Chapter 13
Analysis of Dynamic Circuits by Laplace Transforms
CHAPTER OBJECTIVES
 Expansion of rightsided timedomain waveforms in terms of e^{σt} cos(ωt +ɸ).
 Definition of Laplace transform and inverse integral.
 Various properties of Laplace transform of a real function of time.
 Convergence of inverse integral and region of convergence in splane.
 Basic Laplace transform pairs.
 Solution of differential equations using Laplace transforms.
 Method of partial fractions for inversion of Laplace transforms.
 Sdomain equivalent circuit and applying them for circuit analysis.
 System function H(s) and Laplace transform of impulse response.
 Network functions and polezero plots.
 Graphical interpretation of frequency response function.
INTRODUCTION
We solved firstorder and secondorder circuits using the differential equation model in earlier chapters. Differential equation model can be used to solve multimesh, multinode circuits containing R, L, C, M, linear dependent sources and independent sources too.
A linear timeinvariant circuit containing R, L, C, M, linear dependent sources and a single input source is described by a linear ordinary differential equation with constant coefficients. The differential equation for such a circuit can be expressed in a standard format as below.
The variable y is any circuit voltage or current variable chosen as the describing variable for the circuit and x is the input source function. Standard mesh analysis or nodal analysis technique along with variable elimination will help us to arrive at this equation. However, the variable elimination involved can be considerably tricky in the case of large circuits containing many energy storage elements. This is a serious shortcoming of timedomain analysis by differential equation model.
The order of a circuit is equal to the order of the differential equation that describes it. Order of the circuit will be equal to the total number of independent inductors and capacitors – (number of allinductor nodes + number of allcapacitor loops in the circuit).
The order of a circuit depends also on the kind and location of input. The same circuit will have different order if voltage source input is replaced by current source input.
The coefficients a_{n}_{–}_{1}…a_{0} and b_{m}…b_{0} are decided by the circuit parameters. They are realvalued. a_{n–}_{1}…a_{0} are positive real numbers if the circuit is passive i.e., if it contains only R, L, C and M. They can be zero or negative real numbers if the circuit contains dependent sources. b_{m}…b_{0} can be positive or negative or zero in all circuits.
The format of leftside of the differential equation that describes a circuit is independent of the particular circuit variable chosen as the describing variable in general. That is, a_{n–}_{1}…a_{0} will remain the same even if some other circuit variable is used as the variable y. However, b_{m}…b_{0} will depend on the variable chosen.
This n^{th}order differential equation requires n initial values for solving it if the input function is known only for a range of values of t rather than over entire taxis. The required initial values are
The differential equation can be solved using x(t) for t ≥ 0^{+} and these initial values.
The initial values available in a circuit are the initial current values for all inductors and initial voltage values for all capacitors. It requires considerable effort to translate these values into the required initial values in the case of a circuit containing many energy storage elements. This is another serious shortcoming of the differential equation approach.
Laplace transformation technique converts a linear differential equation with constant coefficients into an algebraic equation. Thus, the task of solving a collection of simultaneous differential equations will be reduced to a much simpler task of solving a set of simultaneous algebraic equations involving Laplace transforms of input signals and Laplace transform of desired output signal. This set of equations may be solved for the Laplace transform of output and the timefunction may be obtained by inverting the transformation. Moreover, we will see later that the initial conditions specified for inductor currents and capacitor voltages can be used directly in Laplace transform method of solving a circuit. This makes it sound as if the technique of Laplace transforms is just a mathematical artifice for solving linear differential equations. Just as Logarithm is not a mathematical artifice for converting multiplication into addition, Laplace transform is not just a mathematical artifice to make solution of differential equations easier.
There is a very compelling reason why we take Laplace transform of a function. The reason is that (i) complex exponential signals are eigen functions of linear timeinvariant circuits, (ii) linear timeinvariant circuits obey superposition principle and (iii) Laplace transform expresses a given arbitrary input function as a sum of complex exponential signals.
Therefore, we commence our study of Laplace transform method of solving a circuit by examining the circuit response to a complex exponential input.
13.1 CIRCUIT RESPONSE TO COMPLEX EXPONENTIAL INPUT
Let the n^{th}order differential equation describing an n^{th}order linear timeinvariant circuit be
where y is some circuit variable identified as the output variable and x is some independent voltage/current source function. Let x(t) = 1 e^{st} be a complex exponential function of unit amplitude and complex frequency s = σ + jω. Let y = A e^{st} be the trial solution, where A is a complex number to be determined. Substituting the trial solution in Eqn. 13.11, we get
Thus, when the input to a linear timeinvariant circuit is a complex exponential function e^{st}, the output is the same complex exponential function multiplied by a complex number. Therefore, the complex frequency of output remains same as that of the input. The output will have a different phase compared to that of input since A is a complex number in general and has an angle. The value of this complex scaling factor depends on the coefficients of circuit differential equation (i.e., on the circuit parameters) and the complex frequency s of the input. In consonance with the symbol H(jω) used for a similar complex number that relates the output to an input of e^{jωt}, we use the symbol H(s) to represent this number A from this point onwards. Therefore,
When x(t) = e^{st} in a linear timeinvariant circuit, y(t) = H(s) e^{st}, where
However, which component of response is this? Since x(t) = 1 e^{st}, the complex exponential function was taken to be applied to the circuit from t = –∞ onwards. Therefore, there is only one component in response and that is the forced response. Therefore, the response given above is the forced response as well as the total response. But if x(t) = 1 e^{st} u(t), then, the above expression yields the forced response component only. The natural response terms in zerostate response and the natural response terms in zeroinput response have to be found out from initial conditions. However, those terms too will be complex exponential functions since natural response terms of a linear timeinvariant circuit are complex exponential functions.
The complex function H(s) of a complex variable s can also be written in polar form as H(s) ∠ θ and in exponential form as H(s) e^{jθ}, where θ is its angle. Therefore, the output y(t) can be expressed as y(t) = H(s) e^{st} ^{+} ^{jθ} = H(s) e^{σ}^{t} e^{j(ωt} ^{+} ^{θ}^{)}. H(s) may be viewed as a generalised frequency response function. Its magnitude gives the ratio between the amplitudes of output complex exponential function and input complex exponential function. Its angle gives the phase angle by which the output complex exponential function leads the input complex exponential function.
The signal e^{st} with a complex value for s goes through a linear timeinvariant circuit and comes out as a scaled replica of itself. The scaling factor is H(s). An input function that goes through a system and comes out as a scaled copy of itself is called an eigen function of the system. The scaling factor is called an eigen value of the system.
Complex exponential signals of e^{st} format with a complexvalued s are eigen functions of linear timeinvariant circuits.
13.2 EXPANSION OF A SIGNAL IN TERMS OF COMPLEX EXPONENTIAL FUNCTIONS
The set of e^{st} format signals for all possible values of s and can be represented as a collection of points in a twodimensional space. The horizontal axis of this plane represents the real part of s and the vertical axis represents the imaginary part of s. Then a point s = σ + jω in this plane will stand for a signal e^{st} = e^{(σ}^{+} ^{jω}^{)}^{t} = e^{σt} e^{jωt}. Such a point which acts as a standin for a complex exponential signal is called a signal point. The complex number representing that point i.e., s, is called the complex frequency of the signal e^{st}. The real part of s has nepers/s as its unit and the imaginary part has radians/s as its unit. Collection of all such signal points – i.e., the plane – is called the signal plane or signal space. The same space is also called the sdomain in circuit studies. The shape of signal for various signal point locations in splane is shown in Fig. 13.21.
Fig. 13.21 Signal point in sdomain versus signal shape
Let v(t) = f(t) u(t) (i.e., right side of some function that is possibly twosided) and v(t) < Me^{αt} for some M and α, then is its Laplace transform, where s = σ + jω is the general complex frequency with σ ≥ α. The Laplace transform exists and the inverse integral converges to v(t) only for those values of s that have Re(s) > α. The region formed by all those values of s in the splane for which the Laplace transform of a timefunction is defined and is convergent is called the region of convergence (ROC) of the Laplace transform. Obviously the ROC of Laplace transform of a rightsided function is the region to the right of Re(s) = α line. This is a vertical straight line parallel to jω axis and crossing σaxis at α.
The timefunction can be obtained from its Laplace transform by carrying out the inversion integral given below.
The Laplace transform defined this way returns the rightside of the underlying function f(t) on inversion. The leftside returned will be zero. In this sense, this Laplace transform may be termed as a unilateral Laplace transform. We deal with only unilateral Laplace transform in this chapter.
Note that the evaluation of inversion integral has to be performed on a line parallel to jωaxis in splane with the line crossing the σaxis within the ROC of the Laplace transform.
Let v(t) be a rightsided function that is bounded by Me^{αt} with some finite value of M and α. Then the Laplace transform pair is defined as
where s = σ + jω is the complex frequency variable standing for the complex exponential function e^{st} with σ ≥ α. The ROC of V(s) is the entire plane to the right of Re(s) = α line.
13.2.1 Interpretation of Laplace Transform
The Laplace transform V(s) is a ‘complex amplitude density function’. Equation 13.22 makes it clear that Laplace transform expresses the given timefunction as a sum of infinitely many complex exponential functions of infinitesimal complex amplitudes. Thus, Laplace transform is an expansion of v(t) in terms of complex exponential functions. The entire ROC is available for evaluating Laplace transform. We will consider an example to clarify this matter.
Example: 13.21
Find the Laplace transform of v(t) = u(t).
Solution
Therefore, V(s) = 1/s with ROC of Re(s) > 0.
Thus, the inversion integral can be evaluated on any vertical straightline on the righthalf in splane. But does not that mean that a steady function like u(t) is being synthesised from oscillations that grow with time? It precisely means that. The synthesis Eqn. 13.22 reveals that infinite growing complex exponential functions of infinitesimal amplitudes, which start at –∞ and go up to + ∞ in time, participate in making the transient timefunction u(t). The contribution from a band of complex frequencies around a complex frequency value s is approximately V(s) × Δs × e^{st}, where Δs is the width of complex frequency band. A similar contribution comes from the band located around s*. These two contributions together will form a growing sinusoidal function as shown below.
Thus, similarly located bands in the two halfsections of the vertical line on which the inversion integral is being evaluated result in a real valued contribution as shown above. Now the inversion integral for 1/s can be written as
Thus, infinitely many exponentially growing sinusoids of frequencies ranging from zero to infinity, each with infinitesimal amplitude, interfere with each other constructively and destructively from t = –∞ to t = + ∞ to synthesise the unit step waveform. Moreover, the exponentially growing sinusoids that participate in this waveform construction process are not unique. The value of σ can be any number > 0. Therefore, each vertical line located in the righthalf of splane yields a distinct set of infinitely many exponentially growing sinusoids which can construct the unit step waveform.
That infinitely many exponentially growing sinusoids interfere with each other to produce a clean zero for all t < 0 and a clean 1 for all t > 0 is indeed counterintuitive and quite surprising when heard first. The inversion integral in Eqn. 13.23 was evaluated using a short computer program for various values of σ and over finite length sections on the vertical line. In effect, the program calculated the partial integral of the form
for various values of σ and ω_{0}. Figure 13.22 shows the resulting waveforms for σ = 0.1 and ω_{0} = 10, 20 and 50.
Fig. 13.22 Partial inversion integral for unit step function for σ = 0.1 and (a) ω_{0} = 10 (b) ω_{0} = 20 (c) ω_{0} = 50
Even a short range of 10 rad/s shows the tendency of the integral to approach step waveform. With ω_{0} = 50 rad/s the integral has more or less yielded step waveform – at least in the range – 1 s to 4 s. We also observe the familiar Gibb’s oscillations at discontinuities. Figure 13.23 shows the results of partial evaluation of inversion integral for σ = 1 and ω_{0} = 10, 20 and 50.
This set of simulation result shows that we have to include more and more components in the partial integral to converge to unit step waveshape in a given timeinterval as we let the components grow at a faster rate, i.e., for higher values of σ. And, keeping σ at a fixed value, we would need to include more and more frequency components when we increase the timerange over which we want convergence. However, we have infinite components at our disposal and it will be possible to include enough of them to recover the u(t) shape up to any finite t however large it may be.
Therefore, Laplace transform expands a transient rightsided timefunction in terms of infinitely many complex exponential functions of infinitesimal amplitudes. The ROC of such a Laplace transform will include portions of righthalf of splane and hence the timedomain waveform gets constructed by growing complex exponential functions though that appears counterintuitive.
Fig. 13.23 Partial inversion integral for step function for σ = 1 and (a) ω_{0} = 10 (b) ω_{0} = 20 (c) ω_{0} = 50
13.3 LAPLACE TRANSFORMS OF SOME COMMON RIGHTSIDED FUNCTIONS
Integral of sum of two functions is sum of integrals of each function. Thus, Laplace transformation is a linear operation. If v_{1}(t) and v_{2}(t) are two rightsided functions and a_{1} and a_{2} are two real numbers, then, a_{1}v_{1}(t) + a_{2}v_{2}(t) ⇔ a_{1}V_{1}(s) + a_{2}V_{2}(s) is a Laplace transform pair. This is called Property of Linearity of Laplace transforms. Now we work out the Laplace transforms for many commonly used rightsided functions using the defining integral and property of linearity.
Let v(t) = e^{sot}u(t) be a rightsided complex exponential function with a complex frequency of s_{o}. Then,
Therefore, e^{sot}u(t) ⇔ 1/ (s – s_{o} ) is a Laplace transform pair with ROC Re(s) > Re(s_{o}).
The special case of v(t) = u(t) is covered by this transform pair with s_{o} = 0.
Therefore, u(t) ⇔ 1/ s is a Laplace transform pair with ROC Re(s) > 0.
The special case of v(t) = cosω_{o}t u(t) is covered by expressing v(t) as = (e^{j}^{ωot} + e^{–jωot}) / 2 by employing Euler’s formula and then applying property of linearity of Laplace transforms.
Therefore, cosω_{o}t u(t) is a Laplace transform pair with ROC Re(s) > 0.
Similarly, sinω_{o}t u(t) is a Laplace transform pair with ROC Re(s) > 0.
Consider v(t) = e^{–}^{αt}cosβt u(t). This can be expressed as = [e^{(e – α + jβ)t} + e^{(e – α– jβ)t} ] / 2 by Euler’s formula. Then,
Therefore, e^{–αt} cosβt u(t) is a Laplace transform pair with ROC of Re(s) > –α.
Similarly, e^{–}^{αt} sin βt u(t) is a Laplace transform pair with ROC of Re(s) > –α.
Now consider v(t)
The Laplace transform of this function can be found from the defining integral as
Now we send v(t) to a limit as Δs → 0.
Therefore, Laplace transform of te^{sot}u(t)
Therefore, te^{sot}u(t) ⇔ 1 / (s – s_{o} )^{2} is a Laplace transform pair with ROC Re(s) > Re(s_{o} ).
The special case of v(t) = t u(t) is covered by this transform pair with s_{o} = 0.
Therefore, tu(t) ⇔ 1/ s^{2} is a Laplace transform pair with ROC Re(s) > 0.
And finally, we consider Thus, δ(t) ⇔ 1 is a Laplace transform pair with ROC of entire splane. It requires all complex exponential functions with equal intensity to synthesise an impulse function in timedomain.
These commonly used Laplace transform pairs are listed in Table 13.31. Some of them have been derived in this section. Others will be taken up later.
Table 13.31 Basic Laplace Transform Pairs
13.4 THE sDOMAIN SYSTEM FUNCTION H(s)
We saw in Section 13.1 that when an input e^{st} is applied to a linear timeinvariant circuit described by an n^{th}order differential equation
the response is given by H(s)e^{st} where
H(s) in this context is the ratio of complex amplitude of forced response component in output to the complex amplitude of input complex exponential function with a complex frequency of s. There is only forced response in this context and forced response itself is the total response.
In Section 13.2, we observed that a rightsided function x(t) can be expressed as a sum of infinitely many complex exponential functions of frequency between σ – j∞ to σ + j∞ with the line Re(s) = σ falling within the ROC of Laplace transform of x(t). We combine these two facts along with superposition principle to arrive at the zerostate response of a linear timeinvariant circuit to a rightsided input function.
Consider a particular value of complex frequency s and a small band of complex frequency Δs centered on it. This band contributes complex exponential functions of frequencies between (s – 0.5Δs) and (s + 0.5Δs). For sufficiently small Δs, we may take all these complex exponential functions to be evolving approximately at the centre frequency of the band, i.e., at s itself. In that case, all the infinitesimal contributions coming from this band may be consolidated into a signal ≈ X(s) Δs e^{st}.
This single complex frequency component with complex amplitude of X(s) Δs will produce a total response component of H(s) X(s) Δs e^{st} in the output. We get the zerostate response of the circuit by adding all such contributions over the line Re(s) = σ falling within the ROC of X(s) and sending the sum to a limit by making Δs → 0. The result will be the following integral.
Compare Eqn. 13.41 with the synthesis equation of Laplace transform given by Eqn. 13.22. It is evident that Eqn. 13.41 is the synthesis equation of the Laplace transform H(s)X(s). But then, a synthesis equation which returns y(t) must be synthesising it from the Laplace transform Y(s) of the timefunction y(t). Therefore, Y(s) = H(s)X(s). This important result requires restatement.
The Laplace transform of zerostate response = Laplace transform of input source function × Ratio of complex amplitude of forced response to the complex amplitude of input complex exponential function at a complex frequency of s.
Now comes a definition. The ratio of Laplace transform of zerostate response to Laplace transform of input source function is defined as the sdomain System Function. And these two are seen to be the same.
∴ The sdomain System Function,
Note carefully that System Function is independent of initial conditions in the circuit since it is the zerostate response to a rightsided input that is employed in its definition. This function is also called a Transfer Function when both x and y are similar quantities, i.e., when x and y are voltages or x and y are currents and is denoted by T(s). It is called an Input Impedance Function and is denoted by Z_{i}(s) if y is the voltage across a terminal pair and x is the current entering the positive terminal. It is called an Input Admittance Function and is denoted by Y_{i}(s) if y is the current into a terminal pair and x is the voltage across the terminal pair. These two, i.e., Z_{i}(s) and Y_{i}(s), together are at times referred to as immittance functions.
If the quantities x and y are voltage–current or current–voltage pair and they refer to different terminal pairs in the circuit, we call the sdomain System Function a Transfer Impedance Function or Transfer Admittance Function, as the case may be. They are represented by Z_{m}(s) and Y_{m}(s), respectively.
We have an expression for H(s) as a ratio of rational polynomials in s in Eqn. 13.42. Rational polynomials are polynomials containing only integer powers of the independent variable. However, there is another more interesting interpretation possible for H(s).
Let us try to find the impulse response of the circuit by this transform technique. We remember that ‘impulse response’ means ‘zerostate response to unit impulse input’ by definition. Hence, we can use the System Function to arrive at the Laplace transform of impulse response as H(s)X(s). But x(t) = δ(t) and therefore X(s) = 1. Hence, for a linear timeinvariant circuit, the following statement holds.
Laplace transform of impulse response = sdomain System Function, and, Impulse response = inverse Laplace transform of sdomain system function This result was anticipated in naming the System Function as H(s).
Once the System Function and Laplace transform of input source function are known, one can obtain the Laplace transform of zerostate response by inverting the product of input transform and System Function. We will take up the task of inverting Laplace transforms in later sections.
13.5 POLES AND ZEROS OF SYSTEM FUNCTION AND EXCITATION FUNCTION
H(s) is the System Function, X(s) is the excitation function and Y(s) is the output function referred to in this section.
We observed that is a ratio of rational polynomials in complex frequency variable s. Further, we observe from Table 13.31 that the excitation functions corresponding to commonly employed input source functions are also in the form of ratio of rational polynomials in s. Thus the output function also turns out to be a ratio of rational polynomials in s. Therefore, we can write
where Q(s) is an m^{th}order polynomial on s and P(s) is an n^{th}order polynomial on s. They are the numerator polynomial and denominator polynomial of System Function, respectively. Similarly, Q_{e}(s) and P_{e}(s) are the numerator and denominator polynomials on s for the excitation function.
Let the n roots of P(s) be represented as p_{1}, p_{2},…, p_{n} and the m roots of Q(s) be represented as z_{1}, z_{2},..., z_{m.} These roots can be complex in general. p_{1}, p_{2},…, p_{n} are the n values of complex frequency s at which the System Function goes to infinity. They are defined as poles of System Function. z_{1}, z_{2},…, z_{m} are the m values of complex frequency s at which the System Function goes to zero value. They are defined as the zeros of System Function.
Similarly the values of s at which X(s) goes to infinity are called the excitation poles and the values of s at which X(s) goes to zero are called the excitation zeros. They are the same as roots of P_{e}(s) and Q_{e}(s), respectively.
Obviously, the System Function poles and excitation function poles together will form the poles of output function. Similarly, the System Function zeros and excitation function zeros together will form the output function zeros. These statements assume that no polezero cancellation takes place.
A diagram that shows the complex signal plane, i.e., the splane, with all poles of a Laplace transform marked by ‘×’ symbol and all zeros marked by ‘o’ symbol is called the polezero plot of that Laplace transform. Some poles and zeros may have multiplicity greater than 1. In that case, the multiplicity is marked near the corresponding pole or zero in the format ‘r = k’ where r indicates the multiplicity and k is the actual value of multiplicity. The default value of r = 1 is not marked.
Example: 13.51
Obtain the polezero plot of the transfer function V_{o}(s)/V_{s}(s), excitation function and the output function in the circuit shown in Fig. 13.51 with v_{s}(t) = 10 e^{–1.5t} cos2t u(t) V.
Fig. 13.51 Circuit for Example: 13.51
Solution
The differential equation describing the second mesh current in this circuit is derived below. The two mesh equations are
Adding the two mesh equations results in
Differentiating the second mesh equation twice with respect to time and using the above equation in the result gives us
v_{o}(t) is numerically equal to i_{2}(t) and hence the differential equation governing v_{o}(t) is
The characteristic equation is s^{3} + s^{2} + 2s + 1 = 0 and its roots are s_{1} = –0.2151 + j1.307, s_{2} = –0.2151– j1.307 and s_{3} = –0.5698. The roots of a polynomial of degree higher than 2 will normally require the help of rootfinding software or numerical methods. There is a pair of complex conjugate roots.
Laplace transform of e^{–1.5t} cos2t u(t) is
Therefore,
We observe that the denominator polynomial is the same as the left side of the characteristic equation of the governing differential equation. This will always be so. Hence poles of System Function (they are also called ‘system poles’) will be the same as the natural frequencies of the circuit for any linear timeinvariant circuit.
Therefore the system poles are p_{1} = –0.2151 + j1.307, p_{2} = –0.2151– j1.307 and p_{3} = –0.5698.
The numerator polynomial of System Function in this case is trivial and there are no ‘system zeros’.
The excitation poles are at p_{e}_{1} = –1.5 + j2 and p_{e}_{1} = –1.5– j2 and excitation zero is at z_{e1} = –1.5.
The polezero plots are shown in Fig. 13.52.
Fig. 13.52 Polezero plots in Example: 13.51 (a) for System Function (b) for excitation function (c) for output function
13.6 METHOD OF PARTIAL FRACTIONS FOR INVERTING LAPLACE TRANSFORMS
Any Laplace transform can be inverted by evaluating the synthesis integral in Eqn. 13.22 on a suitably selected vertical line extending from –∞ to ∞ in the splane within the ROC of the transform being inverted. However, simpler methods based on Residue Theorem in Complex Analysis exist for special Laplace transforms. We do not take up the detailed analysis based on Residue Theorem here. However, the reader has to bear in mind the fact that the ‘method of partial fractions’ for inverting certain special types of Laplace transforms is based on Residue Theorem in Complex Analysis.
Linear timeinvariant circuits are described by linear constantcoefficient ordinary differential equations. All the coefficients are real. Such a circuit will have only realvalued natural frequencies or complexconjugate natural frequencies. Thus, the impulse response of such a circuit will contain only complex exponential functions. Each complex exponential function will have a Laplace transform of the form where s_{o} is the complex frequency of that particular term. Laplace transformation is a linear operation. Hence, Laplace transform of the sum of impulse response terms will be the sum of Laplace transform of impulse response terms. Therefore, Laplace transform of impulse response of a linear timeinvariant circuit will be the sum of finite number of terms of the type. Such a sum will finally become a ratio of rational polynomials in s. The order of denominator polynomial will be the same as the number of firstorder terms of type that entered the sum.
Many of the normally employed excitation functions in linear timeinvariant circuits are also of complex exponential nature. Input functions that can be expressed as linear combinations of complex exponential functions will have Laplace transforms that are ratios of rational polynomials in s as explained in the last paragraph.
Product of Laplace transforms that are ratios of rational polynomials in s will result in a new Laplace transform which will also be a ratio of rational polynomials in s. Hence, the Laplace transform of output of a linear timeinvariant circuit excited by an input source function, that can be expressed as a linear combination of complex exponential functions, will be a ratio of rational polynomials in s.
A Laplace transform that is in the form of a ratio of rational polynomials in s can be inverted by the method of partial fractions.
Let Y(s) = Q(s)/P(s) be such a Laplace transform. Let the degree of denominator polynomial be n and that of numerator be m. The degree of numerator polynomial will usually be less than n. If the Laplace transform of output of a linear timeinvariant circuit shows n ≤ m, it usually implies that the circuit model employed to model physical processes has been idealised too much. We assume that m < n in this section. If m is equal to n or more than n, then Y(s) can be written as and we employ method of partial fractions on only.
Let p_{1}, p_{2},…, p_{n} be the n roots of denominator polynomial. They may be real or complex. If there is a complex root, the conjugate of that root will also be a root of the polynomial. We identify two cases. In the first case all the n roots (i.e., poles of Y(s)) are distinct.
Case1 All the n roots of P(s) are distinct
Then we can express Y(s) as a sum of firstorder factors as below.
Each term in this expansion is a partial fraction. The value of A_{i} appearing in the numerator of i^{th} partial fraction is the ‘residue at the pole p_{i} ’. The problem of partial fractions involves the determination of these residues. Multiply both sides of Eqn. 13.61 by (s – p_{i}), where p_{i} is the pole at which the residue A_{i} is to be evaluated. Remember that Y(s) will contain (s – p_{i}) as a factor in the denominator. Hence, the multiplication by (s – p_{i}) results in cancellation of this factor in Y(s).
Now we evaluate both sides of Eqn. 13.62 at s = p_{i} to get A_{i} = (s – p_{i}_{})Y(s)_{s=pi} . This calculation is repeated for i = 1 to n to complete all the partial fractions.
Each partial fraction of the type can be recognised as the Laplace transform of A_{i}e^{pit} u(t) by consulting relevant entry in Table 13.31. But, though we know that e^{pit }u(t) has a Laplace transform of , how do we know that is the only timefunction that will have as its Laplace transform? It is vital to be sure about that if we want to assert that the timefunction is e^{pit }u(t)whenever we see a Laplace transform The ‘Theorem of Uniqueness of Laplace transforms’ states that a Laplace transform pair is unique. That is, if we have, by some method or other, found out that F(s) is the Laplace transform of f(t), then this theorem assures us that only f(t) will have this F(s) as its Laplace transform and no other function will have F(s) as its Laplace transform. Therefore, whenever we see
a we can write e^{pit }u(t)as its inverse.
Therefore,
Case2 One root of multiplicity r and nr distinct roots for P(s)
In this case the partial fraction expansion is as shown below.
The first root p is assumed to repeat r times. It may be real or complex. The remaining (n – r) roots are designated as p_{r} _{+} _{1}, p_{r} _{+} _{2},…,p_{n}. That Eqn. 13.63 is the partial fraction expansion in this case can be shown by an application of Residue Theorem. We take this as a fact and proceed.
The procedure for evaluating the (n – r) residues at the (n – r) nonrepeating poles of Y(s) is the same as in Case1. Therefore,
We multiply both sides of Eqn. 13.63 by (s – p)^{r} for evaluating the r residues at the repeating pole. The result is
(s – p)^{r} is a factor of denominator of Y(s). Therefore, multiplication of Y(s) by (s – p)^{r} will cancel this factor in the denominator. Now, evaluating both sides with s = p, we get
Now, we differentiate Eqn. 13.64 on both sides with respect to s and substitute s = p to get
Similarly, successive differentiation with respect to s and substitution of s = p leads to
The reader may verify that the partial fraction terms corresponding to the nonrepeating roots will contribute only zero values in all stages of this successive differentiation.
Once all residues are calculated, Eqn. 13.64 is inverted to get the following timefunction:
We have used the Laplace transform pair t^{k}e^{pt}u(t) in arriving at this result. This Laplace transform pair will be proved in a later section.
Example: 13.61
Determine (i) the impulse response (ii) the step response and (iii) the zerostate response when v_{s}(t) = 2e^{–2t} u(t) for v_{o}(t) in the circuit in Fig. 13.61.
Fig. 13.61 Circuit for Example: 13.61
Solution
The mesh equation of the circuit is where i is the current flowing in the mesh. But since i flows in the capacitor and v_{o} (t) is the voltage across capacitor. Therefore, Therefore, the differential equation governing the output voltage is
The System Function
The roots of denominator polynomial are –2.618 and –0.382. The factors of the denominator polynomial are (s + 2.618) and (s + 0.382).
 The impulse response of a linear timeinvariant circuit is the same as the inverse transform of its System Function.
 v_{s}(t) = u(t) ⇒ V_{s}(s) = 1/s. Therefore the Laplace transform of step response Expressing this in partial fractions,
 v_{s}(t) = 2e^{2t}u(t) ⇒ V_{s}(s) = 2/(s + 2). The Laplace transform of zerostate response is given by product of System Function and Laplace transform of input function.
Expressing this in partial fractions,
The resistor value in Fig. 13.61 under Example 13.61 is changed to 2Ω. (i) Find the step response of v_{o}(t) (ii) Determine the zerostate response of v_{o}(t) if v_{s}(t) = e^{–t} u(t) V.
Solution
The differential equation governing the output voltage with 2Ω resistor instead of 3Ω is
The System Function
The roots of denominator polynomial are –1 and –1. The factors of the denominator polynomial are (s + 1) and (s + 1). Therefore, the root at –1 has a multiplicity of 2.
(i) With v_{s}(t) = u(t) the Laplace transform of output is Expressing this in partial fractions, We can find out the residues by applying the expressions developed earlier in this section. Or we may proceed as below.
Now comparing the coefficients of various powers of s in the numerator, we get
Solving these equations, we get, A_{1} = 1, A_{2} = –1 and A_{3} = –1
∴ The step response v_{o}(t) = (1 – te^{–}^{t} – e^{–t}) u(t) V
(ii) With v_{s}(t) = e^{–t} u(t) the Laplace transform of input is the Laplace transform of output is There is no need for partial fractions in this case.
Example: 13.63
Find the impulse response of the circuit in Fig. 13.51 under Example 13.51 in this chapter.
Solution
The System Function was shown to be in Example 13.51.
The characteristic equation is s^{3} + s^{2} + 2s + 1 = 0 and its roots are s_{1} = –0.215 + j1.307, s_{2} = –0.215–j1.307 and s_{3} = –0.57. There is a pair of complex conjugate roots.
Impulse response is obtained by inverting the System Function.
13.7 SOME THEOREMS ON LAPLACE TRANSFORMS
The property of linearity of Laplace transforms was already noted and made use of in earlier sections. We look at other interesting properties of Laplace transform in this section.
13.7.1 TimeShifting Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s) then v_{d}(t) = v(t – t_{d} ) = f(t – t_{d} ) u(t – t_{d} ) has a Laplace transform V_{d}(s) = V(s)e^{–}^{std}.
The shifting operation implied in this theorem is illustrated in Fig. 13.71. Note that there is a difference between f(t – t_{d} ) u(t) and f(t – t_{d} ) u(t – t_{d} _{}).
Fig. 13.71 Illustrating the timeshift operation envisaged in shifting theorem on laplace transforms
This theorem follows from the defining equation for Laplace transforms.
Use variable substitution τ = t – t_{d}
Example: 13.71
Find the zerostate response of a series RC circuit with a time constant of 2 s excited by a rectangular pulse voltage of 10 V height and 2 s duration starting from t = 0. The voltage across the capacitor is the output variable.
Solution
The differential equation governing the voltage across capacitor in a series RC circuit excited by a voltage source is where v is the voltage across the capacitor and v_{s} is the source voltage.
In this case, the equation is Therefore, the System Function is H(s) = 0.5 / (s + 0.5).
The rectangular pulse voltage can be expressed as the sum of 10u(t) and –10u(t – 2). i.e., v_{s} = 10[u(t) – u(t – 2)]. Therefore, its Laplace transform is
We express in partial fractions as and determine A_{1} and A_{2} as
Inverse transform of –10e^{–2s} / s is –10u(t – 2) by Timeshifting Theorem. Similarly, inverse transform of –10e^{–2s } / (s + 0.5) is –10e^{–0.5(t–2)} u(t – 2). Therefore, the output voltage is given by
Figure 13.72 shows the two components of response in dotted curves.
13.7.2 FrequencyShifting Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s), then, v_{d}(t) = v(t)e^{s}^{ot} has a Laplace transform V_{d}(s) = V(s – s_{o} ).
Fig. 13.72 Output response and its components in the circuit in Example 13.71
This theorem follows from the defining equation for Laplace transforms.
13.7.3 TimeDifferentiation Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s), then, has a Laplace transform
V_{d}(s) = sV(s) – v(0^{–}). Note that
The function v(t)e^{–st} will be a decaying function for any value of s in the ROC of V(s). Otherwise, the Laplace transform will not converge for that value of s. Therefore, it will go to zero as t → ∞
Now, by using mathematical induction, we may show that,
Laplace transform of
and that, in general, Laplace transform of
13.7.4 TimeIntegration Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s), then, has a Laplace transform
The function v(t) is stated to possess a Laplace transform. This implies that there is an exponential function Me^{αt} with some positive value of M and some real value for α such that v(t) < Me^{αt}. Otherwise, v(t) would not have a Laplace transform. Therefore, the function will satisfy the inequality and therefore is bounded. Therefore, the Laplace transform of will exist. That is, it is possible to select a value for s such that the function is a decaying function. For such an s, i.e., for a value of s in the ROC of Laplace transform of the value of will go to zero as t → ∞. And, the value of e^{–st} at t = 0^{– }is zero in any case.
Example: 13.72
Find the Laplace transform of t^{n}u(t).
Solution
The function tu(t) is the integral of u(t). Therefore, tu(t) ⇔ 1/ s^{2}. Now the function t^{2}u(t) is 2 times the integral of tu(t). Therefore, t^{2}u(t) ⇔ 2 / s^{3}. Proceeding similarly to the power n, we get, t^{n}u(t) ⇔ n!/ s^{n}^{+1}.
13.7.5 sDomainDifferentiation Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s), then, –tv(t) has a Laplace transform
We show this by determining the Laplace transform of –tv(t) from the defining integral.
We have repeatedly used the limit many times before. We use this limit again with α = –s within the integral.
Since the limiting operation is on s and integration is on t we may interchange the order of these two operations.
13.7.6 sDomainIntegration Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s) and is finite, then, has a Laplace transform
If the integration is carried out in the righthalf of splane (ROC of rightsided functions will have at least a part of righthalf splane in it), then e^{–}^{∞}^{t} in the last step in the equation above will vanish. Then,
13.7.7 Convolution Theorem
If x(t) and y(t) are two rightsided timefunctions with Laplace transforms X(s) and Y(s) respectively and Z(s) = X(s)Y(s), then,
We use the inverse integral to show this.
The integral is called the Convolution Integral between x(t) and y(t) and is denoted by x(t)*y(t). Linear System Theory predicts that convolving the impulse response of a linear timeinvariant circuit with its input source function gives the zerostate response. Hence, the convolution theorem stated here corroborates the fact that Laplace transform of zerostate response is given by the product of Laplace transform of input source function and Laplace transform of impulse response. We had termed the Laplace transform of impulse response as the sdomain System Function.
13.7.8 Initial Value Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s) and exists, then,
We know that the Laplace transform of Therefore,
We assume that s → ∞ with its real part is always positive. This is consistent with the fact that ROC of Laplace transform of rightsided functions will contain the righthalf of splane or at least portions of righthalf splane. Evaluation of the term e^{–st} with t → 0 and s → ∞ simultaneously is to be avoided. Hence, we write the integral as below.
Therefore, . Now we apply the limit s → ∞ with its real part always positive. Then the integral vanishes. Therefore,
13.7.9 Final Value Theorem
If v(t) = f(t) u(t) has a Laplace transform V(s) and exists and all the poles of sV(s) have negative real part, then,
We know that the Laplace transform of Therefore,
One has to be very careful in applying this theorem. This theorem works only if all the poles of sV(s) are in the open lefthalf plane in sdomain. That is, all the poles of sV(s) must have negative real part. Only then will the function v(t) reach a unique and steady final value with time. Otherwise, the value returned by the application of this theorem will not be the final value of v(t). For that matter v(t) may not have a final value at all. Let v(t) be sin ωt. Then V(s) = ω/(s^{2} + ω^{2}) and sV(s) = sω/(s^{2} + ω^{2}). Application of final value theorem says that v(∞) = 0. But there is no unique final value for a sinusoidal waveform. This conflict comes about because of wrong application of the theorem. The sV(s) function in this case has poles on jωaxis and hence the final value theorem is not applicable.
13.8 SOLUTION OF DIFFERENTIAL EQUATIONS BY USING LAPLACE TRANSFORMS
One of the important applications of Laplace transform is in solving linear constantcoefficient ordinary differential equations with initial conditions. The procedure is illustrated below through an example.
Find y(t) for t>0^{+} for x(t) = δ(t) in the given differential equation with y(0^{–}) = 1, y′(0^{–}) = –1 and y′′(0^{–}) = 0.
Solution
Since the differential equation is an equation, both sides of it can be multiplied by e^{–st}. Since the differential equation is satisfied at all instants of time in a given interval, both sides of it can be integrated with respect to time from 0^{– }to ∞. In short, the Laplace transform operation can be carried out on both sides. Laplace transformation is a linear operation and hence Laplace transform of a sum of terms is equal to sum of Laplace transforms of individual terms. Therefore,
Now we apply the ‘Differentiation in Time Theorem on Laplace transforms’ to get
The transform terms that depend only on the input function result in zerostate response. The transform terms that depend only on the initial conditions on output and its derivatives result in zeroinput response.
Since x(t) = δ(t), X(s) = 1 in this example. Substituting the values for initial conditions and Laplace transform of x(t), we get
We need to factorise the denominator polynomial in order to arrive at the partial fraction expansion. It is a thirdorder polynomial with real coefficients. Complex roots, if any, will have to occur in conjugate pairs for such a polynomial. Therefore, a polynomial of odd degree with real coefficients will necessarily possess a realvalued root. We try to locate that real root by the method of bisection.
Try two different values of s such that the polynomial evaluates to a positive and a negative number.
s^{3} + 2.5s^{2} + 2.5s + 1.5 evaluates to 1.5 for s = 0 and –1.5 for s = –2. Therefore, there must be root between 0 and –2. We try the midvalue of –1 and see that the polynomial evaluates to 0.5. Therefore, the root must be between –1 and –2. Hence we try the midvalue –1.5. The polynomial evaluates to 0. Hence the root is s = –1.5. In practice, many iterations may be needed to arrive at a root by this technique.
Now we know that (s + 1.5) is a factor of s^{3} + 2.5s^{2} + 2.5s + 1.5. Therefore, we get the remaining secondorder factor by long division as s^{2} + s + 1. The roots of this factor are –0.5 ± j0.866.
Now we expand each response term in partial fractions. Normally we expand a transform in terms of firstorder partial fractions. However, a secondorder factor with complex conjugate roots may be expanded more conveniently in a form illustrated below.
Comparing the coefficients of various powers of s in the numerator, we get A + B = 0, A + 1.5B + C = 0 and A + 1.5C = 1. Solving for the unknowns, we get A = 0.5715, B = –0.5715 and C = 0.2857.
Therefore, the zerostate response = Inverse of
∴ Zerostate response = [0.5715e^{–1.5t} – 0.5715e^{–0.5t} cos0.866t+0.66e^{–0.5t} sin0.866t]; t ≥ 0^{+}
The zeroinput response is given by the inverse of
∴ Zeroinput response = [e^{–0.5t} cos0.866t – 0.5774e^{–0.5t} cos0.866t]; t ≥ 0^{+}
Total response y(t) is the sum of zeroinput response and zerostate response. Therefore,
If we can solve a differential equation with nonzero initial conditions completely in one stroke using Laplace transforms, then, we can indeed solve linear timeinvariant circuits with nonzero initial conditions for their total response using Laplace transforms. We have to derive the n^{th} order linear constantcoefficient differential equation describing the circuit in terms of a single chosen variable first. In the second step, we have to determine the initial values for that chosen circuit variable and its (n–1) derivatives from the known initial values of inductor currents and capacitor voltages in the circuit. Then we are ready to employ Laplace transform technique to solve for zeroinput response and zerostate response in one step as illustrated in this example.
However, the derivation of differential equation and determination of initial values of chosen variable and its derivatives are the toughest tasks in a circuit analysis problem. Can’t Laplace transform technique help us to simplify these two stages of circuit analysis?
13.9 THE sDOMAIN EQUIVALENT CIRCUIT
Yes, it can. Laplace transform technique tells us that we do not even have to derive the circuit differential equation and initial values required to solve it. Let us see how.
The circuit equations arising out of applying KVL in loops and KCL at nodes are equations that remain true at all values of t. Therefore, such equations can be differentiated and integrated with respect to time without changing their truth content. Moreover, of course, being equations they can be multiplied by the same constant or function on both sides.
We choose to multiply all KCL and KVL equations in a linear timeinvariant circuit by a function e^{–st}, where s is a complex frequency value drawn from splane, with a real part of suitable value such that each term in the equation is converted into an absolutely integrable function of time (so that Laplace transform for that term will converge). Then we choose to integrate the equations from 0^{–} to ∞ in timedomain. We apply the principle that integral of sum of terms is sum of integrals of individual terms.
The result will be a conclusion – (i) the algebraic sum of Laplace transforms of element voltages in any loop in a circuit is zero (ii) the algebraic sum of Laplace transforms of element currents at any node in a circuit is zero.
The Laplace transforms of voltage variables and current variables in a linear timeinvariant circuit obey KVL and KCL, respectively.
Now, suppose we know the relation between the Laplace transform of element voltage and Laplace transform of element current for all circuit elements. Then, we can write the node equations and mesh equations in terms of Laplace transforms of variables straightaway – i.e., we can write the circuit equations in sdomain straightaway instead of writing them in timedomain and transforming them into sdomain at the end of solution process. Hence, we derive the element relationships in sdomain first.
13.9.1 sDomain Equivalents of Circuit Elements
A resistor R is described by the timedomain element equation v_{R}(t) = R i_{R}(t). Multiplying both sides by e^{–s }^{t }^{}and integrating from 0^{–} to ∞, we get V_{R}(s) = R I_{R}(s). Note that we use upper case letters for Laplace transforms. This relation makes it clear that a resistor is represented by a multiplying factor of R that connects the Laplace transform of current through it to the Laplace transform of voltage across it. The ratio of Laplace transform of voltage across an element to the Laplace transform of current through it is defined as its ‘sdomain impedance’. Thus, sdomain impedance of a resistor is R itself.
An inductor L is described by the timedomain element equation where v_{L}(t) and i_{L}(t)are its voltage and current variables as per passive sign convention. Applying Laplace transformation on both sides of this element equation, we get V_{L}(s) = sLI_{L}(s) – Li_{L}_{(0}–_{)} where i_{L}_{(0}–_{)} is the initial current in the inductor at t = 0^{–}. The Laplace transform Li represents an impulse voltage Li_{L}_{(0}–_{)} δ(t) in timedomain and hence it is consistent with the fact that a nonzero initial condition in an inductor can be replaced with an impulse voltage source in series with the inductor. This sdomain equation for an inductor suggests the following sdomain equivalent circuit shown in Fig. 13.91 (a) for an inductor with sL as its sdomain impedance function. The second equivalent circuit shown in Fig. 13.91 (b) follows from and indicates the fact that the sdomain admittance of an inductor is 1/sL and that nonzero initial condition in an inductor can be replaced with a step current source in parallel with it.
Fig. 13.91 sDomain equivalent circuits for an inductor
Note that we use the same graphic symbol for inductor in sdomain as the one we used in timedomain. This will be the case with all other circuit elements too.
A capacitor C is described by the timedomain element equation where v_{C}(t) and i_{C} (t) are its voltage and current variables as per passive sign convention. Applying Laplace transformation on both sides of this element equation, we get I_{C} (s) = sCV_{C} (s) – Cv_{C} _{(0}–_{)} where v_{C} _{(}_{0}–_{)} is the initial voltage across the capacitor at t =0^{–}. The Laplace transform Cv_{C} _{(}_{0}–_{) }represents an impulse current Cv_{C} _{(}_{0}–_{) } δ(t) in timedomain and hence it is consistent with the fact that a nonzero initial condition in a capacitor can be replaced with an impulse current source in parallel with the capacitor. This sdomain equation for a capacitor suggests the following sdomain equivalent circuit shown in Fig. 13.92 (a) for a capacitor with sC as its sdomain admittance function. The second equivalent circuit shown in Fig. 13.92 (b) follows from and indicates the fact that the sdomain impedance of a capacitor is 1/sC and that nonzero initial condition in a capacitor can be replaced with a step voltage source in series with it.
Fig. 13.92 sDomain equivalent circuits for a capacitor
sdomain impedance is assigned the unit of Ω (ohms) and sdomain admittance is assigned the unit of S (Siemens).
Now we can construct the sdomain equivalent circuit for a circuit in timedomain by replacing all sources by their Laplace transforms and replacing all other circuit elements by their sdomain equivalents. The resulting equivalent circuit will have Laplace transforms of voltages and Laplace transforms of currents as the circuit variables instead of the timedomain variables. Each energy storage element will result in an extra independent source representing its initial condition in sdomain equivalent circuit.
Applying KVL and KCL in this circuit will result in algebraic equations involving Laplace transforms of voltages and currents, respectively. Thus the problem of solving a coupled set of integrodifferential equations involving functions of time in the timedomain circuit is translated to solving a coupled set of algebraic equations involving Laplace transforms of variables in the sdomain equivalent circuit. The timefunctions may be determined by inverting the Laplace transforms after they are obtained.
Transforming a timedomain circuit into an sdomain circuit makes it similar to a memoryless circuit with DC excitation since both are described by algebraic equations. Thus, all concepts and techniques developed in the context of analysis of memoryless circuits (and used later in the analysis of phasor equivalent circuits under sinusoidal steadystate) will be directly applicable in the analysis of sdomain equivalent circuits too.
In particular, (i) the concepts of series and parallel equivalent impedances apply without modification (ii) the concepts of input resistance (i.e., drivingpoint resistance) and input conductance (i.e., drivingpoint conductance) apply without modification except that it is ‘input impedance function Z_{i} (s)’ and ‘input admittance function Y_{i} (s)’ in the case of sdomain circuits.
Moreover, the techniques of nodal analysis and mesh analysis can be applied in sdomain circuits. All the circuit theorems, except maximum power transfer theorem, can be applied in the context of sdomain equivalent circuits.
However, Laplace transform of instantaneous power is not equal to product of Laplace transforms of voltage and current. In fact, the sdomain convolution of V(s) and I(s) gives the Laplace transform of p(t) = v(t) i(t). Therefore, dealing with power and energy variables in the sdomain is better avoided. They are better dealt with in timedomain itself.
We observe that the sdomain equivalent circuit makes use of the stated initial values of inductor currents and capacitor voltages right at the start. The sdomain equivalent circuit takes care of these initial values in the form of additional source transforms. Therefore, the circuit solution arrived at by the analysis of sdomain equivalent circuit will contain both zeroinput response components and zerostate response components in one step. Thus, Laplace transform technique yields the total response in a singlestep solution process.
13.9.2 Is sdomain Equivalent Circuit Completely Equivalent to Original Circuit?
No, it is not. It is equivalent to the original circuit only for t ≥ 0^{+}.
Given a nonzero initial value for current at t = 0^{–} in an inductor we have two ways of taking that into account when we try to solve the circuit for t ≥ 0 ^{+} . We may adopt the view that unknown voltage was applied across the inductor in the past which resulted in this initial current. We accept the possibility of other circuit elements that were switched out at t = 0. We can solve for the current for t ≥ 0 + using this initial condition and the known input function. However, we will specify the range of applicability of the circuit solution as t ≥ 0 + specifically.
The second view that we may adopt is that the inductor current was zero at t = 0^{–}; but an impulse voltage of LI_{0} Vs was applied in series with the inductor with suitable polarity. This view will explain the initial current in the inductor becoming I_{0} though it was taken to be zero at t = 0^{–}. Hence, the circuit solution will be the same as the one we obtained by adopting the first point of view.
However, in the second point of view, we are fixing the voltage applied to the inductor in the past at zero value (since the impulse voltage is zero valued for t <0^{–}). Thus, we are assuming that the circuit was the same from t = –∞ onwards, no sources were active in it till t =0^{–}and some impulse sources acted at t = 0 to change the initial energy storage in some elements abruptly. Thus, the circuit and the sources active in the circuit are completely known from t = –∞ onwards in this point of view. Therefore, the circuit solution is valid from t = –∞ In fact, the solution will be zero for (–∞, 0^{–}]. This is usually denoted by multiplying all timefunctions that appear in the circuit solution by u(t). The solution will be correct for all t as far as the fictitious circuit that we assumed in this point of view is concerned. However, that is not the actual circuit that we wanted to solve. The actual circuit that we wanted to solve is the one we described under first point of view. Therefore, timefunctions multiplied by u(t) cannot be the solution in the actual circuit. We cannot solve the actual circuit for t < 0^{– }since the input is really unknown in this timerange. Therefore, only the right side of the solution arrived at from the fictitious circuit, that takes into account initial conditions by bringing in impulse sources, should be accepted as the solution for the actual circuit. Hence, the circuit solution should be specified as timefunctions with the range of applicability specified as t ≥ 0 ^{+}. The solution for t < 0^{– }is left unspecified. It is understood that the circuit cannot be solved for t < 0^{–} with the given data.
sDomain equivalent of a circuit uses the second point of view described above. Hence, it is equivalent to the circuit only for t > 0 ^{+}. Circuit solution from sdomain equivalent circuit is obtained by inverting Laplace Transforms. That will yield timefunctions multiplied by u(t). We have to replace u(t) by ‘for t ≥ 0 ^{+}^{’ }before we accept the solution from sdomain equivalent circuit as the solution for actual timedomain circuit.
However, in practice, this step is skipped often and the solution is left in the ‘timefunction x u(t)’ format itself. This does not lead to errors in practice since we are usually interested in circuit variables for t > 0 ^{+} only. However, in the strict sense, it is a bad practice.
13.10 TOTAL RESPONSE OF CIRCUITS USING SDOMAIN EQUIVALENT CIRCUIT
The application of sdomain equivalent circuit in obtaining the total response of a circuit is illustrated through a set of examples in this section.
Example: 13.101
The inductor L_{1}has an initial current of 1 A and the inductor L_{2} has an initial current of 1 A in the directions marked in the circuit in Fig. 13.101 (i) Find the voltage transfer function V_{o}(s)/V_{s}(s) and the input impedance function V_{s}(s)/I(s) (ii) Determine the total response of v_{o}(t) if v_{S}(t) = 2u(t) V.
Fig. 13.101 Circuit for Example: 13.101
Solution
(i) Transfer functions and immittance functions are defined in the sdomain equivalent circuit. They are defined under zerostate response conditions. They are defined as ratios of Laplace transforms of relevant quantities under zerostate response conditions. Therefore, they are defined under zero initial conditions. The sdomain equivalent circuit with zero initial conditions is shown in Fig. 13.102.
Fig. 13.102 The transformed equivalent circuit for circuit with zero initial conditions
Series–parallel equivalents and voltage–current division principle may be employed to arrive at the required ratios. Input impedance function Z_{i}(s) = V_{s}(s) / I(s).
The transformed current in the second 1Ω resistor may be found out in terms of I(s). Then V_{o}(s) may be obtained from that current by multiplying by 1Ω. Let this current transform be called I_{o}(s). Then,
But
Therefore,
The poles of transfer function are at s = –2.618 and s = –0.382. The zero is at s = 0.
(ii) The total response of v_{o}(t) with v_{S}(t) = 2u(t) V may be solved by mesh analysis or by applying superposition principle. Both methods are illustrated below. The sdomain equivalent circuit with initial conditions accounted and mesh current transforms identified is shown in Fig. 13.103.
The mesh equations are
1 + I_{1}(s)[–s] + I_{2}(s)[2s + 1] –1 = 0
Fig. 13.103 Transformed equivalent circuit of circuit in Fig. 13.101 with initial condition sources included
These are expressed in matrix form as below.
Solving for I_{2}(s), we get
Since
The roots of denominator polynomial (i.e., poles of transfer function) are at s = –2.618 and s = –0.382. The input transform V_{s}(s) = 2/s.
Therefore, zerostate response
The second component of output is expanded in partial fractions as below.
Therefore, zeroinput response
Total response in the actual timedomain circuit is the sum of zerostate response and zeroinput response accepted for only t ≥ 0 ^{+} and is given by
It is not necessary to split the two components of the response this way always. It was done here only to demonstrate how the Laplace transform technique brings out both together in one step. Inverting the total response transform would have resulted in total response straightaway. Indeed
The same solution can be arrived at by using Superposition Theorem. This theorem can be applied only for the zerostate response components due to various inputs. But then, all the initial conditions get translated into sources in the transformed equivalent circuit and hence the circuit analysis problem in the sdomain is always a zerostate response problem. Therefore, superposition principle can be freely applied in transformed equivalent circuits. The solution term due to initial condition sources will be understood as the zeroinput response once we get back to timedomain.
There are three sources in this transformed equivalent circuit. The component circuits required to find out the individual response components are shown in Fig. 13.104.
Fig. 13.104 Component circuits for applying superposition theorem in Example: 13.101
The transfer function of the first circuit is already known as and . Therefore, the output transform in the first circuit is
(s + 1)//1 Ω shares –1 with s Ω in series in the second circuit. Therefore, the voltage transform across (s + 1) Ω is . This voltage transform is divided between s Ω and 1 Ω to yield at the output.
s + 1//s Ω shares 1 in series with 1 Ω to produce across the output.
The total output voltage transform is given by the sum of three output voltage transforms. Therefore, This is the same output transform we obtained in the mesh analysis too.
The timedomain function will be v_{0}(t) = (0.2763e^{–0.382t} + 0.7237e^{–2.618t})u(t) V.
Example: 13.102
The circuit in Fig. 13.105 was in DC steadystate at t = 0. The switch in the circuit closes at t = 0 introducing a new 1 Ω into the circuit. Determine the voltage across the inductor as a function of time.
Solution
The circuit was in DC steadystate prior to switching at t = 0. A capacitor can be modelled by an opencircuit and an inductor by a shortcircuit for DC steadystate analysis. Therefore, the circuit for DC steadystate prior to t = 0 is as shown in Fig. 13.106.
Fig. 13.105 Circuit for Example: 13.102
Fig. 13.106 Circuit under DC steadystate for t < 0
Therefore, the voltage across the capacitor at t = 0^{–} was 20/3 V and current through the inductor at t = 0^{–} was 10/3 A. The circuit solution after t = 0 can be obtained by solving a new circuit with 10u(t) as input, v_{c}(0^{–} ) = 20/3 V and i_{L}(0^{–} ) = 10/3 A.
The new circuit can be analysed by mesh analysis or nodal analysis techniques. We opt for nodal analysis since the desired output is a node voltage variable straightaway. It will be convenient to use a current source in parallel with capacitor and a current source in parallel with inductor to account for initial conditions since we have opted for nodal analysis. The transformed equivalent circuit required is shown in Fig. 13.107.
Fig. 13.107 Transformed equivalent circuit in Example: 13.102
The first source in series with 1 Ω may be replaced by a current source in parallel with 1 Ω. The node equations in matrix form will be
The determinant of Nodal Admittance Matrix is
Solving for V_{3}(s)
Therefore,
The voltage across the inductor is the same as v_{3}(t). Note that the final value of inductor voltage is zero. This is expected since under DC steadystate condition the inductor behaves like a shortcircuit.
Example: 13.103
Verify the initial value theorem and final value theorem on Laplace transforms for i(t) and v_{o} (t) in the initially relaxed circuit shown in Fig. 13.108 when driven by v_{s}(t) = u(t).
Fig. 13.108 Circuit for Example: 13.103
The sdomain equivalent circuit required for analysis is shown in Fig. 13.109.
Fig. 13.109 The sdomain equivalent circuit of the circuit in Fig. 13.108
Two mesh current transforms are identified in the sdomain equivalent circuit. The mesh equations in matrix form is written by inspection by using the rule that diagonal entry is the sum of all impedances in the corresponding mesh and offdiagonal entries are negative of the sum of impedances shared by the two meshes in question.
Solving for I_{1}(s) and I_{2}(s) by Kramer’s rule, we get
Now,
The input is an u(t) function and its transform is 1/s. Therefore,
The poles of sI(s) are in the lefthalf of splane and hence the final value theorem on Laplace transforms is applicable to I(s).
The poles of sV_{o}(s) are in the lefthalf of splane and hence the final value theorem on Laplace transforms is applicable to V_{o}(s).
The initial current at t = 0^{–} through the inductor was zero and initial voltage across the capacitor at that instant was zero. There was no impulse content in voltage at input. Therefore, the inductor current at t = 0 ^{+} remains at zero. There was no impulse current in the circuit. Therefore, the voltage across the capacitor remains at zero at t = 0 ^{+} . The timedomain equivalent circuit at t = 0 ^{+} is shown in Fig. 13.1010 (a). The initial value of current i(t) at 0 ^{+} is clearly zero and the initial value of v_{o}(t) is also zero.
Fig. 13.1010 Equivalent circuits at (a) t = 0 ^{+} and (b) t → ∞
The inductor is replaced by a shortcircuit and the capacitor by an opencircuit under DC steadystate conditions. The resulting circuit is shown in Fig. 13.1010 (b). Hence, the final value of i(t) 1/10 = 0.1 A and the final value of v_{o}(t) is equal to input voltage i.e., 1 V
Hence the initial value theorem and the final value theorem on Laplace transforms are verified for v_{o}(t) and i(t).
Example: 13.104
(a) Find the transfer function in the Opamp circuit shown in
Fig. 13.1011 assuming ideal Opamp. (b) Show its polezero plot for k = 2.9 and k = 3.1 and find its zerostate response to v_{S}(t) = 0.01δ(t) in both cases with R = 10kΩ and c = 1μF. (c) What is the maximum value of k that can be used in the circuit without making it an unstable one?
Fig. 13.1011 The Opamp circuit in Example: 13.104
(a) The Opamp circuit from its noninverting input to its output is a simple noninverting amplifier of gain k and can be replaced with a dependent source as shown in the sdomain equivalent circuit in Fig. 13.1012.
Fig. 13.1012 Transformed equivalent circuit of the circuit in Fig. 13.1011
The node voltage transforms V_{1}(s) and V_{2}(s) are identified in the transformed equivalent circuit. Writing the node equations at these two nodes, we get
Therefore,
Substituting for V_{1}(s) in terms of V_{2}(s) and simplifying, we get
This is the transfer function of the circuit.
(b) The denominator polynomial with R = 10 kΩ, C =1 μF and k = 2.9 is s^{2} + 10s +10000. Therefore, the poles are at s = –5 ± j 99.88. The zero of the transfer function is at s = 0, i.e., at the origin in the splane.
With k = 3.1, the denominator polynomial is s^{2} – 10s + 10000and the poles are at s = 5 ± j 99.88. The zero of the transfer function is again at s = 0. The polezero plots are shown in Fig. 13.1013.
Fig. 13.1013 Polezero plots for the Opamp circuit in Fig. 13.1011 with k = 2.9 and k = 3.1
The zerostate response to v_{S} (t) = 0.01δ(t) is nothing but the impulse response of the circuit scaled by 0.01.
With k = 2.9
The transfer function H(s) Inverse transform of H(s) gives the impulse response of the circuit. We complete the squares in the denominator and identify the inverse transforms as shown below.
This is a stable impulse response since
The zerostate response to 0.01δ(t) = = 2.9e^{–5t} cos(99.88t + 2.86^{0})u(t).
With k = 3.1
The transfer function H(s) Inverse transform of H(s) gives the impulse response of the circuit. We complete the squares in the denominator and identify the inverse transforms as shown below.
This is an unstable impulse response since it is unbounded. The circuit is an unstable one as evidenced by its poles located in the righthalf of splane.
The zerostate response to 0.01δ(t) = = 3.1e^{5t} cos(99.88t – 2.86^{0})u(t).
(c) The transfer function has poles on jωaxis when k = 3. The poles will lie on the righthalf of splane for all values of k > 3. Therefore, k < 3 is the constraint on k value for stability in the circuit.
The circuit is marginally stable at k = 3. It can function as a sinusoidal oscillator with k = 3. But additional circuitry will be needed to stabilize its amplitude of oscillation.
Note that the circuit is a pure RC circuit with one dependent source in it. A passive RC circuit, i.e., a circuit containing only resistors and capacitors and no dependent sources, will have all its poles in the negative real axis. The dependent source is responsible for making the poles complex conjugate in such a circuit. Complex conjugate poles are often necessary in filter circuits to tailor the filter frequency response function suitably to meet filtering specifications.
This circuit is used as a bandpass filter in practice. The value of k will be decided by the bandwidth required in the bandpass filter and will be < 3 at any rate.
Example: 13.105
(a) Obtain the transfer function of the filter circuit shown in Fig. 13.1014 and identify the type of filter. (b) Determine the zerostate response for v_{s}(t) = 0.1 δ(t). R = 10kΩ, C = 1μF and k = 1
Fig. 13.1014 The Opamprc filter circuit in Example: 13.105
Solution
The transformed equivalent circuit is shown in Fig. 13.1015. Initial condition sources are not required since the circuit is needed for determining transfer function and for evaluating zerostate response. The Opamp is assumed to be ideal. There is only one node that has a free node voltage variable. This node and the assigned node voltage transform are also indicated in the equivalent circuit.
Fig. 13.1015 The transformed equivalent circuit of circuit in Fig. 13.1014
Virtual short across the Opamp input terminals and zero input current drawn by the Opamp inverting pin makes the current in the feedback capacitor equal to V(s)/R. This results in the output voltage transform becoming equal to –V(s)/sCR.
Now, we write the node equation at the node1 marked in Fig. 13.1015.
The denominator has its roots in the lefthalf of splane for all positive values of k since a secondorder polynomial with positive coefficients will have both roots in lefthalf of splane. Therefore, the impulse response is stable and will be absolutely integrable. Therefore, the impulse response will have a Fourier transform. If Fourier transform of a timefunction exists, then, the Laplace transform of the same function evaluated on jωaxis is its Fourier transform. Fourier transform of impulse response is the frequency response function. Therefore, the sinusoidal steadystate frequency response function of a stable circuit is given by its Laplace transform evaluated with s = jω.
Therefore, This frequency response function has a magnitude of unity at ω = 0 and 0 at ω = ∞. Therefore, it is a lowpass filter.
Substituting the numerical values, we get
The impulse response of the circuit is obtained by inverting the transfer function. The roots of the denominator are at s = –261.8 and s = –38.2.
A and B can be evaluated as –44.72 and 44.72, respectively.
Therefore, h(t) = 44.72(e^{–38.2t} – e^{–261.8}^{t})u(t)
Therefore, zerostate response for v_{S} (t) = 0.1δ(t) is 4.47(e^{–38.2t} – e^{–261.8}^{t})u(t)V
Example: 13.106
A single Opamp, a resistor and a capacitor can form a good differentiator circuit. This circuit is shown in Fig. 13.1016. (i) Derive the transfer function of the circuit and show that it is a differentiator.
(ii) An ideal Opamp is too good to be true. A practical Opamp suffers from many nonidealities. The particular nonideality that compromises the circuit performance will vary depending on circuit function. For example, we found that integrator circuit is severely compromised by offsets in a practical Opamp. We will see in this example that it is the limited gain and bandwidth of Opamp that compromises the performance of differentiator circuit.
An amplifier contains many capacitors – intentional as well as parasitic – in it and hence represents a highorder dynamic circuit. However, some Opamps like IC 741 can be modelled approximately as a singletime constant amplifier. That is, its gain function is of the form with A ≈ 250,000 and τ ≈ 4 ms. Obtain the transfer function of differentiator circuit with R = 10kΩ and C = 1μF using IC 741 and find its impulse response.
Fig. 13.1016 An Opamp differentiator circuit
(iii) Suggest a method to modify the oscillatory impulse response to critically damped impulse response.
(i) This is essentially an inverting amplifier structure. The transfer function = It is a differentiator since multiplication by s in sdomain is equivalent to differentiation in timedomain according to timedifferentiation theorem on Laplace transforms. It is an inverting differentiator.
Fig. 13.1017 Transformed equivalent circuit for differentiator circuit in Fig. 13.1016
(ii) The Opamp is to be modelled as a dependent source that senses the voltage transform between noninverting pin and inverting pin and produces a voltage transform at its output with respect to ground, where V_{d}(s) is the transform of voltage of noninverting pin with respect to inverting pin. The sdomain equivalent circuit incorporating this model for Opamp is shown in Fig. 13.1017.
Let the node voltage transform at the inverting pin be V_{1}(s). Writing the node equation at inverting pin, we get
Simplifying this equation results in
Therefore,
The DC gain A of any practical Opamp is in thousands and hence A + 1 ≈ A. Therefore,
Note that the order of the circuit is two. The Opamp contributes one extra order to the circuit. Substituting R = 10kΩ, C = 1μF, A = 250000 and τ = 4ms, we get
Step response is the inverse of transfer function multiplied by 1/s.
Of course, if a unit step is really applied to this circuit, the output of Opamp will saturate. But what is to be noted is that the response is highly underdamped. The oscillation is at 12.6 kHz and oscillation period is about 0.08 ms. But the time constant of damping exponential is 40 ms. It takes about 5 time constants for an exponential transient to settle down. That implies that the 12.6 kHz transient oscillations will last for about 200 ms before they die down. That is a bad transient performance.
(iii) The solution is to add a little damping by means of a resistor in series with the input capacitor. This will make the differentiator imperfect however.
13.11 NETWORK FUNCTIONS AND POLEZERO PLOTS
Network function is a ratio of Laplace transforms. It is the ratio of Laplace transform of zerostate response to the Laplace transform of the rightsided input function causing this response. We called it sdomain System Function till now. We use the name ‘Network Function’ synonymously from this section onwards.
But ratio of Laplace transform of zerostate response to Laplace transform of which input function? There should only be one! That is, a network function can be defined and evaluated only in an sdomain equivalent circuit containing one input source transform. Thus, the circuit should have only one independent source active when a network function is evaluated. Hence, there cannot be initial condition sources too. That is why the definition specifies that it is the ratio of Laplace transform of zerostate response to Laplace transform of input function.
The response may be measured across or through any circuit element or combinations of such elements in general. Hence, a variety of network functions are defined in a circuit. In particular, when the response variable and excitation variable pertain to same terminal pair, the network function can only be of drivingpoint impedance or drivingpoint admittance type. The two together are referred as immittance functions.
13.11.1 DrivingPoint Functions and Transfer Functions
Input impedance function or drivingpoint impedance function is where V(s)and I(s) are transforms of voltage and current in a terminal pair as per passive sign convention. Input admittance where V(s)and I(s) are transforms of function or drivingpoint admittance function is voltage and current in a terminal pair as per passive sign convention. They are reciprocals of each other at the same terminal pair. These functions are a special class of network functions.
Transfer impedance function is Z_{m}(s) where V_{ij} (s) is the transform of voltage developed at i^{th} terminal with respect to j^{th} terminal due to a current source I_{pq}(s) delivering current to p^{th} terminal from q^{th} terminal. The terminal pairs p–q and i–j are not the same. Transfer admittance function is where I_{pq}(s) is the transform of current developed from q^{th} terminal to p^{th} terminal due to a voltage source V_{ij} (s) between i^{th} terminal and j^{th} terminal. The terminal pairs p–q and i–j are not the same though they may share a common terminal. These functions are a special class of network functions.
Voltage transfer function is A_{ν}(s) where V_{pq} (s) is the Laplace transform of zerostate voltage response developed across terminal pair p–q due to a voltage source transform V_{ij}(s) applied across terminal pair i–j. The terminal pairs p–q and i–j are not the same though they may share a common terminal.
Current transfer function is A_{i}(s) where I_{pq}(s) is the Laplace transform of zerostate current response developed through terminal pair p–q due to a current source transform I_{ij}(s) applied through terminal pair i–j. The terminal pairs p–q and i–j are not the same though they may share a common terminal.
Thus there are two types of drivingpoint network functions and four types of transfer network functions.
We will use the symbol H(s) when we refer to network functions in general and use Z_{i}(s), Y_{i}(s), Z_{m}(s), Y_{m}(s), A_{v}(s) and A_{i}(s) when we refer to specific network functions.
13.11.2 The Three Interpretations for a Network Function H(s)
The first interpretation is the definition of a network function itself. That is, a network function is the ratio of Laplace transform of zerostate response to the Laplace transform of input source function causing the response.
Two circuit variables are clearly identified in the definition of network function – they are the variable used to measure the zerostate response and the variable that was decided by the input source function. These two variables can be identified in the sdomain equivalent circuit, and circuit analysis in sdomain using nodal analysis or mesh analysis can be performed to arrive at the desired network function. The result will be H(s) in the form of a ratio of rational polynomials in s. Thus, from this point of view, we expect to get an H(s) in the following format. We have chosen to make the coefficient of highest power in s in the denominator 1.
The second interpretation for H(s) comes from the meaning of Laplace transform. Laplace transform of a rightsided function is an expansion of that function in terms of functions of e^{st} type with value of s ranging from σ–j∞ to σ +j∞. The value of σ is such that the expansion converges to the timefunction at all t. The components in expansion, i.e., signals of type e^{st} are from –∞ to ∞ in timedomain. Thus, Laplace transform converts a rightsided input into the sum of everlasting complex exponential inputs. Therefore, the problem of zerostate response with a rightsided input is translated into that of forced response with everlasting complex exponential inputs. And, the ratio of Laplace transform of zerostate response to Laplace transform of input source function must then be the same as the ratio between forced response to input when input is e^{st} (not e^{st} × u(t)).
Forced response to an everlasting complex exponential input of 1e^{st} was seen to be a scaled version e^{st} itself; the scaling factor being a complex number that depends on the complex frequency value s. (Refer Section 13.1)
The timedomain circuit can be analysed using nodal analysis or mesh analysis to arrive at the n^{th}order differential equation relating the response variable (y) to excitation variable (x). The result will be
Then the scaling factor connecting an input of 1e^{st} to the output is
Therefore,
In this expression for H(s), the coefficients come from the coefficients of differential equation governing the circuit. In the expression for H(s) in Eqn. 13.111, the coefficients were the result of circuit analysis in sdomain. We conclude that n' = n, m' = m, all a' values are equal to corresponding a values and all b' values are equal to corresponding b values.
The third interpretation comes from the definition of network function itself. If the input source function is assumed to be δ(t), then H(s) becomes a Laplace transform – it becomes the Laplace transform of impulse response. Thus, H(s) is a ratio of Laplace transforms and a Laplace transform at the same time. It is a Laplace transform when we invert it in order to find the impulse response.
The three faces of H(s)
H(s), the network function, is a Laplace transform if we invert it to find the impulse response. H(s), the network function, is a complex gain if we evaluate it at a particular value of s. In that case, it gives the complex amplitude of the forced response with an input of e^{st} with the value of s same as the value at which H(s) was evaluated. H(s), the network function, functions as a ratio of Laplace transforms when we multiply it by the Laplace transform of input source function and invert the product to determine the zerostate response in timedomain.
13.11.3 Poles and Zeros of H(s) and Natural Frequencies of the Circuit
A network function goes to infinite magnitude at certain values of s. These values are obviously the values of s at which the denominator polynomial evaluates to zero, i.e., at the roots of denominator polynomial. These values of s are called poles of the network function. Thus, poles are roots of denominator polynomial of a network function. Similarly, a network function attains zero magnitude at certain values of s. They are roots of numerator polynomial. They are called zeros of the network function.
A diagram showing the pole points by ‘×’ marking and zero points by ‘o’ marking in complex signal plane (i.e., splane) is called the polezero plot of the network function.
We note from the discussion in the previous subsection that the denominator polynomial of a network function apparently has the same order and same coefficients as that of the characteristic polynomial of differential equation describing the linear timeinvariant circuit. The roots of the characteristic polynomial have been defined as the natural frequencies of the circuit. Does this mean that (i) the degree of denominator polynomial in a network function is the same as the degree of characteristic polynomial (ii) the poles and natural frequencies are the same?
The order of a differential equation is the order of highest derivative of dependent variable. The order of a circuit and order of the describing differential equation are the same. It will also be equal to the total number of independent inductors and capacitors – (number of allcapacitorvoltage source loops + number of allinductorcurrent source nodes).
The order of a network function is the degree of denominator polynomial, i.e., the highest power of s appearing in the denominator polynomial.
Thus we are raising the question – is the order of a network function in a linear timeinvariant circuit same as the order of the circuit?
The characteristic polynomial of a differential equation is quite independent of righthand side of differential equation. But, a network function is very much dependent on the righthand side of the differential equation. Therefore, there exists a possibility of cancellation of some of the denominator factors by numerator factors in the case of a network function. Therefore, the order of a network function can be lower than the order of the circuit. It cannot, however, be higher. This will also imply that the order of two network functions defined within the same network need not be the same.
For instance, let the differential equation describing a linear timeinvariant circuit be
The characteristic equation is s^{2} + 3s + 2 = 0 and the order of circuit is 2. The natural frequencies are s = –1 and s = –2. The zeroinput response can contain e^{–} ^{t} and e^{–}^{2}^{t} terms. But it may contain only one of them for certain combination of initial conditions. Consider y(0) = 1 and y'(0) = –1. Then y(t) = e^{–t} and it will not contain e^{–}^{2}^{t}. Therefore, not all natural response terms need be present in all circuit variables under all initial conditions.
Now consider the network function. It is
The order of network function is 1. It has one pole at s = –2. Therefore, zerostate response to any input will not contain e^{–t} term. This is the effect that a polezero cancellation in a network function has on circuit response. But, note that the same circuit may have other network functions that may not involve such polezero cancellation. It is only this particular circuit variable denoted by y that refuses to have anything to do with the natural response term e^{–t}.
Therefore, we conclude the following:
 The order of a network function and the order of the circuit can be different due to possible polezero cancellations in a particular network function.
 Poles of any network function defined in a linear timeinvariant circuit will be natural frequencies of the circuit.
 However, all natural frequencies need not be present as poles in all network functions defined in that circuit.
 However, all natural frequencies will appear as poles in some network function or other.
 Thus, poles of a network function is a subset of natural frequencies of the circuit and natural frequencies will be unionset of poles of all possible network functions in the circuit.
 A complex frequency that is not a natural frequency of the circuit cannot appear as a pole in any network function in that circuit.
 Both the denominator polynomial and the numerator polynomial of a network function in a linear timeinvariant circuit have real coefficients. Therefore, poles and zeros of a network function either will be realvalued or will occur in complex conjugate pairs.
13.11.4 Specifying a Network Function
A network function H(s) is specified in three ways. In the first method, it is specified as a ratio of rational polynomials in s.
In the second method, it is specified as the ratio of product of firstorder factors in numerator and denominator with a gain factor multiplying the entire ratio.
There are m factors in the numerator and n factors in the denominator. z_{1},z_{2},…,z_{m} are the zeros of the network function and p_{1},p_{2},…,p_{n} are the poles of the network function. Note that though the degree of denominator polynomial is shown as n, which is the order of the circuit, polezero cancellation may take place leaving the denominator polynomial of network function with a degree less than n.
In the third method of specifying a network function, the polezero plot along with the gain factor K is given. The gain factor K may be directly given or indirectly in the form of value of H(s) evaluated at a particular value of s.
Example: 13.111
The circuit shown in Fig. 13.111 is the small signal equivalent circuit of a transistor amplifier for analysis of its behaviour for sinusoidal input at high frequency. Obtain the transfer function between the output voltage and input source voltage.
Fig. 13.111 Smallsignal equivalent circuit of a transistor amplifier in Example: 13.111
Solution
We find the Norton’s equivalent of the circuit to the left of 100pF capacitor first.
Fig. 13.112 Subcircuits for determining Norton’s equivalent
The subcircuits needed for determining this equivalent are shown in Fig. 13.112. The shortcircuit current in the first circuit is The Norton’s equivalent resistance is [(50Ω//2kΩ) + 50Ω] //1kΩ = 89.9 Ω. Thus the required Norton’s equivalent is 9.876×10^{–3}v_{s}(t) A in parallel with 89.9 Ω. The original circuit with this Norton’s equivalent in place is shown in circuit of Fig. 13.113 (a) with R_{1} = 89.9 Ω, R_{2} = 2 kΩ, C_{1} = 100 pF, C_{2} = 5 pF and g_{m} = 0.08. The corresponding sdomain equivalent circuit is shown in circuit (b) of Fig. 13.113.
Fig. 13.113 (a) Reduced version of circuit in Fig. 13.111 and (b) Its sdomain equivalent
The node equations written for the two node voltage transforms V(s) and V_{o}(s) are as follows:
Solving for V_{o}(s) and simplifying the expression, we get
Substituting the numerical values for various parameters, we get
The poles are at s = –10^{9} nepers/s and s = –11.07×10^{6} nepers/s. The zero is at s = 1.6×10^{10 }nepers/s.
Note that compared to the pole at –11.07×10^{6} the other pole and the zero are located two orders away from it. The natural response term contributed by the pole at s = –11.07×10^{6} nepers/s will have a time constant of 90.3 ns whereas the natural response term contributed by the pole at s = –10^{9} nepers/s will have a time constant of 1 ns. Thus the natural response term contributed by the pole at s = –10^{9 }nepers/s will disappear in about 5% of the time constant of the other term. Therefore, the time constant of 90.3 ns is the dominant time constant in this amplifier and the corresponding pole at –11.07×10^{6 }is the dominant pole. The amplifier transfer function can be approximated by neglecting the zero and the nondominant pole to
Example: 13.112
Find (i) Input impedance function and (ii) in the circuit shown in Fig. 13.114.
Fig. 13.114 Circuit for Example: 13.112
Solution
This is a case of cancellation of all poles by zeros leaving a real value for a network function. The input impedance of the circuit is purely resistive at all values of s. This implies that the current drawn by the circuit behaves as in a memoryless circuit. But, this does not mean that the order of the circuit is zero.
The voltage transfer function has a pole at s = –1 and zero at s = 1.
This innocuous circuit challenges our notions on the order of a circuit. It contains two energy storage elements. Hence it must be a secondorder circuit. But no network function defined in this circuit will be secondorder function if the excitation is a voltage source. Even the zeroinput response obtained by shorting the voltage source with initial conditions on inductor and capacitor will contain only e^{–t}. The reader is encouraged to analyse the general situation that develops when many subcircuits with same set of poles in their input admittance functions are connected in parallel and driven by a common voltage source. Similarly, he is encouraged to ponder over the order of a circuit resulting from series connection of many subcircuits with the same set of poles in their input impedance functions driven by a common current source. The reader may also note that the current in the circuit in Fig. 13.114 will have e^{– t} and t e^{– t} terms in zeroinput response due to initial energy storage in inductor and capacitor with input opencircuited (i.e., zeroinput response for current source excitation), thereby confirming it is a secondorder circuit.
13.12 IMPULSE RESPONSE OF NETWORK FUNCTIONS FROM POLEZERO PLOTS
Let be a network function defined in a linear timeinvariant circuit. Then the impulse response of this network function is given by its inverse transform. The transform H(s) × 1 (1 is the Laplace transform of δ(t)) can be expressed in partial fractions as below.
We have assumed that all poles are nonrepeating ones. If there are repeating poles we may assume that the poles are slightly apart by ∆p and evaluate the limit of h(t) as ∆p → 0 after we complete the inversion. We will need a familiar limit for this. This strategy will help us to view all poles as nonrepeating ones at the partial fractions stage.
Thus each pole contributes a complex exponential function to impulse response. The complex frequency of the complex exponential function contributed by a pole to impulse response is the same as the value of the pole frequency itself.
A point s in the complex signal space (i.e., the splane) stands for the complex exponential signal e^{st} for all t. But, when a point s is marked out as a pole of a network function by a ‘×’ mark, that signal point contributes e^{st} u(t) to the impulse response and not e^{st}. Thus, a point in signal space stands for a twosided complex exponential signal in general and stands for a rightsided complex exponential signal when that point is specified as a pole of a network function.
The evaluation of residue A_{i} at the pole p_{i} involves the evaluation of product of terms like (pz_{1})… (p_{i}z_{1}) and (p_{i}p_{1})… (p_{i}p_{i} _{1}) (p_{i}p_{i} _{+} _{1})… (p_{i}p_{n} ). Each of these factors will be a complex number. For instance, consider (p_{i} z_{1}). This is a complex number that can be represented by a directed line drawn from the point s = z_{1} in splane to the point s = p_{i} in the splane with the arrow of the line at s = p_{i} . The length of this line gives the magnitude of the complex number (p_{i}z1) and the angle of the complex number (p_{i}z_{1}) is given by the angle the line makes with the positive real axis in the counterclockwise direction. The magnitude of product of complex numbers is product of magnitudes of individual numbers. The angle of product of complex numbers is the sum of angles of individual complex numbers. Therefore, evaluation of residue A_{i} at the pole p_{i} reduces to determining certain lengths and angles in the polezero plot of the network function.
The reasoning employed in the paragraph above also reveals the roles of poles and zeros of a network function in deciding the impulse response terms. The poles decide the number of terms in impulse response and their complex frequencies. The zeros along with the poles and gain factor K decide the amplitude of each impulse response term.
A network function is a stable one if its impulse response decays to zero with time. This is must be equivalent to stating that its impulse response must be absolutely integrable, i.e., finite. Therefore, a network function is stable if all the impulse response terms are damped ones. That is, all the poles must have negative real values or complex values with negative real parts. Therefore, a network function is stable if and only if all its poles are in the lefthalf of splane excluding the jωaxis.
Note that a stable network function in a linear timeinvariant circuit does not necessarily imply that the circuit itself is stable. A linear timeinvariant circuit is stable only if all the network functions that can be defined in it are stable ones. That is, a stable circuit will have only stable network functions in it. But, an unstable circuit can have both stable and unstable network functions in it.
The graphical interpretation adduced to impulse response coefficients in this section is illustrated in the examples that follow.
Obtain the polezero plots for (i) for positive and negative values of α and sketch the impulse response for α = ± 1.
Solution
These are standard firstorder network functions. They are stable ones for positive values of α and unstable ones for negative values of α. They are important, yet simple, functions.

. Therefore, h(t) = αe^{αt} u(t) The polezero plot and impulse response are shown in Fig.13.121 for α = ±1.
Fig. 13.121 Polezero plot and impulse response for

The polezero plots and impulse responses for are shown in Fig. 13.122.
Fig. 13.122 Polezero plot and impulse response for

. The polezero plots and impulse responses are shown in Fig. 13.123.
Fig. 13.123 Polezero plot and impulse response for
Example: 13.122
A secondorder lowpass network function in standard form is given as where ξ is the damping factor and ω_{n} is the undamped natural frequency as defined in Section 11.6 in Chapter 11. Obtain expressions for impulse response of the network function for positive and negative values of ξ in the range –1<ξ<1.
Solution
The poles are at They are complex conjugate poles for –1<ξ<1. They are located in the righthalf splane for –1<ξ<0 and in the lefthalf splane for 0<ξ<1. They are located on jωaxis at when ξ = 0. The poles have a magnitude of ω_{n} for all values of ξ in the range (–1,1). The pole line of the pole makes an angle of cos^{–1}(ξ) with negative real axis in the case of positive ξ and with positive real axis in the case of negative ξ. Thus the damping factor magnitude is given by cosine of pole angle. See Fig. 13.124.
Fig. 13.124 Polezero plots for a standard secondorder lowpass network function with positive damping
The residue at the pole marked as B is given by ω_{n}^{2} divided by the complex number represented by the line connecting A and B in Fig. 13.124 with arrow towards B. This line is seen to be in length and it makes –90° with positive real axis. Similarly, the residue at the pole marked as B is given by ω_{n}_{}^{2} divided by the complex number represented by the line connecting A and B in Fig. 13.124 with arrow towards B. This line is seen to be in length and it makes 90° with positive real axis.
Therefore,
This response is shown in Fig. 13.125 for ω_{n} = 1 and ξ = 0.7, 0.3, 0.1 and 0.05.
We observe that as the poles get closer and closer to jωaxis, the impulse response oscillations become more and more underdamped and last for many cycles.
The impulse response for ξ = –0.05 in Fig. 13.126 shows the unbounded nature of impulse response of a network function with poles on righthalf splane.
Fig. 13.125 Impulse response of standard secondorder network function for various damping factors
Fig. 13.126 Impulse response of standard secondorder network function for ξ = –0.05
13.13 SINUSOIDAL STEADYSTATE FREQUENCY RESPONSE FROM POLEZERO PLOTS
Let be a network function defined in a linear timeinvariant circuit and let all the poles of this network function be in the lefthalf of splane excluding jωaxis. The zeros can be located anywhere in splane. We know that H(s), the network function, is a complex gain if we evaluate it at a particular value of s. In that case, it gives the complex amplitude of the forced response with an input of e^{st} with the value of s same as the value at which H(s) was evaluated. Sinusoidal steadystate frequency response function of a stable circuit is given by the complex gain offered by the circuit to e^{jωt} signal. Therefore, H(s) evaluated with s = jω gives the frequency response function provided the network function is stable. Thus,
H(jω) is a complex function of a real variable ω. The plots of H(jω) versus ω and H(jω) versus ω yield the frequency response plots of the network function. The first is called the magnitude plot and the second is called the phase plot.
13.13.1 Three Interpretations for H(jω)
We saw that we can interpret the network function H(s) in three ways in Section 13.11. Three interpretations of H(jω) follow from this.
 H(jω) is the ratio of complex amplitudes of output complex exponential and input complex exponential when input complex exponential is of the form Ae^{j}^{ω}^{t}. Equivalently, H(jω) is the complex amplitude of output when input is (not e^{jωt}u(t)). The signal e^{–jωt} is a signal that is different from Hence, H(jω) is a twosided function from this point of view.
If input is e^{jωt}u(t), then, H(jω) e^{jωt} gives the forced response component (same as the steadystate response component).
 H(jω) is the ratio of Laplace transform of zerostate response to Laplace transform of input when input is of the form Ae^{jωt} u(t). The signal Ae^{–jωt} u(t) is not the same as Ae^{jωt} u(t). Hence, H(jω) is a twosided function from this point of view too.
 H(jω) is the expansion of the impulse response h(t) of the circuit in terms of complex exponential signals drawn from jω axis in splane. It expands the timedomain signal into the sum of complex exponential functions of e^{jωt} type (i.e., essentially in terms of sinusoids) with ω value ranging from to –∞ to ∞. Hence, H(jω) is a twosided function from this point of view too.
There are two ways to solve the problem of finding the steadystate output when input variable is cosω_{o}t u(t). The first method is to express cosω_{o}t u(t) as Re(e^{jωot})u(t) and express the output as Re[H(jω_{o} )e^{jωot} ]u(t). This method was called Phasor Method in Chapter 8. This method results in the steadystate response component and is based on the first interpretation of H(jω).
Re[H(jω_{o})e^{jωot}] = Re[ H(jω_{o} )  e^{∠H}^{(}^{jω}o^{)} e^{jωt}] =  H(jω_{o})  cos[ω_{o}t + ∠H(jω_{o})] . Now, there is no harm if H(jω) is thought of as a singlesided function of ω provided we interpret the magnitude of H(jω) as the amplitude of output sinusoidal waveform with input amplitude of 1 and the phase of H(jω) as the phase angle by which the output sinusoidal waveform leads the input sinusoidal waveform.
The second way is to express cosωt u(t) as 0.5 e^{jωt}u(t) + 0.5e^{–jωt} u(t) by applying Euler’s formula and express the steadystate output as 0.5[H(jω_{o} )e^{jωo t} + H(–jω_{o})e^{–jωo t} ] .We have seen in Chapter 7 that H(– jω) = [H(jω)]*.Therefore the steadystate output will be Re[H(jω_{o})]cos ω_{o}t – Im[H(jω_{o})]sin ω_{o}t =  H(jω_{o})  cos[ω_{o}t + ∠H(jω_{o})]; same as in the first method. This method also is based on the first interpretation of H(jω) but uses a twosided version of H(jω).
Frequency response function is not new to us. We had dealt with the frequency response of firstorder circuits and second order circuits in detail earlier in the book. But the observation that H(jω) can be evaluated by evaluating H(s) on jωaxis leads to a graphical interpretation for sinusoidal steadystate frequency response function based on the polezero plot of H(s). This interpretation affords an insight into the variation of magnitude and phase of H(jω) without evaluating it at all values of ω. It helps us to visualise the salient features of frequency response function without extensive calculations.
13.13.2 Frequency Response from PoleZero Plot
Each factor of the form (jω – z_{i}) in the numerator of Eqn. 13.131 is a complex number that can be thought of as a line directed from s = z_{i} to s = jω in splane. The magnitude of (jω – z_{i}) is equal to the length of the line and the angle of (jω – z_{i} ) is the angle that the line makes with the positive real axis in counterclockwise direction. Similar interpretation is valid for factors of the type (jω – p_{i}) in the denominator of Eqn. 13.131 too. Let d_{zi} be the length of the line joining the zero at s = z_{i} to the excitation signal point s = jω on the imaginary axis in splane. Let the angle that the line makes with the positive real axis in counterclockwise be θ_{zt} . Similar quantities for a pole at s = p_{i} are d_{p}_{i} and θ_{ρi} . Then, the frequency response function H(jω) can be expressed in terms of these lengths and angles as
Hence, we can make a rough sketch of the frequency response function by visualising how the various zerodistances and poledistances vary when the ω is taken from –∞ to + ∞ in splane.
Example: 13.131
Obtain the frequency response plots for (i) H(s) = α / (s + α) (ii) H(s) = s / (s + α) using geometrical interpretation of frequency response and obtain expressions for bandwidth in both cases.
Figure 13.131 shows the polezero plot and frequency response function. The pole distance and the pole angle are marked in the Fig. 13.131. Obviously, the gain magnitude goes to 1/√2 times initial gain when ω = α and the phase at that point is –45°. Therefore bandwidth is a rad/s and the function is a lowpass function.
Fig. 13.131 Polezero plot and frequency response of H (s)
Solution (ii)
Figure 13.132 shows the polezero plot and frequency response function. The pole distance and the pole angle are marked in Fig. 13.132. The zero distance is same as the excitation frequency value ω and times the final gain when ω = α and the zero angle is 90°. Obviously, the gain magnitude goes to the phase at that point is 45°. Therefore bandwidth is α rad/s and the function is a highpass function.
Fig. 13.132 Gain and phase plots of
Example: 13.132
The biquadratic network function is a secondorder lowpass function if a = 0, b = 0 and c = ω_{n}^{2}. It is a secondorder highpass function if a = 1 and b = c = 0. It is a bandpass function if a = c = 0 and b = 2ξω_{n} and it is a bandreject function if a = 1, b = 0 and c = ω_{n}^{2}. The frequency response function for lowpass, highpass and bandpass secondorder functions were studied in detail in Section 11.11 in Chapter 11 in the context of frequency response of Series RLC circuit.
Consider the bandpass function and bandreject function and sketch their frequency response plots for ξ << 1.
Solution
The poles of the function are at Let us consider the bandpass function first.
The zero of this function is at s = 0.
The distance of the pole at to the excitation frequency point jω is denoted by d_{1} and the distance of the pole at to jω is d_{2}. The distance of zero at s = 0 to jω is ω itself. The pole angles θ_{1} and θ_{2} are also shown in the polezero diagram in Fig. 13.133.
The magnitude function then is 2ξω_{n}ω/(d_{1}d_{2}) and the phase function is (π/2)–(θ_{1} + θ_{2}). The distances d_{1} and d_{2} are equal to ω_{n} at ω = 0 and the sum of the angles θ_{1} and θ_{2} at that frequency is 360°. Therefore, the gain at zero frequency is 0 (due to the zerodistance of zero) and angle is 90°. As ω → ∞ all the three distances go to ∞ and hence magnitude goes to zero. θ_{1} and θ_{2} go to 90° as ω → ∞ and hence the phase angle of frequency response function goes to –90° as ω → ∞.
As ω increases from 0, the distance d_{1} decreases and the distances d_{2} and d_{3} increase. Consider a pair of ω values equal to i.e., two ω values separated by part of the pole from the imaginary part of the pole. The distance d_{1} undergoes a variation and again to as ω varies from passing through the point The distances d_{2} and d_{3} also vary. However, if ξ << 1, the variation in these two quantities will be negligible over this frequency range and approximation will be satisfactory.
Therefore, the magnitude of frequency response function will vary from and again to as ω varies from passing through the point The imaginary part of poles can be taken as approximately ω_{n} itself for ξ << 1. Therefore, the maximum gain is 1 at ω = ω_{n} and gain goes through The phase angle at ω = ω_{n} is zero. See Fig. 13.133.
Fig. 13.133 Polezero plot and frequency response plot for a second order bandpass function
The center frequency of the narrow bandpass function is seen to be ≈ ω_{n} and the bandwidth is ≈ 2ξω_{n}. Thus the ratio of center frequency to bandwidth of a narrow bandpass secondorder network function is 1/2ξ or Q of the denominator polynomial.
Let us consider the bandreject function now.
The poles are at and zeros are at The distance of the pole at to the excitation frequency point jω is denoted by d_{1} and the distance of the pole at The distance of zero at s = jω_{n} to jω is d_{3} and the distance of zero at s = – jω_{n} to jω is d_{4}. The pole angles θ_{1} and θ_{2} are also shown in the polezero diagram in Fig. 13.134. The zeroangles are –90° and 90° for all ω< ω_{n} and 90° and 90° for all ω> ω_{n} . The gain is given by d_{3}d_{4}/d_{1}d_{2} and starts at 1 at ω = 0 since d_{1} = d_{2} and d_{3} = d_{4}. The gain goes to 1 as ω → ∞ since d_{1}≈ d_{2} and d_{3} ≈ d_{4} under that condition. The gain is zero at ω = ω_{n} since d_{3} is zero under that condition. Therefore, it is a bandreject function.
The polezero plots and frequency response plots are shown in Fig. 13.134. For ξ << 1, it may be shown that the gain crosses and that the phase angles at those frequencies are –45° and 45°. The centre frequency of the narrow bandreject function is seen to be ≈ ω_{n} and the bandwidth is ≈ 2ξω_{n} . Thus the ratio of centre frequency to bandwidth of a narrow bandreject second order network function is 1/2ξ or Q of the denominator polynomial.
Fig. 13.134 Polezero plot and frequency response plot for a secondorder bandpass function
The frequency response of higher order network functions can similarly be sketched. A higher order H(jω) can be expressed as the product of firstorder factors and biquadratic factors. The magnitude response for firstorder factors and biquadratic factors may be sketched separately first and then multiplied together to get the magnitude response curve of H(jω). Phase curves will have to be added.
Poles on negative real axis contribute a monotonically decreasing magnitude response. Poles close to jωaxis render resonant peaks in magnitude response and zeros on jωaxis produce zero gain response at the excitation frequencies equal to the zero locations. Thus graphical interpretation of frequency response function is a valuable aid to a circuit designer who wants to locate poles and zeros of a network function to tailor the frequency response function to meet design specifications.
13.14 SUMMARY
 Let v(t) be a rightsided function that is bounded by Me^{αt} with some finite value of M and α. Then the Laplace transform pair is defined as
where s = σ + jω is the complex frequency variable standing for the complex exponential function e^{st} with σ value > α. The ROC of V(s) is the entire plane to the right of Re(s) = α line.
Table 13.141 Some Impotant Properties of Laplace Transforms
 Laplace transform expands a transient rightsided timefunction in terms of infinitely many complex exponential functions of infinitesimal amplitudes. The ROC of such a Laplace transform will include righthalf of splane and hence the timedomain waveform gets constructed by growing complex exponential functions.
 For a linear timeinvariant circuit described by the ratio of rational polynomials in s defined as
has three interpretations.
 It may be viewed as a generalised frequency response function. Its magnitude gives the ratio between the amplitude of output complex exponential function and input complex exponential function when input is of the form Ae^{st}. Its angle gives the phase angle by which the output complex exponential function leads the input complex exponential function.
 It is also the ratio of Laplace transform of zerostate response to Laplace transform of input source function called ‘the sdomain System Function’.
 Further, it is the Laplace transform of Impulse Response
 Laplace transforms can be inverted by the method of partial fractions.
 Laplace transformation of both sides of a linear constantcoefficient ordinary differential equation converts it into an algebraic equation on transforms. Thus Laplace transform affords a convenient way to solve such differential equations with initial conditions.
 All circuit elements have sdomain equivalents. The sdomain equivalent of the complete circuit can be constructed by replacing each element with its sdomain equivalent. KVL and KCL are directly applicable to the transformed quantities.
 sdomain equivalent circuits may be analysed by nodal analysis or mesh analysis procedures. All circuit theorems developed in the context of memoryless circuits are applicable to sdomain equivalent circuits.
 The sdomain System Function is also called the network function. Immittance functions, transfer functions and transfer immittance functions are three classes of network functions usually employed in circuit analysis. Those complex frequency values at which a network function goes to ∞ are called its poles and those complex frequency values at which the network function goes to zero are called its zeros.
 Polezero plot along with a gain factor K will specify a network function uniquely. The impulse response of the network function may be obtained from its polezero plot. The poles decide the number of terms in impulse response and their complex frequencies. The zeros along with the poles and gain factor K decide the amplitude of each impulse response term.
 A network function is a stable one if its impulse response decays to zero with time. This is must equivalent to stating that its impulse response must be absolutely integrable, i.e., must be finite. Therefore, a network function is stable if all the impulse response terms are damped ones. That is, all the poles must have negative real values or complex values with negative real parts. Therefore, a network function is stable if and only if all its poles are in the lefthalf of splane excluding the jωaxis.
 Sinusoidal steadystate frequency response function H(jω) can be obtained by evaluating H(s) on jωaxis. This evaluation can be carried out in a graphical manner too.
 The DC steadystate gain of a network function is H(0).
13.15 PROBLEMS
 Evaluate the voltage across the capacitor in a series RC circuit with R = 1 kΩ and C = 1000 μF at t = 0 ^{+} and t = 20 sec if the applied voltage to the circuit is (a) 2 e^{0.01}^{t} cost V (b) 2 e^{0.01t}cost u(t) V. Use differential equation approach.
 A current source with i_{S}(t) = f(t) u(t) A is applied to a parallel RL circuit with R = 1 Ω and L = 1 H. The voltage across the combination is found to be = 2e^{–}^{t}sint u(t) V. Find f(t) and the initial current in the inductor. Do not use Laplace transform technique.
 The output variable y in a linear timeinvariant circuit is related to the input variable x by the following differential equation Its zeroinput response is found to contain e term among other terms. If x(t) = 3e^{0.01t} sint, find the instantaneous value of y at t = 10 s. Do not use Laplace transform technique.
 A voltage source of v_{S}(t) = 2e^{–0.2t}cos(t–45°) V is applied to a series RLC circuit with R = 1 Ω, L = 1 H and C = 1 F from t = 0 ^{+} . The circuit is initially relaxed. Determine the total response of current in the circuit by solving the circuit differential equation without using Laplace transforms.
 A bounding exponential Me^{αt} is to be determined for each of the functions listed below. Find the minimum value of α and the corresponding value of M for each. (i) tu(t) (ii) u(t)–u(t–2) (iii) e^{3 t} u(t) (iv) e^{–3 t} u(t) (v) e^{3 t} u(t–3) (vi) e^{–3 t} u(t–2) (vii) e^{–}^{2 t} cos2t u(t) (viii) e^{2 t} cos2t u(t).
 The signals listed in Problem 5 are applied to a parallel RL circuit with R = 2 Ω and L = 1 H as current sources. Find the instantaneous voltage across the combination at t = 2 sec in each case by using Laplace transforms technique.
 The Laplace transform of impulse response of a linear timeinvariant circuit is given by (i) Find the differential equation describing the circuit assuming that the output variable is y and the input variable is x. (ii) Find the total response of the circuit if x(t) = 3 for t >0 + with y(0^{–}) = 1 unit and y'(0^{–}) = 1 unit/s by Laplace transform technique.
 The Laplace transform of current drawn by an initially relaxed dynamic circuit from an unit impulse voltage source is I(s) (i) Find the differential equation relating the current drawn by the circuit to voltage applied. (ii) Find the total response of current if the circuit is initially relaxed and v_{S} (t) = 2e^{–0.5t}cos2t V for t ≥0^{+} is applied to it by using Laplace transforms.
 Find the total response of current in the circuit in Problem 8 if the circuit is initially relaxed at t = 0 and v_{s} (t) = 2e^{–0.5t}cos2t u(t–0.5) V is applied to it.
 Let f(t) be a periodic waveform with a period of T s. Let v(t) = f(t) u(t) and v_{p} (t) = f(t) [u(t)–u(t–T)]. That is, v(t) is the rightside of a periodic waveform and v_{p} (t) is one period of f(t). Develop an expression for Laplace transform of v(t) in terms of Laplace transform of v_{p}(t) using timeshifting theorem. What is the ROC of Laplace transform of v(t) ?
 Find the Laplace transforms of (i) a symmetric square wave of unit amplitude and unit period (ii) a rectangular pulse waveform of unit amplitude with first pulse located between 0 sec and 0.5 sec and pulses repeating every 2 sec using the result derived in Problem 10.
 Solve the system of differential equation
 If and y(0^{–}) = 0, y'(0^{–}) = 1 and y"(0^{–}) = –1, find y(t) by using Laplace transform technique.
 Let x(t) = 3tu(t) and y(t) = 2u(t). Find x(t)* y(t) by inverting X(s)Y(s) and verify by timedomain convolution.
 The impulse response of a linear timeinvariant circuit is 2e^{–0.05t} u(t). Find the zerostate response when input is 3e^{–0}^{.}^{1t} by convolution theorem on Laplace transforms.
 Let x(t) = t[u(t)– u(t–2)] and y(t) = [u(t)– u(t–2)]. Find x(t)* y(t) by inverting X(s)Y(s) and verify by timedomain convolution.
 Find the Laplace transform of (i) cosπt [u(t)– u(t–2)] (ii) 2sinh0.2t [u(t)– u(t–1)] by using shifting theorem.
 Find the Laplace transform of
 Let (i) Find and its inverse transform.(ii) Find and its inverse transform.
 Using Laplace transforms find the value of R in the circuit in Fig. 13.151 such that the damping factor of the circuit for voltage input is 0.2. Find the step response for v_{o}(t) with this R and verify the initial value and final value theorems on step response.
Fig. 13.151
 (i) Obtain the input impedance function and input admittance function for the circuits shown in Fig. 13.152 and prepare polezero plots for these immittance functions. (ii)Determine the zerostate input current as a function of time when input is u(t) V and verify initial value and final value theorems in each case. (iii) Determine the zerostate input voltage as a function of time when input is u(t) A. All circuit elements have unit values.
Fig. 13.152
 (a) Find the voltage transfer function V_{o}(s)/ V_{s} (s) and drivingpoint impedance function in the circuit in Fig. 13.153. (b) Prepare the polezero plot for both network functions. (c) Determine the step response for v_{o}(t) and i_{S}(t) (d) Verify initial value theorem and final value theorem on Laplace transforms in the case of v_{o}(t) and i_{S}(t).
Fig. 13.153
 The initial current in the inductor is 0.5 A and the initial voltage across the capacitor is 1 V in the circuit in Fig. 13.154. A single rectangular pulse of current is applied to the circuit as shown in the figure. Solve for v_{o}(t) by sdomain equivalent circuit method.
Fig. 13.154
 (i) Show that the voltage transfer function in the circuit in Fig. 13.155 is a real number if R_{1}C_{1} = R_{2}C_{2}. (ii) Obtain the input impedance function with R_{1}C_{1} = R_{2}C_{2}.
Fig. 13.155
 The impulse response of i_{x} in the circuit in Fig. 13.156 contains a real exponential term that has a time constant of 1.755 s. (i) Show the polezero plots for I_{x} (s) and V_{o} (s) when v_{S} (t) = u(t) V and the circuit is initially relaxed. (ii) Find i_{x} (t) and v_{o}(t) for t ≥ 0^{+} if v_{S}(t) = u(t) and both capacitors have 1 V across them with the bottom plate positive at t = 0^{–}and inductor has zero current at t = 0^{–}.
Fig. 13.156
 The impulse response of input current in the circuit in Fig. 13.157 contains a (1/6) δ(t) component. (i) Find the value of R. (ii) Find the drivingpoint impedance function and show its polezero plot. (iii) Find the timefunction describing the current delivered by source for t ≥ 0 ^{+} if v_{S} (t) = 2cos(2t + 30°) u(t), i_{1}(0^{–}) = 1A and i_{2}(0^{–}) = –1 A.
Fig. 13.157
 Find the zerostate response for v_{x}(t) by nodal analysis in sdomain in the circuit in Fig. 13.158 if v_{S1}(t) = 2u(t) V and v_{S1}(t) = 2e^{– t} u(t) V
Fig. 13.158
 Find the zeroinput response for v_{x}(t) by mesh analysis in sdomain in the circuit in Fig. 13.158 if first capacitor has 1 V across it at t = 0^{–}with the left plate positive and the second capacitor has 1 V across it with bottom plate positive at t = 0^{–}.
 Polezero plots of some transfer functions are shown in Fig. 13.159. Find the transfer functions and their impulse responses. The DC gain for all the transfer functions is unity.
Fig. 13.159
 Obtain the voltage transfer function in the circuit in Fig. 13.1510 in terms of k and determine the range of values for k such that the transfer function is stable. Use mesh analysis in sdomain.
Fig. 13.1510
 The value of RC product in the circuit in Fig. 13.1511 is 1μs. (i) Derive the voltage transfer function for the circuit and determine the maximum value of A for which the circuit will be stable. (ii) If the value of actually used is 1/10^{th} of this value, calculate the poles and zeros of the voltage transfer function and show the polezero plot. (iii) Sketch the frequency response plots for the above condition by geometrical interpretation of frequency response.
Fig. 13.1511
 The impulse response of a voltage transfer function in a linear timeinvariant circuit is found to contain two waveshapes e^{–0.5t} and e^{–t}^{} sin 2t. The steadystate step response of the same circuit is 0.7 V Find the voltage that must be applied to the circuit if the desired steadystate output is 10 sin (4t + 45°) V by geometrical calculations in splane. Assume that the transfer function has no zeros.
 Mark the polezero plot of the transfer function H (s) and obtain its frequency response plot by geometrical calculations in polezero plot.
 Obtain the voltage transfer function in the Opamp circuit in Fig. 13.1512 and show that the circuit can work as a bandpass filter. Select the values for R_{1}C_{1} and R_{2}C_{2} such that the filter has a centre frequency of 1000 rad/s and bandwidth of 100 rad/s.
Fig. 13.1512
 Sketch the frequency response plots for the transfer functions with polezero plots as in Fig. 13.1513 (a) through (g) approximately by using geometrical interpretation in splane. ‘r’ indicates the multiplicity number. The maximum gain is unity in all cases.
Fig. 13.1513