##### 3.4 *RLC* CIRCUITS

*RLC* circuits behave altogether differently when compared to either *RL* or *RC* circuits. *RLC* circuits are resonant circuits. These can be either series resonant circuits or parallel resonant circuits. A parallel *RLC* circuit is used as a tank circuit in an oscillator to generate oscillations (this is the feedback network that produces the phase shift of 180°). The *RLC* circuit is also used in tuned amplifiers to select a desired frequency band at the output. When a sinusoidal signal is applied as input to a series *RLC* circuit [see Fig. 3.23(a)], the frequency-vs-current characteristic is as shown in Fig. 3.23(b).

At resonance: *X*_{L} = *X*_{C}

**FIGURE 3.23(a)** An *RL*C series circuit with sinusoidal input; (b) the frequency-vs-current characteristic

**FIGURE 3.23(c)** A parallel *RL*C resonant circuit with sinusoidal input; (d) the frequency-vs-voltage characteristic

At resonance, the impedance is minimum, purely resistive and equal to *R*. The current at the resonant frequency, *f _{o}* is maximum, termed

*i*

_{max}. Let us now consider a parallel resonant circuit [see Fig. 3.23(c)] and its frequency-vs-

*v*characteristic, shown in Fig. 3.23(d). In the parallel resonant circuit, the impedance is maximum at resonance and hence, the voltage is maximum at

_{o}*f*. The figure of merit of a tuned circuit, denoted by

_{o}*Q*, is given as:

The larger the value of *Q*, the sharper the response characteristic of the tuned circuit.

#### 3.4.1 The Response of the *RLC* Parallel Circuit to a Step Input

Consider the *RLC* circuit shown in Fig. 3.24(a). Applying Laplace transforms, the above circuit is redrawn shown in Fig. 3.24(b). The impedance of the parallel combination of *Ls* and 1*/Cs* is:

**FIGURE 3.24(a)** *RLC* parallel circuit

**FIGURE 3.24(b)** The Laplace circuit of Fig. 3.24(a)

Multiplying the numerator and denominator by *Cs* we get:

Therefore,

The characteristic equation is:

The roots of this characteristic equation are:

Let *K*, the damping constant, be given by:

From Eq. (3.51), the resonant frequency of the tank circuit is:

From Eqs. (3.56) and (3.57):

Putting Eq. (3.58) in Eq. (3.55):

Therefore,

From Eq. (3.56) we have:

Therefore,

From Eq. (3.53):

For unit step voltage as input:

Applying partial fractions:

Putting *s* = *s*_{1}

Putting *s* = *s*_{2}

Applying inverse Laplace transform to both sides, we get:

From Eq. (3.59):

**(i)** If *K* = 0:

Using Eq. (3.60):

Let

Therefore,

If we substitute the value of *T _{o}* from Eq. (3.57) in Eq. (3.65):

Thus, *K* → 0, as *R* → ∞. Here, *K* can not be zero as assumed ideally, since *R* = ∞ means open-circuiting the resistance *R* in the circuit shown in Fig. 3.24(a), which is absurd because the excitation is not connected to the circuit when *R* = ∞. However, *R* can be made very large, in which case *K* becomes very small, though not zero as expected. The output has a smaller amplitude but is oscillatory in nature, as seen from Eq. (3.66). Thus, when a step is applied as input to the *RLC* circuit in Fig. 3.24(a), with *K* = 0 (practically very small), the response is un-damped oscillations, as shown in Fig. 3.24(c).

**FIGURE 3.24(c)** The response to *K* = 0

**(ii)** If *K <* 1, it is a case of under-damping as shown in Fig. 3.24(d). For this condition, from Eq. (3.59):

Therefore,

Multiply and divide Eq. (3.67) by *K* and substitute *K/T _{o}* = 1/4

*πRC*in it. The resultant equation is:

From Eq. (3.60):

where

The output response is an under-damped sinusoidal waveform. The oscillations die down after a few cycles, as shown in Fig. 3.24(d).

**FIGURE 3.24(d)** The response to *K <* 1

**(iii)** If *K* = 1, it is a case of critical damping. If we substitute the *K* value in the Eq. (3.59), then the roots are *s*_{1} = *s*_{2} = −2*π/T _{o}*. The roots are equal and real.

If the input is a step voltage:

Here,

Therefore,

Applying inverse Laplace transform on both sides:

where *x* = *t/T _{o}*:

Here:

The output response is shown in Fig. 3.24(e).

**FIGURE 3.24(e)** The response to *K* = 1

**(iv)** If *K >* 1, it is a case of over-damping. If *K >* 1, then the roots, from Eq. (3.59), are:

Using the binomial expansion for , we get:

Neglecting the higher order terms, the binomial expansion is reduced to (1 − 1/2*K*^{2}).

As

*K >* 1, (4*K*^{2} − 1) ≈ 4*K*^{2}

As

Therefore,

From Eq. (3.60):

As 4*K*^{2} 1

Therefore,

**FIGURE 3.24(f)** The response to *K >* 1

**FIGURE 3.24(g)** The response of an RLC parallel circuit for different values of *K*

Substituting Eq. (3.73) in Eq. (3.72):

The response is plotted in Fig. 3.24(f).

**(v)** From Fig. 3.24(a), if *R* = 0, the input step *V* is directly available at the output. From Eq. (3.56), *R* = 0 means *K* = ∞. Figure 3.24(g) shows a comparison of the response of an *RLC* parallel circuit for different values of K.

#### 3.4.2 The Response of the *RLC* Series Circuit to a Step Input

Consider a series *RL*C circuit, shown in Fig. 3.25(a). Applying Laplace transforms, the circuit in Fig. 3.25(a) can be redrawn as shown in Fig. 3.25(b). The total impedance *Z(s*) in this circuit is given by the relation:

**FIGURE 3.25(a)** An *RL*C series circuit

**FIGURE 3.25(b)** The Laplace circuit of Fig. 3.25(a)

But

Substituting Eq. (3.77) in Eq. (3.76):

For a step input *V*, from Eq. 3.75

The characteristic equation is:

The roots of this characteristic equation are:

- If either (
*R*/2*L*)^{2}*>*1/*LC*or*R >*, then both the roots are real and different, the circuit is over-damped. - If either (
*R*/2*L*)^{2}= 1/*LC*or*R*= , then both the roots are real and equal, the circuit is critically damped. - If either (
*R*/2*L*)^{2}*<*1/*LC*or*R <*, then both the roots are complex conjugate to each other; the circuit is under-damped. We have from Eq. (3.79):

Putting *s* = *s*_{1}:

Putting *s* = *s*_{2}:

Taking Laplace inverse:

**Case 1**: For over-damped circuit, (*R*/2*L*)^{2} *>* 1/*LC*

Let

From Eq. (3.82), we get:

**Case 2**: For a critically damped circuit, (*R*/2*L*)^{2} = 1/*LC*

Therefore,

Applying inverse Laplace transform on both sides, we get,

**Case 3**: For an under-damped circuit, (*R*/2*L*)^{2} *<* 1/*LC*

Therefore,

where,

**FIGURE 3.25(c)** The step response of an *RL*C series circuit

From Eq. (3.82), we get:

The step response is plotted using Eqs. (3.83), (3.84) and (3.85) in Fig. 3.25(c). For the under-damped condition, we see that there are damped oscillations.

#### 3.4.3 RLC Ringing Circuits

The circuit [see Fig. 3.24(a)], for *K <* 1, gives rise to damped oscillations. This means that the amplitude of the oscillations reduces with every cycle, causing the oscillations to die down ultimately. This circuit is called a ringing circuit. From Eq. (3.68) we have:

where

The amplitude of (*v*_{0}/*V*) is and is at the maximum when *e*^{−2πKx} = 1, when its value is . As *K <* 1, this value is 2*K*. This amplitude reduces to 1/*e* of the maximum value of 2*K* in *n* cycles, when 2*πKx* = 1.

if 2*πKx* = 1

If 2*πKx* = 1, the amplitude decreases to 1/*e* of the initial value. If we know the value of *Q*, we can find the number of cycles for which the circuit rings before the magnitude reduces to 1/*e* of the initial value. Let the amplitude reduce to 1/*e* of the initial value after *n* cycles, at *t*_{1}. At *t* = *t*_{1} = *nT _{o}, x*

_{1}=

*nT*=

_{o}/T_{o}*n*, when 2

*πKx*

_{1}= 1.

But

Therefore,

But

Therefore,

From Eqs. (3.87) and (3.89):

2*KQ* = 1

Also:

2*πKx*_{1} = 1 2*KQ* = 2*πKx*_{1} *Q* = *πx*_{1} = *πn* *n* = *Q/π*

If

The circuit will ring for 6 cycles before the amplitude reduces to 37 per cent of its initial value, as shown in Fig. 3.26.

As *K* becomes smaller, though the amplitude of the output signal decreases, the oscillations continue for more number of cycles. In a sinusoidal oscillator with an *LC* tank circuit, when the power (*V*_{CC}) is switched ON, oscillations develop if *K <* 1 and this condition is sustained as a positive feedback is employed in these oscillators.

**FIGURE 3.26** The output of the ringing circuit for *K <* 1

##### SOLVED PROBLEMS

*Example 3.9:* The periodic ramp with *T*_{1} = *T*_{2} = *τ/*2 shown in Fig. 3.27(a) is applied to a low-pass *RC* circuit. Find the equations to determine the steady-state output waveform. The initial voltage on the condenser is *V*_{1}. Find the maximum and minimum value of the voltage and plot the waveform.

**FIGURE 3.27(a)** The given periodic ramp input; (b) the given *RC* circuit

*Solution:*

From Eq. (3.35), we have the output for a low-pass *RC* circuit to ramp input as:

*v _{o}* (

*t*) = −

*ατ*+

*α t*+

*ατ e*

^{−t/τ}

If there is an initial voltage of *V*_{1} on *C*, Eq. (3.35) gets modified as:

*v _{o}* (

*t*) = −

*ατ*+

*α t*+

*ατ e*

^{−t/τ}+

*V*

_{1}

*e*

^{−t/τ}

For the ramp input, the slope *α* = *V/T*_{1}

Therefore,

The capacitor charges from *V*_{1} to *V*_{2} in time *T*_{1}. During *T*_{2}, when the input is zero, the capacitor discharges from *V*_{2} to *V*_{1}.

Given

*T*_{1} = *T*_{2} =

At

*t* = *T*_{1}, *v _{o}*(

*t*) =

*V*

_{2}

Using (1)

When the ramp input is reduced to zero *V*_{2} decays to *V*_{1}.

Therefore,

Substituting (2) in (3)

From (2)

*V*_{2} = −*V* + (*V*_{1} + 2*V*)*e*^{−0.5} = −*V* + *V*(0.21 + 2)0.606 *V*_{2} = 0.34 V

The minimum value of the output occurs when *dv _{o}*(

*t*)/

*dt*= 0.

*v _{o}*(

*t*) =

*αt*+

*V*

_{1}

*e*

^{−t/τ}−

*α τ*(1 −

*e*

^{−t/τ})

**FIGURE 3.27(c)** The output waveform

At *t* = 0.10*τ, v _{o}* is

*v*

_{o(min)}. Substituting this value in

*v*(

_{o}*t*),

*v*_{o}(min) = *V* × 2 × 0.10 − 2*V*(1 − 0.905) + 0.21*V* × 0.905 = 0.2 V

The output waveform is shown in Fig. 3.27(c).

*Example 3.10:* Prove that an *RC* circuit, shown in Fig. 3.28, behaves as a reasonably good integrator if *RC* 20 *T*, where *T* is the period of an input, *E _{m}* sin

*ω*t.

**FIGURE 3.28** The given low-pass *RC* circuit in which *RC* 20*T*

*Solution:* We have:

If the time constant (*RC*) is very large compared to *T* then:

And the phase angle *θ* = tan^{−1} (*ω/ω*_{2}) = tan − 1 (*τ/T*).

Given that *τ* 20*T*:

At *τ* = 20*T*,

*θ* = tan ^{−1}20 = 87.14°

If *τ* *T, θ* will be nearly 90°. The sinusoidal signal undergoes a phase change of 90°, as required in an integrator.

*Example 3.11:* For the circuit shown in Fig. 3.29, find:

- The expression for the voltage across the capacitor at each stage.
- The rise time of the output in terms of the time constant.
- The ratio of the rise time of the combination of the three sections to the rise time of a single section.

**FIGURE 3.29** Cascaded low-pass *RC* circuits

*Solution:*

- Time constant
*τ*=*RC***Stage 1:**Input =

*v*_{i1}=*V*Output =*v*_{o1}=*V*(1 −*e*^{−t/τ})**Stage 2:**Input =

*v*_{i2}=*v*_{o1}=*V*(1 −*e*^{−t/τ}) Output =*v*_{o2}We have,

**Stage 3:**Input =

*v*_{i3}=*v*_{o2}= Output =*v*_{o3}=*v*_{o}We have,

Therefore,

- Rise time is the time taken for the waveform to rise from 10 per cent to 90 per cent of its max value
*V*. Here, let at*t*=*t*_{2},*v*reach 90 per cent of_{o}*V*Therefore, at 90% of*V, V*= 0.9_{o}/VTherefore

The rise time of the cascaded network is calculated by finding

*t*_{2}and*t*_{1}. As the above equation is transcendental equation,* we should go for trial and error method to get the value of*t*_{2}. The value of*t*_{2}/*τ*at which, LHS value = RHS value, is the value we will take into account to calculate*t*_{2}.Let us start with

*t*_{2}/*τ*= 1*LHS*≠*RHS*If

*t*_{2}/*τ*= 3*LHS*≠*RHS*If

*t*_{2}/*τ*= 5*LHS*≠*RHS*If

*t*_{2}/*τ*= 6It means that the value of

*t*_{2}/*τ*lies between 5 and 6. If*t*_{2}/*τ*= 5.5*LHS*≠*RHS*If

*t*_{2}/*τ*= 5.3*LHS*=*RHS**t*_{2}= 5.3*τ*Let at

*t*=*t*_{1},*v*reach 10% of_{o}*V*10% of

*V*is:The value of

*t*_{1}*/τ*for which LHS = RHS is the value we will take into account to calculate*t*_{1}.Let us start with

*t*_{1}/*τ*= 1*LHS*≠*RHS*If

*t*_{1}/*τ*= 2*LHS*≠*RHS*It means that the value of

*t*_{1}/*τ*lies between 1 and 2.If

*t*_{1}*/τ*= 1.5*LHS*=*RHS*If

*t*_{1}/*τ*= 1.1*LHS*=*RHS*∴

*t*_{1}= 1.1*τ*Rise time =

*t*_{2}−*t*_{1}= 5.3*τ*− 1.1*τ*= 4.2*τ* - The rise time of three stages = 4.2
*τ*Rise timing single stage = 2.2

*τ*Ratio of the rise time of three stages to a single stage is equal to 4.2

*τ/*2.2*τ*= 1.9.

*Example 3.12:* An oscilloscope test probe shown in Fig. 3.30(a) has a cable capacitance of 50 pF. The input impedance of oscilloscope is 1 MΩ in parallel with 5 pF. Find the value of:

**FIGURE 3.30(a)** An attenuator probe

- The capacitor for better performance;
- Attenuation of the probe;
- The input impedance of the compensated probe.

*Solution:* The input resistance of the oscilloscope = *R _{i}* = 1 MΩ

The capacitance of the oscilloscope = *C* = 5 pF

The cable capacitance *C _{b}* = 50 pF

The given circuit can be redrawn as shown in Fig. 3.30(b).

*R*_{2} = 0.28 MΩ || 1 MΩ = 0.2187 MΩ

**FIGURE 3.30(b)** Equivalent circuit redrawn for Fig. 3.30(a)

- In an attenuator the condition for perfect compensation is:
*R*_{1}*C*_{1}=*R*_{2}*C*_{2}The value of capacitor

*C*_{1}for better performance is*C*= 2.559 pF. - The attenuation of the probe =
*α*= = 0.044 - The input impedance
*Z*=_{i}*R*_{1}+*R*_{2}= 4.7 × 10^{6}+ 0.2187 × 10^{6}= 4.9187 × 10^{6}Ω

*Example 3.13:* An ideal pulse generator produces a pulse of 250 *μ*s duration with amplitude of 10 V and the duty cycle is 25 per cent. Find the amplitude, rise time, fall time after passing through the network shown in Fig. 3.31(a).

**FIGURE 3.31(a)** The given attenuator network

*Solution:* The network given is an attenuator with *R*_{1} = 100 kΩ = 0.1 MΩ, *R*_{2} = 1 MΩ, *C*_{1} = 15 pF; *C*_{2} = 3 pF *t _{p}* = 250

*μ*s, per cent duty cycle = 25%.

Duty cycle = = 0.25 *T* = = 1000 × 10^{−6} s

The condition required for perfect compensation is *R*_{1}*C _{P}* =

*R*

_{2}

*C*

_{2}.

*C*_{1} = 15 pF

As *C*_{1} *< C _{P}*, the attenuator is said to be under compensated.

Initial response

Final response

**FIGURE 3.31(b)** The output waveform

Fall time = Rise time = 3.6 × 10^{−6} s

The output waveform is shown Fig. 3.31(b).

##### SUMMARY

- A low-pass circuit transmits low-frequency components and attenuates high frequencies.
- The cut-off frequency of a low-pass or high-pass circuit is the frequency at which the output is 1/ (or 0.707) times the maximum output.
- The upper cut-off frequency of a low-pass circuit,
*f*_{2}is given by 1/2*πRC*. *f*_{2}is the bandwidth of the low-pass circuit.- When a step is applied as input to a low-pass circuit, it takes a finite time for the output to reach the steady-state value. Rise time is defined as the time taken for the output to reach from 10 per cent of its final value to 90 per cent of its final value.
- The rise time
*t*is given as 2.2_{r}*τ (*= 0.35*/f*_{2}). - If
*n*stages of low-pass circuits having rise times*t*_{r1},*t*_{r2}, …,*t*are cascaded, then the rise time of the cascaded stages is given by_{rn}*tr*= . - A low-pass circuit behaves as an integrator if the time constant of the circuit is significantly greater than the time period of the input signal.
- Attenuators are essentially resistive networks employed to reduce the amplitude of the signal.
- An uncompensated attenuator is one in which the output depends on frequency.
- A compensated attenuator is one in which the output is independent of frequency.
- In an ideal attenuator, the output is independent of frequency.
- An attenuator is called a perfectly compensated attenuator if
*v*(0+) =_{o}*v*(∞)._{o} - An attenuator is treated as an over-compensated attenuator if
*v*(0+) >_{o}*v*(∞)._{o} - An attenuator in which
*v*(0+) <_{o}*v*(∞) is called an under-compensated attenuator._{o} - At
*t*= 0+, as capacitors behaves as short circuit,*v*(0+) is calculated by considering capacitors in an attenuator._{o} - At
*t*= ∞, as the capacitors are fully charged, they behave as open circuit for dc. Hence,*v*(∞) is calculated considering only the resistances in an attenuator._{o} - A ringing circuit is an
*RL*C circuit which produces (nearly) undamped oscillations.

##### MULTIPLE CHOICE QUESTIONS

- If two stages of identical low-pass
*RC*circuits are cascaded, the rise time of the cascaded stage is:- 2
*t*_{r} - 0.2
*t*_{r}

- 2
- A step input with certain pulse width is given to a low-pass
*RC*circuit to transmit the same. To minimize the distortion: - The general solution for a single time constant circuit having initial and final values
*v*and_{i}*v*respectively is given by:_{f}*v*=_{o}*v*+ (_{i}*v*−_{f}*v*)_{i}*e*^{−t/τ}*v*=_{o}*v*+ (_{i}*v*−_{i}*v*)_{f}*e*^{−t/τ}*v*=_{o}*v*+ (_{f}*v*−_{i}*v*)_{f}*e*^{−t/τ}*v*=_{o}*v*+ (_{f}*v*−_{f}*v*)_{i}*e*^{−t/τ}

- The following type of attenuator will faithfully reproduce the signal which appears at its input terminals.
- Under-compensated
- Over-compensated
- Perfectly compensated attenuator
- Uncompensated attenuator

- The transmission error,
*e*when a ramp input with_{t}*RC**T*is applied to a low-pass circuit is given by: - Integrators are invariably preferred over differentiators in analogue computer applications due to the following reason;
- The gain of an integrator decreases with frequency
- It is easier to stabilize
- It is more convenient to introduce initial conditions
- All of the above

- The output of an integrator to a square wave input is:
- Triangular wave
- Square wave
- Quadratic response
- Spikes

- The time required for the capacitor to get charged completely is nearly __________ the time constant.
- five times
- one time
- equal to
- two times

- When
*R*_{1}*C*_{1}=*R*_{2}*C*_{2}, the attenuator is said to have achieved:- Perfect compensation
- Over-compensation
- Under-compensation
- No compensation

- When
*R*_{1}*C*_{1}*> R*_{2}*C*_{2}, the attenuator is said to have achieved:- Perfect compensation
- Over-compensation
- Under-compensation
- No compensation

- When
*R*_{1}*C*_{1}*< R*_{2}*C*_{2}, the attenuator is said to have achieved:- Perfect compensation
- Over-compensation
- Under-compensation
- No compensation

##### SHORT ANSWER QUESTIONS

- Obtain the expression for the bandwidth of a low-pass circuit.
- The input to a low-pass
*RC*circuit is a step of*V*. Obtain the expression for the output voltage. - What is meant by the rise time of a pulse? Obtain the expression for the rise time of a low-pass
*RC*circuit. - The input to a low-pass circuit is a pulse of duration
*t*and magnitude_{p}*V*. Plot its output when the time constant is very small and when the time constant is very large. - A symmetric square wave is applied as input to the low-pass circuit. Plot the output waveforms for different time constants.
- A ramp is applied as input to a low-pass circuit. Derive the expression for the transmission error,
*e*._{t} - Show that a low-pass circuit can be used as an integrator if the time constant of the circuit is significantly larger than the time period of the input signal.
- What is an attenuator? Explain the drawbacks of an uncompensated attenuator.
- How can an uncompensated attenuator be modified as a compensated attenuator?
- Compare the responses of perfectly compensated, under-compensated and over-compensated attenuators.
- Illustrate, with a suitable diagram, how a compensated attenuator can be used as a CRO probe.
- Explain under which condition an
*RL*C circuit behaves as a ringing circuit.

##### LONG ANSWER QUESTIONS

- If a symmetric square wave referenced to 0 voltage is the input to a low-pass circuit, derive the expression for the steady-state voltage levels at the output. Plot the typical waveforms.
- A ramp
*v*=_{i}*αt*is applied as input to a low-pass*RC*circuit. Derive the expression for the output voltage. Plot the typical waveforms. - An exponential signal
*v*=_{i}*V*(1 −*e*^{−t/τ1}) is the input to a low-pass*RC*circuit. Derive the expressions for the output when (i)*τ*=*τ*_{1}and (ii)*τ*≠*τ*_{1}. - What is an uncompensated attenuator and its major limitations? Implement a compensated attenuator. Explain the conditions under which this is called a perfectly compensated, over-compensated and under-compensated attenuator.
- Obtain the response of an
*RL*C circuit under critically damped conditions.

##### UNSOLVED PROBLEMS

- A pulse with zero rise time, an amplitude of 10 V and duration 10
*μ*s is applied to an amplifier through a low-pass coupling network shown in Fig. 3p.1. Plot the output waveform to scale under the following conditions:*f*_{2}= 20 MHz,*f*_{2}= 0.2 MHz.

**FIGURE 3p.1**The given coupling network with input - A ramp shown in Fig. 3p.2 is applied to a low-pass
*RC*circuit. Draw to scale the output waveform for the cases:*T*= 0.2*RC*,*T*= 10*RC*.

**FIGURE 3p.2**The given input to the low-pass*RC*circuit - The input to a low-pass
*RC*circuit is periodic and trapezoidal as shown in Fig. 3p.3. Find and sketch the steady-state output if*RC*= 10*T*_{1}= 10*T*_{2}.**FIGURE 3p.3**The given input to the low-pass*RC*circuit - A 1 kHz symmetrical square wave of peak-to-peak voltage 20 V, as shown in Fig. 3p.4, is applied to a low-pass
*RC*circuit with*R*= 100 Ω,*C*= 1*μF*. Sketch the output waveform to scale by determining the corner voltages.**FIGURE 3p.4**The given input to the low-pass*RC*circuit - The input waveform shown in Fig. 3p.5 is applied to a low-pass
*RC*network. Sketch the waveform of the voltage across the capacitor to scale for two cycles. The time constant of the*RC*circuit is 0.11 ms.**FIGURE 3p.5**The given input to a low-pass*RC*circuit - The periodic waveform shown in Fig. 3p.6 is applied to an
*RC*integrating network whose time constant is 15*μ*s. Sketch the output and calculate the maximum and minimum values of output voltage with respect to the ground.**FIGURE 3p.6**The given input to a low-pass*RC*circuit - The waveform shown in Fig. 3p.7 is applied to a low-pass
*RC*circuit. Sketch the output waveform to scale by determining the corner voltages for the following cases: (1)*RC*= 20*T*, (2)*RC*=*T*/20.**FIGURE 3p.7**The given input to the low-pass*RC*circuit - A square wave extends ±0.5V with respect to ground. The duration of the positive section is 0.1s and the negative duration is 0.2 s, as shown in Fig. 3p.8. This waveform is applied as a input to a low-pass
*RC*circuit whose time constant is 0.2 s, sketch the steady-state output waveform to scale, and find the maximum and minimum values of the output.**FIGURE 3p.8**The given input to the low-pass*RC*circuit - A pulse of amplitude 10 V and pulse width of 10
*μ*s, as shown in Fig. 3p.9, is applied to an*RC*circuit with*R*= 100 kΩ and*C*= 0.1*μ*F. Sketch the capacitor voltage waveform to scale.**FIGURE 3p.9**The given input to the low-pass*RC*circuit - An oscilloscope has an input impendence of 10 MΩ in parallel with 25 pF. Design a 10:1 voltage divider for the oscilloscope. Determine the values of the input resistance and the capacitance for the compensated divider.
- The attenuator shown Fig. 3p.10 has
*R*_{1}=*R*_{2}= 1 MΩ,*C*_{2}= 50 pF. Find the magnitudes of the initial and final responses and draw the output waveform to scale for*C*_{1}= 50 pF, 75 pF and 25 pF.**FIGURE 3p.10**The given attenuator circuit - The periodic ramp with
*T*_{1}=*T*_{2}= 2*τ*, shown in Fig. 3p.11, is applied to a low-pass*RC*circuit. Find the equations to determine the steady-state output waveform. The initial voltage on the condenser is*V*_{1}. Find the maximum and minimum value of the voltage and plot the waveform.**FIGURE 3p.11**The given periodic ramp input and the low-pass circuit - For a low-pass
*RC*circuit, it is desired to pass a 3 ms sweep for a ramp input with less than 0.4 per cent transmission error. Calculate the upper3-dB frequency. - A step input of 20 V is applied to an
*RC*integrating circuit. Calculate the upper 3-dB frequency and the value of resistance, if the rise time and capacitor values are 100*μ*s and 0.28*μ*F, respectively.