13. Filters and Attenuators – Network Analysis and Synthesis

13. Filters and Attenuators

CHAPTER OBJECTIVES

After carefully studying this chapter, you should be able to do the following:

Explain the basic function of a filter circuit.

Distinguish between a passive filter and an active filter.

Classify passive filters and explain function of each type of filter.

Explain the parameters of a filter.

Draw and explain basic filter networks in T and π sections.

Carryout analysis of filter networks in both T-section and π-section.

Make analysis of constant K-type or proto-type filters.

Solve problems on constant K-type filters.

Make comparison of parameters of constant K-type low-pass and high-pass filters.

State the limitations of constant K-type filters.

Modify constant K-type filters to obtain m-derived filters.

Analyse all types of m-derived filters.

Develop composite filters using constant K-type and m-derived filters.

13.1 INTRODUCTION

A filter blocks unwanted signals or noise signals and passes wanted or desired signals. A filter is basically frequency selective network that allows signals of a particular band of frequencies and rejects or attenuates signals of other frequencies. For example, a low-pass filter (LPF) passes low frequency signals and attenuates all frequencies above a selected cut-off frequency.

A high-pass filter (HPF) network will pass only those input signals whose frequencies are above the selected cut-off frequency.

A band-pass filter (BPF) passes signals of a selected frequency band, while a band-stop filter (BSF) blocks frequencies within its band. The band of frequencies passed by a filter is called pass band and the band of frequency that separates pass band and stop band of a filter is called cut-off frequency. Figure 13.1 shows various types of filters, their input and output wave forms and also their frequency response.

An LPF passes low frequency signals and blocks or attenuates signals that have frequencies above a given cut-off frequency (fc). The input wave form for an LPF, as shown in Figure 13.1(a), is composed of a low frequency signal and high frequency unwanted signal. The filter will allow the low frequency signal. This low frequency signal will appear at the output of the filter. However, the high frequency unwanted signal (noise signal) will be stopped or drastically reduced at the output.

Figure 13.1(a) Low-Pass Filter

The gain versus frequency (f) response graph for the filter and the attenuation versus frequency response have been shown in the Figure.

An HPF allows high frequency input signals and stops or attenuates the low frequency signals as shown in Figure 13.1(b).

Figure 13.1(b) High-Pass Filter

Wave forms of BPF and BSF have been shown in Figure 13.1(c) and (d), respectively.

Figure 13.1(c) Band-Pass Filter

Figure 13.1(d) Band-Stop Filter

13.1.1 Measurement in Decibels

The change in the output from a filter is measured in decibels (dB). The decibel is one-tenth of a bel. Since bel is large, the unit of decibel is used. Let the output voltage of a filter unit changes from v1 to v2 as the frequency changes. The change in power is expressed as the log of the ratio of power that changes from p1 to p2. The change in power (ΔP) is expressed as follows:

If power is assumed to be dissipated in the load resistance RL, then ΔP is given as in the following:

Assuming power as i2 RL, ΔP can be expressed as follows:

The change in output power, voltage and current can be measured using the above three expressions respectively.

13.2 TYPES OF FILTERS

Basically filters are of two types: Active filters and Passive filters.

Active filters are the filters having active elements like OP-AMP and transistor, in addition to resistor and capacitor. An active filter not only passes or stops a particular band of frequency but also amplifies the signal that passes through it.

Passive filters are made up of only passive components like inductor and capacitor. Such filters cannot amplify the signal that passes through them.

In this chapter, we will describe passive filters only.

13.3 CLASSIFICATION OF PASSIVE FILTERS

On the basis of functions they perform, passive filters are classified as follows:

1.Low-Pass Filters (LPF)

2.High-Pass Filters (HPF)

3.Band-Pass Filters (BPF)

4.Band-Stop Filters (BSF)

13.3.1 Low-Pass Filters

These are the filters that pass all the frequencies lower than the selected cut-off frequency fc and attenuate/stop/suppress the signals whose frequency is greater than fc. Attenuation characteristic for an ideal LPF is shown in Figure 13.2. From the attenuation characteristic of the LPF, it is clear that attenuation is zero in pass band and attenuation of signal is maximum in stop band.

It is to be noted that the characteristic shown is for an ideal LPF. For the low-pass filter the pass-band is from 0 to fc and stop-band is from fc to ∞.

Figure 13.2 Attenuation Characteristic of an LPF

13.3.2 High-Pass Filters

It is a filter that passes all the signals whose frequency is higher than the cut-off frequency and stops the signals whose frequency is less than fc.

The attenuation characteristics of HPF is shown in Figure 13.3.

Therefore, for a high-pass filter (HPF) the pass band is from fc to ∞ and the stop-band is from 0 to fc.

Figure 13.3 Attenuation Characteristic of an HPF

13.3.3 Band-Pass Filters

It is a filter that passes a particular band of frequencies and stops all other frequencies.

It has two cut-off frequencies: lower cut-off frequency (f1) and higher cut-off frequency (f2). This filter passes all those signals whose frequency lies inside the band f1 to f2 and stops all other frequency signals. The attenuation characteristics for BPF is shown in Figure 13.4. Therefore, for a BPF, the following are given:

Pass band: f1 to f2

Stop band: 0 to f1 and f2 to ∞.

Figure 13.4 Attenuation Characteristic of a BPF

13.3.4 Band-Stop or Band-Elimination Filter

It is a filter that stops particular band of frequencies and passes all other frequencies. It is just opposite to that of a band-pass filter (BPF). The attenuation characteristics for a band-stop filter is shown in Figure 13.5.

Figure 13.5 Attenuation Characteristic of a Band-Stop Filter

Therefore, for a BPF, the following are given:

Pass band: 0 to f1 and f2 to ∞

Stop band: f1 to f2.

13.4 PARAMETERS OF A FILTER

There are four important parameters that are necessary to analyse the performance of a filter network. They are as follows:

1.Propagation constant γ

2.Attenuation constant α

3.Phase shift constant β

4.Characteristic impedance Z0

13.4.1 Propagation Constant (γ)

For any two-port network terminated by characteristic impedance Z0, as shown in Figure 13.6, we can write the following:

Figure 13.6 A Two-port Network Terminated by Characteristics Impedance Z0

where γ is known as propagation constant. Propagation constant determines the propagation performance of any two-port network.

Moreover,

where α is real part of γ and is known as attenuation constant of the filter, and β is imaginary part of γ and is known as phase constant.

13.4.2 Attenuation Constant

Whenever a signal passes through a passive network/filter, it gets attenuated, because passive components like capacitors and inductors consume some of the signal energy. The attenuation constant determines the attenuation of the signal when it passes through the filter.

Units of Attenuation

Attenuation can be expressed in decibels or nepers.

Neper: It is defined as the natural log of the ratio of input current or voltage or power to the output current or voltage or power.

Decibel: It is defined as the ten times the common log of the ratio of input current/voltage/power and output current/voltage/power.

Decibel (D) can be written as follows:

Relation Between Nepers and Decibel

13.4.3 Phase Shift Constant (β)

When the signal passes through the filter, it gets some shift in phase. Phase shift constant signifies the phase shift in the signal when it passes through the filter.

Units of Phase Shift

The unit of phase shift is radians or degrees. The relation between radians and degrees can be written as follows:

13.4.4 Characteristic Impedance (Z0)

Characteristic impedance is the image impedance of a two-port network. For symmetric networks, the image impedance at port 1-1′ is equal to the image impedance at port 2-2′. They are equal to the characteristic impedance Z0.

13.5 FILTER NETWORKS

A filter is constructed from reactive elements such as inductors and capacitors. Filter produces no attenuation to pass band or transmission band and provides complete attenuation to all other frequencies called attenuation band or stop band. Filters are used in the communication systems.

Filters are made of symmetrical T- or π-Sections. T- and π-Sections are the combination of asymmetrical L networks, as shown in Figures 13.7 and 13.8, respectively.

13.5.1 Formation of Symmetrical T-Network

Figure 13.7 Symmetrical and T-section Network is Made from Asymmetrical L-networks

13.5.2 Formation of Symmetrical π-Network

Figure 13.8 Symmetrical π-Network Made from Asymmetrical L-networks

A ladder network is a cascade or series connection of many T- and π-Sections. There are two types of ladder networks.

These are shown in Figures 13.9 and 13.10, respectively.

It is formed by connecting many T-sections in series, as shown in Figure 13.9.

It is formed by connecting several sections of π-networks in cascade, as shown in Figure 13.10.

13.6 ANALYSIS OF FILTER NETWORKS

In this section, we will explain how the basic parameters of T- and π-networks are determined.

13.6.1 Symmetrical T-network

We know that the basic parameters of a filter network are its characteristic impedance, propagation constant, attenuation constant and phase shift constant. These are calculated as follows.

Characteristic Impedance (Z0)

If a two-port network is symmetrical, the image impedance Zi1 at port 1-1′ is equal to the image impedance Zi2 at port 2-2′ and that image impedance is called the characteristic impedance Z0.

Figure 13.11 A T-network

When the network is terminated with Z0 (characteristic impedance), then input impedance

Consider the T-network shown in Figure 13.11.

The input impedance of the T-network shown in figure is written as follows:

Now, substitute Zin = Z0 (using equation (13.7))

The expression is for characteristic impedance of a T-network.

Z0 (characteristic impedance) can also be expressed in terms of ZOC (open-circuit impedance) and ZSC (short-circuit impedance).

Let us find open-circuit impedance of T-network. From Figure 13.12, it is observed that the open-circuit impedance is written as follows:

Figure 13.12 Open-circuit Impedance of a T-network

From Figure 13.13, the short-circuit impedance of the T-network is calculated as follows:

Figure 13.13 Short-circuit Impedance of a T-network

Multiplying equations (13.9) and (13.10), we get the following:

Therefore, the characteristic impedance for a T-network, ZOT is given as follows:

Figure 13.14(a) A Symmetrical T-network Connected to a Load Impedance Z0

Propagation Constant (γ)

As shown in Figure 13.14, I1, I2, V1 and V2 are input current, output current, input voltage and output voltage, respectively.

Applying KVL in mesh II, as shown in Figure 13.14(a), we get the following:

Now, by definition, we get the following form:

Substituting this value in equation (13.12), the following form can be obtained:

Substituting this value in the equation, we get the following form:

Squaring both sides, the equation can be written as follows:

Dividing both sides by 2, we get the following:

This is the required expression for propagation constant of symmetrical T-network.

Attenuation Constant (α) and Phase Shift Constant (β)

From equation (13.14), propagation constant is a complex function represented as γ = α + . The real part α is a measure of change in the magnitude of the current or the voltage in the network and the imaginary part β is a measure of difference in the phase between the input and output currents or voltages.

Substituting γ = α+, in the equation, we get the following:

Case I: When α = 0, equation (13.15) can be written as follows:

Considering the magnitude,

Case II: To get expression for α, substitute β = π in equation (i)

Cut-off Frequency (fc)

We have

For pass band, α = 0

Substituting γ=jβ in the equation, we get the following form:

Now, the limits of cos θ are ±1

Therefore, in pass band, the equation can be written as follows:

Cut-off frequency can be obtained by substituting the following:

Figure 13.14(b)

The graphical representation of the equations is shown in the following Figure 13.14(b).

Summary of Equations of T-network

1. : characteristic impedance

2. : propagation constant

3. : attenuation constant

4. : phase shift

5.Z1 + 4Z2 = 0 : equation to obtain cut-off frequency.

13.6.2 Analysis of π-Network

Let us find the characteristic impedance for π-network shown in Figure 13.15.

Figure 13.15 A π-network

From the figure, the input impedance of the network is given as follows:

By definition, Zin = Z0, substituting this value in the equation, we get the following:

Multiplying and dividing by Z1, we get the following:

Now, we will prove the following:

where ZOC and ZSC are, respectively, the open-circuit and short-circuit impedance of the π-Network.

Open-circuit Impedance (Zoc) of π-Network

Figure 13.16 Determination of Open-circuit Impedance of a π-Network

Let us find short-circuit impedance of π-network:

From Figure 13.17 and its equivalent circuit shown in Figure 13.18, we get,

Figure 13.17 Determination of Short-circuit Impedance of a π-Network

Multiplying equation (13.17) and (13.18), we get the following:

Figure 13.18 Equivalent Circuit of Figure 13.17

Multiplying and dividing by Z1 we write the equation as follows:

The relation between the characteristic impedances of T-network ZOT and π-network Z are given as in the following:

We have

Substituting the value of from equation (13.19) in equation (13.20), we get the following form:

Hence, the product of characteristic impedance of T-network and π-network is equal to the product of series and shunt impedances.

Here, Z1 is equal to the total series impedance of the filter network and Z2 is equal to the total shunt impedance of the filter network.

13.6.3 Summary of Parameters of Filter Network

Parameters of filter network are given in Table 13.1.

Table 13.1 Parameters of T-network and π-Network

13.7 CLASSIFICATION OF FILTERS

So far, we have seen four types of filters, that is, LPF, HPF, BPF and BSF. When a number of signals are transmitted along a line, filters need to be used to separate them. For example, a speech channel using a carrier frequency of 50 kHz requires a bandwidth of say, 46 to 54 kHz. A band-pass filter (BPF) should pass freely signal of any frequency within this band of frequencies, that is, 46 kHz to 54 kHz and reject all other signals outside this range.

Thus, the output of a filter varies with the frequency. This is called the frequency response of a filter. If the range of frequency or the range of variation of the signal amplitude is large, logarithmic scale is used to plot the frequency response.

We have also known that filters are of two basic types: passive filters and active filters. In both cases, filters are designed to select or reject a band of frequencies. This is achieved by using series or parallel combination of R, L and C. In active filters, operational amplifiers or transistors and R, L and C are used.

Filters may be classified to be of constant K-type or m-Derived type.

Depending upon the relationship between Z1 and Z2, filters are classified as follows:

1.Constant K-type or prototype filters.

2.m-Derived filters

These are described in this section.

13.8 CONSTANT K-TYPE OR PROTOTYPE FILTERS

A constant K-type filter is a filter that satisfies the following relationship:

where Z1 is the series arm impedance.

Z2 is the shunt arm impedance.

K is the design impedance or nominal impedance or zero characteristic impedance.

Constant K-type filters can be of T-type or π-type. These filters are also called prototype filters because other complex type of filters can be derived from constant K-type filters. Constant K-type filters may be of low-pass type, high-pass type, band-pass type or band-stop type. These are discussed in the following sections.

13.8.1 Constant K-type Low-Pass Filters (LPF)

In LPF, the series element is inductor and shunt element is capacitor. T- and π-Section for constant K-type LPF are shown in Figure 13.19.

Figure 13.19 Constant K-type LPF

Expressions for Different Parameters of Constant K-type LPF

Design Impedance (K). In constant K-type LPF,

Total series impedance, Z1 = jωL and

Total shunt impedence,

Further, for constant K-type filters, we have, Z1Z2 = K2

Substituting the values of Z1 and Z2 in the equation, we get the following:

Cut-off Frequency (fc). We have limits of pass band as given in the following:

Now, we have fc = 0 and the condition .

Substituting j2 = −1, we have

Attenuation Constant (α). To find attenuation constant, we have the following relation:

Substituting Z1 = jωL and in the equation, we get the following:

Substitute ω = 2πf

Now, from equation (13.23), we have the following form:

Substituting this value in equation (13.24), we get the following forms:

and attenuation in pass band α = 0.

Phase Constant β. In stop band, β = π and in pass band,

The performance characteristic of a constant K-type LPF has been shown in Figure 13.20.

Figure 13.20 Performance Characteristic of a Constant K-type LPF

Characteristic Impedance (Z0). We have

Substituting Z1 = jωL and , we get the following:

Further, for LPF

and

Substituting the value of Z1 = jωL

Using equation (13.26), we get the following form:

Design Parameters. We have

Equating the two equations, that is, equation (iii) and (iv), we get the following form:

Substituting this value in equation (iii), the equation can be written as follows:

Summary of Constant K-type LPF

1.Design impedance:

2.Design parameters:

3.Cut-off frequency:

4.Attenuation:

5.Phase constant:

6.Characteristic impedance:

Solved Numerical on Constant K-type LPF

Example 13.1 Design a low-pass T-section filter having a cut-off frequency of 1.5 kHz to operate with a design impedance of 600 Ω.

Figure 13.21(a)

Figure 13.21(b)

Figure 13.21(c)

Solution: Given

Now, for constant K-type LPF we write the equation as follows:

The required design is shown in Figure 13.21a.

Example 13.2 Figure 13.21(b) shows a passive filter section. Find its cut-off frequency and characteristic impedance at f = 0.

Solution: Given filter is T-section LPF of Constant K-type. By comparing Figure 13.21(b) with Figure 13.21(c), we get,

Now, for LPF, fc can be written as follows:

Example 13.3 Design a constant K-type LPF having a cut-off frequency of 2000 Hz and a zero-frequency characteristic impedance of 200 Ω. Draw T- and π-Section of the filter.

Solution: Given

Now, for an LPF, L and C can be calculated as follows:

Required design is given in Figure 13.22.

Figure 13.22

Example 13.4 A constant K-type LPF composed of T-section has 63.6 mH inductance in each series arm and 0.088 μF in the shunt arm. Find (1) cut-off frequency and (2) attenuation in β at 5000 Hz.

Solution: The circuit is given in Figure 13.23. For comparison, the general design has also been shown.

Figure 13.23

1.Cut-off frequency

2.Attenuation at 5000 Hz, that is, f = 5000 Hz and we have

Example 13.5 Each arm of a symmetrical T-section LPF consists of 6 mH inductor, while the shunt arm is a 0.03 μF capacitor. Find the design impedance and cut-off frequency.

Solution: Given circuit is shown in Figure 13.24

Now for LPF, the design impedance

and

Figure 13.24

Example 13.6 Design the T- and π-Section of a constant K-type LPF having a cut-off frequency of 10 kHz and design impedance of 450 Ω. Further, find its characteristic impedance and phase constant at 5 kHz as well as determine the attenuation at 12 kHz.

Solution: Given

Now for LPF, L and C can be calculated as follows:

The constant X-type LPF for both T- and π-Sections are shown in Figure 13.25.

Figure 13.25

Characteristic impedance at f = 5 kHz = 1500 Hz

We now calculate Phase shift at f = 5 kHz = 5000 Hz as

Attenuation α at f = 12 kHz, is calculated as below

For constant K-type LPF, we have the value of α as,

13.8.2 Constant K-type High-Pass-Filters (HPF)

In an HPF, the series element is a capacitor and the shunt arm element is an inductor, that is, in case of HPF.

The circuit configuration of constant K-type HPF, both T-type and π-type, have been shown in Figure 13.26.

Figure 13.26 Circuit Configurations of HPF

Expression of parameters for HPF are given as follows.

Design Impedance (K)

We know that for constant K-type HPF, the following can be written as follows:

Substituting the values of Z1 and Z2, we get the following:

Cut-off Frequency (fc)

In pass band, we get the following:

That is,

Attenuation Constant (α)

In pass band, α = 0 and in stop band

We get

Now for HPF, we have the following:

Substituting this value in equation (13.31), we get:

Phase Constant (β)

In pass band, β = π

In stop band,

Performance of HPF

The variation of α with frequency and variation of β with frequency have been shown in Figure 13.27 and Figure 13.28, respectively.

Figure 13.27 Variation of Attenuation with Frequency

Characteristic Impedance (Z0)

We have the following:

Figure 13.28 Variation Phase Shift with Frequency

Substituting

Design Parameters

We have derived the following for HPF:

and

From equations (13.33) and (13.34), it is clear that C can be written as follows:

Substituting this value in equation (13.33), we get the value of L as follows:

Summary of Constant K-type HPF

Attenuation:

Phase constant:

Characteristic impedance:

13.8.3 Comparison of Constant K-type LPF and HPF

We have, so far, discussed the constant K-type low-pass and high-pass filters in both π-section and T-section. The design parameters of these filters are shown in a consolidated manner in Table 13.2.

Table 13.2 Comparative Table of Design Parameters of Constant K-type LPF and HPF

Solved Numericals on Constant K-type HPF

Example 13.7 Design a constant K-type HPF having a cut-off frequency of 5500 Hz and a design impedance of 750Ω. Draw T-section filter and π-Section filter.

Solution: Given

Now for an HPF, L and C can be calculated as follows:

Therefore, the T-section and π-section filters are as shown in Figure 13.29.

Figure 13.29

Example 13.8 A T-section HPF has a cut-off frequency of 3000 Hz and infinite frequency characteristic impedance of 500 Ω. Find characteristic impedance at 5000 Hz.

Solution: Given

Now for HPF, we get the following:

Example 13.9 Figure 13.30 shows an HPF section. Find the cutoff frequency and characteristic impedance at f = ∞.

Figure 13.30

Solution: Given circuit is a T-type HPF and we know the general circuit configuration for T-type HPF is shown in Figure 13.31.

Figure 13.31

Therefore, given

Characteristic impedance at f = ∞

Example 13.10 Figure 13.32 shows a high-pass filter section. Find cut-off frequency and characteristic impedance at f = ∞.

Figure 13.32

Solution: The general circuit configuration of π-Section high-pass filter is as shown in Figure 13.33.

By comparing the general configuration with the given π-Section we get

and

Characteristic impedance at f equal to ∞ is calculated as

Figure 13.33

Example 13.11 Design the T- and π-Section of a constant K-type high-pass filter having cutoff frequency of 20 kHz and design impedance of 450 Ω. Also, find its characteristic impedance and phase constant at 25 kHz as well as determine the attenuation at 4 kHz.

Solution: Given

Now for an HPF, we get the following:

Therefore, the designs are shown in Figure 13.34.

Figure 13.34

The characteristic impedance at f = 25,000 Hz

The phase constant at 25 kHz, that is, at f = 25,000 Hz.

We have

The attenuation at f = 4 kHz, that is, at 400 Hz

For constant K-type HPF, we have the following:

13.8.4 Constant K-type Band-Pass Filter

A band-pass filter can be obtained by connecting a LPF and a HPF in cascade as shown in Figure 13.35.

The circuit configuration of constant K-type BPF has been shown in Figure 13.36.

Figure 13.35 Block Diagram of a Constant K-type Band-pass Filter

T-section

After connecting the T-sections in cascade (series arm is made up of series resonant circuit and short arm is made up of parallel resonant circuit), we get a band-pass filter as shown in Figure 13.37.

Figure 13.36 A Band-pass Filter Obtained by Connecting in Series an LPF and an HPF

Figure 13.37 A Band-pass Filter in T-section Obtained by Connecting in Series an LPF and an HPF

π-Section

After connecting π-sections in cascade (series), we get band-pass filter in π-section as shown in Figure 13.38.

Figure 13.38 A Band-pass Filter in π-Section Obtained by Connecting in Series a π-Section of Constant K-type LPF and a Constant K-type HPF

Analysis of Constant K-type BPF

For series arm, resonant frequency = and for shunt arm, resonant frequency =

In band-pass filter, the resonant frequency of series arm is equal to the resonant frequency of shunt arm.

Design Impedance (K)

and Z2 (total shunt arm impedance)

Now, for constant K-type filters, Z1 Z2 = K2

Substituting the value of Z1 and Z2 in the equation, we get the following:

Using equation (13.35), that is, L1C1 = L2C2, we get the following:

Using equation (13.36), that is, using ,

Cut-off Frequencies

In BPF, there are two cut-off frequencies: low cut-off frequency (f1) and higher cut-off frequency (f2). Let us derive the expressions for f1 and f2 for pass band,

Cut-off frequency can be obtained by the following equations:

Multiplying both sides by Z1, we get

Case 1: Take Z1 = j2K.

Therefore,

This is the one cut-off frequency of BPF.

Case II: When Z1 = − j2K

Then

Taking the negative value, we get the following:

This is the second cut-off frequency of BPF.

Attenuation (α)

Substituting the values of Z1 and Z2 from equations (13.37) and (13.38)

Phase Constant (β)

Resonant Frequency (fo)

We know cut-off frequencies occur when Z1 + 4Z2 = 0 or, Z1 = −4Z2

Multiplying both sides by Z1, we get the following:

That is, Z1 at f1 = Z1 at f2 [∴ Magnitude of Z1 is same at f1 and f1, the difference is only of equation]

Now, when ω0 is resonant free

Characteristic Impedance (Z0)

For a band-pass filter, the following equations can be written as,

Design Parameters

Now, at f1, that is, higher cut-off frequency

Therefore, we get

where f1 is upper cut-off frequency and f2 is lower cut-off frequency.

Further, we have to calculate the value of K.

Substituting the value of C1 in the equation, we get the following form:

Further, we know

Substituting the value of C1 in the equations, we get the value of L1.

and we have

Summary of Band-Pass-Filter:

Circuit Configuration.

Figure 13.39 Band-pass Filter in T-section and in π-section

Design Impedance.

Cut-off Frequencies.

Resonant Frequency.

Phase Constant (β).

Characteristic Impedances.

Design Parameters.

Performance of Constant K-type BPF.

Figure 13.40 α Versus f Characteristic

Figure 13.41 β Versus f Characteristic

Numericals on BPF

Example 13.12 Design the T-section and π-section of a constant K-type BPF that has a pass band from 1500 to 5500 Hz and characteristic resistance of 200 Ω. Further, find resonant frequency of series and shunt arms.

Solution: Given f1 (lower cut-off frequency) = 1500 Hz; f2 (upper cut-off frequency) = 5500 Hz

and K = 200 Ω

Now, for a constant K-type BPF,

Required design is given as shown in Figure 13.42.

Figure 13.42

Now, resonant frequency of series arm

13.8.5 Constant K-type Band-Stop/Band-Elimination Filter

By interchanging the series and shunt arms of the band-pass-filter, we can obtain the band-stop-filter as has been shown in Figure 13.43(a) and (b).

Circuit Configuration

Figure 13.43 Band-stop Filter Developed from Band-pass Filter (a) T-section; (b) π-section

That is, series elements have been connected in parallel and shunt elements have been placed in series.

Further, in this filter, the components, that is, L1, C1, L2 and C2 are so selected that their resonant frequencies are same. This frequency is known as the resonant frequency or centre frequency or rms frequency of the filter. So, we will have

For a band-stop filter, Z1 and Z2 can be calculated as:

and

The various parameters of a band-stop filter are calculated as,

Design impedance (K): Know that for constant K-type filters, Z1 Z2 = K2

Substituting the value of Z1 and Z2 from equations (13.40) and (13.41), we get the following form:

Substituting, L1C1 = L2C2 from equation (13.39), in the above, we get

From equation (13.39), L1C1 = L2C2 or

Substituting this value in the equation, we get the value of K.

Cut-off frequency: Cut-off frequencies can be obtained from the equation

Multiplying both sides by Z1, we get

Case I: When Z1 = j2K (let us assume that it is at f1)

Then, using equation (13.40), we have the following form:

or Expression for lower cut-off frequency of BPF,

Case II: When Z1 = − j2K (let us assume that it is at f2), then the following equation can be obtained

Considering the positive value only, we get f2 as follows:

Expression for higher cut-off frequency,

Attenuation (α): We have

Using equations (13.40) and (13.41), we get

Putting, from equation (13.39), we get

Phase shift (β):

Resonant frequency (f0):

Characteristic impedance (Z0): We have

Using equations (13.40) and (13.41), we get the following

Design parameters:

Putting we get

Further, we have or

Summary of Band-Stop Filter

Circuit configuration:

Figure 13.44 Band-stop Filter in T and p-sections

Design impedance:

Cut-off frequencies:

Attenuation (α):

Phase shift (β):

Resonant frequency(f0):

Characteristic impedance (Z0):

Design parameters:

Example 13.13 Design a passive constant K-type BSF having a design impedance of 200 Ω and cut-off frequency 2000 Hz and 6000 Hz.

Solution: Given K= 200 Ω; f1 = 2000 Hz; f2 = 6000 Hz

Now, for a constant K-type BSF, we have the following form:

Required design is shown as in Figures 13.45 and 13.46.

Figure 13.45

Figure 13.46

13.8.6 Comparison of Constant K-type Filters

After discussing all the constant K-type filters, we now present their circuit parameters in a consolidated way in Table 13.3.

Table 13.3 Parameters of Constant K-type Filters

13.8.7 Limitations of Constant K-type Filters

The constant K-type filters have mainly the following two limitations:

1.The characteristic impedance of the filter circuit does not remain constant over the pass band and it is a function of frequency, that is, it varies with frequency.

We know filter gives ideal performance only if it is terminated by a resistance equal to the characteristic impedance.

Since in constant K-type filters, characteristic impedance is different at different frequencies, a mismatch occurs.

2.Its attenuation does not rise abruptly beyond the cut-off frequency as shown in Figure 13.47 and Figure 13.48 for a low-pass filter.

Figure 13.47 α Versus f Characteristic (Actual Case)

Figure 13.48 α Versus f Characteristic (Ideal Case)

To overcome these limitations of constant K-type filters, m-derived filters are used.

13.9 m-DERIVED FILTERS

In m-derived filters, prototype or constant K-type filter is used.

Following modifications are done to obtain m-derived filters from constant K-type filters.

1.The series impedance is multiplied by a factor m

2.The shunt impedance is divided by a factor m

3.An additional impedance of the opposite sign is added either in series or in parallel

The resonant frequency of the shunt arm is so chosen such that it is slightly higher than that of the constant K-type filter.

At the frequency f, the impedance of shunt branch is zero, that is, the attenuation is infinite. This can be seen from the expression for attenuation as

At f = f, Z2 = 0 and hence α = ∞. This produces very sharp cut-off or attenuation.

Here, m is a constant and 0 < m < 1.

13.9.1 m-Derived T-section

As mentioned earlier, m-derived T-section as shown in Figure 13.50 can be obtained from constant K-type filter shown in Figure 13.49 by making the following modifications:

Figure 13.49 Constant K-type

Figure 13.50 m-Derived T-Section

1.Multiply series impedance, that is, by m.

2.Impedance Z2 is replaced by , such that the value of Z0 (characteristic impedance) should be same for both the cases.

Now, for constant K-type, T-section, we have the following:

For m-derived T-section, we have the following:

Now, equating the characteristic impedances,

The m-derived T-section is as shown in Figure 13.51. Also transformation from constant K-type to m-derived type has been illustrated in Figure 13.52.

Figure 13.51 Final m-Derived T-Section

Figure 13.52 Transformation from Constant K-type to m-Derived Type Filter

13.9.2 m-Derived π-Section

The constant K-type filter and the corresponding m-derived filter in π-sections are shown in Figure 13.53.

Figure 13.53 Constant K-type Filter and its m-Derived in π-section

Therefore,

So, is the parallel combination of mZ1 and .

Therefore, final m-derived π-section is redrawn as shown in Figure 13.54.

Figure 13.54 m-Derived π-section Filter

13.9.3 m-Derived Low-Pass Filter

In a Low-pass filter, series element is inductor, that is, Z1 = jωL and shunt element is the capacitor, that is,

Representing Z1 as L and Z2 as the circuit configurations are drawn as shown in Figures 13.55 and 13.56.

Circuit Configurations

Figure 13.55 m-Derived Low-pass Filter in T-section

Figure 13.56 π-Section for m-Derived Low-Pass Filter

Analysis of m-derived LPF is given in the following:

Frequency of infinite attenuation (f):

We know (cut-off frequency for LPF)

Equating both sides, equation (13.46) we get

Attenuation (α):

For m-derived LPF,

Now, for m-derived LPF,

Substituting the value of (1-m2),

Alteration Constant, where fc > f >

Phase shift (β) is calculated as,

Now, substituting the value of from equation (13.48) in the equation, we get

Variation of attenuation (α) and phase shift (β ) with frequency are shown in Figure 13.57.

Figure 13.57 Shows the Variation of Attenuation and Phase Shift of m-Derived Filter

Therefore, m-derived LPF produces very sharp attenuation but the attenuation falls off beyond f. This is the disadvantage of m-derived filters.

Characteristic impedance: For T-section, the following equation can be obtained:

Using equation (13.47), we have the following form:

For π-Section, we get the following form:

13.9.4 Summary of m-Derived Low-Pass Filter

Circuit configurations:

Figure 13.58 m-Derived Low-pass Filters in T and π-Sections

Expression for m:

Cut-off frequency:

Design parameters:

Attenuation (α):

Phase shift ( β):

Characteristic impedance:

13.9.5 m-Derived High-Pass Filter

Circuit configurations:

Figure 13.59 m-Derived High-pass Filters in T and π-sections

Expression for f: At f, the transmission through filter is zero and attenuation is infinite.

Now,

Now, for constant K-type HPF, we get the following form:

That is, f < fC Expression for m: Now, we have

Attenuation (α):

Firstly, let us find from T-section:

Therefore,

Therefore, attenuation

Phase shift (β): We have

Using equation (13.50), we get the value of β.

The variations of α and β with frequency have been shown in Figure 13.60.

Figure 13.60 Variation of Attenuation and Phase-shift with Frequency

Characteristic impedance (Z0): We have the following form:

Using equation (13.49), we have

Now, Z1Z2 = K2

13.9.6 Summary of m-Derived HPF

Circuit configurations:

Figure 13.61 m-Derived High-pass Filters in T and π-sections

Expression for m:

Cut-off frequencies:

Component values:

Attenuation (α):

Phase shift (β):

Characteristic impedance:

13.9.7 Comparison of m-Derived LPF and HPF

After discussing m-Derived LPF and HPF, we are now in a position to present these filters with their design parameters in a consolidated way as in Table 13.4.

Table 13.4 Design Parameters of m-derived Low-pass and High-pass Filters

13.9.8 m-Derived Band-Pass Filter

Circuit configuration:

Figure 13.62 m-Derived Band-pass Filters Obtained by Cascading m-Derived LPF and HPF in T-sections

After connecting in cascade T-section of m-derived low-pass filter and T-section of m-derived high-pass filter, we will obtain the m-derived band-pass filter as has been shown in Figure 13.63.

Figure 13.63 T-Section of m-Derived BPF

After cascading T-section of m-derived low-pass filter with π-section of m-derived low-pass filter as shown in Figure 13.64(a), we get a π-section band-pass filter as has been shown in Figure 13.64(b).

Figure 13.64 A Band-pass Filter in π-section as in Figure (b) is Formed by Cascading T-section and π-section of m-Derived Low-pass Filters as Shown in Figure (a).

Various useful results for m-derived BPF are as follows:

Expression for m:

where f2 (upper cut-off frequency) and f1 (lower cut-off frequency)

are the frequency of infinite attenuation.

Component values:

13.9.9 m-Derived Band-Stop Filter

m-derived band-stop filters can be made from m-derived band-pass filters both in T-section and π-section after changing their series and shunt arms as shown in Figure 13.65 and Figure 13.66, respectively.

Figure 13.65 m-Derived Band-stop Filter in T-section Obtained from T-section Band-pass Filter

Figure 13.66 m-Derived Band-stop Filter in π-section Obtained from π-section of m-Derived Band-pass Filter

Some useful results for m-derived BSF are as follows:

Component values:

13.10 COMPOSITE FILTERS

We have just studied constant K-type and m-derived filters. It has been observed that a constant K-type filter does not give sharp cut-off/attenuation. We have also seen that an m-derived filter has a sharp-cut-off or attenuation, but its attenuation decreases for frequencies beyond f. The decrease in attenuation beyond f is a limitation of m-derived filters that can be overcome by cascading the m-derived filter with that of constant K-type filter having a rising attenuation beyond cut-off.

Attenuation characteristics of constant K-type filter cascaded with m-derived filter are shown in Figure 13.67.

Figure 13.67 Attenuation Characteristics (α versus f) of Constant K-type Filters

Therefore, the composite filter is a filter that consists of a number of m-derived and prototype sections with two terminating half sections. Terminating half sections are used for proper impedance matching. Terminating half sections are used at both source end and load end with m = 0.6.

A composite filter consists of the following:

1.One or more constant K-type sections to produce cut-off between the pass band and the stop band at a specified frequency fc

2.One or more m-derived sections to given infinite attenuation at a frequency f near to fc

3.Two terminating m-derived half section with m = 0.6 to provide matching with the source and the load.

Block diagram representation of a composite filter is shown in Figure 13.68.

Figure 13.68 A Composite Filter Shown in Block Diagram Form

13.10.1 Composite Low-Pass Filter

Figure 13.69 shows a composite low-pass filter T-section which consists of two terminating half sections at both source and load, one constant K-type section, and one m-derived section. Figure 13.70 shows a similar composition of a composite low-pass filter in π-section.

Figure 13.69 Composite LPF in T-section

Figure 13.70 Composite LPF in π-section

Procedure to Design Composite LPF

Step 1: Design constant K-type section.

For this, find and

and design T- or π-section using following configurations as shown in Figure 13.71.

Figure 13.71 Constant K-section of the Composite Filter in T and π-sections

Step 2: Design m-derived sections.

For this, firstly find , and then, find the component values and hence design T- or π-Section as shown in Figure 13.72.

Figure 13.72 m-Derived Sections of the Composite Low-pass Filter in T and π-sections

Step 3: Design terminating half section using m = 0.6 as shown in Figure 13.73.

Figure 13.73 Terminating Half sections of Composite Low-pass Filter in T and π-sections

Step 4: Assemble all the sections of the designed composite filter.

13.10.2 Composite High-Pass Filter

On the same line of designing composite LPF, a high-pass composite filter can be developed in both T and π-sections as shown in Figures 13.74 and 13.75, respectively.

Figure 13.74 A Composite High-pass Filter in π-section

Figure 13.75 A Composite High-pass Filter in π-section

13.11 ADDITIONAL SOLVED NUMERICALS ON FILTERS

13.11.1 Problems on m-Derived Low-pass Filters

Example 13.14 An m-derived LPF has a cut-off frequency of 2000 Hz. If m = 0.3, find the frequency of infinite attenuation.

Solution: Given

Now, for m-derived LPF, we get the following equation:

Substituting values,

Example 13.15 An m-derived LPF has fc = 1500 Hz and f = 2000 Hz. Find the parameter.

Solution: Given

Now, for m-derived LPF, we have the following equation:

Example 13.16 Design the T-section and π-section of m-derived LPF having a design impedance of 500 Ω, cut-off frequency 2200 Hz and frequency of infinite attenuation of 2500 Hz.

Solution: Given

Now, for m-derived LPF, we can write the equation as follows:

Now, for LPF, we can calculate L and C as follows:

Now, the design of T-type m-derived LPF and component values are as follows:

The required T-section is shown in Figure 13.76

Figure 13.76

The design of π-type m-derived LPF and the component values are given as follows:

The required π-Section is shown in Figure 13.77.

Figure 13.77

Example 13.17 Design an m-derived LPF with T-section having cut-off frequency of 7.2 kHz and infinite attenuation at 7.5 kHz and design impedance of 500 Ω.

Solution: Given

Now, for LPF, we calculate the value of L and C as follows:

and

Now, let us find the component values of T-section m-derived LPF as in the following:

The required design of m-derived low-pass filter in T-section is shown in Figure 13.78.

Figure 13.78

13.11.2 Problems on m-Derived High-pass Filters

Examples 13.18 An m-derived HPF has fc = 1400 Hz and f = 1300 Hz. Find the value of m.

Solution: Given

Now, for m-derived HPF, m can be calculated as follows:

Example 13.19 An m-derived HPF has fc = 1700 Hz and m = 0.6. Find the frequency of infinite attenuation.

Solution: Given

Now, for HPF, f can be calculated as follows:

Example 13.20 Design m-derived HPF having a design impedance of 300 W, cut-off frequency of 2000 Hz and frequency of infinite attenuation of 1700 Hz.

Solution: Given

Now, for m-derived HPF, the following can be obtained:

T-section:

Component values for T-section are as follows:

Therefore, the required design is shown in Figure 13.79. π-Section:

Figure 13.79

Firstly, let us find component values as in the following:

The required design is shown in Figure 13.80.

Figure 13.80

13.11.3 Problems on Composite Filters

Example 13.21 Design a composite LPF to work into 150Ω with cut-off frequency of 1.5 kHz and very high attenuation at 2 kHz.

Solution: Given

Step 1: Let us design constant K-type filter

Now, for constant K-type LPF, component values are as follows:

Therefore, designs are as shown in Figure 13.81:

Figure 13.81

Step 2: Let us design m-derived LPF and for m-derived LPF.

T-section: Component values are as follows.

The circuit designed is shown in Figure 13.82. π-Section:

Figure 13.82

Component values of the π-section as in Figure 13.83 are as follows:

Figure 13.83

Step 3: Let us design terminating half sections.

Terminating half section for T-type as shown in Figure 13.84: For terminating half sections, m = 0.6

Figure 13.84

Terminating half section for π-type as shown in Figure 13.85: Now,

Figure 13.85

Step 4: Now assemble the composite low-pass filter in T-section as shown in Figure 13.86 and in π-section as shown in Figure 13.87.

Figure 13.86

Figure 13.87

Example 13.22 Design and draw the T-section of a composite LPF having the following specifications:

1.Required cut-off frequencies: 2 kHz; 2. Design impedances: 500 Ω

Assume suitable value of m for m-derived full section and half section.

Solution: Given

Step 1 : Let us design constant K-type LPF.

For constant K-type LPF, the value of L and C can be calculated as follows:

Therefore, required T-section of constant K-type LPF is shown in Figure 13.88

Figure 13.88

Step 2: Design m-derived LPF (T-section)

For m-derived full section, if value of m’ is not given to us, then it is assumed as 0.4.

Therefore, let m’ = 0.4

Now,

Therefore, required T-section of m-derived LPF filter is shown in Figure 13.89.

Figure 13.89

Step 3: Design the terminating half section of T-section m-derived LPF with m = 0.6 as shown in Figure 13.90. Component values are as follows:

Figure 13.90

Step 4: Assemble the designed composite filter as shown in Figure 13.91.

Figure 13.91

Example 13.23 Design a composite HPF to work into 500 Ω resistances with cut-off frequency of 1500 Hz and with high attenuation of 1400 Hz.

Solution: Given

Now, for m-derived FPF, the following can be calculated:

Therefore, clearly, for full m-derived section, we will use m = 0.359 and for terminating half section of m-derived filter, we will use m = 0.6

Step 1 : Design constant K-type HPF.

For HPF, L and C can be calculated as follows:

Therefore, the required design of constant K-type HPF in T-section and in π-section are shown, respectively, in Figures 13.92 and 13.93.

Figure 13.92

Step 2: Design m-derived HPF.

For this, we will use m = 0.359.

T-section: Component values are as follows.

The design is shown in Figure 13.94.

π-Section: Component values are as follows.

Figure 13.93

Figure 13.94

Figure 13.95

The design is shown in Figure 13.95.

Step 3: Design of terminating half section with m = 0.6.

For T-section: Component values can be calculated as follows.

The design is shown in Figure 13.96.

Figure 13.96

For π-section: The calculation of component values are given as follows.

The design is shown in Figure 13.97.

Figure 13.97

Design of Composite HPF

Now we are able to make the composite high-pass filter with two terminating half sections, the constant-K section and a m-derived section in both T-section and π-section with the calculated values as has been shown in Figure 13.98 and Figure 13.99, respectively.

Figure 13.98 Composite High-pass Filter Network in T-section

Figure 13.99 Composite High-pass Filter Network in π-section

13.12 ATTENUATORS

Attenuators are used to reduce the signal level required in various applications in the field of electronics. For example, attenuators may be used as volume controllers in a radio station or to obtain a small voltage required in a testing of laboratory, etc.

13.12.1 Introduction

Attenuator is a network, normally a two-port resistive network that reduces the signal level to a desired value. It is inserted between a source and a load to reduce current, voltage and power.

Attenuators may be symmetrical or asymmetrical. In this section, we will discuss only symmetrical resistance attenuators. A resistance attenuator is a network consisting of resistors and is designed to reduce the voltage, current or power by known amount. The block diagram of an attenuator has been shown in Figure 13.100.

Figure 13.100

Units of Attenuation Produced by Attenuator

Attenuation produced by an attenuator may be expressed in decibels (dB) or in nepers (N)

Relation between decibels and nepers can be given as follows:

Basically, there are four types of attenuators. They are as follows:

1.T-type Attenuator;

2.π-Type Attenuator

3.Lattice Attenuator;

4.Bridged T-type Attenuator

These four types of attenuators are described in the following sections.

13.12.2 T-type Attenuator

Symmetrical T-type attenuator is shown in Figure 13.101.

Here, R0 is the design impedance.

Let us find some parameters of T-type attenuator.

Figure 13.101 T-type Attenuator

Attenuation in Nepers (N)

By definition,

Now, applying KVL in mesh II, we get the following:

Characteristic Impedance (R0)

(input resistance of T-type network from Figure)

By definition,

Design Parameters

From equation (13.53), we have the following:

Now, to find the value of R2, let us consider equation (13.52),

Now, substituting the value of R1 from equation (13.53) in the equation, we get the following:

Summary of T-type Attenuator

1.

2.

3.Design parameters are as follows:

Example 13.24 Find the characteristic impedance for the T-network shown in Figure 13.102.

Figure 13.102

Solution: Given a T-attenuator, let us compare it with the general network of T-type attenuator as shown in Figure 13.103.

Figure 13.103

We get

Now, for T-type attenuator, we have the following.

By definition,

Example 13.25 Design a T-type actuator to given attenuation of 20 dB and to work in a line of 200 Ω impedance.

Solution: Given D = 20 dB; R0 = 200 Ω

Now, we have

Further, for T-type attenuator, we get the following forms:

and

Therefore, the required attenuator is shown in Figure 13.104.

Figure 13.104

13.12.3 π-Type Attenuator

Circuit Configuration

Now, we should know that for an attenuator shown in Figure 13.105, the following can be written

Figure 13.105

To find the values of R1 and R2, firstly, let us bisect the π-network as shown in Figure 13.106.

Figure 13.106

Now,

Open-circuit impedance of half section

Now, we have general equations (13.55) and (13.56)

Substituting the value of R2 from equation (13.58) in equation (13.59), we get the following:

From equation (13.58), we have the following:

Summary

Therefore, the design equations for π-type attenuators are as follows:

Example 13.26 Design a π-type attenuator to give an attenuation of 3 N and to work in a line impedance of 375 Ω.

Solution: Given

Now, for π-type attenuator, we get

Substituting the values of N and R0, we get

Substituting N = 3 and R0 = 375 Ω, the following are calculated.

Therefore, the required π-type attenuator is shown in Figure 13.107.

Figure 13.107

13.12.4 Lattice Attenuator

Circuit configuration for symmetric lattice attenuator is shown in Figure 13.108.

Figure 13.108

Now, the network shown in Figure 13.108 can also be drawn as shown in Figure 13.109.

Figure 13.109

Figure 13.110

We calculate can the open-circuit impedance ROC from Figure 13.110.

Figure 13.111

To find short-circuit impedance, we refer to Figure 13.111

Further, from Figure 13.109,

Now, we have

Substituting the values of ROC and RSC from equations (13.62) and (13.63) in the equation, we get the following:

Now, from equation (13.64), we get the following:

Applying components and dividend, we get

Now, from equation (13.65), R2 can be calculated as follows:

Here, substituting the values of R1, we get the following equation:

Summary

Therefore, for lattice attenuator, design equations are as follows:

Example 13.27 Design a symmetrical lattice attenuation to give an attenuation of 20 dB and to work in a line of 200 Ω impedance.

Solution: Given

Now, for lattice attenuation, we get the following equation:

Substituting the value of N and R0 in the equation, we have the following equation:

Therefore, required lattice attenuator is shown in Figure 13.112.

Figure 13.112

13.12.5 Bridged T-type Attenuator

Circuit configuration: The circuit configuration of a bridged T-type attenuator has been shown in Figure 13.113.

Figure 13.113 Bridged T-network

Bisected half section of the bridged T-network is shown in Figure 13.114.

Figure 13.114 Bisected Half Section of Bridged T-network

Now, from Figure 13.114,

Circuit to find RSC is shown in Figure 13.115.

Now, from equation (13.67) and the circuit shown in Figure 13.115, we can write

Figure 13.115

Figure 13.116

Further, from equation (13.66), we get the following equation:

Therefore, we get the following set of equations:

Therefore, for bridged T-type attenuator, the equation is written as follows:

Example 13.28 Design a bridged T-type attenuator for attenuation of 10 dB and load resistance of 200 Ω.

Solution: Given

Now, a bridged T-type attenuator we can write as

Therefore, the required bridged T-type attenuator is shown in Figure 13.117.

Figure 13.117

13.13 MORE SOLVED PROBLEMS ON FILTERS AND ATTENUATORS

Example 13.28 Given two capacitors of 1 μF each and coil L of 10 mH, compute the following:

1.Cut-off frequency and characteristic impedance at infinite frequency for an HPF.

2.Cut-off frequency and characteristic impedance at zero frequency for an LPF.

Draw the constructed sections of filters using these elements.

Solution: Given HPF is shown in Figure 13.118.

Figure 13.118

ZOT at infinite frequency = R0

Given

LPF from the given capacitors and inductor can be drawn as shown in Figure 13.119.

Figure 13.119

Example 13.29 For the network shown in Figure 13.120, find the characteristic impedance

Figure 13.120

Solution: To find Z0:

Now, we know the following:

Example 13.30 A symmetrical T-section has the following data:

Determine the T-section parameters and represent the two-port network.

Solution: We have to find Z1 and Z2 of the network shown Figure 13.121

Figure 13.121

From the expression of ZSC and ZOC as above, we get

Substituting the given values, we can calculate the value of Z2. as

From equation (13.67), we obtain the following form:

Substituting the values of Z0C and Z2 in the equation, we get the following set of equations:

Therefore, required T-section parameters are as follows:

Example 13.31 Design a T-type symmetrical attenuation network that offers 40 dB attenuation with a load of 400 Ω.

Solution: Given attenuation is 40 dB

Series arm resistance can be given as follows:

Shunt arm resistance can be written as follows:

Figure 13.122

The designed network in shown in Figure 13.122.

Example 13.32 Design a constant K-type BPF section having a cut-off frequency of 2 kHz and 5 kHz and a nominal impedance of 600 Ω. Draw the configuration of the filter.

Solution: Now, for constant K-type BPF, we get the value of L1, L2, C1 and C2 as follows:

Therefore, the required configuration is shown in Figure 13.123.

Figure 13.123

Example 13.33 Design the symmetrical bridge T-type attenuator with an attenuation of 40 dB and an impedance of 600 Ω.

Solution: Given

Therefore,

We have the following form;

Therefore,

Therefore, the required symmetrical bridge T-type attenuator is shown in Figure 13.124.

Figure 13.124

Example 13.34 Design an m-derived T-section (high-pass) filter with a cut-off frequency fc = 20 kHz, f = 16 kHz and a design impedance R0 = 600 Ω.

Solution: We know for m-derived filter, the following equation can be obtained:

For HPF, L and C can calculated as follows:

The components of T-section (m-derived HPF) are given as in the following:

Therefore, required filter is shown in Figure 13.125.

Figure 13.125

Example 13.35 Design a symmetrical T-section having parameters of ZOC = 1000 Ω & ZSC = 600Ω.

Solution: Given

Now, for a symmetrical T-section, we have the following:

Therefore, required symmetrical T-section is shown in Figure 13.126.

Figure 13.126

Example 13.36 Design an m-derived LPF (T- and π-Section) having a design impedance of 500Ω and cut-off frequency 1500 Hz and an infinite attenuation frequency of 2000 Hz.

Solution: Given

We know, for an LPF, m can be calculated as follows:

Now, components of T-section (m-derived LPF) can be given as follows:

The values of elements of π-section (m-derived LPF) can be written as follows:

Therefore, configurations in T-section and in π-section are shown in Figure 13.127.

Figure 13.127

Example 13.37 Design a symmetrical bridged T-type attenuator to provide attenuation of 60 dB and to work into a line of characteristic impedance of 600 Ω.

Solution: Given

Therefore, required attenuator is shown in Figure 13.128.

Example 13.38 In a symmetrical T-network, if the ratio of input and output power is 6.76. Calculate the attenuation in neper and dB. Further, design this attenuator operating between source and load resistance of 100 Ω.

Figure 13.128

Solution: Given

Series arm resistance is given as follows:

Shunt arm resistance is calculated as in the following:

Therefore, the required symmetrical T-type network is shown in Figure 13.129.

Figure 13.129

1.Design an m-derived LPF (T- and π-Section) having a design resistance of R0 = 500 Ω and the cut-off frequency (fc) of 1500 Hz and an infinite attenuation frequency of 2000 Hz.

[Ans. As shown in Figure 13.130]

Figure 13.130

2.Design a symmetrical bridged T-type attenuator to provide attenuation of 60 dB and to work into a line of characteristic impedance 600 Ω.

[Ans. As shown in Figure 13.131]

Figure 13.131

3.Design a symmetrical bridge T-type attenuator with an attenuation of 40 dB and an impedance of 600 Ω

[Ans. As shown in Figure 13.132]

Figure 13.132

4.Design a m-derived T-section (HPF) with a cut-off frequency fc = 20 kHz, f = 16 kHz and a design impedance 600 Ω

[Ans. As shown in Figure 13.133]

Figure 13.133

5.Design a symmetrical T-section having parameters of ZOC = 1000 Ω and ZSC = 600 Ω

[Ans. As shown in Figure 13.134]

Figure 13.134

6.Design a T-type symmetrical attenuator that offers 40 dB attenuation with a load of 400 Ω

[Ans. As shown in Figure 13.135]

Figure 13.135

7.Design a constant K-type BPF section having cut-off frequencies of 2 kHz and 5 kHz and a nominal impedance of 600 Ω. Draw the configuration of the filter.

[Ans. As shown in Figure 13.136]

Figure 13.136

8.Given the network for HPF

Figure 13.137

Find the cut-off frequency and characteristic impedance at infinity frequency for an HPF.

[Ans. 1.12 kHz, 141.4 Ω]

9.Given π-Section LPF

Figure 13.138

Compute the cut-off frequency and characteristic impedance at zero frequency for an LPF.

[Ans. 4.5 kHz, 707 Ω]

10.Design a constant K-type BPF T-section having cut-off frequencies 2 kHz and 5 kHz and a nominal impedance of 600 Ω. Draw the configuration of the filter.

[Ans. As shown in Figure 13.139]

Figure 13.139

11.Design a prototype LPF, assuming cut-off frequency fc

13.What is an attenuator? Classify and state its applications.

14.Derive the relationships between neper and decibel units

15.Derive the expression for characteristic impedance of a symmetrical bridged T-type network

16.What are the disadvantages of the prototype filters? How are they overcome in composite filters?

17.Discuss the characteristics of a filter.

18.Differentiate between the attenuator and the amplifier. List the practical applications of attenuators.

19.With the help of frequency response curves, give the classification of filters.

20.Derive an expression for design impedance of a symmetrical T-type attenuator.

21.Explain how the reactance and impedance of an HPF varies with frequency.

22.What is an attenuator? Define the terms decibel and neper. Derive the relation between the two.

23.Write short notes on the following

1.LPF and its design

2.T- and π-Section attenuators.

24.What are the requirements of an ideal filter? Distinguish between an LPF and an HPF.

25.What is the propagation constant of a symmetrical T-section and symmetrical π-Section? What is its significance?

26.How do you choose the resonant frequency of an m-derived LPF and m-derived HPF.

27.What is a band-elimination filter?

MULTIPLE CHOICE QUESTIONS

1.A 26 dB output in watts is equal to

(a)2.4 w

(b)0.26 w

(c)0.156 w

(d)0.4 w

2.The characteristic impedance of an LPF in attenuation band is

(a)Purely imaginary

(b)Zero

(c)Complex frequency

(d)Real value

3.The real part of the propagation constant shows:

(a) Variation of voltage and current on basic unit

(b) Variation of phase shift/position of voltage

(c) Reduction in voltage, current value of signal amplitude.

(d) Reduction of only voltage amplitude.

4.The purpose of an attenuator is to

(a)Increase signal strength

(b)Decrease reflections

(c)Decrease value of signal strength

(d)Provide impedance matching

5 In a transmission line terminated by characteristic impedance, Z0

(a) There is no reflection of the incident wave

(b) The reflection is maximum due to termination.

(c) There are a large number of maximum and minimum on the line.

(d) The incident current is zero for any applied signal.

6.An attenuator is a

(a)Resistive network

(b)R-L network

(c)R-C network

(d)L-C network.

7.If ‘α’ attenuation in nepers, then

(a)Attenuation in dB =

(b)Attenuation in dB = 8.686α

(c)Attenuation in dB = 0.1α

(d)Attenuation is dB = 0.01α

8.For a constant K-type high-pass π-Section filter, characteristic impedance Z0 for f < fc is

(a)Resistance

(b)Inductive

(c)Capacitive

(d)Inductive or capacitive

9.An ideal filter should have

(a)Zero attenuation in the pass band

(b)Zero attenuation in the attenuation band

(c)Infinite attenuation in the pass band

(d)Infinite attenuation in the attenuation band

10.For an m-derived HPF, the cut-off frequency is 4 kHz and the filter has an infinite attenuation at 3.6 kHz, the value of m is

(a)0.436

(b)4.36

(c)0.34

(d)0.6

11.If ZOC = 120 Ω and ZSC = 30 Ω, the characteristic impedance is

(a)30 Ω

(b)60 Ω

(c)120 Ω

(d)150 Ω

12.If ZOC = 100 Ω and ZSC = 64 Ω, the characteristic impedance is

(a)400 Ω

(b)60 Ω

(c)80 Ω

(d)170 Ω

13.The frequency of infinite attenuation (f) of a low-pass m-derived section is

(a)Equal to fc

(b)f = ∞

(c)Close to but greater than fc

(d)Close to but less than fc of the filter

14.A constant K-type BPF has pass band from 1000 Hz to 4000 Hz. The resonance frequency of shunt and series arm is a

(a)2500 Hz

(b)500 Hz

(c)2000 Hz

(d)3000 Hz

15.A constant K-type low-pass T-section filter has Z0 = 600 Ω at zero frequency. At f=fc, the characteristic impedance is

(a)600 Ω

(b)0 Ω

(c)∞ Ω

(d)more than 600 Ω

16.In m-derived terminating half section, m is equal to

(a)0.1

(b)0.3

(c)0.6

(d)0.9

17.In a symmetrical T-type attenuator with attenuation N and characteristic impedance R0, the resistance of each series arm is equal to

(a)R0

(b)(N − 1) R0

(c)

(d)

18.For a transmission line, open-circuit and short-circuit impedance are 20 Ω and 5 Ω. The characteristic impedance of the line is

(a)10 Ω

(b)20 Ω

(c)25 Ω

(d)100 Ω

19.For a prototype LPF, the phase constant β in the attenuation band is

(a)α

(b)0

(c)

(d)

20.In the m-derived HPF, the resonant frequency is to be chosen so that it is

(a)Above the cut-off frequency

(b)Below the cut-off frequency

(c)Equal to the cut-off frequency

(d)Infinitely high

21.In a symmetrical π-section attenuator with attenuation N and characteristic impedance R0, the resistance of each shunt arm is equal to

(a)R0

(b)(N − 1) R0

(c)

(d)

22.Bridged T-type network can be used as

(a)Attenuator

(b)LPF

(c)HPF

(d)BPF

23.One neper is equal to

(a)0.8686 dB

(b)8.686 dB

(c)0.115 dB

(d)86.86 dB

24.Terminating half sections used in composite filters are built with the following value of m

(a)0.3

(b)0.8

(c)0.1

(d)0.6

25.A transmission line works as

(a)Attenuator

(b)LPF

(c)HPF

(d)None of (a), (b) and (c).

1.d

2.a

3.c

4.c

5.a

6.a

7.b

8.d

9.a

10.a

Because for m-derived HPF

11.b

Because

12.c

13.c

14.c

15.b

16.c

17.c

18.a

19.a

20.b

21.d

22.a

23.c

24.d

25.b