Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
6th Edition
ISBN: 9780078028229
Author: Charles K Alexander, Matthew Sadiku
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 14, Problem 57P

Determine the center frequency and bandwidth of the band-pass filters in Fig. 14.88.

Chapter 14, Problem 57P, Determine the center frequency and bandwidth of the band-pass filters in Fig. 14.88. , example  1

Chapter 14, Problem 57P, Determine the center frequency and bandwidth of the band-pass filters in Fig. 14.88. , example  2

(a)

Expert Solution
Check Mark
To determine

Find the center frequency and bandwidth of the band-pass filter shown in Figure 14.88(a).

Answer to Problem 57P

The value of the center frequency (ω0) and the bandwidth (B) of the band-pass filter shown in Figure 14.88(a) is 1rads and 3rads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.88(a) in the textbook.

Formula used:

Write the expression to calculate the impedance of the passive elements resistor, inductor and capacitor in s-domain.

ZR(s)=R        (1)

ZL(s)=sL        (2)

ZC(s)=1sC        (3)

Here,

R is the value of the resistor,

L is the value of the inductor, and

C is the value of the capacitor.

Calculation:

The given circuit is drawn as Figure 1.

Fundamentals of Electric Circuits, Chapter 14, Problem 57P , additional homework tip  1

Use equation (1) to find ZR1(s).

ZR1(s)=R1

Use equation (1) to find ZR2(s).

ZR2(s)=R2

Use equation (3) to find ZC1(s).

ZC1(s)=1sC1

Use equation (3) to find ZC2(s).

ZC2(s)=1sC2

The s-domain circuit of the Figure 1 is drawn as Figure 2.

Fundamentals of Electric Circuits, Chapter 14, Problem 57P , additional homework tip  2

Write the general expression to calculate the transfer function of the circuit in Figure 2.

H(s)=VoVs        (4)

Here,

Vo is the output response of the system, and

Vs is the input response of the system.

Refer to Figure 2, the series connected impedances ZC2(s) and ZR2(s) are connected in parallel with the impedance ZC1(s) and this parallel combination is connected in series with the impedance ZR1(s).

Therefore, the equivalent impedance is calculated as follows.

Zeq(s)=ZR1(s)+(ZC1(s)(ZC2(s)+ZR2(s)))

Substitute R1 for ZR1(s), R2 for ZR2(s), 1sC1 for ZC1(s) and 1sC2 for ZC2(s) in above equation to find Zeq(s).

Zeq(s)=R1+(1sC1(1sC2+R2))        (5)

Refer to the given data. The value of the resistors R1 and R2 is same and the value of the capacitors C1 and C2 is same. Therefore, R1=R2=R and C1=C2=C.

Substitute R for R1 and R2, C for C1 and C2 in equation (5) to find Zeq(s).

Zeq(s)=R+(1sC(1sC+R))=R+(1sC)(1sC+R)(1sC+1sC+R)=R+(1sC)(1+sRCsC)(2sC+R)=R+(1+sRCsC)sC(2+sRCsC)

Simplify the above equation to find Zeq(s).

Zeq(s)=R+(1+sRC)sC(2+sRC)=R+(1+sRC)sC(2+sRC)=2sRC+s2R2C2+1+sRCsC(2+sRC)=1+3sRC+s2R2C2sC(2+sRC)

The reduced circuit of Figure 2 is drawn as Figure 3.

Fundamentals of Electric Circuits, Chapter 14, Problem 57P , additional homework tip  3

Refer to the Figure 3, the current through the circuit is expressed as,

I=VsZeq(s)

Apply current division rule on Figure 2 to find I1.

I1=ZC1(s)ZC1(s)+ZC2(s)+ZR2(s)I

Substitute VsZeq(s) for I, R2 for ZR2(s), 1sC1 for ZC1(s) and 1sC2 for ZC2(s) in above equation to find I1.

I1=(1sC1)(1sC1)+(1sC2)+R2(VsZeq(s))

Substitute 1+3sRC+s2R2C2sC(2+sRC) for Zeq(s), R for R2, C for C1 and C2 in above equation to find I1.

I1=((1sC)(1sC)+(1sC)+R)(Vs(1+3sRC+s2R2C2sC(2+sRC)))=((1sC)2sC+R)(Vs(sC(2+sRC)1+3sRC+s2R2C2))=((1sC)2+sRCsC)(Vs(sC(2+sRC)1+3sRC+s2R2C2))=Vs2+sRC(sC(2+sRC)1+3sRC+s2R2C2)

Refer to Figure 2, the output voltage Vo is calculated as,

Vo=I1R2

Substitute Vs2+sRC(sC(2+sRC)1+3sRC+s2R2C2) for I1 and R for R2 in above equation to find Vo.

Vo=(Vs2+sRC(sC(2+sRC)1+3sRC+s2R2C2))(R)=(R2+sRC(sC(2+sRC)1+3sRC+s2R2C2))Vs

Rearrange the above equation to find VoVs.

VoVs=R(2+sRC)(sC(2+sRC)1+3sRC+s2R2C2)=(sRC1+3sRC+s2R2C2)

Substitute (sRC1+3sRC+s2R2C2) for VoVs in equation (4) to find H(s).

H(s)=(sRC1+3sRC+s2R2C2)=sRCR2C2(s2+3sRCR2C2+1R2C2)=(33)sRC(s2+3RCs+1R2C2)=13[(3RC)ss2+(3RC)s+(1R2C2)]

Compare the denominator factor of above equation with the standard quadratic equation (s2+ω0Qs+ω02).

ω02=1R2C2        (6)

ω0Q=3RC        (7)

Substitute 1Ω for R and 1F for C in equation (6) to find ω02.

ω02=1(1Ω)2(1F)2=11Ω2F2=11Ω2(s2Ω2) {1F=1s1Ω}=11s2

Take square root on both sides of the above equation to find ω0.

ω02=11s2ω0=1rads

Write the expression to calculate the bandwidth of the band-pass filter.

B=ω0Q

Substitute B for ω0Q, 1Ω for R and 1F for C in equation (7) to find B.

B=3(1Ω)(1F)=31Ω(sΩ) {1F=1s1Ω}=31s=3rads

Conclusion:

Thus, the value of the center frequency (ω0) and the bandwidth (B) of the band-pass filter shown in Figure 14.88(a) is 1rads and 3rads respectively.

(b)

Expert Solution
Check Mark
To determine

Find the center frequency and bandwidth of the band-pass filter shown in Figure 14.88(b).

Answer to Problem 57P

The value of the center frequency (ω0) and the bandwidth (B) of the band-pass filter shown in Figure 14.88(b) is 1rads and 3rads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.88(b) in the textbook.

Calculation:

The given circuit is drawn as Figure 4.

Fundamentals of Electric Circuits, Chapter 14, Problem 57P , additional homework tip  4

Use equation (1) to find ZR1(s).

ZR1(s)=R1

Use equation (1) to find ZR2(s).

ZR2(s)=R2

Use equation (2) to find ZL1(s).

ZL1(s)=sL1

Use equation (2) to find ZL2(s).

ZL2(s)=sL2

The s-domain circuit of the Figure 4 is drawn as Figure 5.

Fundamentals of Electric Circuits, Chapter 14, Problem 57P , additional homework tip  5

Write the general expression to calculate the transfer function of the circuit in Figure 5.

H(s)=VoVs        (8)

Refer to Figure 5, the series connected impedances ZR2(s) and ZL2(s) are connected in parallel with the impedance ZR1(s) and this parallel combination is connected in series with the impedance ZL1(s).

Therefore, the equivalent impedance is calculated as follows.

Zeq(s)=ZL1(s)+(ZR1(s)(ZR2(s)+ZL2(s)))

Substitute R1 for ZR1(s), R2 for ZR2(s), sL1 for ZL1(s) and sL2 for ZL2(s) in above equation to find Zeq(s).

Zeq(s)=sL1+(R1(R2+sL2))        (9)

Refer to the given data. The value of the resistors R1 and R2 is same and the value of the inductors L1 and L2 is same. Therefore, R1=R2=R and L1=L2=L.

Substitute R for R1 and R2, L for L1 and L2 in equation (9) to find Zeq(s).

Zeq(s)=sL+(R(R+sL))=sL+R(R+sL)(R+R+sL)=sL+R2+sRL(2R+sL)=2sRL+s2L2+R2+sRL2R+sL

Simplify the above equation to find Zeq(s).

Zeq(s)=R2+3sRL+s2L22R+sL

The reduced circuit of Figure 5 is drawn as Figure 6.

Fundamentals of Electric Circuits, Chapter 14, Problem 57P , additional homework tip  6

Refer to the Figure 6, the current through the circuit is expressed as,

I=VsZeq(s)

Apply current division rule on Figure 5 to find I1.

I1=ZR1(s)ZR1(s)+ZR2(s)+ZL2(s)I

Substitute VsZeq(s) for I, R1 for ZR1(s), R2 for ZR2(s) and sL2 for ZL2(s) in above equation to find I1.

I1=R1R1+R2+sL2(VsZeq(s))

Substitute R2+3sRL+s2L22R+sL for Zeq(s), L for L2, R for R1 and R2 in above equation to find I1.

I1=RR+R+sL(Vs(R2+3sRL+s2L22R+sL))=R2R+sL(Vs(2R+sLR2+3sRL+s2L2))=RVsR2+3sRL+s2L2

Refer to Figure 5, the output voltage Vo is calculated as,

Vo=sL2I1

Substitute RVsR2+3sRL+s2L2 for I1 and L for L2 in above equation to find Vo.

Vo=s(L)(RVsR2+3sRL+s2L2)=(sRLR2+3sRL+s2L2)Vs

Rearrange the above equation to find VoVs.

VoVs=(sRLR2+3sRL+s2L2)

Substitute (sRLR2+3sRL+s2L2) for VoVs in equation (8) to find H(s).

H(s)=sRLR2+3sRL+s2L2=sRLL2(s2+3sRLL2+R2L2)=(33)sRL(s2+3RLs+R2L2)=13((3RL)s(s2+3RLs+R2L2))

Compare the denominator factor of above equation with the standard quadratic equation (s2+ω0Qs+ω02).

ω02=R2L2        (10)

ω0Q=3RL        (11)

Substitute 1Ω for R and 1H for L in equation (10) to find ω02.

ω02=(1Ω)2(1H)2=1Ω21H2=1Ω21Ω2s2 {1H=1Ω1s}=11s2

Take square root on both sides of the above equation to find ω0.

ω02=11s2ω0=1rads

Substitute B for ω0Q, 1Ω for R and 1H for L in equation (11) to find B.

B=3(1Ω)1H=3Ω1Ωs {1H=1Ω1s}=31s=3rads

Conclusion:

Thus, the value of the center frequency (ω0) and the bandwidth (B) of the band-pass filter shown in Figure 14.88(b) is 1rads and 3rads respectively.

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Chapter 14 Solutions

Fundamentals of Electric Circuits

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