Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89. Figure 14.89

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Answer 1

Textbook Question

Chapter 14, Problem 59P

Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89.

Chapter 14, Problem 59P, Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89. Figure 14.89

Figure 14.89

Expert Solution & Answer

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer to Problem 59P

The value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.89 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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , 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 (2) to find ZL(s).

ZL(s)=sL

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

ZC(s)=1sC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 2

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

H(s)=VoVi (4)

Here,

Vo is the output response of the system, and

Vi is the input response of the system.

Refer to Figure 2, the series connected impedances 1sC and sL are connected in parallel with the impedance R2.

Therefore, the equivalent impedance is calculated as follows.

Z=R2∥(1sC+sL)=R2∥(1+s2LCsC)=R2(1+s2LCsC)R2+(1+s2LCsC)=(R2(1+s2LC)sC)(sR2C+1+s2LCsC)

Simplify the above equation to find Z.

Z=R2(1+s2LC)1+sR2C+s2LC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 3

Apply voltage division rule on Figure 3 to find Vo.

Vo=ZZ+R1Vi

Rearrange the above equation to find VoVi.

VoVi=ZZ+R1

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find VoVi.

VoVi=(R2(1+s2LC)1+sR2C+s2LC)(R2(1+s2LC)1+sR2C+s2LC)+R1=(R2(1+s2LC)1+sR2C+s2LC)(R2+s2R2LC+R1+sR1R2C+s2R1LC1+sR2C+s2LC)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC

Substitute R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC for VoVi in equation (4) to find H(s).

H(s)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC (5)

Refer to Figure 3, the input impedance is expressed as,

Zin=R1+Z

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find Zin.

Zin=R1+R2(1+s2LC)1+sR2C+s2LC=R1+R2+s2R2LC1+sR2C+s2LC=R1+sR1R2C+s2R1LC+R2+s2R2LC1+sR2C+s2LC

Substitute jω for s in above equation to find Zin(ω).

Zin(ω)=R1+(jω)R1R2C+(jω)2R1LC+R2+(jω)2R2LC1+(jω)R2C+(jω)2LC=R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC1+jωR2C+j2ω2LC=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC1+jωR2C−ω2LC {∵j2=−1}=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C

Simplify the above equation to find Zin(ω).

Zin(ω)=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C×(1−ω2LC)−jωR2C(1−ω2LC)−jωR2C=((1−ω2LC−jωR2C)(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC))(1−ω2LC)2−(jωR2C)2 {∵a2−b2=(a+b)(a−b)}=(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC−ω2R1LC−jω3R1R2LC2+ω4R1L2C2−ω2R2LC+ω4R2L2C2−jωR1R2C−j2ω2R1R22C2+jω3R1R2LC2−jωR22C+jω3R22LC2)(1−ω2LC)2−j2(ωR2C)2=(R1+R2−2ω2R1LC−2ω2R2LC+ω2R1R22C2+ω4R1L2C2+ω4R2L2C2+j(−ωR22C+ω3R22LC2))(1−ω2LC)2+(ωR2C)2 {∵j2=−1}

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

−ω0R22C+ω03R22LC2(1−ω02LC)2+(ω0R2C)2=0−ω0R22C+ω03R22LC2=0ω03R22LC2=ω0R22Cω02LC=1

Simplify the above equation to find ω02.

ω02=1LC

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

ω02=1LCω0=1LC

Substitute 1 mH for L and 4 μF for C in above equation to find ω0.

ω0=1(1 mH)(4 μF)=1(1×10−3)(4×10−6) H⋅F {∵1 m=10−3, 1 μ=10−6}=1(4×10−9) (s2F⋅F) {∵1 H=1 s21 F}=15.811×103 rads

Simplify the above equation to find ω0.

ω0=15.811 krads {∵1 k=103}

Substitute jω for s in equation (5) to find H(ω).

H(ω)=R2(1+(jω)2LC)R1+jωR1R2C+(jω)2R1LC+R2+(jω)2R2LC=R2(1+j2ω2LC)R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC

H(ω)=R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC {∵j2=−1} (6)

Substitute 0 for ω in equation (6) to find the minimum value H(0).

H(0)=R2(1−(0)2LC)R1+j(0)R1R2C−(0)2R1LC+R2−(0)2R2LC=R2(1−0)R1+0−0+R2−0=R2R1+R2

Substitute ∞ for ω in equation (6) to find the maximum value H(∞).

H(∞)=limω→∞(R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC)=limω→∞(ω2R2(1ω2−LC)ω2(R1+R2ω2+jωR1R2Cω2−LC(R1+R2)))=limω→∞(R2(1ω2−LC)(R1+R2ω2+jR1R2Cω−LC(R1+R2)))=R2(1(∞)2−LC)(R1+R2(∞)2+j(R1R2C∞)−LC(R1+R2))

Simplify the above equation to find H(∞).

H(∞)=R2(0−LC)(0+j(0)−LC(R1+R2))=−R2LC−LC(R1+R2)=R2R1+R2

At corner frequency ω1 and ω2, the transfer function is expressed as,

|H(ω)|=12H(∞)

Substitute R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC for H(ω) and R2R1+R2 for H(∞) in above equation.

|R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC|=12(R2R1+R2)|R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))+jωR1R2C|=R22(R1+R2)(R2(1−ω2LC))2+02(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)

Simplify the above equation.

R2(1−ω2LC)(R1+R2)(R2)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12(R1+R2)(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12

Substitute 6 Ω for R1, 4 Ω for R2, 1 mH for L and 4 μF for C in above equation.

(6+4)(1−ω2(1 m)(4 μ))(6+4−ω2(1 m)(4 μ)(6+4))2+(ω(6)(4)(4 μ))2=12(6+4)(1−ω2(1× 10−3)(4× 10−6))(6+4−ω2(1× 10−3)(4× 10−6)(6+4))2+(ω(6)(4)(4× 10−6))2=12 {∵1 m=10−3, 1 μ=10−6}(10)(1−ω2(4× 10−9))(10−ω2(4× 10−8))2+(ω(96× 10−6))2−12=02(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(2)(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=0

Simplify the above equation.

2(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=02(10−(4× 10−8)ω2)=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2

Square on both sides of the above equation to simplify it.

(2(10−(4× 10−8)ω2))2=((10−(4× 10−8)ω2)2+((96× 10−6)ω)2)22(10−(4× 10−8)ω2)2=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(96× 10−6)2ω2+1(10−(4× 10−8)ω2)2−2(10−(4× 10−8)ω2)2=0(96× 10−6)2ω2−1(10−(4× 10−8)ω2)2=0

Simplify the above equation.

(96× 10−6)2ω2−(10−(4× 10−8)ω2)2=0(9.216× 10−9)ω2−(102+((4× 10−8)ω2)2−2(10)((4× 10−8)ω2))=0(9.216× 10−9)ω2−(100+(1.6× 10−15)ω4−(8× 10−7)ω2)=0(9.216× 10−9)ω2−100−(1.6× 10−15)ω4+(8× 10−7)ω2=0

Simplify the above equation.

−(1.6× 10−15)ω4+(8.092× 10−7)ω2−100=0

(1.6× 10−15)ω4−(8.092× 10−7)ω2+100=0 (7)

Assume s=ω2 and substitute in the equation (7) to simplify it.

(1.6× 10−15)s2−(8.092× 10−7)s+100=0 (8)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (9)

Compare the equation (8) with the quadratic equation (as2+bs+c=0) to obtain the following values.

a=(1.6× 10−15)b=−(8.092× 10−7)c=100

Substitute (1.6× 10−15) for a, −(8.092× 10−7) for b, and 100 for c in equation (9) to find s1,2.

s1,2=−(−(8.092× 10−7))±(−(8.092× 10−7))2−4(1.6× 10−15)(100)2(1.6× 10−15)=(8.092× 10−7)±(1.48× 10−14)(3.2× 10−15)=(8.092× 10−7)±(1.2167× 10−7)(3.2× 10−15)= 2.1471×108, 2.9109×108

Substitute the roots of characteristic equation in equation (8).

(s1−(2.1471×108))(s2−(2.9109×108))=0

Substitute ω2 for s in above equation.

(ω12−(2.1471×108))(ω22−(2.9109×108))=0 (10)

Simplify the equation (10) to find ω12.

ω12−(2.1471×108)=0ω12=(2.1471×108)

Take square root on both sides of the above equation to find ω1.

ω12=(2.1471×108)ω1=14.653×103 radsω1=14.653 krads {∵1 k=103}

Simplify the equation (10) to find ω22.

ω22−(2.9109×108)=0ω22=(2.9109×108)

Take square root on both sides of the above equation to find ω2.

ω22=(2.9109×108)ω2=17.061×103 rads=17.061 krads {∵1 k=103}

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

B=ω2−ω1 (11)

Here,

ω1 is the lower corner frequency, and

ω2 is the upper corner frequency.

Substitute 14.653 krads for ω1 and 17.061 krads for ω2 in equation (11) to find B.

B=17.061 krads−14.653 krads=2.408 krads

Conclusion:

Thus, the value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

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Answer 2

Textbook Question

Chapter 14, Problem 59P

Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89.

Chapter 14, Problem 59P, Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89. Figure 14.89

Figure 14.89

Expert Solution & Answer

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer to Problem 59P

The value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.89 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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , 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 (2) to find ZL(s).

ZL(s)=sL

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

ZC(s)=1sC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 2

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

H(s)=VoVi (4)

Here,

Vo is the output response of the system, and

Vi is the input response of the system.

Refer to Figure 2, the series connected impedances 1sC and sL are connected in parallel with the impedance R2.

Therefore, the equivalent impedance is calculated as follows.

Z=R2∥(1sC+sL)=R2∥(1+s2LCsC)=R2(1+s2LCsC)R2+(1+s2LCsC)=(R2(1+s2LC)sC)(sR2C+1+s2LCsC)

Simplify the above equation to find Z.

Z=R2(1+s2LC)1+sR2C+s2LC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 3

Apply voltage division rule on Figure 3 to find Vo.

Vo=ZZ+R1Vi

Rearrange the above equation to find VoVi.

VoVi=ZZ+R1

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find VoVi.

VoVi=(R2(1+s2LC)1+sR2C+s2LC)(R2(1+s2LC)1+sR2C+s2LC)+R1=(R2(1+s2LC)1+sR2C+s2LC)(R2+s2R2LC+R1+sR1R2C+s2R1LC1+sR2C+s2LC)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC

Substitute R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC for VoVi in equation (4) to find H(s).

H(s)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC (5)

Refer to Figure 3, the input impedance is expressed as,

Zin=R1+Z

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find Zin.

Zin=R1+R2(1+s2LC)1+sR2C+s2LC=R1+R2+s2R2LC1+sR2C+s2LC=R1+sR1R2C+s2R1LC+R2+s2R2LC1+sR2C+s2LC

Substitute jω for s in above equation to find Zin(ω).

Zin(ω)=R1+(jω)R1R2C+(jω)2R1LC+R2+(jω)2R2LC1+(jω)R2C+(jω)2LC=R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC1+jωR2C+j2ω2LC=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC1+jωR2C−ω2LC {∵j2=−1}=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C

Simplify the above equation to find Zin(ω).

Zin(ω)=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C×(1−ω2LC)−jωR2C(1−ω2LC)−jωR2C=((1−ω2LC−jωR2C)(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC))(1−ω2LC)2−(jωR2C)2 {∵a2−b2=(a+b)(a−b)}=(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC−ω2R1LC−jω3R1R2LC2+ω4R1L2C2−ω2R2LC+ω4R2L2C2−jωR1R2C−j2ω2R1R22C2+jω3R1R2LC2−jωR22C+jω3R22LC2)(1−ω2LC)2−j2(ωR2C)2=(R1+R2−2ω2R1LC−2ω2R2LC+ω2R1R22C2+ω4R1L2C2+ω4R2L2C2+j(−ωR22C+ω3R22LC2))(1−ω2LC)2+(ωR2C)2 {∵j2=−1}

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

−ω0R22C+ω03R22LC2(1−ω02LC)2+(ω0R2C)2=0−ω0R22C+ω03R22LC2=0ω03R22LC2=ω0R22Cω02LC=1

Simplify the above equation to find ω02.

ω02=1LC

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

ω02=1LCω0=1LC

Substitute 1 mH for L and 4 μF for C in above equation to find ω0.

ω0=1(1 mH)(4 μF)=1(1×10−3)(4×10−6) H⋅F {∵1 m=10−3, 1 μ=10−6}=1(4×10−9) (s2F⋅F) {∵1 H=1 s21 F}=15.811×103 rads

Simplify the above equation to find ω0.

ω0=15.811 krads {∵1 k=103}

Substitute jω for s in equation (5) to find H(ω).

H(ω)=R2(1+(jω)2LC)R1+jωR1R2C+(jω)2R1LC+R2+(jω)2R2LC=R2(1+j2ω2LC)R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC

H(ω)=R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC {∵j2=−1} (6)

Substitute 0 for ω in equation (6) to find the minimum value H(0).

H(0)=R2(1−(0)2LC)R1+j(0)R1R2C−(0)2R1LC+R2−(0)2R2LC=R2(1−0)R1+0−0+R2−0=R2R1+R2

Substitute ∞ for ω in equation (6) to find the maximum value H(∞).

H(∞)=limω→∞(R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC)=limω→∞(ω2R2(1ω2−LC)ω2(R1+R2ω2+jωR1R2Cω2−LC(R1+R2)))=limω→∞(R2(1ω2−LC)(R1+R2ω2+jR1R2Cω−LC(R1+R2)))=R2(1(∞)2−LC)(R1+R2(∞)2+j(R1R2C∞)−LC(R1+R2))

Simplify the above equation to find H(∞).

H(∞)=R2(0−LC)(0+j(0)−LC(R1+R2))=−R2LC−LC(R1+R2)=R2R1+R2

At corner frequency ω1 and ω2, the transfer function is expressed as,

|H(ω)|=12H(∞)

Substitute R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC for H(ω) and R2R1+R2 for H(∞) in above equation.

|R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC|=12(R2R1+R2)|R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))+jωR1R2C|=R22(R1+R2)(R2(1−ω2LC))2+02(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)

Simplify the above equation.

R2(1−ω2LC)(R1+R2)(R2)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12(R1+R2)(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12

Substitute 6 Ω for R1, 4 Ω for R2, 1 mH for L and 4 μF for C in above equation.

(6+4)(1−ω2(1 m)(4 μ))(6+4−ω2(1 m)(4 μ)(6+4))2+(ω(6)(4)(4 μ))2=12(6+4)(1−ω2(1× 10−3)(4× 10−6))(6+4−ω2(1× 10−3)(4× 10−6)(6+4))2+(ω(6)(4)(4× 10−6))2=12 {∵1 m=10−3, 1 μ=10−6}(10)(1−ω2(4× 10−9))(10−ω2(4× 10−8))2+(ω(96× 10−6))2−12=02(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(2)(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=0

Simplify the above equation.

2(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=02(10−(4× 10−8)ω2)=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2

Square on both sides of the above equation to simplify it.

(2(10−(4× 10−8)ω2))2=((10−(4× 10−8)ω2)2+((96× 10−6)ω)2)22(10−(4× 10−8)ω2)2=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(96× 10−6)2ω2+1(10−(4× 10−8)ω2)2−2(10−(4× 10−8)ω2)2=0(96× 10−6)2ω2−1(10−(4× 10−8)ω2)2=0

Simplify the above equation.

(96× 10−6)2ω2−(10−(4× 10−8)ω2)2=0(9.216× 10−9)ω2−(102+((4× 10−8)ω2)2−2(10)((4× 10−8)ω2))=0(9.216× 10−9)ω2−(100+(1.6× 10−15)ω4−(8× 10−7)ω2)=0(9.216× 10−9)ω2−100−(1.6× 10−15)ω4+(8× 10−7)ω2=0

Simplify the above equation.

−(1.6× 10−15)ω4+(8.092× 10−7)ω2−100=0

(1.6× 10−15)ω4−(8.092× 10−7)ω2+100=0 (7)

Assume s=ω2 and substitute in the equation (7) to simplify it.

(1.6× 10−15)s2−(8.092× 10−7)s+100=0 (8)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (9)

Compare the equation (8) with the quadratic equation (as2+bs+c=0) to obtain the following values.

a=(1.6× 10−15)b=−(8.092× 10−7)c=100

Substitute (1.6× 10−15) for a, −(8.092× 10−7) for b, and 100 for c in equation (9) to find s1,2.

s1,2=−(−(8.092× 10−7))±(−(8.092× 10−7))2−4(1.6× 10−15)(100)2(1.6× 10−15)=(8.092× 10−7)±(1.48× 10−14)(3.2× 10−15)=(8.092× 10−7)±(1.2167× 10−7)(3.2× 10−15)= 2.1471×108, 2.9109×108

Substitute the roots of characteristic equation in equation (8).

(s1−(2.1471×108))(s2−(2.9109×108))=0

Substitute ω2 for s in above equation.

(ω12−(2.1471×108))(ω22−(2.9109×108))=0 (10)

Simplify the equation (10) to find ω12.

ω12−(2.1471×108)=0ω12=(2.1471×108)

Take square root on both sides of the above equation to find ω1.

ω12=(2.1471×108)ω1=14.653×103 radsω1=14.653 krads {∵1 k=103}

Simplify the equation (10) to find ω22.

ω22−(2.9109×108)=0ω22=(2.9109×108)

Take square root on both sides of the above equation to find ω2.

ω22=(2.9109×108)ω2=17.061×103 rads=17.061 krads {∵1 k=103}

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

B=ω2−ω1 (11)

Here,

ω1 is the lower corner frequency, and

ω2 is the upper corner frequency.

Substitute 14.653 krads for ω1 and 17.061 krads for ω2 in equation (11) to find B.

B=17.061 krads−14.653 krads=2.408 krads

Conclusion:

Thus, the value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

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Answer 3

Textbook Question

Chapter 14, Problem 59P

Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89.

Chapter 14, Problem 59P, Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89. Figure 14.89

Figure 14.89

Expert Solution & Answer

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer to Problem 59P

The value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.89 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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , 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 (2) to find ZL(s).

ZL(s)=sL

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

ZC(s)=1sC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 2

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

H(s)=VoVi (4)

Here,

Vo is the output response of the system, and

Vi is the input response of the system.

Refer to Figure 2, the series connected impedances 1sC and sL are connected in parallel with the impedance R2.

Therefore, the equivalent impedance is calculated as follows.

Z=R2∥(1sC+sL)=R2∥(1+s2LCsC)=R2(1+s2LCsC)R2+(1+s2LCsC)=(R2(1+s2LC)sC)(sR2C+1+s2LCsC)

Simplify the above equation to find Z.

Z=R2(1+s2LC)1+sR2C+s2LC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 3

Apply voltage division rule on Figure 3 to find Vo.

Vo=ZZ+R1Vi

Rearrange the above equation to find VoVi.

VoVi=ZZ+R1

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find VoVi.

VoVi=(R2(1+s2LC)1+sR2C+s2LC)(R2(1+s2LC)1+sR2C+s2LC)+R1=(R2(1+s2LC)1+sR2C+s2LC)(R2+s2R2LC+R1+sR1R2C+s2R1LC1+sR2C+s2LC)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC

Substitute R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC for VoVi in equation (4) to find H(s).

H(s)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC (5)

Refer to Figure 3, the input impedance is expressed as,

Zin=R1+Z

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find Zin.

Zin=R1+R2(1+s2LC)1+sR2C+s2LC=R1+R2+s2R2LC1+sR2C+s2LC=R1+sR1R2C+s2R1LC+R2+s2R2LC1+sR2C+s2LC

Substitute jω for s in above equation to find Zin(ω).

Zin(ω)=R1+(jω)R1R2C+(jω)2R1LC+R2+(jω)2R2LC1+(jω)R2C+(jω)2LC=R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC1+jωR2C+j2ω2LC=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC1+jωR2C−ω2LC {∵j2=−1}=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C

Simplify the above equation to find Zin(ω).

Zin(ω)=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C×(1−ω2LC)−jωR2C(1−ω2LC)−jωR2C=((1−ω2LC−jωR2C)(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC))(1−ω2LC)2−(jωR2C)2 {∵a2−b2=(a+b)(a−b)}=(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC−ω2R1LC−jω3R1R2LC2+ω4R1L2C2−ω2R2LC+ω4R2L2C2−jωR1R2C−j2ω2R1R22C2+jω3R1R2LC2−jωR22C+jω3R22LC2)(1−ω2LC)2−j2(ωR2C)2=(R1+R2−2ω2R1LC−2ω2R2LC+ω2R1R22C2+ω4R1L2C2+ω4R2L2C2+j(−ωR22C+ω3R22LC2))(1−ω2LC)2+(ωR2C)2 {∵j2=−1}

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

−ω0R22C+ω03R22LC2(1−ω02LC)2+(ω0R2C)2=0−ω0R22C+ω03R22LC2=0ω03R22LC2=ω0R22Cω02LC=1

Simplify the above equation to find ω02.

ω02=1LC

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

ω02=1LCω0=1LC

Substitute 1 mH for L and 4 μF for C in above equation to find ω0.

ω0=1(1 mH)(4 μF)=1(1×10−3)(4×10−6) H⋅F {∵1 m=10−3, 1 μ=10−6}=1(4×10−9) (s2F⋅F) {∵1 H=1 s21 F}=15.811×103 rads

Simplify the above equation to find ω0.

ω0=15.811 krads {∵1 k=103}

Substitute jω for s in equation (5) to find H(ω).

H(ω)=R2(1+(jω)2LC)R1+jωR1R2C+(jω)2R1LC+R2+(jω)2R2LC=R2(1+j2ω2LC)R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC

H(ω)=R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC {∵j2=−1} (6)

Substitute 0 for ω in equation (6) to find the minimum value H(0).

H(0)=R2(1−(0)2LC)R1+j(0)R1R2C−(0)2R1LC+R2−(0)2R2LC=R2(1−0)R1+0−0+R2−0=R2R1+R2

Substitute ∞ for ω in equation (6) to find the maximum value H(∞).

H(∞)=limω→∞(R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC)=limω→∞(ω2R2(1ω2−LC)ω2(R1+R2ω2+jωR1R2Cω2−LC(R1+R2)))=limω→∞(R2(1ω2−LC)(R1+R2ω2+jR1R2Cω−LC(R1+R2)))=R2(1(∞)2−LC)(R1+R2(∞)2+j(R1R2C∞)−LC(R1+R2))

Simplify the above equation to find H(∞).

H(∞)=R2(0−LC)(0+j(0)−LC(R1+R2))=−R2LC−LC(R1+R2)=R2R1+R2

At corner frequency ω1 and ω2, the transfer function is expressed as,

|H(ω)|=12H(∞)

Substitute R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC for H(ω) and R2R1+R2 for H(∞) in above equation.

|R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC|=12(R2R1+R2)|R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))+jωR1R2C|=R22(R1+R2)(R2(1−ω2LC))2+02(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)

Simplify the above equation.

R2(1−ω2LC)(R1+R2)(R2)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12(R1+R2)(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12

Substitute 6 Ω for R1, 4 Ω for R2, 1 mH for L and 4 μF for C in above equation.

(6+4)(1−ω2(1 m)(4 μ))(6+4−ω2(1 m)(4 μ)(6+4))2+(ω(6)(4)(4 μ))2=12(6+4)(1−ω2(1× 10−3)(4× 10−6))(6+4−ω2(1× 10−3)(4× 10−6)(6+4))2+(ω(6)(4)(4× 10−6))2=12 {∵1 m=10−3, 1 μ=10−6}(10)(1−ω2(4× 10−9))(10−ω2(4× 10−8))2+(ω(96× 10−6))2−12=02(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(2)(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=0

Simplify the above equation.

2(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=02(10−(4× 10−8)ω2)=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2

Square on both sides of the above equation to simplify it.

(2(10−(4× 10−8)ω2))2=((10−(4× 10−8)ω2)2+((96× 10−6)ω)2)22(10−(4× 10−8)ω2)2=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(96× 10−6)2ω2+1(10−(4× 10−8)ω2)2−2(10−(4× 10−8)ω2)2=0(96× 10−6)2ω2−1(10−(4× 10−8)ω2)2=0

Simplify the above equation.

(96× 10−6)2ω2−(10−(4× 10−8)ω2)2=0(9.216× 10−9)ω2−(102+((4× 10−8)ω2)2−2(10)((4× 10−8)ω2))=0(9.216× 10−9)ω2−(100+(1.6× 10−15)ω4−(8× 10−7)ω2)=0(9.216× 10−9)ω2−100−(1.6× 10−15)ω4+(8× 10−7)ω2=0

Simplify the above equation.

−(1.6× 10−15)ω4+(8.092× 10−7)ω2−100=0

(1.6× 10−15)ω4−(8.092× 10−7)ω2+100=0 (7)

Assume s=ω2 and substitute in the equation (7) to simplify it.

(1.6× 10−15)s2−(8.092× 10−7)s+100=0 (8)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (9)

Compare the equation (8) with the quadratic equation (as2+bs+c=0) to obtain the following values.

a=(1.6× 10−15)b=−(8.092× 10−7)c=100

Substitute (1.6× 10−15) for a, −(8.092× 10−7) for b, and 100 for c in equation (9) to find s1,2.

s1,2=−(−(8.092× 10−7))±(−(8.092× 10−7))2−4(1.6× 10−15)(100)2(1.6× 10−15)=(8.092× 10−7)±(1.48× 10−14)(3.2× 10−15)=(8.092× 10−7)±(1.2167× 10−7)(3.2× 10−15)= 2.1471×108, 2.9109×108

Substitute the roots of characteristic equation in equation (8).

(s1−(2.1471×108))(s2−(2.9109×108))=0

Substitute ω2 for s in above equation.

(ω12−(2.1471×108))(ω22−(2.9109×108))=0 (10)

Simplify the equation (10) to find ω12.

ω12−(2.1471×108)=0ω12=(2.1471×108)

Take square root on both sides of the above equation to find ω1.

ω12=(2.1471×108)ω1=14.653×103 radsω1=14.653 krads {∵1 k=103}

Simplify the equation (10) to find ω22.

ω22−(2.9109×108)=0ω22=(2.9109×108)

Take square root on both sides of the above equation to find ω2.

ω22=(2.9109×108)ω2=17.061×103 rads=17.061 krads {∵1 k=103}

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

B=ω2−ω1 (11)

Here,

ω1 is the lower corner frequency, and

ω2 is the upper corner frequency.

Substitute 14.653 krads for ω1 and 17.061 krads for ω2 in equation (11) to find B.

B=17.061 krads−14.653 krads=2.408 krads

Conclusion:

Thus, the value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

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Answer 4

Textbook Question

Chapter 14, Problem 59P

Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89.

Chapter 14, Problem 59P, Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89. Figure 14.89

Figure 14.89

Expert Solution & Answer

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer to Problem 59P

The value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.89 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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , 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 (2) to find ZL(s).

ZL(s)=sL

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

ZC(s)=1sC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 2

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

H(s)=VoVi (4)

Here,

Vo is the output response of the system, and

Vi is the input response of the system.

Refer to Figure 2, the series connected impedances 1sC and sL are connected in parallel with the impedance R2.

Therefore, the equivalent impedance is calculated as follows.

Z=R2∥(1sC+sL)=R2∥(1+s2LCsC)=R2(1+s2LCsC)R2+(1+s2LCsC)=(R2(1+s2LC)sC)(sR2C+1+s2LCsC)

Simplify the above equation to find Z.

Z=R2(1+s2LC)1+sR2C+s2LC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 3

Apply voltage division rule on Figure 3 to find Vo.

Vo=ZZ+R1Vi

Rearrange the above equation to find VoVi.

VoVi=ZZ+R1

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find VoVi.

VoVi=(R2(1+s2LC)1+sR2C+s2LC)(R2(1+s2LC)1+sR2C+s2LC)+R1=(R2(1+s2LC)1+sR2C+s2LC)(R2+s2R2LC+R1+sR1R2C+s2R1LC1+sR2C+s2LC)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC

Substitute R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC for VoVi in equation (4) to find H(s).

H(s)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC (5)

Refer to Figure 3, the input impedance is expressed as,

Zin=R1+Z

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find Zin.

Zin=R1+R2(1+s2LC)1+sR2C+s2LC=R1+R2+s2R2LC1+sR2C+s2LC=R1+sR1R2C+s2R1LC+R2+s2R2LC1+sR2C+s2LC

Substitute jω for s in above equation to find Zin(ω).

Zin(ω)=R1+(jω)R1R2C+(jω)2R1LC+R2+(jω)2R2LC1+(jω)R2C+(jω)2LC=R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC1+jωR2C+j2ω2LC=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC1+jωR2C−ω2LC {∵j2=−1}=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C

Simplify the above equation to find Zin(ω).

Zin(ω)=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C×(1−ω2LC)−jωR2C(1−ω2LC)−jωR2C=((1−ω2LC−jωR2C)(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC))(1−ω2LC)2−(jωR2C)2 {∵a2−b2=(a+b)(a−b)}=(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC−ω2R1LC−jω3R1R2LC2+ω4R1L2C2−ω2R2LC+ω4R2L2C2−jωR1R2C−j2ω2R1R22C2+jω3R1R2LC2−jωR22C+jω3R22LC2)(1−ω2LC)2−j2(ωR2C)2=(R1+R2−2ω2R1LC−2ω2R2LC+ω2R1R22C2+ω4R1L2C2+ω4R2L2C2+j(−ωR22C+ω3R22LC2))(1−ω2LC)2+(ωR2C)2 {∵j2=−1}

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

−ω0R22C+ω03R22LC2(1−ω02LC)2+(ω0R2C)2=0−ω0R22C+ω03R22LC2=0ω03R22LC2=ω0R22Cω02LC=1

Simplify the above equation to find ω02.

ω02=1LC

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

ω02=1LCω0=1LC

Substitute 1 mH for L and 4 μF for C in above equation to find ω0.

ω0=1(1 mH)(4 μF)=1(1×10−3)(4×10−6) H⋅F {∵1 m=10−3, 1 μ=10−6}=1(4×10−9) (s2F⋅F) {∵1 H=1 s21 F}=15.811×103 rads

Simplify the above equation to find ω0.

ω0=15.811 krads {∵1 k=103}

Substitute jω for s in equation (5) to find H(ω).

H(ω)=R2(1+(jω)2LC)R1+jωR1R2C+(jω)2R1LC+R2+(jω)2R2LC=R2(1+j2ω2LC)R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC

H(ω)=R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC {∵j2=−1} (6)

Substitute 0 for ω in equation (6) to find the minimum value H(0).

H(0)=R2(1−(0)2LC)R1+j(0)R1R2C−(0)2R1LC+R2−(0)2R2LC=R2(1−0)R1+0−0+R2−0=R2R1+R2

Substitute ∞ for ω in equation (6) to find the maximum value H(∞).

H(∞)=limω→∞(R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC)=limω→∞(ω2R2(1ω2−LC)ω2(R1+R2ω2+jωR1R2Cω2−LC(R1+R2)))=limω→∞(R2(1ω2−LC)(R1+R2ω2+jR1R2Cω−LC(R1+R2)))=R2(1(∞)2−LC)(R1+R2(∞)2+j(R1R2C∞)−LC(R1+R2))

Simplify the above equation to find H(∞).

H(∞)=R2(0−LC)(0+j(0)−LC(R1+R2))=−R2LC−LC(R1+R2)=R2R1+R2

At corner frequency ω1 and ω2, the transfer function is expressed as,

|H(ω)|=12H(∞)

Substitute R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC for H(ω) and R2R1+R2 for H(∞) in above equation.

|R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC|=12(R2R1+R2)|R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))+jωR1R2C|=R22(R1+R2)(R2(1−ω2LC))2+02(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)

Simplify the above equation.

R2(1−ω2LC)(R1+R2)(R2)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12(R1+R2)(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12

Substitute 6 Ω for R1, 4 Ω for R2, 1 mH for L and 4 μF for C in above equation.

(6+4)(1−ω2(1 m)(4 μ))(6+4−ω2(1 m)(4 μ)(6+4))2+(ω(6)(4)(4 μ))2=12(6+4)(1−ω2(1× 10−3)(4× 10−6))(6+4−ω2(1× 10−3)(4× 10−6)(6+4))2+(ω(6)(4)(4× 10−6))2=12 {∵1 m=10−3, 1 μ=10−6}(10)(1−ω2(4× 10−9))(10−ω2(4× 10−8))2+(ω(96× 10−6))2−12=02(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(2)(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=0

Simplify the above equation.

2(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=02(10−(4× 10−8)ω2)=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2

Square on both sides of the above equation to simplify it.

(2(10−(4× 10−8)ω2))2=((10−(4× 10−8)ω2)2+((96× 10−6)ω)2)22(10−(4× 10−8)ω2)2=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(96× 10−6)2ω2+1(10−(4× 10−8)ω2)2−2(10−(4× 10−8)ω2)2=0(96× 10−6)2ω2−1(10−(4× 10−8)ω2)2=0

Simplify the above equation.

(96× 10−6)2ω2−(10−(4× 10−8)ω2)2=0(9.216× 10−9)ω2−(102+((4× 10−8)ω2)2−2(10)((4× 10−8)ω2))=0(9.216× 10−9)ω2−(100+(1.6× 10−15)ω4−(8× 10−7)ω2)=0(9.216× 10−9)ω2−100−(1.6× 10−15)ω4+(8× 10−7)ω2=0

Simplify the above equation.

−(1.6× 10−15)ω4+(8.092× 10−7)ω2−100=0

(1.6× 10−15)ω4−(8.092× 10−7)ω2+100=0 (7)

Assume s=ω2 and substitute in the equation (7) to simplify it.

(1.6× 10−15)s2−(8.092× 10−7)s+100=0 (8)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (9)

Compare the equation (8) with the quadratic equation (as2+bs+c=0) to obtain the following values.

a=(1.6× 10−15)b=−(8.092× 10−7)c=100

Substitute (1.6× 10−15) for a, −(8.092× 10−7) for b, and 100 for c in equation (9) to find s1,2.

s1,2=−(−(8.092× 10−7))±(−(8.092× 10−7))2−4(1.6× 10−15)(100)2(1.6× 10−15)=(8.092× 10−7)±(1.48× 10−14)(3.2× 10−15)=(8.092× 10−7)±(1.2167× 10−7)(3.2× 10−15)= 2.1471×108, 2.9109×108

Substitute the roots of characteristic equation in equation (8).

(s1−(2.1471×108))(s2−(2.9109×108))=0

Substitute ω2 for s in above equation.

(ω12−(2.1471×108))(ω22−(2.9109×108))=0 (10)

Simplify the equation (10) to find ω12.

ω12−(2.1471×108)=0ω12=(2.1471×108)

Take square root on both sides of the above equation to find ω1.

ω12=(2.1471×108)ω1=14.653×103 radsω1=14.653 krads {∵1 k=103}

Simplify the equation (10) to find ω22.

ω22−(2.9109×108)=0ω22=(2.9109×108)

Take square root on both sides of the above equation to find ω2.

ω22=(2.9109×108)ω2=17.061×103 rads=17.061 krads {∵1 k=103}

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

B=ω2−ω1 (11)

Here,

ω1 is the lower corner frequency, and

ω2 is the upper corner frequency.

Substitute 14.653 krads for ω1 and 17.061 krads for ω2 in equation (11) to find B.

B=17.061 krads−14.653 krads=2.408 krads

Conclusion:

Thus, the value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer to Problem 59P

The value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.89 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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , 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 (2) to find ZL(s).

ZL(s)=sL

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

ZC(s)=1sC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 2

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

H(s)=VoVi (4)

Here,

Vo is the output response of the system, and

Vi is the input response of the system.

Refer to Figure 2, the series connected impedances 1sC and sL are connected in parallel with the impedance R2.

Therefore, the equivalent impedance is calculated as follows.

Z=R2∥(1sC+sL)=R2∥(1+s2LCsC)=R2(1+s2LCsC)R2+(1+s2LCsC)=(R2(1+s2LC)sC)(sR2C+1+s2LCsC)

Simplify the above equation to find Z.

Z=R2(1+s2LC)1+sR2C+s2LC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 3

Apply voltage division rule on Figure 3 to find Vo.

Vo=ZZ+R1Vi

Rearrange the above equation to find VoVi.

VoVi=ZZ+R1

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find VoVi.

VoVi=(R2(1+s2LC)1+sR2C+s2LC)(R2(1+s2LC)1+sR2C+s2LC)+R1=(R2(1+s2LC)1+sR2C+s2LC)(R2+s2R2LC+R1+sR1R2C+s2R1LC1+sR2C+s2LC)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC

Substitute R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC for VoVi in equation (4) to find H(s).

H(s)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC (5)

Refer to Figure 3, the input impedance is expressed as,

Zin=R1+Z

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find Zin.

Zin=R1+R2(1+s2LC)1+sR2C+s2LC=R1+R2+s2R2LC1+sR2C+s2LC=R1+sR1R2C+s2R1LC+R2+s2R2LC1+sR2C+s2LC

Substitute jω for s in above equation to find Zin(ω).

Zin(ω)=R1+(jω)R1R2C+(jω)2R1LC+R2+(jω)2R2LC1+(jω)R2C+(jω)2LC=R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC1+jωR2C+j2ω2LC=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC1+jωR2C−ω2LC {∵j2=−1}=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C

Simplify the above equation to find Zin(ω).

Zin(ω)=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C×(1−ω2LC)−jωR2C(1−ω2LC)−jωR2C=((1−ω2LC−jωR2C)(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC))(1−ω2LC)2−(jωR2C)2 {∵a2−b2=(a+b)(a−b)}=(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC−ω2R1LC−jω3R1R2LC2+ω4R1L2C2−ω2R2LC+ω4R2L2C2−jωR1R2C−j2ω2R1R22C2+jω3R1R2LC2−jωR22C+jω3R22LC2)(1−ω2LC)2−j2(ωR2C)2=(R1+R2−2ω2R1LC−2ω2R2LC+ω2R1R22C2+ω4R1L2C2+ω4R2L2C2+j(−ωR22C+ω3R22LC2))(1−ω2LC)2+(ωR2C)2 {∵j2=−1}

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

−ω0R22C+ω03R22LC2(1−ω02LC)2+(ω0R2C)2=0−ω0R22C+ω03R22LC2=0ω03R22LC2=ω0R22Cω02LC=1

Simplify the above equation to find ω02.

ω02=1LC

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

ω02=1LCω0=1LC

Substitute 1 mH for L and 4 μF for C in above equation to find ω0.

ω0=1(1 mH)(4 μF)=1(1×10−3)(4×10−6) H⋅F {∵1 m=10−3, 1 μ=10−6}=1(4×10−9) (s2F⋅F) {∵1 H=1 s21 F}=15.811×103 rads

Simplify the above equation to find ω0.

ω0=15.811 krads {∵1 k=103}

Substitute jω for s in equation (5) to find H(ω).

H(ω)=R2(1+(jω)2LC)R1+jωR1R2C+(jω)2R1LC+R2+(jω)2R2LC=R2(1+j2ω2LC)R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC

H(ω)=R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC {∵j2=−1} (6)

Substitute 0 for ω in equation (6) to find the minimum value H(0).

H(0)=R2(1−(0)2LC)R1+j(0)R1R2C−(0)2R1LC+R2−(0)2R2LC=R2(1−0)R1+0−0+R2−0=R2R1+R2

Substitute ∞ for ω in equation (6) to find the maximum value H(∞).

H(∞)=limω→∞(R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC)=limω→∞(ω2R2(1ω2−LC)ω2(R1+R2ω2+jωR1R2Cω2−LC(R1+R2)))=limω→∞(R2(1ω2−LC)(R1+R2ω2+jR1R2Cω−LC(R1+R2)))=R2(1(∞)2−LC)(R1+R2(∞)2+j(R1R2C∞)−LC(R1+R2))

Simplify the above equation to find H(∞).

H(∞)=R2(0−LC)(0+j(0)−LC(R1+R2))=−R2LC−LC(R1+R2)=R2R1+R2

At corner frequency ω1 and ω2, the transfer function is expressed as,

|H(ω)|=12H(∞)

Substitute R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC for H(ω) and R2R1+R2 for H(∞) in above equation.

|R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC|=12(R2R1+R2)|R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))+jωR1R2C|=R22(R1+R2)(R2(1−ω2LC))2+02(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)

Simplify the above equation.

R2(1−ω2LC)(R1+R2)(R2)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12(R1+R2)(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12

Substitute 6 Ω for R1, 4 Ω for R2, 1 mH for L and 4 μF for C in above equation.

(6+4)(1−ω2(1 m)(4 μ))(6+4−ω2(1 m)(4 μ)(6+4))2+(ω(6)(4)(4 μ))2=12(6+4)(1−ω2(1× 10−3)(4× 10−6))(6+4−ω2(1× 10−3)(4× 10−6)(6+4))2+(ω(6)(4)(4× 10−6))2=12 {∵1 m=10−3, 1 μ=10−6}(10)(1−ω2(4× 10−9))(10−ω2(4× 10−8))2+(ω(96× 10−6))2−12=02(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(2)(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=0

Simplify the above equation.

2(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=02(10−(4× 10−8)ω2)=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2

Square on both sides of the above equation to simplify it.

(2(10−(4× 10−8)ω2))2=((10−(4× 10−8)ω2)2+((96× 10−6)ω)2)22(10−(4× 10−8)ω2)2=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(96× 10−6)2ω2+1(10−(4× 10−8)ω2)2−2(10−(4× 10−8)ω2)2=0(96× 10−6)2ω2−1(10−(4× 10−8)ω2)2=0

Simplify the above equation.

(96× 10−6)2ω2−(10−(4× 10−8)ω2)2=0(9.216× 10−9)ω2−(102+((4× 10−8)ω2)2−2(10)((4× 10−8)ω2))=0(9.216× 10−9)ω2−(100+(1.6× 10−15)ω4−(8× 10−7)ω2)=0(9.216× 10−9)ω2−100−(1.6× 10−15)ω4+(8× 10−7)ω2=0

Simplify the above equation.

−(1.6× 10−15)ω4+(8.092× 10−7)ω2−100=0

(1.6× 10−15)ω4−(8.092× 10−7)ω2+100=0 (7)

Assume s=ω2 and substitute in the equation (7) to simplify it.

(1.6× 10−15)s2−(8.092× 10−7)s+100=0 (8)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (9)

Compare the equation (8) with the quadratic equation (as2+bs+c=0) to obtain the following values.

a=(1.6× 10−15)b=−(8.092× 10−7)c=100

Substitute (1.6× 10−15) for a, −(8.092× 10−7) for b, and 100 for c in equation (9) to find s1,2.

s1,2=−(−(8.092× 10−7))±(−(8.092× 10−7))2−4(1.6× 10−15)(100)2(1.6× 10−15)=(8.092× 10−7)±(1.48× 10−14)(3.2× 10−15)=(8.092× 10−7)±(1.2167× 10−7)(3.2× 10−15)= 2.1471×108, 2.9109×108

Substitute the roots of characteristic equation in equation (8).

(s1−(2.1471×108))(s2−(2.9109×108))=0

Substitute ω2 for s in above equation.

(ω12−(2.1471×108))(ω22−(2.9109×108))=0 (10)

Simplify the equation (10) to find ω12.

ω12−(2.1471×108)=0ω12=(2.1471×108)

Take square root on both sides of the above equation to find ω1.

ω12=(2.1471×108)ω1=14.653×103 radsω1=14.653 krads {∵1 k=103}

Simplify the equation (10) to find ω22.

ω22−(2.9109×108)=0ω22=(2.9109×108)

Take square root on both sides of the above equation to find ω2.

ω22=(2.9109×108)ω2=17.061×103 rads=17.061 krads {∵1 k=103}

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

B=ω2−ω1 (11)

Here,

ω1 is the lower corner frequency, and

ω2 is the upper corner frequency.

Substitute 14.653 krads for ω1 and 17.061 krads for ω2 in equation (11) to find B.

B=17.061 krads−14.653 krads=2.408 krads

Conclusion:

Thus, the value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Answer 8

Expert Solution & Answer

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer to Problem 59P

The value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Explanation of Solution

Given data:

Refer to Figure 14.89 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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , 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 (2) to find ZL(s).

ZL(s)=sL

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

ZC(s)=1sC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 2

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

H(s)=VoVi (4)

Here,

Vo is the output response of the system, and

Vi is the input response of the system.

Refer to Figure 2, the series connected impedances 1sC and sL are connected in parallel with the impedance R2.

Therefore, the equivalent impedance is calculated as follows.

Z=R2∥(1sC+sL)=R2∥(1+s2LCsC)=R2(1+s2LCsC)R2+(1+s2LCsC)=(R2(1+s2LC)sC)(sR2C+1+s2LCsC)

Simplify the above equation to find Z.

Z=R2(1+s2LC)1+sR2C+s2LC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 3

Apply voltage division rule on Figure 3 to find Vo.

Vo=ZZ+R1Vi

Rearrange the above equation to find VoVi.

VoVi=ZZ+R1

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find VoVi.

VoVi=(R2(1+s2LC)1+sR2C+s2LC)(R2(1+s2LC)1+sR2C+s2LC)+R1=(R2(1+s2LC)1+sR2C+s2LC)(R2+s2R2LC+R1+sR1R2C+s2R1LC1+sR2C+s2LC)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC

Substitute R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC for VoVi in equation (4) to find H(s).

H(s)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC (5)

Refer to Figure 3, the input impedance is expressed as,

Zin=R1+Z

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find Zin.

Zin=R1+R2(1+s2LC)1+sR2C+s2LC=R1+R2+s2R2LC1+sR2C+s2LC=R1+sR1R2C+s2R1LC+R2+s2R2LC1+sR2C+s2LC

Substitute jω for s in above equation to find Zin(ω).

Zin(ω)=R1+(jω)R1R2C+(jω)2R1LC+R2+(jω)2R2LC1+(jω)R2C+(jω)2LC=R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC1+jωR2C+j2ω2LC=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC1+jωR2C−ω2LC {∵j2=−1}=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C

Simplify the above equation to find Zin(ω).

Zin(ω)=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C×(1−ω2LC)−jωR2C(1−ω2LC)−jωR2C=((1−ω2LC−jωR2C)(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC))(1−ω2LC)2−(jωR2C)2 {∵a2−b2=(a+b)(a−b)}=(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC−ω2R1LC−jω3R1R2LC2+ω4R1L2C2−ω2R2LC+ω4R2L2C2−jωR1R2C−j2ω2R1R22C2+jω3R1R2LC2−jωR22C+jω3R22LC2)(1−ω2LC)2−j2(ωR2C)2=(R1+R2−2ω2R1LC−2ω2R2LC+ω2R1R22C2+ω4R1L2C2+ω4R2L2C2+j(−ωR22C+ω3R22LC2))(1−ω2LC)2+(ωR2C)2 {∵j2=−1}

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

−ω0R22C+ω03R22LC2(1−ω02LC)2+(ω0R2C)2=0−ω0R22C+ω03R22LC2=0ω03R22LC2=ω0R22Cω02LC=1

Simplify the above equation to find ω02.

ω02=1LC

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

ω02=1LCω0=1LC

Substitute 1 mH for L and 4 μF for C in above equation to find ω0.

ω0=1(1 mH)(4 μF)=1(1×10−3)(4×10−6) H⋅F {∵1 m=10−3, 1 μ=10−6}=1(4×10−9) (s2F⋅F) {∵1 H=1 s21 F}=15.811×103 rads

Simplify the above equation to find ω0.

ω0=15.811 krads {∵1 k=103}

Substitute jω for s in equation (5) to find H(ω).

H(ω)=R2(1+(jω)2LC)R1+jωR1R2C+(jω)2R1LC+R2+(jω)2R2LC=R2(1+j2ω2LC)R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC

H(ω)=R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC {∵j2=−1} (6)

Substitute 0 for ω in equation (6) to find the minimum value H(0).

H(0)=R2(1−(0)2LC)R1+j(0)R1R2C−(0)2R1LC+R2−(0)2R2LC=R2(1−0)R1+0−0+R2−0=R2R1+R2

Substitute ∞ for ω in equation (6) to find the maximum value H(∞).

H(∞)=limω→∞(R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC)=limω→∞(ω2R2(1ω2−LC)ω2(R1+R2ω2+jωR1R2Cω2−LC(R1+R2)))=limω→∞(R2(1ω2−LC)(R1+R2ω2+jR1R2Cω−LC(R1+R2)))=R2(1(∞)2−LC)(R1+R2(∞)2+j(R1R2C∞)−LC(R1+R2))

Simplify the above equation to find H(∞).

H(∞)=R2(0−LC)(0+j(0)−LC(R1+R2))=−R2LC−LC(R1+R2)=R2R1+R2

At corner frequency ω1 and ω2, the transfer function is expressed as,

|H(ω)|=12H(∞)

Substitute R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC for H(ω) and R2R1+R2 for H(∞) in above equation.

|R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC|=12(R2R1+R2)|R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))+jωR1R2C|=R22(R1+R2)(R2(1−ω2LC))2+02(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)

Simplify the above equation.

R2(1−ω2LC)(R1+R2)(R2)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12(R1+R2)(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12

Substitute 6 Ω for R1, 4 Ω for R2, 1 mH for L and 4 μF for C in above equation.

(6+4)(1−ω2(1 m)(4 μ))(6+4−ω2(1 m)(4 μ)(6+4))2+(ω(6)(4)(4 μ))2=12(6+4)(1−ω2(1× 10−3)(4× 10−6))(6+4−ω2(1× 10−3)(4× 10−6)(6+4))2+(ω(6)(4)(4× 10−6))2=12 {∵1 m=10−3, 1 μ=10−6}(10)(1−ω2(4× 10−9))(10−ω2(4× 10−8))2+(ω(96× 10−6))2−12=02(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(2)(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=0

Simplify the above equation.

2(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=02(10−(4× 10−8)ω2)=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2

Square on both sides of the above equation to simplify it.

(2(10−(4× 10−8)ω2))2=((10−(4× 10−8)ω2)2+((96× 10−6)ω)2)22(10−(4× 10−8)ω2)2=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(96× 10−6)2ω2+1(10−(4× 10−8)ω2)2−2(10−(4× 10−8)ω2)2=0(96× 10−6)2ω2−1(10−(4× 10−8)ω2)2=0

Simplify the above equation.

(96× 10−6)2ω2−(10−(4× 10−8)ω2)2=0(9.216× 10−9)ω2−(102+((4× 10−8)ω2)2−2(10)((4× 10−8)ω2))=0(9.216× 10−9)ω2−(100+(1.6× 10−15)ω4−(8× 10−7)ω2)=0(9.216× 10−9)ω2−100−(1.6× 10−15)ω4+(8× 10−7)ω2=0

Simplify the above equation.

−(1.6× 10−15)ω4+(8.092× 10−7)ω2−100=0

(1.6× 10−15)ω4−(8.092× 10−7)ω2+100=0 (7)

Assume s=ω2 and substitute in the equation (7) to simplify it.

(1.6× 10−15)s2−(8.092× 10−7)s+100=0 (8)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (9)

Compare the equation (8) with the quadratic equation (as2+bs+c=0) to obtain the following values.

a=(1.6× 10−15)b=−(8.092× 10−7)c=100

Substitute (1.6× 10−15) for a, −(8.092× 10−7) for b, and 100 for c in equation (9) to find s1,2.

s1,2=−(−(8.092× 10−7))±(−(8.092× 10−7))2−4(1.6× 10−15)(100)2(1.6× 10−15)=(8.092× 10−7)±(1.48× 10−14)(3.2× 10−15)=(8.092× 10−7)±(1.2167× 10−7)(3.2× 10−15)= 2.1471×108, 2.9109×108

Substitute the roots of characteristic equation in equation (8).

(s1−(2.1471×108))(s2−(2.9109×108))=0

Substitute ω2 for s in above equation.

(ω12−(2.1471×108))(ω22−(2.9109×108))=0 (10)

Simplify the equation (10) to find ω12.

ω12−(2.1471×108)=0ω12=(2.1471×108)

Take square root on both sides of the above equation to find ω1.

ω12=(2.1471×108)ω1=14.653×103 radsω1=14.653 krads {∵1 k=103}

Simplify the equation (10) to find ω22.

ω22−(2.9109×108)=0ω22=(2.9109×108)

Take square root on both sides of the above equation to find ω2.

ω22=(2.9109×108)ω2=17.061×103 rads=17.061 krads {∵1 k=103}

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

B=ω2−ω1 (11)

Here,

ω1 is the lower corner frequency, and

ω2 is the upper corner frequency.

Substitute 14.653 krads for ω1 and 17.061 krads for ω2 in equation (11) to find B.

B=17.061 krads−14.653 krads=2.408 krads

Conclusion:

Thus, the value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer 12

Answer to Problem 59P

The value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Answer 13

Explanation of Solution

Given data:

Refer to Figure 14.89 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.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , 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 (2) to find ZL(s).

ZL(s)=sL

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

ZC(s)=1sC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 2

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

H(s)=VoVi (4)

Here,

Vo is the output response of the system, and

Vi is the input response of the system.

Refer to Figure 2, the series connected impedances 1sC and sL are connected in parallel with the impedance R2.

Therefore, the equivalent impedance is calculated as follows.

Z=R2∥(1sC+sL)=R2∥(1+s2LCsC)=R2(1+s2LCsC)R2+(1+s2LCsC)=(R2(1+s2LC)sC)(sR2C+1+s2LCsC)

Simplify the above equation to find Z.

Z=R2(1+s2LC)1+sR2C+s2LC

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

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 14, Problem 59P , additional homework tip 3

Apply voltage division rule on Figure 3 to find Vo.

Vo=ZZ+R1Vi

Rearrange the above equation to find VoVi.

VoVi=ZZ+R1

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find VoVi.

VoVi=(R2(1+s2LC)1+sR2C+s2LC)(R2(1+s2LC)1+sR2C+s2LC)+R1=(R2(1+s2LC)1+sR2C+s2LC)(R2+s2R2LC+R1+sR1R2C+s2R1LC1+sR2C+s2LC)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC

Substitute R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC for VoVi in equation (4) to find H(s).

H(s)=R2(1+s2LC)R1+sR1R2C+s2R1LC+R2+s2R2LC (5)

Refer to Figure 3, the input impedance is expressed as,

Zin=R1+Z

Substitute R2(1+s2LC)1+sR2C+s2LC for Z in above equation to find Zin.

Zin=R1+R2(1+s2LC)1+sR2C+s2LC=R1+R2+s2R2LC1+sR2C+s2LC=R1+sR1R2C+s2R1LC+R2+s2R2LC1+sR2C+s2LC

Substitute jω for s in above equation to find Zin(ω).

Zin(ω)=R1+(jω)R1R2C+(jω)2R1LC+R2+(jω)2R2LC1+(jω)R2C+(jω)2LC=R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC1+jωR2C+j2ω2LC=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC1+jωR2C−ω2LC {∵j2=−1}=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C

Simplify the above equation to find Zin(ω).

Zin(ω)=R1+jωR1R2C−ω2R1LC+R2−ω2R2LC(1−ω2LC)+jωR2C×(1−ω2LC)−jωR2C(1−ω2LC)−jωR2C=((1−ω2LC−jωR2C)(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC))(1−ω2LC)2−(jωR2C)2 {∵a2−b2=(a+b)(a−b)}=(R1+jωR1R2C−ω2R1LC+R2−ω2R2LC−ω2R1LC−jω3R1R2LC2+ω4R1L2C2−ω2R2LC+ω4R2L2C2−jωR1R2C−j2ω2R1R22C2+jω3R1R2LC2−jωR22C+jω3R22LC2)(1−ω2LC)2−j2(ωR2C)2=(R1+R2−2ω2R1LC−2ω2R2LC+ω2R1R22C2+ω4R1L2C2+ω4R2L2C2+j(−ωR22C+ω3R22LC2))(1−ω2LC)2+(ωR2C)2 {∵j2=−1}

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

−ω0R22C+ω03R22LC2(1−ω02LC)2+(ω0R2C)2=0−ω0R22C+ω03R22LC2=0ω03R22LC2=ω0R22Cω02LC=1

Simplify the above equation to find ω02.

ω02=1LC

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

ω02=1LCω0=1LC

Substitute 1 mH for L and 4 μF for C in above equation to find ω0.

ω0=1(1 mH)(4 μF)=1(1×10−3)(4×10−6) H⋅F {∵1 m=10−3, 1 μ=10−6}=1(4×10−9) (s2F⋅F) {∵1 H=1 s21 F}=15.811×103 rads

Simplify the above equation to find ω0.

ω0=15.811 krads {∵1 k=103}

Substitute jω for s in equation (5) to find H(ω).

H(ω)=R2(1+(jω)2LC)R1+jωR1R2C+(jω)2R1LC+R2+(jω)2R2LC=R2(1+j2ω2LC)R1+jωR1R2C+j2ω2R1LC+R2+j2ω2R2LC

H(ω)=R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC {∵j2=−1} (6)

Substitute 0 for ω in equation (6) to find the minimum value H(0).

H(0)=R2(1−(0)2LC)R1+j(0)R1R2C−(0)2R1LC+R2−(0)2R2LC=R2(1−0)R1+0−0+R2−0=R2R1+R2

Substitute ∞ for ω in equation (6) to find the maximum value H(∞).

H(∞)=limω→∞(R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC)=limω→∞(ω2R2(1ω2−LC)ω2(R1+R2ω2+jωR1R2Cω2−LC(R1+R2)))=limω→∞(R2(1ω2−LC)(R1+R2ω2+jR1R2Cω−LC(R1+R2)))=R2(1(∞)2−LC)(R1+R2(∞)2+j(R1R2C∞)−LC(R1+R2))

Simplify the above equation to find H(∞).

H(∞)=R2(0−LC)(0+j(0)−LC(R1+R2))=−R2LC−LC(R1+R2)=R2R1+R2

At corner frequency ω1 and ω2, the transfer function is expressed as,

|H(ω)|=12H(∞)

Substitute R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC for H(ω) and R2R1+R2 for H(∞) in above equation.

|R2(1−ω2LC)R1+jωR1R2C−ω2R1LC+R2−ω2R2LC|=12(R2R1+R2)|R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))+jωR1R2C|=R22(R1+R2)(R2(1−ω2LC))2+02(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)R2(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=R22(R1+R2)

Simplify the above equation.

R2(1−ω2LC)(R1+R2)(R2)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12(R1+R2)(1−ω2LC)(R1+R2−ω2LC(R1+R2))2+(ωR1R2C)2=12

Substitute 6 Ω for R1, 4 Ω for R2, 1 mH for L and 4 μF for C in above equation.

(6+4)(1−ω2(1 m)(4 μ))(6+4−ω2(1 m)(4 μ)(6+4))2+(ω(6)(4)(4 μ))2=12(6+4)(1−ω2(1× 10−3)(4× 10−6))(6+4−ω2(1× 10−3)(4× 10−6)(6+4))2+(ω(6)(4)(4× 10−6))2=12 {∵1 m=10−3, 1 μ=10−6}(10)(1−ω2(4× 10−9))(10−ω2(4× 10−8))2+(ω(96× 10−6))2−12=02(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(2)(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=0

Simplify the above equation.

2(10−(4× 10−8)ω2)−(10−(4× 10−8)ω2)2+((96× 10−6)ω)2=02(10−(4× 10−8)ω2)=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2

Square on both sides of the above equation to simplify it.

(2(10−(4× 10−8)ω2))2=((10−(4× 10−8)ω2)2+((96× 10−6)ω)2)22(10−(4× 10−8)ω2)2=(10−(4× 10−8)ω2)2+((96× 10−6)ω)2(96× 10−6)2ω2+1(10−(4× 10−8)ω2)2−2(10−(4× 10−8)ω2)2=0(96× 10−6)2ω2−1(10−(4× 10−8)ω2)2=0

Simplify the above equation.

(96× 10−6)2ω2−(10−(4× 10−8)ω2)2=0(9.216× 10−9)ω2−(102+((4× 10−8)ω2)2−2(10)((4× 10−8)ω2))=0(9.216× 10−9)ω2−(100+(1.6× 10−15)ω4−(8× 10−7)ω2)=0(9.216× 10−9)ω2−100−(1.6× 10−15)ω4+(8× 10−7)ω2=0

Simplify the above equation.

−(1.6× 10−15)ω4+(8.092× 10−7)ω2−100=0

(1.6× 10−15)ω4−(8.092× 10−7)ω2+100=0 (7)

Assume s=ω2 and substitute in the equation (7) to simplify it.

(1.6× 10−15)s2−(8.092× 10−7)s+100=0 (8)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=−b±b2−4ac2a (9)

Compare the equation (8) with the quadratic equation (as2+bs+c=0) to obtain the following values.

a=(1.6× 10−15)b=−(8.092× 10−7)c=100

Substitute (1.6× 10−15) for a, −(8.092× 10−7) for b, and 100 for c in equation (9) to find s1,2.

s1,2=−(−(8.092× 10−7))±(−(8.092× 10−7))2−4(1.6× 10−15)(100)2(1.6× 10−15)=(8.092× 10−7)±(1.48× 10−14)(3.2× 10−15)=(8.092× 10−7)±(1.2167× 10−7)(3.2× 10−15)= 2.1471×108, 2.9109×108

Substitute the roots of characteristic equation in equation (8).

(s1−(2.1471×108))(s2−(2.9109×108))=0

Substitute ω2 for s in above equation.

(ω12−(2.1471×108))(ω22−(2.9109×108))=0 (10)

Simplify the equation (10) to find ω12.

ω12−(2.1471×108)=0ω12=(2.1471×108)

Take square root on both sides of the above equation to find ω1.

ω12=(2.1471×108)ω1=14.653×103 radsω1=14.653 krads {∵1 k=103}

Simplify the equation (10) to find ω22.

ω22−(2.9109×108)=0ω22=(2.9109×108)

Take square root on both sides of the above equation to find ω2.

ω22=(2.9109×108)ω2=17.061×103 rads=17.061 krads {∵1 k=103}

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

B=ω2−ω1 (11)

Here,

ω1 is the lower corner frequency, and

ω2 is the upper corner frequency.

Substitute 14.653 krads for ω1 and 17.061 krads for ω2 in equation (11) to find B.

B=17.061 krads−14.653 krads=2.408 krads

Conclusion:

Thus, the value of the bandwidth (B) and the center frequency (ω0) of the band-stop filter shown in Figure 14.89 is 2.408 krads and 15.811 krads respectively.

Answer 14

Ch. 14.2 - Obtain the transfer function VoVs of the RL...Ch. 14.2 - Prob. 2PP Ch. 14.4 - Draw the Bode plots for the transfer function...Ch. 14.4 - Sketch the Bode plots for H()=50j(j+4)(j+10)2 Ch. 14.4 - Construct the Bode plots for H(s)=10s(s2+80s+400)Ch. 14.4 - Obtain the transfer function H() corresponding to...Ch. 14.5 - A series-connected circuit has R = 4 and L = 25...Ch. 14.6 - A parallel resonant circuit has R = 100 k, L = 50...Ch. 14.6 - Calculate the resonant frequency of the circuit in...Ch. 14.7 - For the circuit in Fig. 14.40, obtain the transfer...

Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89. Figure 14.89

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Answer to Problem 59P

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