ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<
ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<
9th Edition
ISBN: 9781260540666
Author: Hayt
Publisher: MCG CUSTOM
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Chapter 14, Problem 81E

(a)

To determine

Design a circuit which produces a transfer function of H(s)=5(s+1).

(a)

Expert Solution
Check Mark

Explanation of Solution

Given data:

The given transfer function is,

H(s)=VoutVin=5(s+1)

Calculation:

The transfer function of the circuit is,

H(s)=VoutVin=5(s+1)

The above transfer function has a zero at s=1.

The Figure 14.39 (b) in the textbook, that shows a cascade two stages of the circuit with a zero at s=1R1C1 is redrawn as shown in Figure 1.

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<, Chapter 14, Problem 81E , additional homework tip  1

For a single zero,

s=1R1C1

Substitute 1 for s in the above equation.

1=1R1C1

1=1R1C1        (1)

Consider the value of R1=1kΩ.

Substitute 1k for R in equation (1) as follows,

1=1(1×103)C1{1k=103}C1=1mF{1m=103}

Transfer function:

The input impedance of the cascaded circuit in Figure 1 is,

Z1=R1(1C1s)Z1=R1(1C1s)R1+1C1sZ1=R1R1C1(s+1R1C1)

Z1=(1C1s+1R1C1)

Then, write the Formula for the transfer function for the cascaded two stage amplifier.

H(s)=ZfZ1

Substitute (1C1s+1R1C1) for Z1 and Rf for Zf in above equation to find H(s).

H(s)=Rf(1C1s+1R1C1)

Thus, the transfer function for H(s) is,

H(s)=RfC1(s+1R1C1)

Substitute 1 for 1R1C1 and 1m for C in the above equation to find H(s).

H(s)=Rf(1×103)(s+1){1m=103}        (2)

Completing the design by letting Rf=5kΩ in the above equation as follows,

H(s)=(5×103)(1×103)(s+1){1k=103}=5(s+1)

If the input will be inverted, add an inverting amplifier with gain of 1 to provide the transfer function as follows.

H(s)=5(s+1)

Thus, the final design of the circuit is,

Rf=5kΩ, R1=1kΩ, and C1=1mF.

Conclusion:

Thus, a circuit is designed which produces a transfer function of H(s)=5(s+1).

(b)

To determine

Design a circuit which produces a transfer function of H(s)=5(s+1).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given data:

The given transfer function is,

H(s)=VoutVin=5(s+1)

Calculation:

The transfer function of the circuit is,

H(s)=VoutVin=5(s+1)

The above transfer function has pole at s=1.

The Figure 14.39 (a) in the textbook, that shows a cascade two stages of the circuit with pole at s=1RfCf is redrawn as shown in Figure 2.

ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<, Chapter 14, Problem 81E , additional homework tip  2

For pole s=1,

s=1RfACfA

Substitute 1 for s in the above equation.

1=1RfACfA

1=1RfACfA        (3)

Let arbitrarily consider RfA=5kΩ

Substitute 5k for RfA in equation (3) as follows,

1=1(5×103)CfA{1k=103}CfA=2×104×102×102CfA=200×106CfA=200μF{1μ=106}

Transfer function:

Find the feedback impedance of the cascaded circuit in Figure 2.

Zf=Rf1CfsZf=Rf(1Cfs)Rf+1CfsZf=1Cfs+1CfRf

Write the formula for the transfer function of the cascaded circuit in Figure 2 as follows

H(s)=ZfZ1

Substitute 1Cfs+1CfRf for Zf and R1 for Z1 in above equation to find the transfer function of the cascaded circuit in Figure 2.

H(s)=(1Cfs+1CfRf)R1H(s)=(1Cfs+1CfRf)R1

Therefore, consider the transfer function H(s) is,

H(s)=1R1ACfA(s+1RfACfA)

Substitute 1 for 1RfACfA and 200μ for CfA in the above equation to find H(s).

H(s)=1R1A(200×106)(s+1){1μ=106}        (4)

Completing the design by letting R1A=1kΩ in equation (4) as follows,

H(s)=1(1×103)(200×106)(s+1){1k=103}=5(s+1)

If the input will be inverted, add an inverting amplifier with gain of 1 to provide the transfer function as follows.

H(s)=5(s+1)

Thus, the final design of the circuit is,

RfA=5kΩ, CfA=200μF, and R1A=1kΩ.

Conclusion:

Thus, a circuit is designed which produces a transfer function of 5(s+1).

(c)

To determine

Design a circuit which produces a transfer function of H(s)=5(s+1)(s+2).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given data:

The given transfer function is,

H(s)=VoutVin=5(s+1)(s+2)

Calculation:

The transfer function of the circuit is,

H(s)=VoutVin=5(s+1)(s+2)

For the above transfer function, it has a zero at s=1

Refer to Figure 1 in Part (a), that shows a cascade two stages of the circuit with a zero at s=1R1C1.

For a single zero,

s=1R1C1

Substitute 1 for s in the above equation.

1=1R1C1

1=1R1C1        (5)

Let arbitrarily consider R1=1kΩ

Substitute 1k for R in equation (5) as follows,

1=1(1×103)C1{1k=103}C1=1mF{1m=103}

Consider the same circuit shown in Figure 1 and the transfer function as in a cascaded circuit,

H1(s)=RfC1(s+1R1C1)

Substitute 1 for 1R1C1 and 1m for C in the above equation to find H1(s).

H1(s)=Rf(1×103)(s+1){1m=103}        (6)

Completing the design by letting Rf=5kΩ in equation (6) as follows,

H1(s)=(5×103)(1×103)(s+1){1k=103}=5(s+1)

Thus, the final design of the circuit is,

Rf=5kΩ, R1=1kΩ, and C1=1mF.

The given transfer function has a pole at s=2.

Refer to Figure 2 in Part (b), that shows a cascade two stages of the circuit with pole at s=1RfCf.

For pole s=2,

s=1RfACfA

Substitute 2 for s in the above equation.

2=1RfACfA

2=1RfACfA        (7)

Let arbitrarily consider RfA=5kΩ

Substitute 5k for RfA in equation (7) as follows,

2=1(5×103)CfA{1k=103}CfA=12×5×103CfA=1×104×102×102CfA=100×106

The above equation becomes,

CfA=100μF{1μ=106}

Consider the same circuit shown in Figure 2 and the transfer function as in a cascaded circuit,

H2(s)=1R1ACfA(s+1RfACfA)

Substitute 2 for 1RfACfA and 100μ for CfA in the above equation to find H2(s).

H2(s)=1R1A(100×106)(s+2){1μ=106}        (8)

Completing the design by letting R1A=10kΩ in equation (8) as follows,

H2(s)=1(10×103)(100×106)(s+2){1k=103}=1(s+2)

Thus, the final design of the circuit is,

RfA=5kΩ, CfA=100μF, and R1A=10kΩ.

Therefore, the overall transfer function of the cascaded circuit is,

H(s)=H1(s)H2(s)

Substitute 5(s+1) for H1(s) and 1(s+2) for H2(s) in the above equation as follows,

H(s)=(5(s+1))(1(s+2))=5(s+1)(s+2)

Conclusion:

Thus, a circuit is designed which produces a transfer function of 5(s+1)(s+2).

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ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<

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