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• Syed Ali Haider

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C HAPTER 9 Section 9-1: Transfer Function Problem 9.1 Determine the resonant frequency of the circuit shown in Fig. P9.1, given that R = 100 Ω, L = 5 mH, and C = 1 µ F. L Zi

C

R

Figure P9.1: Circuit for Problem 9.1.

Solution: 

1 Zi = R k jω C



+ jω L

=

R jω C R + jω1C

=

R + jω L 1 + jω RC

=

R(1 − ω 2 LC) + jω L 1 − jω RC × 1 + jω RC 1 − jω RC

+ jω L

[R(1 − ω 2 LC) + ω 2 RLC] + j[ω L − ω R2C(1 − ω 2 LC)] 1 + ω 2 R2C2 R + jω [L − R2C + ω 2 R2 LC2 ] = . 1 + ω 2 R2C2 =

At resonance, the imaginary part of Zi is zero. Thus,

ω [L − R2C + ω 2 R2 LC2 ] = 0, which gives two solutions, a trivial resonance at ω = 0, and r 1 1 ω0 = − 2 2 LC R C r 1 1 = − 4 = 104 rad/s. −3 −6 5 × 10 × 10 10 × 10−12

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©2009 National Technology and Science Press

Problem 9.2 Determine the resonant frequency of the circuit shown in Fig. P9.2, given that R = 100 Ω, L = 5 mH, and C = 1 µ F. C Zi

L

R

Figure P9.2: Circuit for Problem 9.2.

Solution: 1 jω C 1 jω LR + = R + jω L jω C

Zi = (R k jω L) +

=

−ω 2 RLC + R + jω L . −ω 2 LC + jω RC

After a few steps of algebra, the expression simplifies to Zi =

ω 4 RL2C2 − jω [ω 2 (L2C − R2 LC2 ) + R2C] . ω 4 L2C2 + ω 2 R2C2

At resonance, the imaginary part of Zi is zero. Thus,

ω [ω 2 (L2C − R2 LC2 ) + R2C] = 0, which gives ω0 = 0 (trivial resonance) and s R2 ω0 = R2 LC − L2 s =

104 104 × 5 × 10−3 × 10−6 − 25 × 10−6

= 2 × 104 rad/s.

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Problem 9.3 For the circuit shown in Fig. P9.3, determine (a) the transfer function H = Vo /Vi , and (b) the frequency ωo at which H is purely real. C

L2

V1

+

+ L1

Vi

_

R

Vo

_

Figure P9.3: Circuit for Problem 9.3.

Solution: (a) KCL at node V1 gives: V1 − Vi V1 V1 + + = 0, ZC Z L1 R + Z L2 where ZC = 1/ jωC, ZL1 = jω L1 , and ZL2 = jω L2 . Also, voltage division gives Vo =

V1 R . R + j ω L2

Solving for the transfer function gives H=

Vo −ω 2 RL1C = . Vi R(1 − ω 2 L1C) + jω (L1 + L2 − ω 2 L1 L2C)

(b) We need to rationalize the expression for H: H=

−ω 2 RL1C R(1 − ω 2 L1C) + jω (L1 + L2 − ω 2 L1 L2C) ×

=

R(1 − ω 2 L1C) − jω (L1 + L2 − ω 2 L1 L2C) R(1 − ω 2 L1C) − jω (L1 + L2 − ω 2 L1 L2C)

−ω 2 R2 L1C(1 − ω 2 L1C) + jω 3 RL1C(L1 + L2 − ω 2 L1 L2C) . R2 (1 − ω 2 L1C)2 + ω 2 (L1 + L2 − ω 2 L1 L2C)2

The imaginary part of H is zero if ω = 0 (trivial solution) or if r L1 + L2 ω0 = . L1 L2C

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Problem 9.4 For the circuit shown in Fig. P9.4, determine (a) the transfer function H = Vo /Vi , and (b) the frequency ωo at which H is purely real. R1 Vi

+

+ _

C

L

R2

Vo

_

Figure P9.4: Circuit for Problem 9.4.

Solution: (a) KCL at mode V0 : Vo − Vi + Vo R1



1 1 + jω C + jω L R2



= 0.

Solution leads to H=

Vo jR2 ω L = . 2 Vi R1 R2 (1 − ω LC) + jω L(R1 + R2 )

(b) Rationalizing the expression for H: H= =

jR2 ω L R1 R2 (1 − ω 2 LC) − jω L(R1 + R2 ) × R1 R2 (1 − ω 2 LC) + jω L(R1 + R2 ) R1 R2 (1 − ω 2 LC) − jω L(R1 + R2 )

ω 2 R2 L2 (R1 + R2 ) + jω LR1 R22 (1 − ω 2 LC) . R21 R22 (1 − ω 2 LC)2 + ω 2 L2 (R1 + R2 )2

The imaginary part of H is zero when ω = 0 (trivial resonance) or when 1 ω0 = √ . LC

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Problem 9.6 The corner frequency of the highpass filter circuit shown in Fig. P9.6 is approximately 1 Hz. Scale the circuit up in frequency by a factor of 105 , while keeping the values of the inductors unchanged. 1Ω Vs

+ _

1F

0.3 F

0.04 H

0.04 H

1F

+ 1Ω

Vo

_

Figure P9.6: Circuit for Problem 9.6.

Solution: L′ Km =1= , L Kf Kf = 105

=⇒

Km = 105 .

C′ 1 = = 10−10 . C Km Kf R′ = Km = 105 . R 100 pF 100 kΩ Vs

+ _

30 pF

100 pF

0.04 H

100 kΩ

+ 0.04 H

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Vo

_

©2009 National Technology and Science Press

Problem 9.9 Circuit (b) in Fig. P9.9 is a scaled version of circuit (a). The scaling process may have involved magnitude or frequency scaling, or both simultaneously. If R1 = 1 kΩ gets scaled to R′1 = 10 kΩ, supply the impedance values of the other elements in the scaled circuit. j10 Ω L1

j20 Ω −j5 Ω L2

C1 R1

C2 R2

1 kΩ

−j50 Ω 2 kΩ

(a) Original circuit

L1

L2

C1 R1

10 kΩ

C2 R2

(b) Scaled circuit Figure P9.9: Circuits for Problem 9.9.

Solution: Km = From the original circuit,

R′1 = 10. R1

ZL1 = jω L1 = j10 Ω.

For the scaled circuit, ZL′1 = jω ′ L′1 = jKf ω ·

Km L1 = jKm ω L1 = j10 × 10 = j100 Ω, Kf

ZL′2 = j200 Ω, ZC1′ =

1 − jKm Kf − j10 = = = − j50 Ω, jω ′C1′ Kf ωC1 ωC1

ZC2′ = − j500 Ω,

R′2 = Km R2 = 10 × 2 kΩ = 20 kΩ. j100 Ω

j200 Ω −j50 Ω 10 kΩ

−j500 Ω 20 kΩ

Scaled circuit

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©2009 National Technology and Science Press