HIGHER-ORDER ACTIVE CIRCUITS

LC ¼ 0 A solution is of the form

8.11 HIGHER-ORDER ACTIVE CIRCUITS

Application of circuit laws to circuits which contain op amps and several storage elements produces, in general, several first-order differential equations which may be solved simultaneously or be reduced to a higher-order input-output equation. A convenient tool for developing the equations is the complex frequency s (and generalized impedance in the s-domain) as used throughout Sections 8.5 to 8.10.

Again, we assume ideal op amps (see Section 7.16). The method is illustrated in the following examples.

EXAMPLE 8.13 FindHðsÞ ¼V₂=V₁in the circuit of Fig. 8-41 and show that the circuit becomes a noninverting integrator if and only ifR₁C₁¼R₂C₂.

Apply voltage division, in the phasor domain, to the input and feedback paths to find the voltages at the terminals of the op amp.

At terminal A: V_A¼ 1 1þR₁C₁sV₁ At terminal B: V_B¼ R₂C₂s

1þR₂C₂sV₂ ButV_A¼V_B. Therefore,

V₂

V₁¼ 1þR₂C₂s ð1þR₁C₁sÞR₂C₂s Only ifR₁C₁¼R₂C₂¼RCdo we get an integrator with a gain of 1=RC

V₂ V₁¼ 1

RCs; v₂¼ 1 RC

ðt 1

v₁dt

EXAMPLE 8.14 The circuit of Fig. 8-42 is called an equal-component Sallen-Key circuit. Find HðsÞ ¼V₂=V₁ and convert it to a differential equation.

Write KCL at nodes A and B.

At node A: V_AV₁

R þV_AV_B

R þ ðV_AV₂ÞCs¼0 At node B: V_BV_A

R þV_BCs¼0

Let 1þR₂=R₁¼k, thenV₂¼kV_B. EliminatingV_AandV_B between the above equations we get V₂

V₁¼ k

R²C²s²þ ð3kÞRCsþ1 R²C²d²v₂

dt² þ ð3kÞRC dv₂

dt þv₂¼kv₁

EXAMPLE 8.15 In the circuit of Fig. 8-42 assumeR¼2k,C¼10nF, andR₂¼R₁. Findv₂ifv₁¼uðtÞ.

By substituting the element values inHðsÞfound in Example 8.14 we obtain V₂

V₁¼ 2

410¹⁰s²þ210⁵sþ1 d²v₂

dt² þ510⁴dv₂

dt þ2510⁸v₂¼510⁹v₁ The response of the preceding equation fort>0 tov₁¼uðtÞis

v₂¼2þe^tð2 cos!t2:31 sin!tÞ ¼2þ3:055e^tcosð!tþ130:98Þ where¼25 000 and!¼21 651 rad/s.

EXAMPLE 8.16 Find conditions in the circuit of Fig. 8-42 for sustained oscillations inv₂ðtÞ(with zero input) and find the frequency of oscillations.

In Example 8.14 we obtained

V₂

V₁¼ k

R²C²s²þ ð3kÞRCsþ1

For sustained oscillations the roots of the characteristic equation in Example 8.14 should be imaginary numbers.

This happens whenk¼3 orR₂¼2R₁, in which case!¼1=RC.

Solved Problems

8.1 A seriesRLC circuit, withR¼3 k,L¼10 H, andC¼200mF, has a constant-voltage source, V¼50 V, applied at t¼0. (a) Obtain the current transient, if the capacitor has no initial charge. (b) Sketch the current and find the time at which it is a maximum.

¼ R

2L¼150 s¹ !²₀¼ 1

LC¼500 s² ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ²!²₀ q

¼148:3 s¹ ðaÞ

The circuit is overdampedð > !₀Þ.

s₁¼ þ¼ 1:70 s¹ s₂¼ ¼ 298:3 s¹

i¼A₁e^1:70tþA₂e^298:3t and

Since the circuit contains an inductance,ið0^þÞ ¼ið0Þ ¼0; also,Qð0^þÞ ¼Qð0Þ ¼0. Thus, att¼0^þ, KVL gives

0þ0þLdi

dt ^0þ¼V or di dt ^0þ¼V

L¼5 A=s

Applying these initial conditions to the expression fori, 0¼A₁ð1Þ þA₂ð1Þ

5¼ 1:70A₁ð1Þ 298:3A₂ð1Þ from whichA₁¼ A₂¼16:9 mA.

i¼16:9ðe^1:70te^298:3tÞ ðmAÞ (b) For the time of maximum current,

di

dt¼0¼ 28:73e^1:70tþ5041:3e^298:3t Solving by logarithms,t¼17:4 ms. See Fig. 8-18.

Fig. 8-18

8.2 A seriesRLCcircuit, withR¼50;L¼0:1 H;andC¼50mF, has a constant voltageV¼100 V applied att¼0. Obtain the current transient, assuming zero initial charge on the capacitor.

¼ R

2L¼250 s¹ !²₀¼ 1

LC¼2:010⁵s² ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ²!²₀ q

¼j370:8 rad=s

This is an oscillatory caseð < !₀Þ, and the general current expression is i¼e^250tðA₁cos 370:8tþA₂sin 370:8tÞ The initial conditions, obtained as in Problem 8.1, are

ið0^þÞ ¼0 di dt ₀þ

¼1000 A=s and these determine the values:A₁¼0,A₂¼2:70 A. Then

i¼e^250tð2:70 sin 370:8tÞ ðAÞ

8.3 Rework Problem 8.2, if the capacitor has an initial chargeQ₀¼2500mC.

Everything remains the same as in Problem 8.2 except the second initial condition, which is now 0þLdi

dt ₀þ

þQ₀

C ¼V or di dt ₀þ

¼100 ð2500=50Þ

0:1 ¼500 A=s The initial values are half those in Problem 8.2, and so, by linearity,

i¼e^250tð1:35 sin 370:8tÞ ðAÞ

8.4 A parallelRLC network, withR¼50:0, C¼200mF, andL¼55:6 mH, has an initial charge Q₀¼5:0 mC on the capacitor. Obtain the expression for the voltage across the network.

¼ 1

2RC¼50 s¹ !²₀¼ 1

LC¼8:9910⁴s² Since!²₀> ², the voltage function is oscillatory and so!_d¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

!²₀² q

¼296 rad/s. The general voltage expression is

v¼e^50tðA₁cos 296tþA₂sin 296tÞ

WithQ₀¼5:010³C,V₀¼25:0 V. Att¼0,v¼25:0 V. Then,A₁¼25:0.

dv

dt¼ 50e^50tðA₁cos 296tþA₂sin 296tÞ þ296e^50tðA₁sin 296tþA₂cos 296tÞ Att¼0,dv=dt¼ V₀=RC¼!_dA₂A₁, from whichA₂¼ 4:22. Thus,

v¼e^50tð25:0 cos 296t4:22 sin 296tÞ ðVÞ

8.5 In Fig. 8-19, the switch is closed att¼0. Obtain the current i and capacitor voltage v_C, for t>0.

As far as the natural response of the circuit is concerned, the two resistors are in parallel; hence, ¼R_eqC¼ ð5Þð2mFÞ ¼10ms

By continuity,v_Cð0^þÞ ¼v_Cð0Þ ¼0. Furthermore, ast! 1, the capacitor becomes an open circuit, leav- ing 20in series with the 50 V. That is,

ið1Þ ¼50

20¼2:5 A v_Cð1Þ ¼ ð2:5 AÞð10Þ ¼25 V

Knowing the end conditions onv_C, we can write

v_C¼ ½v_Cð0^þÞ v_Cð1Þe^t=þv_Cð1Þ ¼25ð1e^t=10Þ ðVÞ whereintis measured inms.

The current in the capacitor is given by i_C¼Cdv_C

dt ¼5e^t=10 ðAÞ and the current in the parallel 10-resistor is

i₁₀¼ v_C

10¼2:5ð1e^t=10Þ ðAÞ i¼i_Cþi₁₀¼2:5ð1þe^t=10Þ ðAÞ Hence,

The problem might also have been solved by assigning mesh currents and solving simultaneous differ- ential equations.

8.6 For the time functions listed in the first column of Table 8-2, write the corresponding amplitude and phase angle (cosine-based) and the complex frequencys.

See columns 2 and 3 of the table.

8.7 For each amplitude and phase angle in the first column and complex frequencysin the second column in Table 8-3, write the corresponding time function.

See column 3 of the table.

Fig. 8-19

Table 8-2

Time Function A 8 s

iðtÞ ¼86:6 A 86:6 08 A 0

iðtÞ ¼15:0e^2103t ðAÞ 15:0 08 A 210³Np/s vðtÞ ¼25:0 cosð250t458Þ ðVÞ 25:0 458 V j250 rad/s vðtÞ ¼0:50 sinð250tþ308Þ ðVÞ 0:50 608 V j250 rad/s iðtÞ ¼5:0e^100tsinð50tþ908Þ ðAÞ 5:0 08 A 100j50 s¹ iðtÞ ¼3 cos 50tþ4 sin 50t ðAÞ 5 53:138 A j50 rad/s

Table 8-3

A 8 s Time Function

10 08 þj120 10 cos 120 t

2 458 j120 2 cosð120 tþ458Þ 5 908 2 j50 5e^2tcosð50t908Þ 15 08 5000 j1000 15e^5000tcos 1000t

100 308 0 86.6

8.8 An amplitude and phase angle of 10 ffiffiffi p2

458V has an associated complex frequency s¼ 50þj100 s¹. Find the voltage att¼10 ms.

vðtÞ ¼10 ffiffiffi 2 p

e^50tcosð100tþ458Þ ðVÞ Att¼10²s, 100t¼1 rad¼57:38, and so

v¼10 ffiffiffi 2 p

e^0:5cos 102:38¼ 1:83 V

8.9 A passive network contains resistors, a 70-mH inductor, and a 25-mF capacitor. Obtain the respective s-domain impedances for a driving voltage (a) v¼100 sinð300tþ458Þ ðVÞ, (b) v¼100e^100tcos 300tðVÞ.

(a) Resistance is independent of frequency. Ats¼j300 rad/s, the impedance of the inductor is sL¼ ðj300Þð7010³Þ ¼j21

and that of the capacitor is

1

sC¼ j133:3 (b) Ats¼ 100þj300 s¹,

sL¼ ð100þj300Þð7010³Þ ¼ 7þj21 1

sC¼ 1

ð100þj300Þð2510⁶Þ¼ 40j120

8.10 For the circuit shown in Fig. 8-20, obtainvat t¼0:1 s for source current ðaÞ i¼10 cos 2t(A), (b) i¼10e^tcos 2t(A).

Z_inðsÞ ¼2þ2ðsþ2Þ

sþ4 ¼ ð4Þsþ3 sþ4 (a) Ats¼j2 rad/s,Z_inðj2Þ ¼3:22 7:138. Then,

V¼IZ_in¼ ð10 08Þð3:22 7:138Þ ¼32:2 7:138 V or v¼32:2 cosð2tþ7:138Þ ðVÞ andvð0:1Þ ¼32:2 cosð18:598Þ ¼30:5 V.

(b) Ats¼ 1þj2 s¹,Z_inð1þj2Þ ¼3:14 11:318. Then

V¼IZ_in¼31:4 11:318 V or v¼31:4e^tcosð2tþ11:318Þ ðVÞ andvð0:1Þ ¼31:4e^0:1cos 22:778¼26:2 V.

8.11 Obtain the impedance Z_inðsÞ for the circuit shown in Fig. 8-21 at (a) s¼0, (b) s¼j4 rad/s, (c) jsj ¼ 1.

Fig. 8-20 Fig. 8-21

Z_inðsÞ ¼2þ

2ðsþ1Þ 4 s 2ðsþ1Þ þ4

s

¼ ð2Þs²þ3sþ4 s²þsþ2

(a) Z_inð0Þ ¼4, the impedance offered to a constant (dc) source in the steady state.

Z_inðj4Þ ¼2ðj4Þ²þ3ðj4Þ þ4

ðj4Þ²þj4þ2 ¼2:33 29:058 ðbÞ

This is the impedance offered to a source sin 4tor cos 4t.

(c) Z_inð1Þ ¼2. At very high frequencies the capacitance acts like a short circuit across theRLbranch.

8.12 Express the impedance ZðsÞ of the parallel combination of L¼4 H and C¼1 F. At what frequenciessis this impedance zero or infinite?

ZðsÞ ¼ ð4sÞð1=sÞ 4sþ ð1=sÞ¼ s

s²þ0:25

By inspection,Zð0Þ ¼0 andZð1Þ ¼0, which agrees with our earlier understanding of parallelLCcircuits at frequencies of zero (dc) and infinity. ForjZðsÞj ¼ 1,

s²þ0:25¼0 or s¼ j0:5 rad=s

A sinusoidal driving source, of frequency 0.5 rad/s, results in parallel resonance and an infinite impedance.

8.13 The circuit shown in Fig. 8-22 has a voltage source connected at terminalsab. The response to the excitation is the input current. Obtain the appropriate network functionHðsÞ.

HðsÞ ¼ response excitation¼ IðsÞ

VðsÞ 1 ZðsÞ ZðsÞ ¼2þð2þ1=sÞð1Þ

2þ1=sþ1¼8sþ3

3sþ1 from which HðsÞ ¼ 1

ZðsÞ¼3sþ1 8sþ3

8.14 ObtainHðsÞfor the network shown in Fig. 8-23, where the excitation is the driving currentIðsÞ and the response is the voltage at the input terminals.

Applying KCL at junctiona,

IðsÞ þ2IðsÞ ¼s

5V⁰ðsÞ or V⁰ðsÞ ¼15 s IðsÞ

Fig. 8-22 Fig. 8-23

At the input terminals, KVL gives

VðsÞ ¼2sIðsÞ þV⁰ðsÞ ¼ 2sþ15 s

IðsÞ

HðsÞ ¼VðsÞ

IðsÞ ¼2s²þ15 Then s

8.15 For the two-port network shown in Fig. 8-24 find the values ofR₁, R₂, and C, given that the voltage transfer function is

H_vðsÞ V_oðsÞ

V_iðsÞ¼ 0:2 s²þ3sþ2

The impedance looking intoxx⁰is

Z⁰¼ ð1=sCÞðR₁þR₂Þ

ð1=sCÞ þR₁þR₂¼ R₁þR₂ 1þ ðR₁þR₂ÞCs Then, by repeated voltage division,

V_o V_i¼ V_o

V_xx⁰

V_xx⁰ V_i

¼ R₂ R₁þR₂

Z⁰ Z⁰þs1

¼ R₂=ðR₁þR₂ÞC s²þ 1

ðR₁þR₂ÞCsþ1 C

Equating the coefficients in this expression to those in the given expression forH_vðsÞ, we find:

C¼1

2F R₁¼3

5 R₂¼ 1

15

8.16 Construct the pole-zero plot for the transfer admittance function HðsÞ ¼ I_oðsÞ

V_iðsÞ¼s²þ2sþ17 s²þ3sþ2 In factored form,

HðsÞ ¼ðsþ1þj4Þðsþ1j4Þ ðsþ1Þðsþ2Þ Poles exist at1 and2; zeros at1 j4. See Fig. 8-25.

Fig. 8-24

8.17 Obtain the natural frequencies of the network shown in Fig. 8-26 by driving it with a conveniently located current source.

The response to a current source connected atxx⁰ is a voltage across these same terminals; hence the network functionHðsÞ ¼VðsÞ=IðsÞ ¼ZðsÞ. Then,

1 ZðsÞ¼1

1þ 1 2=sþ 1

2þ4s¼ 1 2

s²þ2:5sþ1:5 sþ0:5 ZðsÞ ¼ ð2Þ sþ0:5

s²þ2:5sþ1:5¼ ð2Þ sþ0:5 ðsþ1Þðsþ1:5Þ Thus,

The natural frequencies are the poles of the network function,s¼ 1:0 Np=s¼2 ands¼ 1:5 Np/s.

8.18 Repeat Problem 8.17, now driving the network with a conveniently located voltage source.

The conductor at yy⁰ in Fig. 8-26 can be opened and a voltage source inserted. Then, HðsÞ ¼IðsÞ=VðsÞ ¼1=ZðsÞ:

The impedance of the netework at terminalsyy⁰ is ZðsÞ ¼2þ4sþ 1ð2=sÞ

1þ2=s¼ ð4Þs²þ2:5sþ1:5 sþ2 HðsÞ ¼ 1

ZðsÞ¼ 1 4

sþ2 s²þ2:5sþ1:5 Then;

Fig. 8-25

Fig. 8-26

The denominator is the same as that in Problem 8.17, with the same roots and corresponding natural frequencies.

8.19 A 5000-rad/s sinusoidal source, V¼100 08V in phasor form, is applied to the circuit of Fig. 8-27. Obtain the magnitude-scaling factorK_mand the element values which will limit the current to 89 mA (maximum value).

At!¼5000 rad/s,

Z_in¼j!L₁þ

ðj!L₂Þ Rþ 1 j!C

j!L₂þRþ 1 j!C

¼j0:250þðj0:500Þð0:40j0:80Þ

0:40j0:30 ¼1:124 69:158

ForjVj ¼100 V,jIj ¼100=1:124¼89:0 A. Thus, to limit the current to 8910³A, the impedance must be increased by the factorK_m¼10³.

The scaled element values are as follows: R¼10³ð0:4Þ ¼400, L₁¼10³ð50mHÞ ¼50 mH, L₂¼10³ð100mHÞ ¼100 mH, andC¼ ð250mFÞ=10³¼0:250mF.

8.20 Refer to Fig. 8-28. Obtain HðsÞ ¼V_o=V_i for s¼j410⁶rad/s. Scale the network with K_m¼10³ and compareHðsÞfor the two networks.

At!¼410⁶rad/s,X_L¼ ð410⁶Þð0:510³Þ ¼2000. Then, HðsÞ ¼V_o

V_i¼ j2000

2000þj2000¼ 1 ffiffiffi2 p 458

After magnitude scaling, the inductive reactance is 10³ð2000Þ ¼2 and the resistance is 10³ð2 kÞ ¼2. Thus

HðsÞ ¼ j2 2þj2¼ 1

ffiffiffi2 p 458 Fig. 8-27

Fig. 8-28

The voltage transfer function remains unchanged by magnitude scaling. In general, any dimensionless transfer function is unaffected by magnitude scaling; a transfer function having unitsis multiplied byK_m; and a function having units S is multiplied by 1=K_m.

8.21 A three-element series circuit containsR¼5, L¼4 H, andC¼3:91 mF. Obtain the series resonant frequency, in rad/s, and then frequency-scale the circuit with K_f ¼1000. Plot jZð!Þj for both circuits.

Before scaling,

!₀¼ 1 ffiffiffiffiffiffiffi LC

p ¼8 rad=s and Zð!₀Þ ¼R¼5 After scaling,

R¼5 L¼ 4 H

1000¼4 mH C¼3:91 mF

1000 ¼3:91mF

!₀¼1000ð8 rad=sÞ ¼8000 rad=s Zð!₀Þ ¼R¼5

Thus, frequency scaling by a factor of 1000 results in the 5-impedance value being attained at 8000 rad/s instead of 8 rad/s. Any other value of the impedance is likewise attained, after scaling, at a frequency 1000 times that at which it was attained before scaling. Consequently, the two graphs ofjZð!Þjdiffer only in the horizontal scale—see Fig. 8-29. (The same would be true of the two graphs of_Zð!Þ.)

Supplementary Problems

8.22 In theRLCcircuit of Fig. 8-30, the capacitor is initially charged toV₀¼200 V. Find the current transient after the switch is closed att¼0. Ans: 2e^1000tsin 1000t ðAÞ

8.23 A seriesRLCcircuit, withR¼200,L¼0:1 H, andC¼100mF, has a voltage source of 200 V applied at t¼0. Find the current transient, assuming zero initial charge on the capacitor.

Ans: 1:055ðe^52te^1948tÞ ðAÞ

8.24 What value of capacitance, in place of the 100mF in Problem 8.23, results in the critically damped case?

Ans: 10mF

Fig. 8-29

8.25 Find the natural resonant frequency,jj, of a seriesRLCcircuit withR¼200,L¼0:1 H,C¼5mF.

Ans: 1000 rad/s

8.26 A voltage of 10 V is applied att¼0 to a seriesRLCcircuit withR¼5,L¼0:1 H,C¼500mF. Find the transient voltage across the resistance. Ans: 3:60e^25tsin 139t ðVÞ

8.27 In the two-mesh circuit shown in Fig. 8-31, the switch is closed att¼0. Findi₁andi₂, fort>0.

Ans: i₁¼0:101e^100tþ9:899e^9950t ðAÞ;i₂¼ 5:05e^100tþ5:00þ0:05e^9950t ðAÞ

8.28 A voltage has thes-domain representation 100 308V. Express the time function for (a) s¼ 2 Np/s, (b) s¼ 1þj5 s¹. Ans: ðaÞ 86:6e^2t ðVÞ; ðbÞ 100e^tcosð5tþ308Þ ðVÞ

8.29 Give the complex frequencies associated with the currentiðtÞ ¼5:0þ10e^3tcosð50tþ908Þ ðAÞ.

Ans: 0;3j50 s¹

8.30 A phasor current 25 408A has complex frequency s¼ 2þj3 s¹. What is the magnitude of iðtÞ at t¼0:2 s? Ans: 4:51 A

8.31 Calculate the impedance ZðsÞ for the circuit shown in Fig. 8-32, at (a) s¼0; ðbÞ s¼j1 rad/s, (c) s¼j2 rad/s, (d) jsj ¼ 1. Ans: ðaÞ 1; ðbÞ 1:58 18:438; ðcÞ 1:84 12:538; ðdÞ 2

8.32 The voltage source in thes-domain circuit shown in Fig. 8-33 has the time-domain expression v_iðtÞ ¼10e^tcos 2t ðVÞ

Obtaini_oðtÞ. Ans: 7:07e^tcosð2tþ98:138Þ ðAÞ

8.33 In the time domain, a series circuit ofR,L, andChas an applied voltagev_iand element voltagesv_R,v_L, and v_C. Obtain the voltage transfer functions (a) V_RðsÞ=V_iðsÞ, (b) V_CðsÞ=V_iðsÞ:

Fig. 8-30 Fig. 8-31

Fig. 8-32 Fig. 8-33

Ans: ðaÞ Rs=L s²þR

Lsþ 1 LC

; ðbÞ 1=LC s²þR

Lsþ 1 LC

8.34 Obtain the network functionHðsÞfor the circuit shown in Fig. 8-34. The response is the voltageV_iðsÞ.

Ans: ðsþ7j2:65Þðsþ7þj2:65Þ ðsþ2Þðsþ4Þ

8.35 Construct thes-plane plot for the transfer function of Problem 8.34. EvaluateHðj3Þfrom the plot.

Ans: See Fig. 8-35.

ð7:02Þð9:0Þ 2:86þ38:918

ð3:61Þð5:0Þ 56:318þ36:878¼3:50 51:418

8.36 ObtainHðsÞ ¼V_iðsÞ=I_iðsÞfor the circuit shown in Fig. 8-36 and construct the pole-zero plot.

Ans: HðsÞ ¼sðs²þ1:5Þ

s²þ1 : See Fig. 8-37.

8.37 Write the transfer functionHðsÞwhose pole-zero plot is given in Fig. 8-38.

Ans: HðsÞ ¼k s²þ50sþ400 s²þ40sþ2000

Fig. 8-34

Fig. 8-35

8.38 The pole-zero plot in Fig. 8-39 shows a pole ats¼0 and zeros ats¼ 50j50. Use the geometrical method to evaluate the transfer function at the test pointj100.

Ans: Hðj100Þ ¼223:6 26:578

8.39 A two-branch parallel circuit has a resistance of 20in one branch and the series combination ofR¼10 andL¼0:1 H in the other. First, apply an excitation,I_iðsÞ, and obtain the natural frequency from the denominator of the network function. Try different locations for applying the current source. Second, insert a voltage source,V_iðsÞ, and obtain the natural frequency. Ans: 300 Np/s in all cases

8.40 In the network shown in Fig. 8-40, the switch is closed att¼0. Att¼0^þ,i¼0 and di

dt¼25 A=s

Obtain the natural frequencies and the complete current,i¼i_nþi_f. Ans: 8:5 Np/s,23:5 Np/s;i¼ 2:25e^8:5t0:25e^23:5tþ2:5 ðAÞ

Fig. 8-36 Fig. 8-37

Fig. 8-38 Fig. 8-39