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This is solution manual for problems related Electrical Circuit Analysis course. It was provided at National Institute of Industrial Engineering by Prof. Sanjay Das. It includes: Active, Filter, Circuits, Prototype, Butterworth, High, Pass, Transfer, Functions, Currents, Input, Node
Typology: Exercises
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AP 15.1 H(s) = −(R 2 /R 1 )s s + (1/R 1 C)
1 R 1 C = 1 rad/s; R 1 = 1 Ω,. .· C = 1 F
..^ · Hprototype(s) = −s s + 1
AP 15.2 H(s) =
s + (1/R 2 C)
s + 5000
1 R 1 C = 20,000; C = 5 μF
15–2 CHAPTER 15. Active Filter Circuits
AP 15.3 ωc = 2πfc = 2π × 104 = 20, 000 π rad/s
..^ · kf = 20, 000 π = 62, 831. 85
kf km
kf km
..^ · km = 1 (0. 5 × 10 −^6 )(62, 831 .85)
AP 15.4 For a 2nd order prototype Butterworth high pass filter
H(s) = s^2 s^2 +
2 s + 1
For the circuit in Fig. 15.
H(s) =
s^2 s^2 +
( (^2) R 2 C
) s +
( (^1) R 1 R 2 C^2
)
Equate the transfer functions. For C = 1F,
2 R 2 C
AP 15.5 Q = 8, K = 5, ωo = 1000 rad/s, C = 1 μF
For the circuit in Fig 15.
H(s) =
) s
s^2 +
) s +
( R 1 + R 2 R 1 R 2 R 3 C^2
)
Kβs s^2 + βs + ω^2 o
β =
βC
β = ωo Q
= 125 rad/s
15–4 CHAPTER 15. Active Filter Circuits
P 15.1 Summing the currents at the inverting input node yields
0 − Vi Zi
0 − Vo Zf
..^ · Vo Zf
Vi Zi
..^ · H(s) = Vo Vi
Zf Zi
P 15.2 [a] Zf = R 2 (1/sC 2 ) [R 2 + (1/sC 2 )]
R 2 C 2 s + 1
=
s + (1/R 2 C 2 ) Likewise
Zi =
s + (1/R 1 C 1 )
..^ · H(s) = −(1/C^2 )[s^ + (1/R^1 C^1 )] [s + (1/R 2 C 2 )](1/C 1 )
= −
[s + (1/R 1 C 1 )] [s + (1/R 2 C 2 )]
[b] H(jω) =
[ jω + (1/R 1 C 1 ) jω + (1/R 2 C 2 )
]
H(j0) =
[c] H(j∞) = −
( j j
[d] As ω → 0 the two capacitor branches become open and the circuit reduces to a resistive inverting amplifier having a gain of −R 2 /R 1. As ω → ∞ the two capacitor branches approach a short circuit and in this case we encounter an indeterminate situation; namely vn → vi but vn = 0 because of the ideal op amp. At the same time the gain of the ideal op amp is infinite so we have the indeterminate form 0 · ∞. Although ω = ∞ is indeterminate we can reason that for finite large values of ω H(jω) will approach −C 1 /C 2 in value. In other words, the circuit approaches a purely capacitive inverting amplifier with a gain of (− 1 /jωC 2 )/(1/jωC 1 ) or −C 1 /C 2.
Problems 15–
P 15.3 [a] Zf =
s + (1/R 2 C 2 )
Zi = R 1 +
sC 1
s [s + (1/R 1 C 1 )]
H(s) = −
[s + (1/R 2 C 2 )]
s R 1 [s + (1/R 1 C 1 )]
= −
s [s + (1/R 1 C 1 )][s + (1/R 2 C 2 )]
[b] H(jω) = −
(^ jω jω + (^) R 11 C 1
) ( jω + (^) R 21 C 2
)
H(j0) = 0 [c] H(j∞) = 0 [d] As ω → 0 the capacitor C 1 disconnects vi from the circuit. Therefore vo = vn = 0. As ω → ∞ the capacitor short circuits the feedback network, thus Zf = 0 and therefore vo = 0.
P 15.4 [a] K = 10(10/20)^ = 3.16 =
ωcC
(2π)(10^3 )(750 × 10 −^9 )
[b]
P 15.5 [a] R 1 =
ωcC
(2π)(8 × 103 )(3. 9 × 10 −^9 )
= 5. 10 kΩ
..^ · R 2 = 5. 01 R 1 = 25. 57 kΩ
Problems 15–
P 15.7 For the RC circuit
H(s) = Vo Vi
s s + (1/RC)
R′^ = kmR; C′^ =
kmkf
kf
kf
= kf
H′(s) = s s + (1/R′C′)
s s + kf
(s/kf ) (s/kf ) + 1
For the RL circuit
H(s) = s s + (R/L)
R′^ = kmR; L′^ = kmL kf
R′ L′^ = kf
) = kf
H′(s) = s s + (R′/L′)
s s + kf
(s/kf ) (s/kf ) + 1
P 15.8 H(s) = (R/L)s s^2 + (R/L)s + (1/LC)
βs s^2 βs + ω^2 o For the prototype circuit ωo = 1 and β = ωo/Q = 1/Q. For the scaled circuit
H′(s) = (R′/L′)s s^2 + (R′/L′)s + (1/L′C′)
where R′^ = kmR; L′^ = km kf L; and C′^ =
kf km
′ L′^
kmR km kf L^
= kf
) = kf β
kf km km kf LC^
k f^2 LC = k^2 f
15–8 CHAPTER 15. Active Filter Circuits
ω′ o β′^
kf ωo kf β
therefore the Q of the scaled circuit is the same as the Q of the unscaled circuit. Also note β′^ = kf β.
..^ · H′(s) =
( (^) kf Q
) s s^2 +
( (^) kf Q
) s + k f^2
H′(s) =
( (^1) Q
) ( (^) s kf
) [( s kf
) 2
( (^) s kf
)
]
P 15.9 [a] L = 1 H; C = 1 F
R =
[b] kf = ω o′ ωo = 40,000; km =
Thus, R′^ = kmR = (0.05)(100,000) = 5 kΩ
L′^ = km kf
kmkf
= 250 pF
[c]
P 15.10 [a] Since ω^2 o = 1/LC and ωo = 1 rad/s,
C =
[b] H(s) = (R/L)s s^2 + (R/L)s + (1/LC)
H(s) = (1/Q)s s^2 + (1/Q)s + 1
15–10 CHAPTER 15. Active Filter Circuits
km =
= 40,000; kf = ω o′ ωo
Thus,
R′^ = kmR = 40 kΩ; L′^ = km kf
(0.04) = 32 mH;
kmkf
= 12. 5 nF
[b]
P 15.12 For the scaled circuit
H′(s) =
s^2 +
( (^1) L′C′
)
s^2 +
( (^) R′ L′
) s +
( (^1) L′C′
)
km kf
kmkf
k^2 f LC
; R′^ = kmR
′ L′^
= kf
)
It follows then that
H′(s) =
s^2 +
( (^) k 2 f LC
)
s^2 +
( (^) R L
) kf s + k
(^2) f LC
=
( (^) s kf
) 2
( (^1) LC
) [( s kf
) 2
( (^) R L
) ( (^) s kf
)
( (^1) LC
)]
= H(s)|s=s/kf
P 15.13 For the circuit in Fig. 15.
H(s) =
s^2 +
( (^1) LC
)
s^2 + (^) RCs +
( (^1) LC
)
Problems 15–
It follows that
H′(s) = s^2 + (^) L′^1 C′ s^2 + (^) Rs′C′ + (^) L′^1 C′
where R′^ = kmR; L′^ = km kf
kmkf
k^2 f LC
1 R′C′^
kf RC
H′(s) =
s^2 +
( (^) k 2 f LC
)
s^2 +
( (^) kf RC
) s + k^2 f LC
=
( (^) s kf
) 2
) 2
( (^1) RC
) ( (^) s kf
)
P 15.14 [a] For the circuit in Fig. P15.14(a)
H(s) =
Vo Vi
s +
s 1 Q
s
s^2 + 1 s^2 +
( (^1) Q
) s + 1
For the circuit in Fig. P15.14(b)
H(s) = Vo Vi
Qs + Qs 1 + Qs + Qs
= Q(s^2 + 1) Qs^2 + s + Q
H(s) = s^2 + 1 s^2 +
( (^1) Q
) s + 1
Problems 15–
P 15.17 From the solution to Problem 14.24, ωo = 10^6 rad/s and β = 2π(10.61) krad/s. Calculate the scale factors:
kf = ω′ o ωo
km = kf L′ L
Thus,
R′^ = kmR = (0.2)(750) = 150 Ω C′^ =
kmkf
= 2 μF
Calculate the bandwidth:
β′^ = kf β = (0.05)[2π(10. 61 × 103 )] = 3333 rad/s
To check, calculate the quality factor:
ωo β
2 π(10. 61 × 103 )
ω′ o β′^
= 15 (Checks)
P 15.18 [a] km =
= 1000; kf =
kmC′^
km kf
(1) = 200 mH
[b]
V − 10 /s 1000
1000 + (5 × 106 /s)
s
s 1000 s + 5 × 106
100 s
V = 10(s + 5000) 2 s^2 + 10, 000 s + 25 × 106
5(s + 5000) s^2 + 5000s + 12. 5 × 106
15–14 CHAPTER 15. Active Filter Circuits
Io =
25(s + 5000) s(s^2 + 5000s + 12. 5 × 106 )
=
s
s + 2500 − j 2500
s + 2500 + j 2500 K 1 = 0.01; K 2 = − 0. 005
io(t) = 10 − 10 e−^2500 t^ cos 2500t mA
Since km = 1000 and the source voltage didn’t change, the amplitude of the current is reduced by a factor of 1000. Since kf = 5000 the coefficients of t are multiplied by 5000.
P 15.19 km =
= 100; kf = ω′ o ωo
kmkf
= 8 nF
50 Ω → 5 kΩ; 700 Ω → 70 kΩ
km kf
vφ = 5 × 10 −^4 vφ
The original expression for the current:
io(t) = 1728 + 2880e−^20 t^ cos(15t + 126. 87 ◦) mA
The frequency components will be multiplied by kf = 5000:
20 → 20(5000) = 10^5 ; 15 → 15(5000) = 75, 000
The magnitudes will be reduced by km = 100:
1728 → 1728 /100 = 17.28; 2880 → 2880 /100 = 28. 80
The expression for the current in the scaled circuit is thus,
io(t) = 17.28 + 28. 80 e−^105 t^ cos(75, 000 t + 126. 87 ◦) mA
15–16 CHAPTER 15. Active Filter Circuits
[b] H(s) = −Ks (s + 1)
[c] H′(s) = − (K (s/kf ) s kf + 1
) (^) = −Ks (s + kf )
P 15.22 [a] Hhp = −s s + 1
; kf = ω′ o ω
1000(2π) 1
= 2000π
..^ · H′ hp = −s s + 2000π 1 RH CH
= 2000π;. .· RH =
(2000π)(0. 1 × 10 −^6 )
= 1. 59 kΩ
Hlp =
s + 1 ; kf = ω o′ ω
5000(2π) 1 = 10, 000 π
..^ · H′ lp = −^10 ,^000 π s + 10, 000 π 1 RLCL = 10, 000 π;. .· RL =
(10, 000 π)(0. 1 × 10 −^6 )
[b] H′(s) = −s s + 2000π
− 10 , 000 π s + 10, 000 π
= 10 , 000 πs (s + 2000π)(s + 10, 000 π)
[c] ωo =
ωc 1 ωc 1 =
√ (2000π)(10, 000 π) = 1000π
20 rad/s
H′(jωo) = (10, 000 π)(j 1000 π
(2000π + j 1000 π
20)(10, 000 π + j 1000 π
j 10
(2 + j
20)(10 + j
Problems 15–
[d] G = 20 log 10 (0.8333) = − 1. 58 dB [e]
P 15.23 [a] For the high-pass section:
kf = ω′ o ω
4000(2π) 1 = 8000π
H′(s) = −s s + 8000π
..^ · 1 R 1 (10 × 10 −^9 )
= 8000π; R 1 = 3. 98 kΩ. .· R 2 = 3. 98 kΩ
For the low-pass section:
kf = ω′ o ω
400(2π) 1
= 800π
H′(s) = − 800 π s + 800π
..^ · 1 R 2 (10 × 10 −^9 ) = 800π; R 2 = 39. 8 kΩ. .· R 1 = 39. 8 kΩ
0 dB gain corresponds to K = 1. In the summing amplifier we are free to choose Rf and Ri so long as Rf /Ri = 1. To keep from having many different resistance values in the circuit we opt for Rf = Ri = 39. 8 kΩ.
Problems 15–
[f]
P 15.24 [a] H(s) = (1/sC) R + (1/sC)
s + (1/RC)
H(jω) =
jω + (1/RC)
|H(jω)| =
ω^2 + (1/RC)^2
|H(jω)|^2 =
ω^2 + (1/RC)^2 [b] Let Va be the voltage across the capacitor, positive at the upper terminal. Then Va − Vi R 1
Solving for Va yields
Va = (R 2 + sL)Vi R 1 LCs^2 + (R 1 R 2 C + L)s + (R 1 + R 2 ) But
Vo = sLVa R 2 + sL Therefore Vo = sLVi R 1 LCs^2 + (L + R 1 R 2 C)s + (R 1 + R 2 )
H(s) = sL R 1 LCs^2 + (L + R 1 R 2 C)s + (R 1 + R 2 )
H(jω) = jωL [(R 1 + R 2 ) − R 1 LCω^2 ] + jω(L + R 1 R 2 C)
15–20 CHAPTER 15. Active Filter Circuits
|H(jω)| = √ ωL [R 1 + R 2 − R 1 LCω^2 ]^2 + ω^2 (L + R 1 R 2 C)^2
|H(jω)|^2 = ω^2 L^2 (R 1 + R 2 − R 1 LCω^2 )^2 + ω^2 (L + R 1 R 2 C)^2
= ω^2 L^2 R^21 L^2 C^2 ω^4 + (L^2 + R 12 R^22 C^2 − 2 R 12 LC + 2R 1 R 2 LC)ω^2 + (R 1 + R 2 )^2 [c] Let Va be the voltage across R 2 positive at the upper terminal. Then Va − Vi R 1
Va R 2
(0 − Va)sC + (0 − Va)sC + 0 − Vo R 3
..^ · Va = R^2 Vi 2 R 1 R 2 Cs + R 1 + R 2
and Va = − Vo 2 R 3 Cs It follows directly that
H(s) = Vo Vi
− 2 R 2 R 3 Cs 2 R 1 R 2 Cs + (R 1 + R 2 )
.. H^ · (jω) = −^2 R^2 R^3 C(jω) (R 1 + R 2 ) + jω(2R 1 R 2 C)
|H(jω)| = √^2 R^2 R^3 Cω (R 1 + R 2 )^2 + ω^24 R^21 R 22 C^2
|H(jω)|^2 = 4 R^22 R^23 C^2 ω^2 (R 1 + R 2 )^2 + 4R^21 R^22 C^2 ω^2
P 15.25 ωo = 2πfo = 400π rad/s
β = 2π(1000) = 2000π rad/s
..^ · ωc 2 − ωc 1 = 2000π
√ ωc 1 ωc 2 = ωo = 400π
Solve for the cutoff frequencies:
ωc 1 ωc 2 = 16 × 104 π^2
