CHAPTER SIXTEEN SOLUTIONS Engineering Circuit Analysis, 6 th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved 1. We have a parallel RLC with R = 1 kΩ, C = 47 µF and L = 11 mH. (a) Q o = R(C/L) ½ = 65.37 (b) f o = ω o / 2π = (LC) -½ / 2π = 221.3 Hz (c) The circuit is excited by a steady-state 1-mA sinusoidal source: The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC = C(s 2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s 2 + s/RC + 1/LC) and Z(jω) = (1/C) (jω) / (1/LC – ω 2 + jω/RC). Since V = 10 -3 Z, we note that |V| > 0 as ω → 0 and also as ω → ∞. 10 -3 ∠0 o A jω ω ωL -j/ ω ω ωC
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CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
1. We have a parallel RLC with R = 1 kΩ, C = 47 µF and L = 11 mH. (a) Qo = R(C/L)½ = 65.37
(b) fo = ωo/ 2π = (LC)-½ / 2π = 221.3 Hz (c) The circuit is excited by a steady-state 1-mA sinusoidal source: The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC = C(s2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s2 + s/RC + 1/LC) and Z(jω) = (1/C) (jω) / (1/LC – ω2 + jω/RC).
Since V = 10-3 Z, we note that |V| > 0 as ω → 0 and also as ω → ∞.
10-3∠∠∠∠ 0o A jωωωωL -j/ ωωωωC
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
2. (a) R = 1000 Ω and C = 1 µF. Qo = R(C/L)½ = 200 so L = C(R/ Qo)2 = 25 µH (b) L = 12 fH and C = 2.4 nF R = Qo (L/ C)½ = 447.2 mΩ (c) R = 121.7 kΩ and L = 100 pH C = (Qo / R)2 L = 270 aF
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
3. We take the approximate expression for Q of a varactor to be
Q ≈ ωCjRp/ (1 + ω2 Cj2 Rp Rs)
(a) Cj = 3.77 pF, Rp = 1.5 MΩ, Rs = 2.8 Ω (b) dQ/dω = [(1 + ω2 Cj
2 Rp Rs)(Cj Rp) - ωCjRp(2ωCj2 Rp Rs)]/ (1 + ω2 Cj
2 RpRs) Setting this equal to zero, we may subsequently write
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
4. Determine Q for (dropping onto a smooth concrete floor): (a) A ping pong ball: Dropped twice from 121.1 cm (arbitrarily chosen). Both times, it bounced to a height of 61.65 cm.
Q = 2πh1/ (h1 – h2) = 12.82 (b) A quarter (25 ¢). Dropped three times from 121.1 cm. Trial 1: bounced to 13.18 cm Trial 2: bounced to 32.70 cm Trial 3: bounced to 16.03 cm. Quite a bit of variation, depending on how it struck. Average bounce height = 20.64 cm, so
Qavg = 2πh1/ (h1 – h2) = 7.574 (c) Textbook. Dropped once from 121.1 cm. Didn’t bounce much at all- only 2.223 cm. Since the book bounced differently depending on angle of incidence, only one trial was performed.
Q = 2πh1/ (h1 – h2) = 6.4
All three items were dropped from the same height for comparison purposes. An interesting experiment would be to repeat the above, but from several different heights, preferrably ranging several orders of magnitude (e.g. 1 cm, 10 cm, 100 cm, 1000 cm).
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
5.
2 2
80Np/s, 1200 rad/s, Z( 2 ) 400
1200 80 1202.66 rad/s Q 7.5172
( )( ) ( )( 2 )Now, Y( ) C Y( 2 ) C2
80( 80 2400)Y( 160 1200) C Y( 160 1200)160 1200
d d
oo o
d d dd
d
j
s j s j js js j
jj jj
α ω α ω
ωωα
α ω α ω α α ωα ωα ω
= = − + = Ω
= + = ∴ = =
+ − + + − − += ∴ − + =− +
− − +∴ − + = ∴ − + =− +
2
1 1 3080C400 2 15
1 229 1 1C 15.775 F; L 43.88mH; R 396.732,000 901 C 2o
jj
Cµ
ω α−
− +=− +
∴ = = = = = = Ω
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
34. (a) (b)
(c) 6
6 1010 rad/s, Q stays the same, B 21.40 krad/s46.73
ω = ∴ = =o o
3 6 4
44 3 4
,8 , 2
4 3
2 2, ,4 6
4 6
1/ (2 8)10 10 10 rad/s
10Q 10 / 8 10 10 125 R 0.64125
10 10 102 8 10mH Q 156.250.64
1R 0.64 156.25 15.625 ; Q 100, R 100 1 1010 10
R 20 15.625 10 4.673 Q 10 10 4
− −
−
−
−
−
ω + =
= × = ∴ = = Ω
× ×+ = ∴ = =
∴ = × = Ω = = = × = Ω×
∴ = = Ω ∴ = × ×
Bo
L L S
L
L P C C P
P o
k k
k 3.673 10 46.73× =
6 4K 10 /10 100, K 1 R s stay the same; 2 mH 20 H, 8mH 80 H,1 F 10nF′= = = ∴ → µ → µ µ →f m
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
35. (a) (b)
3
0.1K 250, K 400 0.1F 1 F250 400
2 2505 1250 , 2H 1.25H, 4 I 10 I400
= = ∴ → = µ×
×Ω → Ω → = →
m f
x x
3 61250
33
6 3 6 3
33 3
110 . Apply 1 V I 10 , I1250
1 101000 I 10 I1.25
1 0.8 0.8I 10 (1 10 ) 10 ; 1012500.8 10 1 1000I 10 0.2 10 Z 5 V 0
I 0.2
−
−−
− − −
−− −
ω = ∴ = ↓ =
−∴ = ∴→ =
∴ = + + − = + =
×∴ = + = × ∴ = = = − Ω =
x
x L
in
in th ocin
s
sss
s s s s js s
j j j kj j
1 µF
1.25 H
1250 Ω 103
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
36. (a) I 2 0 A, 50 V 60 25 V= ∠ ° ω = ∴ = ∠ °s out (b) I 2 40 A, 50 V 60 65 V= ∠ ° ω = ∴ = ∠ °s out (c) I 2 40 A, 200, OTSK= ∠ ° ω = ∴s (d) K 30, I 2 40 A, 50 V 1800 65 V= = ∠ ° ω = ∴ = ∠ °m S out (e) K 30, K 4, I 2 40 A, 200 V 1800 65 V= = = ∠ ° ω = ∴ = ∠ °m f s out
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
42. (a) (b) (c)
2
2 2
2
2
20 400 20 400H( ) 1
1 2 0.5( / 20) ( / 20)400
20, 0.520 log 400 52dBCorrection at is 20 log 2 0 dB
o
o
s sss s s
s ss
ω ζ
ω ζ
+ += + + =
+ × +=
∴ = ==
=
5 : H 52 2 20 log 5 24.0dB(plot)
H 20 log 1 16 4 23.8dB (exact)
100 : H 0dB (plot)
H 20 log 1 0.04 0.2 0.170 dB (exact)
dB
dB
dB
dB
j
j
ω
ω
= = − × =
= − + =
= =
= − + = −
Hdb
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
43. (a)
(c) 0.520, H( 20) H 15.68 dB H( 20) 80.541 4 0.5 dB
jj jj
ω = = ∴ = − ∠ = − °− +
225
V 25 25 0.025H( )V 10 25 1000 / 10 25 1000 11 2
8 10 10110, 1/ 8 correction 20 log 2 12 dB8
0.025 32 dB
= = = =+ + + + + +
∴ = = ∴ = − × = → −
R
o
s sss s s s s s
ω ζ(b)
HdB ang(H)
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
44. 3 6
1 2 3 3 6
1/(50 10 10 ) 201st two stages, H ( ) H ( ) 10; H ( )1/(200 10 10 ) 5
20 400H( ) ( 10)( 10)5 1 / 5
400 52 dB
s s ss s
ss s
−
−
− × × −= = − = =+ × × +
− − ∴ = − − = + + − →
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
45. (a) 20log10(0.1) = -20 dB (b) (c)
51 1 1
15 51
5 6
5 6
1st stage: C 1 F, R , R 10 H (S) R C 0.11/ R C
2nd stage: R 10 , R 10 , C 1 F H ( )1/ R C
1/(10 10 ) 10H ( )1/(10 10 ) 10
3rd stage: same as 2nd1H( ) ( 0.1 )
A A fA A fA A
B fBB fB fB B
fB fB
B
s s
ss
ss s
s s
µ
µ
−
−
= = ∞ = ∴ = − = −
−= = = ∴ =
+
×∴ = = −+ × +
−∴ = − 2
0 10 0.110 10 (1 /10)
ss s s
− = − + + +
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
46. An amplifier that rejects high-frequency signals is required. There is some ambiguity in the requirements, as social conversations may include frequencies up to 50 kHz, and echolocation sounds, which we are asked to filter out, may begin below this value. Without further information, we decide to set the filter cutoff frequency at 50 kHz to ensure we do not lose information. However, we note that this decision is not necessarily the only correct one.
Our input source is a microphone modeled as a sinusoidal voltage source having a peak amplitude of 15 mV in series with a 1-Ω resistor. Our output device is an earphone modeled as a 1-kΩ resistor. A voltage of 15 mV from the microphone should correspond to about 1 V at the earphone according to the specifications, requiring a gain of 1000/15 = 66.7.
If we select a non-inverting op amp topology, we then need 65.7 1- 66.7 1
==RR f
Arbitrarily choosing R1 = 1 kΩ, we then need Rf = 65.7 kΩ. This completes the amplification part. Next, we need to filter out frequencies greater than 50 kHz.
Placing a capacitor across the microphone terminals will “short out” high frequencies.
We design for ωc = 2πfc = 2π(50×103) = filtermicCR
1 . Since Rmic = 1 Ω, we require
Cfilter = 3.183 µF.
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
47. We choose a simple series RLC circuit. It was shown in the text that the “gain” of the
circuit with the output taken across the resistor is ( )[ ] 2
122222 -1
RC CRLC
AV
ωω
ω
+= .
This results in a bandpass filter with corner frequencies at
LCLCCR-RC
Lc 24
22 ++=ω and LC
LCCRRCHc 2
4 22 ++=ω
If we take our output across the inductor-capacitor combination instead, we obtain the opposite curve- i.e. a bandstop filter with the same cutoff frequencies. Thus, we want
2π(20) = LC
LCCR-RC2
4 22 ++ and 2π(20×103) =
LCLCCRRC
24
22 ++
Noting that
Hcω – Lcω = R/L = 125.5 krad/s, we arbitrarily select R = 1 kΩ, so that L =
7.966 mH. Returning to either cutoff frequency expression, we then find C = 7.950 µF
PSpice verification. The circuit performs as required, with a lower corner frequency of about 20 Hz and an upper corner frequency of about 20 kHz.
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
48. We choose a simple RC filter topology:
Where RC1
1 VV
in
out
ωj+= and hence
( )2in
out
RC1
1 VV
ω+= . We desire a cutoff
frequency of 1 kHz, and note that this circuit does indeed act as a low-pass filter (higher frequency signals lead to the capacitor appearing more and more as a short circuit). Thus,
( ) 21
RC1
1 2
c
=+
=ω
where ωc = 2πfc = 2000π rad/s.
A small amount of algebra yields 1 + [2π(1000)RC]2 = 2 or 2000πRC = 1. Arbitrarily setting R = 1 kΩ, we then find that C = 159.2 nF. The operation of the filter is verified in the PSpice simulation below:
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
49. We are not provided with the actual spectral shape of the noise signal, although the reduction to 1% of its peak value (a drop of 40 dB) by 1 kHz is useful to know. If we place a simple high-pass RC filter at the input of an op amp stage, designing for a pole at 2.5 kHz should ensure an essentially flat response above 25 kHz, and a 3 dB reduction at 2.5 kHz. If greater tolerance is required, the 40 dB reduction at 1 kHz allows the pole to be moved to a frequency even closer to 1 kHz. The PSpice simulation below shows a
filter with R = 1 kΩ (arbitrarily chosen) and C = nF 63.66 )1000)(105.2(2
13 =
×π.
At a frequency of 25 kHz, the filter shows minimal gain reduction, but at 1 kHz any signal is reduced by more than 8 dB.
We therefore design a simple non-inverting op amp circuit such as the one below, which
with Rf = 100 kΩ and R1 = 1 kΩ, has a gain of 100 V/V. In simulating the circuit, a gain of approximately 40 dB at 25 kHz was noted, although the gain dropped at higher frequencies, reaching 37 dB around 80 kHz. Thus, to completely assess the suitability of design, more information regarding the frequency spectrum of the “failure” signals would be required.
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
50. We select a simple series RLC circuit with the output taken across the resistor to serve as a bandpass filter with 500 Hz and 5000 Hz cutoff frequencies. From Example 16.12, we know that
(500)2 4LC CR2LC
12LR- 22 πω =++=
Lc
and
(5000)2 4LC CR2LC
12LR 22
Hπω =++=c
With -
LH cc ωω = 2p(5000 – 500) = R/L, we (arbitrarily) select R = 1 kΩ, so that L = 35.37 mH. Substituting these two values into the equation for the high-frequency cutoff, we find that C = 286.3 nF. We complete the design by selecting R1 = 1 kΩ and Rf = 1 kΩ for a gain of 2 (no value of gain was specified). As seen in the PSpice simulation results shown below, the circuit performs as specified at maximum gain (6 dB or 2 V/V), with cutoff frequencies of approximately 500 and 5000 KHz and a peak gain of 6 dB.
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
51. For this circuit, we simply need to connect a low-pass filter to the input of a non- inverting op amp having Rf/R1 = 9 (for a gain of 10). If we use a simple RC filter, the cutoff frequency is
(3000)2 RC1 πω ==c
Selecting (arbitrarily) R = 1 kΩ, we find C = 53.05 nF. The PSpice simulation below shows that our design does indeed have a bandwidth of 3 kHz and a peak gain of 10 V/V (20 dB).
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
52. We require four filter stages, and choose to implement the circuit using op amps to isolate each filter sub- circuit. Selecting a bandwidth of 1 rad/s (no specification was given) and a simple RLC filter as suggested in the problem statement, a resistance value of 1 Ω leads to an inductor value of 1 H (bandwidth for this type of filter = ωH – ωL = R/L). The capacitance is found by designing each filter’s respective resonant frequency ( LC1 ) at the desired “notch” frequency. Thus, we require CF1 = 10.13 µF, CF2 = 2.533 µF, CF3 = 1.126 µF and CF4 = 633.3 nF. The Student Version of PSpice® will not permit more than 64 nodes, so that the total solution must be simulated in two parts. The half with the filters for notching out 50 and 100 Hz components is shown below; an additional two op amp stages are required to complete the design.
CHAPTER SIXTEEN SOLUTIONS
Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved
53. Using the series RLC circuit suggested, we decide to design for a bandwidth of 1 rad/ s (as no specification was provided). With ωH – ωL = R/ L, we arbitrarily select R = 1 Ω so that L = 1 H. The capacitance required is obtained by setting the resonant frequency of the circuit ( LC1 ) equal to 60 Hz (120π rad/s). This yields C = 7.04 µF.