370
P13.4-16 P13.4-17
We require
p
p
k C
Then R R and C
2001
5001
34 50
1 5 50 0 04
1
1 1
2
2 2
1 2
1 1 2 2
= =
= =
= = == = = =
C R
C R
dB R
Pick C F k k Fµ µ. , . .Ω Ω
H jj
j j j( )
( )
( )( )( )ω ω
ω ω ω = +
+ + +
10 50 1
2 1 50 1 80 1
Pick the appropriate circuit from Table 13.4-2.
Hk
where
R
pR
( )ωω ω
=
1+ jp
1+ jp
k = C R
p 1
C
= 1
C
1 2
1 2
11
2
−
=1
22
371
P13.4-18 P13.4-19
ϕ ω ω ω ω = − − −− − − −tan tan tan tan1 1 1 150 2 20 80
At
ω
ω
=
=+
=+
=
⇒ = = −
−
−
−
10 000
1
1
1
1 10 10 1010
20 20 10 40
22
4 3 62
2
2
,
log log
HR C
H dB
H ( )( )
( ),
ω ωω ω ω= = −
+= −
+
V
V
R R
j R C j
o
s
2 1
2110
110 000
HR R
j R CR R
j R CTotal1 22 1
21 22
2 12
2
2
11
1, , ( ) H H = −+
⇒ = =+
ω ω
At low frequency
H R R
Now also R C let C F
R k
T T
,
,
,
( ) ( )
for H need R R
so
R
= ∴ = == = =
∴ = = =−
2 12
1 2
1 2
1 26
1
1000 1 1
1 1000 10 1
ω µ
Ω
(a)
(b)
(a)
(b) 10=20dB
(c) 10,000 rad/sec
372
P13.4-20 P13.4-21
Assume V then V V V
ji o
Use s
~ ,− = ==
+ −0
ω
Voltage divider yields : V V V
o
s o o
sCR sC
sRC V
KCL at V V V mR V V R V V snC
=+
⇒ = +
− + − + − =
1 1 0
1 1 1 1
11
1
0
/
: ( )
⇒ = =− +
=
=+
H j VV j Q mnRC
mn
m
o
so o
( )ωω ω ω ω
ω where
Q
1
1
1
1
2 0
Plugging V into above yields
1
VmR
sCsC
ms nRC
V
mR
V
V s m RC nmR C s
os
o
s
1
1
1 1
2 2
2 2 2
+ + +
=
∴ =+ + +( )
373
Section 13-5: Resonant Circuits P13.5-1
Vj C
Rj C
V
V V
Rj C V V
j C R j C R C R V Vo a
a sa o
o o s
( ) ( )
( ) ( )( ( ) ( ))
( )( )
ω ω
ω
ω
ω ω ω ω ω
ω ω ω
V j
=+
= − + −
⇒ + + = +
1
1
0
1 1
2
2
11
1 1 2 2 1 1
T(V
V C R j C C R R j
der poles with
o
s
o
ω ωω ω ω ω ω
ω δ
)( )
.
.
Second or and
= =+ −
=− + +
= =
1
1
1
8 1
0 42 2
21 2 1 2
2
ω06
1 1
1120
130
10
60
x
k rad= =
=−LC
sec
Q CL
Q Qk
Q Qk
BW RC
= =×
=
= − +
+ =
= +
+ =
= =×
=
−
−
R
rad s
rad s
k rad s
10 000
130
10
1120
20
2 25852
2 26152
1 1
100001
3010
3
6
10 0
2
02
20 0
2
02
6
,
.
.
ω ω ω ω
ω ω ω ω
Notice that BW = = Q2
0ω ω ω− 1
374
P13.5-2 P13.5-3 P13.5-4 P13.5-5 P13.5-6
R H
rad s
Hk
Q
At rad s H
Q
= = =⋅
=
=
=
+ −
= = =
=
+ −
⇒ =
−
−
k
, so
Q
( )
( )
. ( ).
,
..
ω
ω
ωωω
ωω
ω ω
0 3
0
2
0
02
3
22
8
20 10400
1000
1
897 64
2010200
200400
1897 61000
1000897 6
8
Ω
Now
LC
CL
11000
400 8
0
Q
L = 50 mHC = 20 F
= =
= =
⇒ω
µ
ω05 41
101
10 10 rad s Q BW rad s= = = = = =LC R
L
C
R
L, ,
ω04 31
101
10 10 Q BW = = = = = =LC
rad sR
L
C
R
Lrad s, ,
R
CF
LmH
= =
= = ⇒ =
⋅= = ⇒ =
−
Z
BW C
L
ω
µ
ω
0
60
100
1100
500 20
1
20 102500 8
Ω
R Y
LL H
CF
= =
= = ⇒ =
= = ⇒ =
BW
C
1100
100500 0 2
1
0 22500 0 8
0
0
ω
ω µ
Ω
Ω.
..
375
P13.5-7 P13.5-8
Y C R j L R
R R CLR j L CR R
R R j L
R j L
R j L
R R R CLR L L CR R j R LCR R j L R R CLR
R R L
ω ωω
ω ω
ω
ω
ω
ω ω ω ω ω
ω
j
= ++
+
=+ − + +
+× −
−
=+ − − + + − + −
−
1 1
1 2
1 22
2 1 2
2 1
1
1
1 1 22
22
1 2 1 1 2 1 22
2
2 12 2
( )
( )
ω ω ω
ω
ω
is the fr which the imaginary part of is zero
rad sec
=
− + − =
= − =
0
1 1 2 1 2 02
2
02 1
22
22
0
12 9
equency at Y
R LCR R L R R CLR
LR CR R
CL RM
:
.
( )( )( )
( )( ) ( )100 100
5100 2 135 10 2 135100 100 1000 0 1000 0 1000 90
100 100 100 2 135 100 50 2 135100100 100
| | 1000
j
jVo j jj
j
V Vo
−° °∠− ∠−− °° °= ∠ = ∠ = = ∠° °− ∠− + ∠−+
−
∴ =
(a) Using voltage divider
(b) Do a source transformation to obtain
376
P13.5-9 P13.5-10
So have an RLC resonant circuit with ω0 1 400= =LC
Z j LG j C
R
R G LC j LG CR
G j C
~; R G
= + ++
=
=+ − + +
+
1
2
2 2
1 22
2 1
2
11
1
ωω
ω ω ω
ω
or LG CR
R G LC
CG
thus LG CR
R G LC
CG
C LG
LCC G L
tan tan
&
− −++ −
=
++ −
= ⇒ = − >
1 2 1
1 22
1
2
2 1
1 22
2
2 22
2 22
1
1
ω ωω
ω
ω ωω
ω ω
with R rad s
C L
LCC mF L mH
and C G L checks
1 2 0
02 4
2
22
1 100
10 10 5
R and
if
= = =
= = − = ⇒ =
>
Ω ω
ω ,
Z j
in
in
LR j C
R j C
ZR RLC j L
j RC
= ++
=− +
+
ω ωω
ω ω
ω
/1
1
2
∴ =− +
+ | |Z
R RLC L
RCin
ω ω
ω
2 2 2
21
at resonan Zce = Z 0~
∠ °
(a)
Therefore the circuit is operated at resonance.This means that the L - C parallel combination has an overall Z =
and hence I = I . When R is suddenly changed from 100 to 1 k , due to the resonance condition, V R I
suddenly increases by a factor of 10. This in turn causes very large (equal & opposite ) currents in the L& C as the
capacitor voltage is forced to abruptly change. Thus very large currents radiate electromagnetic radition and
thus see sparks. Clearly we need a variable capacitor that varies as R varies such that when R 100 ,
and thus have V = R I constant while both R and I change abruptly.
s o L o L o
L L
o o L o L o
∞=
≠≠
Ω Ω
ΩΩ Ω
377
P13.5-11
at ω = =+
11
1 2LC Z
C L R C Lin, | |
( )
Let V
R
ω ω
ωωω
ω ω
ωθ θ
ωθ θ
A A and V B
Y I
V
V V
V
A B
AR
A Bcos j Bsin
AR
Y A Bcos Bsin
AR
2
2
= ∠θ = = ∠θ
= =
−
= − ∠θ
= − −
=− +
°
| |
2 2
(b)
(c)
378
PSpice Problems SP 13-1
3 600 10 0.6
w 2 3.1823 20 sec
g
at the frequency
radπ
−= × =
= × =
This simulation shows that the gain is
379
SP 13-2
The phase shift is 45
2 3.1831 20 sec
as required.
at the frequency
radω π
°−
= × =
Here is a simulation of the circuit when R = 50k
380
SP 13-3
This simulation shows that the gain of the circuit is 0.1 at 10Hz and 0.995 at 10,000 Hz.
To satisfy the specifications on the corner frequencies, the gain must be ≤ 1.414 at 200 Hz and ≥ 7.07 at 2000 Hz.
Both conditions are met.
The circuit satisfies the specifications.
381
SP 13-4
16 420
dB dB
decadedB decade
− − =( )
This simulation shows
• The high-frequency gain is 33.928 ≅ 34dB
• The slope of the low-frequency asymptote is
382
SP 13-5
The peak of the frequency response is 72dB = 4000 at 2.25Hz = 14130rad / sec. So k = 4000 and
and are ident
ωω ω
ω
ω ω
0
1 2
0
2 1
0
2 1
14130
72 3 69
2250
2332 4 2172114
=− =
=−
=−
=−
=
ified as the frequencies where the gain is dB
Q f
f f . .
383
SP 13-6
R3 and the dependent source E1 model an ideal op amp. V(R is v V is v V is the an4 1 7 1 60: ) ( ). ( : ) . ( : )s t R t V t swer given in this solution manual. After the transient part dies out, V(R7:1) is identical to V(V6:t). The answer is
correct.
384
SP 13-7 SP 13-8
Is 1 0 ac 1 R1 2 0 600 C1 1 0 114n C2 2 0 37.9n L1 1 2 36.37m .ac dec 50 100 10k .probe .end
V1 1 0 ac 1 R2 1 2 1.35k L3 2 3 500m R4 3 0 47k E5 4 0 3 0 10 R6 4 6 1 R7 4 5 213m C8 5 6 353u C9 6 0 3.53m .ac dec 50 20 20k .probe .End
385
When 630 rad / sec, T which agrees with the tabulated values of T( ) corresponding to
= 200 and 400 rad sec.
ω ω ω
ω
< − ~ 1
10
SP 13-9 Verification Problems VP 13-1
VP 13-2
BWQ
= = = ≠ω0 10 00070
143 714 rad s,
.
When / sec, T which agrees with the tabulated values of |T( | corresponding to
= 12600, 25000, 50000 and 100000 rad sec
ω ω ωω
> −6300 rad 1 ~ )
.
Is 1 5 ac 1m C1 1 5 2.53n R1 1 2 1 L1 2 5 10u L2 3 0 10u R2 3 4 1 C2 4 0 2.53n K1 L1 L2 0.06 Rdummy 5 0 1000k
.ac dec 200 900k 1100k
.probe
.end
At ω = 630 we expect |T(ω)| = -3dB = 0.707. This agrees with the tabulated value of |T(ω)| corresponding
to ω = 6310.
At ω=630 we expect |T(ω)| = -20 + 3=-17dB = 0.14 which agrees with the tabulated values of |T(ω)|
corresponding to ω =400 and 795 rad/s.
This data does seem reasonable.
This report is not correct.
386
VP 13-3 VP 13-4 Design Problems DP 13-1
ω01
10 159
120
500 79 6
rad kHz
rad Hz
= = =
= =
= = =
LCk s
Q R
LC
BWRL
s
.
.
H
Z
ω
ω
ω = −+
+
=
=
=
k
jz
jp
where
k R
R
C R
p C R
1
1
1
1
2
1
1 1
2 2
The reported results are correct.
The network function indicates a zero at 200 rad/s and a pole at 800 rad/s. In contrast, the Bode plot indicates a
pole at 200 rad/s and a zero at 800 rad/s.
The Bode plot and network function don’t correspond to each other.
Pick an appropriate circuit from Table 13.4-2.
387
DP 13-2
2 1
1 2
1 1 2 2
1
11 2 1 2
1
The specifications indicate that
2 k , 5 k ,
1 12 1000 2 10,000
2 2000 rad . 0.05 .
1 1.592 k , 2 3.183 , 0.0
R p C
R z C
z and pC R C R
Try z s Pick C F Then
CR R R k C
pC z kk
π π
π µ
= = = =
⋅ < = ⋅ > =
= ⋅ =
= = Ω = = Ω = =
2 2
1
1: p 31.42 rad s 2 10, 000
F
Check kC R
µ
π= = < ⋅
HV
V
j CR
o
s
( )( )
( )ω ω
ω
ω
=
= 1
HV
V
j CR
j Lj C
R
R
j CR
j LR
j CR
LC
jRC LC
Pick LC
s
H LC
LCj
LC RC LC
o
s
( )( )
( )
||
||
( ) .
( )
ω ω
ω
ω
ωω
ω
ωω
ω ω
ω π ω ω
ω
= =
+
= +
++
=− + +
= = ⋅ =
=− + +
1
1
1
1
1
1 1
12 100 10
1
1 1 1 1
2
03
0
0
rad When
So
HCL
We require
dB HCL
CL
Finally
LC
CL
C nF
L mH
( )
. ( )
( )
.
.
.
ω
ω
π
0
0
3
3 0 707 1000
12 100 10
0 707 1000
113
2 26
R
R
=
− = = = =
= ⋅
=
⇒=
=
388
DP 13-3
R k
R k
R k
R M
R M
R k
C F
C F
1
2
3
4
5
6
1
2
10
866
8 06
1
2 37
499
0 47
01
========
ΩΩΩ
ΩΩ
Ω
.
.
.
.
µµ
⇒ = − +
V sin f
z
RV
R
V
R31
1( )
⇒ = − = −+
V o sf
isV
Z
ZV
R
R C R s5
4 1 5
11
2( )
⇒ = − V Vfs
oC R s1
36
( )
Solving (1) (3) for VV yields
VV
n the values for the resistors & capacitors, can draw
o
in→
=+ +
o
in
R
R R Cs
sR C
sR
R R R C C
plugging i
3
1 4 1
2
5 1
3
2 4 6 1 2
1
ckt A is inverting summer
ckt B is first order LPF
Continued ckt C is an integrator
V
V
dB)
0
in
(
389
DP 13-4
where
K R C , pC R1 2 1 1
1 1
1= =
We require
10
2001
5001
1 2 2 14
3
11 1
22 4
K K R CR
R
pR C
pC R
= − =
= =
= =
where
KR
R p
C R24
32
2 4
1= − =,
H Kj
j1 1
1
1( )ω ω
ω=−+
p
HK
jP
22
2
1( )ω ω=
+
Pick C F Then R p C
Pick C F Then R p C
Next R
R
R
R
Let R and R
1 1
1 1
2 4
2 2
2
3
6 3 2
3
2 3
11
5
011
20
10 10 20 10 500
500 1
= = =
= = =
= ⋅ ⇒ =
= =
−
µ
µ
.
. .
( )( )
k
k
k k
Ω
Ω
Ω Ω
390
DP 13-5
Pick C Fp C
k
Pick C Fp C
k
Next
R
R
R
R
Let R k k
1 1
1 1
2 4
2 2
2
3
6 3 2
3
2 3
201
500
11
10
10 20 10 10 10 50
200 4
= = =
= = =
= ⋅ ⋅ ⇒ =
= =
−
µ
µ
.
.
( )( )
Then R
Then R
and R
Ω
Ω
Ω Ω
H Kj
jp
HK
jp
where where
K R CC R
KR
R C R
1 1
1
22
2
1 2 1 11 1
24
32
2 2
1 1
1 1
( ) ( )
, ,
ω ωω ω ω= −
+=
+
= = = − = p p
We require
dB K K CR
R
pR C
pR C
20 10
011
1001
1 2 2 14
3
11 1
24 2
= = − =
= =
= =
R
.
391
DP 13-6 DP 13-7 DP 13-8
T
R
R
j R C( )ω
ω=
+
+
1
1
2
3
1
θ ω = − −tan 11R C
G
R
R
R C
R
R=+
+=
+
+
1
1
1
1
3
2
12
3
2
2( ) tanω θ
RC1 = −tan( )θ
ω
R G R
s R k k F3
22
1 2 3
1 1
45 2 1000 10 10 18 284 01
= ⋅ + − ⋅=− = = ⇒ = = = =
( (tan ) )
deg, , , , . , .
θθ ω µ G rad R R k CΩ Ω Ω
8001
0 5 102 5
32 40 100
2001 1
200 1000 05
20 100 5 0 5
0 05
16 1
2
1
2
2
z x
k
R k
pR C
C k
F
dB kp
z
F
F
F
= = ⇒ =
= = ⇒ =
= = ⇒ = =
= = = =
−RR
dB R
R
C
( . ).
( )( ).
. .
.
Ω
Ω
Ωµ
µ µ
µ
HR
j C
j CR
j CR
CR
( )
tan
( )
( )( . ).
( )
ω
ω
ωω
ω
ω
R tan
x k
lim H R
RR R k
w
= −
+= −
+
= + −
⇒ = − =
= = ⇒ = =
° −
−
→∞
2 2
1
11
1 6
2
12 1
11 1
195 180 90
270 195
1000 01 1037 3
10 10 373
Ω
Ω
The network function of this circuit is
The phase shift of this network function is
The gain of this network function is
Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is calculated from
Next pick R2 = 10kΩ (a convenient value) and calculated R3 using
From Table 13.4-2 and the Bode plot:
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