Circuit Analysis Exercise 6
Exercise 7 Frequency Response, Bode Plot, and Resonance
Problem 1. (Hambley 6.25)
Problem 2. (Hambley 6.28)
1
Circuit Analysis Exercise 6
Problem 3. (Hambley 6.30)
2
Circuit Analysis Exercise 6
Problem 4. (Hambley 6.32)
3
Circuit Analysis Exercise 6
Problem 5. (Hambley 6.45)
Problem 6. (Hambley 6.54)
Problem 8. (Hambley 6.59)
4
Circuit Analysis Exercise 6
Problem 9. (Hambley 6.65)
Problem 10. (Hambley 6.68)
Problem 11. (Hambley 6.72)
5
Circuit Analysis Exercise 6
Problem 12. (Hambley 6.77)
Problem 13. (Hambley 6.81)
Problem 14. (Hambley 6.86)
6
Circuit Analysis Exercise 6
Problem 15. (Rizzoni 6.3)
Repeat Problem 6.1 for the circuit of Figure P6.2.
a. Determine the frequency response Vout ( j ) / Vin ( j ) for the circuit of Figure P6.2.
b. Plot the magnitude and phase of the circuit for frequencies between 10 and 10 7 rad/s on graph
paper, with a linear scale for frequency.
Figure P6.2
Solution:
Known quantities:
Resistance and capacitance values, in the circuit of Figure P6.3.
Find:
a) The frequency response for the circuit of Figure P6.3.
b) Plot magnitude and phase of the circuit using a linear scale for frequency.
c) Repeat part b., using semilog paper.
d) Plot the magnitude response using semilog paper with magnitude in dB.
Analysis:
First, we find the Thévenin equivalent circuit seen by the capacitor:
RT 2000200010002000
and
vOC 


a)
2000
v
vin  in
20002000
2
1
vo u t
j C
1
1



vOC RT  1
1 j RTC 1 j 0.02
j C
v out
1

4 2
vOC
 1 + 4  1 0 
v out 1 v out
0.5


v in
2 vOC
1 + 4  1 04  2
 () arctan 0.02

b)
The plots obtained using Matlab are shown below:

7
Circuit Analysis Exercise 6
c)
d)
8
Circuit Analysis Exercise 6
Problem 16. (Rizzoni 6.8)
In the circuit shown in Figure P6.8, if L  190mH R1  2.3k
C  55nF R2  1.1k
a. Determine how the input impedance behaves at extremely high or low frequencies.
b. Find an expression for the input impedance in the form
 1  jf 1 ( ) 
Z ( j )  Z 0 

1  jf 2 ( ) 
Z 0  R1 
L
R2 C
f1 ( ) 
R1 LC  R1  R2
 ( R1 R2 C  L)
f 2 ( ) 
 2 LC  1
CR2
c. Determine the four frequencies at which f1(w)=+1 or -1 and f2(w)=+1 or -1.
d. Plot the impedance(magnitude and phase) versus frequency.
e.
Figure P6.8
Problem 17. (Rizzoni 6.24)
Are the filters shown in Figure P6.24 low-pass, high-pass, badpass, or bandstop (notch) filters?
9
Circuit Analysis Exercise 6
Figure P6.24
Solution:
Known quantities:
With reference to Figure P6.8
L = 1 9 0 mH R1 = 2 .3 k
C = 5 5nF
R2 = 1 .1 k
Find:

V  j 
a) How the input impedance, Z  j  = i
, behaves at extremely high or low frequencies.
I i  j 
b) An expression for the input impedance in the form:
1 j f 
1
Z  j  Z o 



1 j f 2 
f 1  

 2 R1 LC  R1  R 2
R1 R 2 C  L 
Z o  R1 
f 2  
L
R2C
 2 LC 1
C R 2
c) Determine the four cutoff frequencies at which f1[] = +1 or -1 and f2[] = +1 or -1.
d) Determine the resonant frequency of the circuit.
e) Plot the impedance vs. frequency.
Analysis:
As   ,
a)
As   0,
Z L    Open, Z C  0
 Shor t Z  R1
Z C    Open, Z L  0
 Shor t  Z  R1  R2

10
Circuit Analysis Exercise 6
b)
V[ j ]
[
+ ZL ]
= Z R1 + Z C Z R 2
 R1 +
I[ j ]
ZC + [ Z R2 + Z L ]
Z[ j ] =

R C  j  LC 1
R1R2C  L  j  R1LC  R1  R2
2
 Z  j  
2

jC

jC
(
[ 1 -  2 LC ] + R 2 ) + j (  R1 R 2 C + L )  j 
R 2 + j L
 R1


2
1 -  LC + j  R 2 C
[ 1 -  2 LC ] + j  R 2 C
 j
= R1 +

1
] [ R 2 + jL ]
jC
1
+ [ R 2 + jL ]
jC
[
 R R C  L 
1 j
1 2
R2C
2
 2 R1LC  R1  R2
R1R2C  L
1 j
 2 LC 1
R2C
c)
Both f1[] and f2[] can be positive or negative, and therefore equal to plus or minus one
depending on the frequency; therefore, both cases must be considered.
 c [ R1 R 2 C + L ] =  1
R1 [ 1-  2c LC ] + R 2
1
+ R2
R
]  c - R1
= 0
 2c  [ 2 +
L
C
R1
R1 LC
f 1 [  c] =

R2 + 1
L
R1 C
=
1100
1
+
0.19
[ 2300 ] [ 5510-9 ]
3400
R1 + R 2 =
R1 LC
[ 2300] [ 0.19 ] [ 55
10-9 ]
= 13.69 k
rad
s
rad
= 141.46 M 2
s
Where only
1
1
[  13.69103 ] 
( [  13.69103 ]2 - 4[1][ - 141.5106 ] )1/2
2
2
c = 
=  6.845103  13.724103
  c1 = 6.879 k
rad
s
rad
s
 c4 = 20.569 k
the positive answers are physically valid, i.e., a negative frequency is physically impossible.
 2c L C - 1 =  1
c R2C

f 2 [ c] =

1100
rad
R2
=
= 5 .7 9 k
L
0 .1 9
s
c = 

  c2
  2c  [
1
R2
] c = 0
L
LC
1
1
rad
=
= 9 5 .6 9 M2
-9
LC
[ 0 .1 9 ] [ 5
150 ]
s
1
1
[  5 7 9 0 ]
( [  5 7 9 0]2  4 [1 ][ 9 5 .6 91 06 ] )1 / 2 =  2 8 9 5 1 0 2 0 1
2
2
rad
rad
= 7 .3 1 k
 c3 = 1 3 .0 9 k
s
s
Again, the negative roots were rejected because they are physically impossible.
d)
Magnitude and phase response in semilogarithmic frequency plots:
11
Circuit Analysis Exercise 6
Problem 18. (Rizzoni 6.27)
In the filter circuit shown in Figure P6.27:
RS  100 RL  5k
Rc  2400 L  1mH
C  0.5nF
Compute and plot the frequency response function. What type of filter is this?
Figure P6.27
Solution:
Known quantities:
The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.27.
Find:
Compute and plot the frequency response function. What type of filter is this?
Analysis:
First, we find the Thévenin equivalent circuit seen by the capacitor:

1

1
1
ZT  ZC || Z Rs  Z L  || Z Rc  Z L    jC 


RS  jL RC  jL  


RS  jLRC  jL
jC RS  jLRC  jL  RS  RC  j2L
12

Circuit Analysis Exercise 6
and, by node analysis,
VOC 

ZC | |Z Rs  Z L  | |Z Rc  Z L 
RC  jL
vi n 
vi n
Z Rs  Z L
jC RS  jLRC  jL  RS  RC  j 2L
Z RL
Vo u t
RL
1




VOC ZT  Z RL ZT  RL 1 ZT
RL
jC RS  jLRC  jL  RS  RC  j2L
jC RS  jLRC  jL  RS  RC  j2L  RS  jLRC  jL / RL
Therefore,

Vout
RC  jL

Vi n jCRS  jLRC  jL RS  RC  j2L RS  jLRC  jL/RL
Substituting the numerical values:






8 101 7  j 2 101 2 
Vo u t

Vin  j3   j2 9 105  j 4.24101 2  1.016101 8
The corresponding Bode diagrams are shown in the
figure.
The magnitude of the voltage transfer function is
highest at the resonant frequency and decreases at
higher and lower frequencies. Therefore, this is a
band-pass filter.
However, it is not a particularly good filter since the voltage gain [or actually insertion loss] is not
very different at the resonant and lower frequencies. This is due to the large inductor losses
modeled here as the equivalent resistance Rc. This causes a low "Q" circuit.
Problem 19. (Rozzpni 6.31)
In the filter circuit shown in Figure P6.31:
RS  500 RL  5k
Rc  4k
C  5 pF
L  1mH
Computer and plot the voltage frequency response function
V ( j )
H ( j )  o
Vi ( j )
What type of filter is this?
13
Circuit Analysis Exercise 6
Figure P6.31
Solution:
Known quantities:
The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.31.
Find:
Compute and plot the voltage frequency response function. What type of filter is this?
Analysis:
First, we find the Thévenin equivalent circuit seen by the capacitor:

1

1
ZT  Z Rs  ZC || Z Rc  Z L   RS   jC 

RC  jL  

 RS 
RC  jL  RC  jL  RS 1 jC RC  jL
jC RC  jL 1
jC RC  jL 1
and VOC  Vin
Z RL
Vout
RL
1




Vin
ZT  Z RL ZT  RL 1 ZT
RL


Therefore,

jC RC  jL 1
jC RC  jL 1 RC  jL  RS 1 jC RC  jL/ RL

2

 R  R 
S 
1 C

RL 

1 jCR C   j LC

 R  L 
 R 
2
jCR C 1 S 
  j LC 1 S 
 RL  RL 
 RL 

Substituting the numerical values:



1 j 2 108    j 5101 5
Vo u t

Vin  j2 5.5101 5  j 2.22107  1.9

2


The corresponding Bode diagrams are
shown in the figure.
The magnitude of the voltage transfer
function is lowest at the resonant
frequency and increases at higher and
lower frequencies. Therefore, this is a
band stop or "notch" filter.
14
Circuit Analysis Exercise 6
At its resonant frequency, a parallel resonant circuit has a high equivalent resistance that is resistive.
Connected here in series with the load, this high impedance reduces the magnitude of the voltage
transfer function [or voltage gain or insertion loss] at the resonant frequency.
The loading due to the inductor losses, modeled here as an equivalent "coil" resistance, is fairly
small giving a substantially lower gain at the resonant frequency compared with the gain at higher
or lower frequencies. Therefore this is a high "Q" circuit with good performance and selectivity.
The inductor losses also affect only slightly the resonant frequency.
The cutoff frequencies are difficult [but not impossible] to determine in circuits containing a
parallel resonant circuit which includes inductor losses, so no attempt was made to do so.
Problem 20. (Rizzoni 6.43)
In the circuit shown in Figure P6.43:
a. Determine the frequency response function
V ( j )
H ( j )  o
Vi ( j )
b. Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog
paper.
Show the factored polynomial and all the steps in constructing the plot. Clearly show the break
frequencies on the  axis. Hint: To factor the denominator polynomial, you may find it helpful
to use the MATLABTM command “roots”.
c. Use MATLABTM and the Bode command to generate the same plot, and verify that your answer
is indeed correct. Assume
R1  R2  1k, C1  1F , C2  1mF, L  1H .
Figure P6.43
Solution:
Known quantities:
The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.43:
R1 =R2 1 k C
L1 H
1 =1 F C2 = 1 mF
Find:

a) The frequency response function H v  j  =
Vo u t j 
for the circuit of Figure P6.43.
Vin  j 
15
Circuit Analysis Exercise 6
b) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog
paper.
c) Use Matlab and the Bode command to generate the same plot.
Analysis:
First, we find the Thévenin equivalent circuit seen by the capacitor:
1

1
1
ZT  Z R2  Z C1 || Z L || Z R1  R2   jC1 
  
jL R1 

a)



R1 
 j 2 LC1  j
L
1
R1
 j 2 LC1R2  jL1  R2   R2
and
j
VO C 
Vout
VOC
L
R1
 j 2 LC1  j
L
1
R1
Vin
 j 2 LC1  j
L
1
j C2
R1
Thus,








ZT  1
R
L
j C2  j 3 LC1C2 R2   j 2 LC1  C2 1  2   j  C2 R2   1
 R 
R

1 

 1


Vo u t

Vi n
1
j
L
R1





R1 
 L

 C 2 R2   1
 R1

 j 3 LC 1C 2 R2   j 2 L C1  C 2 1  R2   j 
b)
Substituting the numerical values and expressing the frequency response function in
factored form, we have:
H v  j   10 3
c)
j
j
j


 1
 1
 968 .361  1031 .638 
The sketch plots and the ones obtained using Matlab are shown below:
 j  1
16