Download HW5 solution

Problem 4.37
Solution:
Known quantities:
Functions.
Find:
The phasor form.
Analysis:
In phasor form:
a) V(jw)  155-25 V





b) 5 sin 1000t  40o  5 cos 1000t  40o  90o  V(jw)  5-130 V
I(jw)  1063  15-42  10 cos 63  j10 sin 63  15 cos -42  j15 sin -42 
c)
4.54  j8.91  11.15  j 10.04  15.69  j1.13  15.73-4.12 A
d) I(jw)  460-25 - 220-75  416.90 - j194.40  56.94-j 212.50  359.96  j18.10  360.42.88 A
Problem 4.38
Solution:
Known quantities:
Complex number.
Find:
The polar form.
Analysis:
a) 4  j 4=4 245=5.6645 
b) -3  j 4  5 126.9
c) j  2  j 4  3  -1 -j 3  3.16-108.4
Problem 4.39
Solution:
Known quantities:
Complex number.
Find:
The polar form.
Analysis:
a) 50  j104  j8  (50.9911.30)(8.9463.43)  456.174.7
50  j104  j8  200  j 400  j 40  j 2 80  120  j 440  456.174.7
b)  j 2  24  j52  j 7   (2.82135)(6.4051.34)(7.2874.05)  131.8260.4  131.8-99.6
 j 2  24  j52  j 7   -36 - j126 - j 4 - j 2 14  -22 - j130  131.8-99.6
Problem 4.51
Solution:
Known quantities:
The values of the impedance, Z1  5.97 k ,
o
Z 2  2.30 o  , Z 3  1711o  and the voltages applied to the circuit shown in Figure P4.51,
v s1t   v s2 t   170 cos377t  V .
Find:
The current through Z 3 .
Analysis:
KVL:
Vs1  Vs2  I 3Z 3  0
Vs1  Vs2  1700 o V  170  j0 V
Vs1  Vs2 1700 o V 1700 o V 3400 o V


 20 11o A
Z3
1711o 
1711o 


rad
i 3t   20 cos377
 t 11o  A


s
I3 
Problem 4.54
Solution:
Known quantities:
The values of the impedance and the voltage applied to the
circuit shown in Figure P4.54.
Find:
The current in the circuit.
Analysis:
Assume clockwise currents:
rad
3
, VS  120 o V
s
1
ZC 
  j  , Z L  jL  j 9   Z total  3 j9  j  3 j8 
jC
12
12
A  1.4045  69.44 o A , i t   1.4 cos t  69.4 o
I

3  j8 8.5469.44 o


A
Problem 4.65
Solution:
Known quantities:
The circuit shown in Figure P4.65, the values of resistance,
inductor and capacitor.
Find:
The frequency causes the equivalent impedance to appear to be purely resistive
Analysis:
1
1 

 15  j  0.001ω  6 
jC
10 ω 

This impedance is purely resistive if its imaginary part (the reactance) is zero.
1
Therefore, we solve for the frequency, , at which 0.001ω 
. So   31,622.77 rad/s
6
10 ω
The series impedance of the circuit is Z eq  R  jL 
Problem 6.2
Solution:
Known quantities:
Resistance and capacitance values, in the circuit of Figure
P6.2.
Find:
a)
b)
c)
d)
The frequency response for the circuit of Figure P6.2.
Plot magnitude and phase of the circuit using a linear scale for frequency.
Repeat part b., using semilog paper.
Plot the magnitude response using semilog paper with magnitude in dB.
Analysis:
First, we find the Thévenin equivalent circuit seen by the capacitor:
RT  500 500  250 
and
v
500
vT 
vin  in
500  500
2
1
vout
v
j C
1
1
1 vout



 out 
a)
vT
1  j RT C 1  j 0.05 
vin
2 vT
RT  1
j C
vout
vT

1
1 + 0.0025 2
vout
1 vout

vin
2 vT

1
4 + 0.01 2
 ()   arctan0.05
b)
The plots obtained using Matlab are shown below:
c)
d)
Problem 6.5
Solution:
Known quantities:
Resistance, inductance and capacitance values, in the circuit
of Figure P6.5
Find:
a) The frequency response for the circuit of Figure P6.5
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:
R1
jC1
R1
ZT  Z R 2  ZC1 || Z R1  R2 
 R2 
1
1  jC1R1
 R1
jC1
and
Z R1
R1
jC1 R1
vT 
vin 
vin 
vin
1
Z R1  Z C1
1  jC1 R1
R1 
jC1
1
vout
ZC2
j C 2
1



a)
vT
ZT  Z C 2 
 1


R1
R1
 R2 
 
 jC 2
1   R2 
j C 2

j
C
R

1
j
C
R


1
1 1
1 1


Therefore,
v out
jC1R1
jC1R1
1



2


v in 1  jC1R1
R1
1  j C1R1  C 2 R1  R2   j  C1C 2 R1R2
1  R2 
j C 2
1  jC1R1 

Substituting the numerical values:
j 2
v out

v in
1   2  j 2.6

b)
c)

The plots obtained using Matlab are shown below:
d)
Problem 6.12
Solution:
Known quantities:
The values of the resistors and of the capacitor in the circuit
of Figure P6.12:
R1 = 100 
R L = 100  R 2 = 50 
C = 80 nF
Find:
Compute and plot the frequency response function.
Analysis:
Using voltage division:
Z eq =
Z R2 ZC
Z R2+ ZC
1
jC
1
R2 +
jC
R2
=
Voltage Divider : H v [j ] =
jC
j C
=
R2
1 + j  R2C
Z RL
V o [j ]
=
=
[j
]

+
Vi
Z R1 Z eq + Z RL
R1 +
1 + j R 2 C
RL

R2
+ R L 1 + j R 2 C
1 + jR 2 C
1+ j  R 2 C
R L 1 + j R 2 C 
RL
=
 +  C
R1 + R 2 + R L + j R1 + R L   R 2 C
R1 + R 2 + R L 
1+ j R1 R L R 2 
R1 + R 2 + R L 

Plotting the response in a Bode Plot:
=
Magnitude of the voltage divider at zero and infinite raidan frequencies are
Problem 6.20
Solution:
Known quantities:
Frequency response H v  j of the circuit of Example 6.3.
Find:
The frequency at which the output of the circuit is attenuated by 10 percent.
Analysis:
The frequency response of the circuit is: H v  j  
From Example 6.3:  0 
1
1  jCR
1
 2,128rad /sec
CR
The attenuation introduced by the circuit is: H v  j  
1
1  /  0 
2
Thus,
H v  j 
1
1 / 2128
2
 0.9

 1 2 2
 1  1031 rad/s
0.9 
  2128 