ENGR 223
HW# 3 Solution
14:1 Find:
a. cutoff frequency in Hz (fc )
b. H(jω) @ ω = ωc , 0.125ωc , 8ωc
c. Steady state of vo (t) when ω = ωc , 0.125ωc , 8ωc
d. Draw Bode plot of H(s)
Solution:
fc = 3.82kHz
ω (rad/s)
H(jω)
vo (t) (V)
3
◦
ωc = 24 × 10
0.707∠ − 45
14.1 cos(24000t − 45◦ )
.125ωc = 3 × 103 0.992∠ − 7.12◦
19.8 cos(3000t − 7.12◦ )
◦
3
0.124∠ − 82.9 2.48 cos(192000t − 82.9◦ )
8ωc = 192 × 10
1
ENGR 223
HW# 3 Solution
14:3 Find:
a. H(s) = Vo /Vi
b. ω when H(jω) is at its maximum
c. |H(jω)|max
√
d. ω when |H(jω)| = |H(jω)|max / 2
e. ωc , H(j0), H(jωc ), H(j0.2ωc ), and H(j5ωc ) when Rl = 300Ω
Solution:
a.
R/L
H(s) =
s + (R + Rl )/L
b.
ω = 0 rad/s
c.
|H(jω)|max =
d.
ωc =
R
R + Rl
R + Rl
L
e.
ωc = 30, 000 rad/s
ω (rad/s)
H(jω)
0
0.8∠0◦
ωc = 30000 0.565∠ − 45◦
.2ωc = 6000 0.784∠ − 11◦
5ωc = 150000 0.156∠ − 79◦
2
ENGR 223
HW# 3 Solution
14:9 Find:
a. ωc
b. R
c. draw schematic with output voltage labeled
d. H(s)
e. H(s) if filter is loaded with R of same value as found in part b
f. ωc of loaded filter
g. Gain in passband of loaded filter
Solution:
a. ωc = 628 rad/s
b. R=339Ω
R
+
vi(t)
c.
d.
+
339Ω
-
4.7uF
C
vo(t)
-
H(s) =
628
1/RC
=
s + 1/RC
s + 628
H(s) =
1/RC
628
=
s + 2/RC
s + 1260
e. Loaded
f. Loaded ωc = 1260 rad/s
628
g. |H(j0)| = 1260
= 0.500
14:11 Find:
a. fc
b. H(jω) @ ω = ωc , 0.125ωc , 8ωc
c. Steady state of vo (t) when ω = ωc , 0.125ωc , 8ωc
Solution:
fc = 99.5Hz
ω (rad/s)
H(jω)
vo (t) (V)
◦
ωc = 625
0.707∠45
53.0 cos(625t + 45◦ )
◦
.125ωc = 78.1 0.124∠83
9.38 cos(78.1t + 83◦ )
8ωc = 5000
0.992∠7.13◦ 74.4 cos(5000t + 7.13◦ )
3
ENGR 223
HW# 3 Solution
4
14:13 Find:
a. Design HPF where fc = 300Hz and C=100nF (Don’t forget to draw the schematic)
b. fc if filter loaded with 47kΩ resistor
Solution:
C
+
vi(t)
a.
+
100nF
5.31kΩ
vo(t)
R
-
RL
-
Load resistor is not part of actual filter. I have shown it for reference.
b. Loaded fc = 334Hz
14:17 Find:
a. Design HPF where ωc = 1500krad/s and L=100uH. R must be a standard value (Don’t
forget to draw the schematic)
b. Find smallest standard value load resistor so ωc > 1200krad/s
Solution:
R
+
vi(t)
a.
-
+
150Ω
100uH
L
vo(t)
RL
-
Load resistor is not part of actual filter. I have shown it for reference.
b. RL > 680Ω
14:25 Find:
a. R and L
b. cutoff frequencies in Hz (f1 and f2 )
c. Bandwidth (β) in Hz
Solution:
Ko βs
(1/RC)s
= 2
H(s) = 2
s + (1/RC)s + (1/LC)
s + βs + ωo2
a. L=50mH, R=5kΩ
b. f1 = 2.88kHz and f2 = 3.52kHz
c. β = 637Hz
ENGR 223
HW# 3 Solution
5
14:31 Find:
a. wo
b. β
c. Q
d. steady state vo (t)
20
for RL in kΩ. It helps to use numbers for everything but RL
e. prove Q = 1+(100/RL)
f. Plot Q vs RL for 20kΩ ≤ RL ≤ 2M Ω
g. Bode plot of H(s)
Solution:
H(s) =
a.
b.
c.
d.
e.
f.
(1/RC)s
Ko βs
50 × 103 s
=
=
s2 + (R + RL)s/(R · RL · C) + (1/LC)
s2 + βs + ωo2
s2 + 62.5 × 103 s + 1 × 1012
ωo = 1M rad/s
β = 62.5krad/s
Q = 16
vo (t) = 200cos(1 × 106 t)mV
You have the answer, work out the math and practice algebra.
ENGR 223
HW# 3 Solution
6
ENGR 223
HW# 3 Solution
7
14:37 Find:
a. Qualitative analysis to prove circuit is band reject
b. H(s) in terms of RLC
c. derive wo from H(s)
d. derive wc1 and wc2 in terms of RLC
e. derive β in terms of RLC
f. derive Q in terms of RLC
Solution:
a. at low frequency w = 0 L is SC so vo = vi
at high frequency w = inf C is SC so vo = vi
R
vi < vi
at mid frequencies vo = ZL //Z
C ·R
∴ it rejects mid band more than other bands and is band reject
b.
s2 + (1/LC)
H(s) = 2
s + (1/RC)s + (1/LC)
c. Show all work of how you come to solution. Hint: H(jωo ) = 0
ωo = √
d. Show all work. Hint: |H(jωc )| =
1
LC
√1 |H(jω)|max
2
1
ωc1 = −
+
2RC
1
+
ωc2 =
2RC
s
s
1
2RC
1
2RC
2
+
2
+
1
LC
1
LC
2
2
There are actually 4 answers for wc but two of them are negative so they are not
possible
e.
β=
f.
1
RC
RC
Q= √
LC
ENGR 223
HW# 3 Solution
8
14:51 Find:
a. Design Series RLC filter to select Low Frequency DTMF tone. β = 941 − 697 Hz,
R=600Ω
b. |H(s)| at each low-band frequency
c. |H(s)| at the lowest high-band frequency (1209Hz)
Solution:
C
+
+
vi(t)
98.7nF
391mH
R
600Ω
-
f (Hz)
697
941
770
852
1209
L
vo(t)
-
|H(j2πf )
.707
.707
.946
.946
.343
low-band or high-band
low
low
low
low
high