9.14 Problem Generate Bode magnitude and phase plots for the following voltage
transfer functions:
4 104 (60 + j6 )
(a) H( ) =
(4 + j2 )(100 + j2 )(400 + j4 )
(1 + j0.2 )2 (100 + j2 )2
(b) H( ) =
( j )3 (500 + j )
8 102 (10 + j10 )
(c) H( ) =
j (16 2 + j4 )
4 104 2 (100 2 + j50 )
(d) H( ) =
(5 + j5 )(200 + j2 )3
j5 103 (20 + j2 )
(e) H( ) =
(2500 2 + j20 )
512(1 + j )(4 + j40 )
(f) H( ) =
(256 2 + j32 )2
j(10 + j ) 108
(g) H( ) =
(20 + j )2 (500 + j )(1000 + j )
Solution:
(a)
H( ) =
=
=
4 104 (60 + j6 )
(4 + j2 )(100 + j2 )(400 + j4 )
4 104 60(1 + j /10)
4 100 400(1 + j /2)(1 + j /50)(1 + j /100)
15(1 + j /10)
.
(1 + j /2)(1 + j /50)(1 + j /100)
Constant term 15
=
23.5 dB
Simple pole with c = 2 rad/s
Simple zero with c = 10 rad/s
Simple pole with c = 50 rad/s
Simple pole with c = 100 rad/s
dB
40
20 log |1 + j/10|
23.5
20
23.5 dB
M [dB]
3.5
0
0.1
1
10
1000
100
(rad/s)
20 log |1 + j/100|
20
20 log |1 + j/2|
20 log |1 + j/50|
Magnitude
90o
(1 + j/10)
0
0.1
1
10
100
1000
1
1 + j/100
90o
1
1 + j/50
1
1 + j/2
()
180o
Phase
(b)
H( ) =
(1 + j0.2 )2 (100 + j2 )2
( j )3 (500 + j )
(rad/s)
=
=
Constant term 20
j104 (1 + j /5)2 (1 + j /50)2
500 3 (1 + j /500)
j20(1 + j /5)2 (1 + j /50)2
3 (1 + j /500)
=
26 dB
Pole of order 3 @ origin
Simple zero with c = 5 rad/s, of order 2
Simple zero of order 2 with c = 50 rad/s
Simple pole with c = 500 rad/s
dB
80
60
40 log |1 + j/5|
40 log |1 + j/50|
40
26
20
26 dB
0
0.1
1
10
100
1000
10,000
(rad/s)
16
20
20 log 3
M [dB]
20 log |1 + j/500|
36
40
Magnitude
450o
360o
()
270o
(1 + j/5)2
180o
(1 + j/50)2
j
90o
0
0.1
1
10
100
1
1 + j/500
90o
Phase
(c)
H( ) =
8 102 (10 + j10 )
j (16 2 + j4 )
1000
10,000
(rad/s)
=
=
Constant factor 5 102
j8 102 10(1 + j )
16 [1 ( /4)2 + j /4]
j5 102 (1 + j )
[1 ( /4)2 + j /4]
=
Zero factor with c = 1 rad/s
Quadratic pole with c = 4 rad/s
Pole @ origin
26 dB
dB
20 log |1 + j|
20
20 log
0
0.1
6
1
10
100
1000
20
26
(rad/s)
26 dB
40
60
66
Quadratic pole
M [dB]
80
Magnitude
90o
(1 + j)
0
0.1
1
10
100
1000
(rad/s)
1
1 (/4)2 + j/4
()
j
90o
180o
Phase
(d)
H( =
=
4 ) 104 2 (100 2 + j50 )
(5 + j5 )(200 + j2 )3
4 104 100 2 [1 ( /10)2 + j /2] 0.1 2 [1 ( /10)2 + j /2]
=
5 (200)3 (1 + j )(1 + j /100)3
(1 + j )(1 + j /100)3
Constant factor 0.1
=
20 dB
Zero at origin of order 2
Simple pole with c = 1 rad/s
Quadratic zero with c = 10 rad/s
Third order pole with c = 100 rad/s
dB
40
Quadratic zero
with c = 10 rad/s
20
0
0.1
1
10
20
(rad/s)
20 dB
Third order pole
@ c = 100 rad/s
M [dB]
40
1000
100
Pole with
c = 1 rad/s
60
Magnitude
180o
Quadratic zero
90o
0
0.1
1
10
100
1000
(rad/s)
Simple pole
90o
180o
()
Third order pole
270o
Phase
(e)
H( ) =
j5 103 (20 + j2 )
j5 103 20 (1 + j /10)
=
(2500 2 + j20 )
2500[1 ( /50)2 + j /125)
=
Constant term 40
=
j40 (1 + j /10)
[1 ( /50)2 + j /125)
32 dB
Zero @ origin
Simple zero with c = 10 rad/s
Quadratic pole with c = 50 rad/s
dB
M [dB]
80
52
20 log |1 + j/10|
40
32
32 dB
20 log
0
1
5
10
50 100
(rad/s)
500 1000
Quadratic @ 50 rad/s
40
Magnitude
180o
()
90o
j
Zero @ 10 rad/s
0
1
5
10
50 100
500 1000
Quadratic
180o
Phase
(f)
H( ) =
512(1 + j )(4 + j40 )
(256 2 + j32 )2
(rad/s)
=
=
Constant term 1/32
512 4(1 + j )(1 + j /0.1)
2562 [1 ( /16)2 + j /8]2
(1 + j /0.1)(1 + j )
32[1 ( /16)2 + j /8]2
=
30.1 dB
Simple zero with c = 0.1 rad/s
Simple zero with c = 1 rad/s
Quadratic pole of order 2 with c = 16 rad/s
dB
40
20
0
16
0.01
0.1
1
10
160
100
1000
(rad/s)
20
30.1 dB
40
M [dB]
60
80 dB/decade
80
Magnitude
180o
90o
0
1.6
0.01
0.1
160
1
10
100
1000
90o
()
180o
270o
Quadratic pole,
second order
360o
Phase
(g)
H( ) =
j(10 + j ) 108
(20 + j )2 (500 + j )(1000 + j )
(rad/s)
=
=
j10 108 (1 + j /10)
202 500 1000(1 + j /20)2 (1 + j /500)(1 + j /1000)
(1 +
j5(1 + j /10)
j /500)(1 + j /1000)
j /20)2 (1 +
Constant term 5
=
14 dB
Simple zero with c = 10 rad/s
Simple pole of order 2 with c = 20 rad/s
Simple pole with c = 500 rad/s
Simple pole with c = 1000 rad/s
dB
20
14
0
14 dB
1
10
100
(rad/s)
10000
1000
20
40
60
M [dB]
Pole of order 2
80
Magnitude
180o
j
90o
0
50
1
10
500
100
90o
1000
10000
()
180o
270o
Phase
(rad/s)