Exercises
E8.1
Give11 ~ , I I Ptransfer fluliction
we
rlc t,erminc: that
The polar plot is shown in Figure E8.1.
FIGURE E8.1
4
Polar plot for U ( s j = i-,+l)l
The phase angle is campuled via
At
L,J
= 28.3, we determine that
= -90'
- 70.5' - 35.;1" + 15.8' = 180"
Computing the magllitude yields
when w = 28.3. Wc call also rewrite G(s) as
372
CHAPTER 8
Frequency Response Methods
Put S(jω);
∴
0.5
jω (1 + jω) (1 + 0.5jω)
0.5
q
|G (jω) H (jω)| = √
2
ω 1 + ω × 1 + (0.5jω)2
G (jω) H (jω) =
∠G (jω) H (jω) = −90 − tan −1 ω − tan−1 0.5ω
(b)
−180◦ = −90 − tan−1 ω − tan−1 0.5ω
tan−1 ω + tan−1 0.5ω = 90◦
tan tan−1 ω + tan−1 0.5ω = tan 90
1
ω + 0.5ω
=
1 − ω × 0.5ω
0
1 − 0.5ω 2 = 0
0.5ω 2 = 1, ω = 1.414 rad/sec
(c) |G(jω)H(jω)| =
E8.3
0.5
p
= 0.617
1.414 1 + 1.4142 × 1 + (0.5 × 1.414)2
√
The loop transfer function is
Gc (s)G(s) =
300(s + 100)
.
s(s + 10)(s + 40)
The phase angle is computed via
φ(ω) = −90o − tan−1
ω
ω
ω
− tan−1
+ tan−1
.
10
40
100
At ω = 28.3, we determine that
φ = −90o − 70.5o − 35.3o + 15.8o = 180o .
Computing the magnitude yields
1
|Gc G(jω)| =
ω 2 2
300(100)(1 + ( 100
) )
1
1
ω 2 2
ω 2 2
ω10(1 + ( 10
) ) 40(1 + ( 40
) )
= 0.75 ,
373
Exercises
when ω = 28.3. We can also rewrite Gc (s)G(s) as
Gc (s)G(s) =
s
75( 100
+ 1)
.
s
+ 1)( 40
+ 1)
s
s( 10
Then, the magnitude in dB is
ω 2
ω
) ) − 10 log 10 (1 + ( )2 )
100
10
ω 2
− 10 log 10 (1 + ( ) ) − 20 log 10 ω = −2.5 dB ,
40
20 log 10 |Gc G| = 20 log 10 (75) + 10 log 10 (1 + (
at ω = 28.3.
E8.4
(a) Transfer function,
Y (s)
R2
=
R1 × 1
U (s)
R2 + R + cs1
1
cs
Y (s)
R2 (1 + R1 cs)
=
U (S)
R1 R2 cs + R2 + R1
h
i
R1 cR2 s + R11 c
h
i
=
2)
R1 cR2 s + (RR11+R
cR2
s+
=
s+
R2
(R1 +R2 )
s + T1
Y (s)
=
1
U (s)
s + αT
where T = R1 C and α =
(b)
1
1
R1 c
1
R1 c
R2
R1 + R 2
T = R1 C = 10 × 103 × 0.1 × 10−3 = 1 ms
R2
5 × 103
=
= 0.33
R1 + R 2
(5 + 10) × 103
s + 101−3
Y (s)
G(s) =
=
1
U (s)
s + 0.33×
−3
α=
374
CHAPTER 8
Frequency Response Methods
Time constant form,
0.33 (0.001s + 1)
(0.00033s + 1)
0.33 (1 + 0.001jω)
G1 (jω) =
(1 + 0.00033jω)
q
0.33 1 + (0.001ω)2 ∠ tan−1 (0.001ω)
|G1 (jω)| = q
1 + (0.00033ω)2 ∠ tan−1 (0.00033ω)
G1 (s) =
FIGURE E8.4
E8.5
The lower portion for ω < 2 is
20 log
at ω = 8. Therefore,
20 log
which occurs when
K
= 0 dB ,
ω
K
= 0 dB
8
K=8.
We have a zero at ω = 2 and another zero at ω = 4. The zero at ω = 4
yields
a = 0.25 .
375
Exercises
We also have a pole at ω = 8, and a second pole at ω = 24. The pole at
ω = 24 yields
b = 1/24 .
Therefore,
8(1 + s/2)(1 + s/4)
.
s(1 + s/8)(1 + s/24)(1 + s/36)
G(s) =
E8.6
The loop transfer function is
L(s) =
10
.
s(s/6 + 1)(s/100 + 1)
The Bode diagram is shown in Figure E8.6. When 20 log 10 |L| = 0 dB, we
have
ω = 6.7 rad/sec .
Bode Diagram
50
Magnitude (dB)
0
−50
−100
−150
−200
−90
Phase (deg)
−135
−180
−225
−270
−1
10
0
FIGURE E8.6
Bode Diagram for L(s) =
E8.7
1
10
2
10
10
Frequency (rad/sec)
3
10
10
.
s(s/6+1)(s/100+1)
The transfer function is
T (s) =
(s2
4
.
+ s + 1)(s2 + 0.4s + 4)
4
10
377
Exercises
Bode Diagram
20
Magnitude (dB)
10
0
−10
−20
−30
180
Phase (deg)
135
90
45
0
−45
−90
−2
10
−1
10
FIGURE E8.8
Bode Diagram for Gc (s)G(s) =
1
2
10
3
10
100(s−1)
.
s2 +25s+100
The Bode diagram for G(s) is shown in Figure E8.9.
40
n dB
20
ëé ê
0
-20
-40
10-1
se deg
E8.9
0
10
10
Frequency (rad/sec)
íìê
100
àá â
ã?ä2â
å210æçMè á ad/sec)
1
102
103
100
àá â
ã?ä?âå?10æçMè á ad/sec])
102
103
50
0
-50
10-1
FIGURE E8.9
Bode Diagram for G(s) =
(s/5+1)(s/20+1)
.
(s+1)(s/80+1)
1
?'lie node plot of T ( s ) is S ~ ~ O WillI Figure
~
Ed. 13, where i d g = 4.9 rudlsec.
Bode Diagram
-'l
1 o0
1 o'>
lo1
103
Frequency (md/sec)
FIGURE €0.13
Bode Diagram for GI81 =
1 OU
.*7.klly~+20s+llo
(a) Wllsll G g ( s ) is out 01 the iool,, the chal.acteristir eclllntioll is
or s2 -I-0.6s
+ G -- 0 . Tlius. C - o . G , / (JC)
~
(A) 11'1th GJ(sj, tlrc cbal.rtcter.:stic eclua,tioii is
- 0.12.
393
Problems
(b)
Gc (s) · Gp (s) =
s (s2
K (s + 5)
+ 2s + 2) (s + 10)
Time constant form,
0.25 K (1 + 0.25)
s (0.5s2 + s + 1) (1 + 0.1s)
0.25 × 40 (1 + 0.2s)
=
s (0.5s2 + s + 1) (1 + 0.1s)
10 (1 + j0.2ω)
∴ G (jω) =
jω (1 + jω − 0.2ω 2 ) (1 + j0.1ω)
G (s) =
(c) Frequency corresponding to ‘0’ db gain = 2.78 rad/sec
P8.6
(a) The transfer function is
GH(s) =
3.98(1 + s/1)
.
s(1 + s/10)2
We have a zero at ω = 1 and two poles at ω = 10.0. The low frequency
approximation is K/s and at ω = 1 we have
K
20 log
= 12dB .
ω
Therefore,
K = 3.98
at ω = 1 (an approximation). The phase plot is shown in Figure P8.6a.
(b) The transfer function is
GH(s) =
s
.
(1 + s/10)(1 + s/50)
The poles are located by noting that the slope is ±20 dB/dec. The
low frequency approximation is Ks, so
20 log Kω = 0dB .
At ω = 1 we determine that
K=1.
The phase plot is shown in Figure P8.6b.
394
CHAPTER 8
Frequency Response Methods
(a)
-40
-60
se deg
-80
-100
‡ † ˆ -120
-140
-160
-180
-2
10
10
-1
10
0
10
1
10
2
‘“ ”
100
se deg
50
‡† ˆ
0
-50
-100
-1
10
10
FIGURE P8.6
Phase plots for (a) G(s) =
1
‰ Š ‹Œ‹Ž 10
’‘ Š
3.98(s/1+1)
.
s(s/10+1)2
10
2
10
ad/sec)
(b) G(s) =
s
.
(s/10+1)(s/50+1)
The loop transfer function is
L(s) =
Kv
.
s(s/π + 1)2
(a) Set Kv = 2π. The Bode plot is shown in Figure P8.7a.
n dB
40
žŸ
20
0
-20
10-1
100
• – —˜™2—š2› œ’ –
101
ad/sec)
-80
-100
se deg
P8.7
0
¢¡ Ÿ
-120
-140
-160
-180
10-1
FIGURE P8.7
(a) Bode plot for L(s) =
100
• – —˜™2—š2› œ’ –
Kv
,
s(s/π+1)
101
ad/sec)
where Kv = 2π.
3
395
Problems
(b) The logarithmic magnitude versus the phase angle is shown in Figure P8.7b.
40
30
n dB
20
¨
¦§
10
0
-10
-20
-170
-160
-150
-140
-130
£X¤*¥
-120
-110
-100
-90
se deg
FIGURE P8.7
CONTINUED: (b) Log-magnitude-phase curve for L(jω).
P8.8
The transfer function is
T (s) =
s2
K
.
+ 7s + K
(a) When P.O. = 10%, we determine that ζ = 0.59 by solving
√ 2
0.10 = e−πζ/ 1−ζ .
So, 2ζωn = 7 implies that ωn = 5.93, hence K = ωn2 = 35.12. Also,
p
Mpω = (2ζ 1 − ζ 2 )−1 = 1.05 .
(b) For second-order systems we have
p
ωr = ωn 1 − 2ζ 2 = 0.55ωn = 3.27
when ζ = 0.59 and ωn = 5.93.
(c) We estimate ωB to be
ωB ≈ (−1.19ζ + 1.85)ωn = 1.14ωn = 6.8 rad/s .