Transcript Slide 1
Chapter 5
Sinusoidal Input
1
Chapter 5
Processes are also subject to periodic, or cyclic, disturbances.
They can be approximated by a sinusoidal disturbance:
0 for t 0
U sin t
A sin t for t 0
where:
A = amplitude, ω = angular frequency
A
U sin ( s ) 2
s 2
Examples:
1. 24 hour variations in cooling water temperature.
2. 60-Hz electrical noise (in USA!)
2
For a sine input to the 1st order process:
K
U ( s) A 2
and G s
2
s
s 1
Chapter 5
output is...
0
1s
2
K
A
Y ( s)
2
2
2
2
2
s 1 s s 1 s s 2
By partial fraction decomposition,
AK 2
0 2 2
1
AK
1 2 2
1
AK
2 2 2
1
3
Inverting,
this term dies out for large t
AK t
AK
e
sin(t )
2 2
2
2
1
1
AK
y
sin(t ) B sin(t )
2 2
1
where
B
K
amplitude ratio
2
2
A
1
Chapter 5
y (t )
arctan( ) phase angle
Note that the amplitude ratio and phase angle is not a
function of t but of τ and ω. For large t, y(t) is also
sinusoidal, output sine is attenuated by…
1
2 2 1
4
Figure 13.1 Attenuation and time shift between input and output
sine waves (K= 1). The phase angle φ of the output signal is given
φ t / P 360 , where t is the (period) shift and P
by
is the period of oscillation.
Frequency Response Characteristics of a
First-Order Process
ˆ sin ωt φ as t where :
For x(t ) A sin t , y t A
Aˆ
KA
ω2 τ 2 1
and φ tan 1 ωτ
1. The output signal is a sinusoid that has the same frequency, ,
as the input signal, x t Asin t
2. The amplitude of the output signal, Aˆ , is a function of the
frequency and the input amplitude, A:
Aˆ
KA
ω2 τ 2 1
Dividing both sides by the input signal amplitude A yields the
amplitude ratio (AR)
Aˆ
K
AR
A
ω2 τ 2 1
which can, in turn, be divided by the process gain to yield the
normalized amplitude ratio (ARN) (or magnitude ratio):
AR N
1
ω2 τ 2 1
3. The output has a phase shift, φ, relative to the input. The
amount of phase shift depends on frequency, i.e.,
t
2 tan 1
P
Basic Theorem of Frequency
Response
The frequency response of a system can be found
by substituting j for s in the system transfer function,
i.e.
Laplace domain
G s
s j
ferquency domain
G j
8
Properties
G j AR amplitude ratio
G j phase angle
Note,
G j R jI G j e jG j
G j R I
2
G j tan
1
2
I
R
9
Shortcut Method for Finding
the Frequency Response
The shortcut method consists of the following steps:
Step 1. Set s=j in G(s) to obtain G jω .
Step 2. Rationalize G(j); We want to express it in the form.
G(j)=R + jI
where R and I are functions of . Simplify G(j) by
multiplying the numerator and denominator by the
complex conjugate of the denominator.
Step 3. The amplitude ratio and phase angle of G(s) are given
by:
AR R 2 I 2
Memorize
tan 1 ( R / I )
Example 13.1
Find the frequency response of a first-order system, with
1
G s
τs 1
(13-16)
Solution
First, substitute s jω in the transfer function
1
1
G jω
τjω 1 jωτ 1
(13-17)
Then multiply both numerator and denominator by the complex
conjugate of the denominator, that is, jωτ 1
jωτ 1
jωτ 1
G jω
2 2
jωτ 1 jωτ 1 ω τ 1
1
ω τ 1
2 2
j
ωτ
ω τ 1
2 2
R jI
(13-18)
R
where:
I
1
(13-19a)
ω τ 1
ωτ
2 2
(13-19b)
ω τ 1
2 2
From Step 3 of the Shortcut Method,
2
1
ωτ
AR R 2 I 2 2 2 2 2
ω τ 1 ω τ 1
or
2
2 2
ω τ 1
1 ω2 τ 2
AR
Also,
φ tan
1
1
ω2 τ 2 1
2
(13-20a)
I
1
1
tan
ωτ
tan
ωτ (13-20b)
R
Complex Transfer Functions
Consider a complex transfer G(s),
Ga s Gb s Gc s
G s
G1 s G2 s G3 s
Substitute s=j,
Ga jω Gb jω Gc jω
G jω
G1 jω G2 jω G3 jω
(13-22)
(13-23)
From complex variable theory, we can express the magnitude and
angle of G jω as follows:
G jω
Ga jω Gb jω Gc jω
(13-24a)
G1 jω G2 jω G3 jω
G jω Ga jω Gb jω Gc jω
[G1 jω G2 jω G3 jω
]
(13-24b)
Transfer Functions in Series
n
Y s
G s Gi s
X s
i 1
s j
n
G j Gi j
i 1
n
Gi j
i 1
G j e
Gi j e
Gi j e
i 1
i 1
n
n
ln G j jG j ln Gi j j Gi j
i 1
i 1
G j
n
Gi j
n
n
ln G j ln Gi j
i 1
n
G j Gi j
i 1
14
Transfer Functions in Series
Gi s
Kp
e s
zi
s 1
1
pi s 1
1
2 2
i s 2 i i s 1
0 i 1
15
Bode Diagrams
• A special graph, called the Bode diagram or
Bode plot, provides a convenient display of the
frequency response characteristics of a transfer
function model. It consists of plots of AR and
phase angle as a function of frequency.
• Ordinarily, frequency is expressed in units of
radians/time.
16
Bode Plot of A First-order System
Recall:
AR N
1
ω2 τ 2 1
and φ tan 1 ωτ
At low frequencies ( ω 0 and ω
AR N 1 and
At high frequencies ( ω 0 and ω
1) :
1) :
AR N 1/ ωτ and
17
Figure 13.2 Bode diagram for a first-order process.
• Note that the asymptotes intersect at ω ωb 1/ τ, known as the
break frequency or corner frequency. Here the value of ARN
from (13-21) is:
AR N ω ωb
1
0.707
11
(13-30)
• Some books and software defined AR differently, in terms of
decibels. The amplitude ratio in decibels ARd is defined as
AR d 20 log AR
(13-33)
Negative Zeros
Consider a process zero term,
G s K ( sτ 1)
Substituting s=j gives
G jω K ( jωτ 1)
Thus:
AR G jω K ω2 τ 2 1
φ G jω tan 1 ωτ
Note: In general, a multiplicative constant (e.g., K) changes
the AR by a factor of K without affecting φ .
20
Negative Zeros
1
AR N ω τ 1 and φ tan
ωτ
At low frequencies ( ω 0 and ω
1) :
2 2
AR N 1 and
At high frequencies ( ω 0 and ω
1) :
AR N ωτ and
21
Positive Zeros
G s 1 s
AR N ω2 τ 2 1 and φ tan 1 ωτ
At low frequencies ( ω 0 and ω
AR N 1 and
1) :
At high frequencies ( ω 0 and ω
1) :
AR N ωτ and
22
Examples
1
2
1
1s 1 2 s 1
z1s 1
p1
s 1 p 2 s 1
23
n-th Power of s
G s sn
n 1, 2,
G j j
n
AR
n
n
Im G j
tan
Re G j
1
24
Integrating Elements
The transfer function for an integrating element
was given in Chapter 5:
Y s
1
G s
U s s
1
1
AR G j
j
φ G jω K 90
25
Differentiating Element
G s s
AR G j
φ G jω tan
1
0
90
26
Second-Order System
G s
K
τ 2 s 2 2ζτs 1
G j
K 1 2 2
1
2 2
2
4
j
2 2 2
2 K
1
2 2
2
4 2 2 2
1
ARN
1
tan
2 2
2
2
2
2
1 2 2
1
27
Second-Order System
Low-frequency asymptote (slope=0):
0 ARN 1 and 0
High-frequency asymptote (slope=-2):
ARN
1
2
2
and
Beaking frequency:
1
1
b 1 ARN b
and tan
2
2
28
Resonant Peak
d ARN
1
0
d
2
2 2 1 2 2 0
3
2
2 2 1 2 2 8 2
1 2 2
1
2
*
1
2
1
AR*
2 1 2 2
Notice that, since and must both be real,
1
( 0.707)
2
29
Figure 13.3 Bode diagrams for second-order processes.
Time Delay
Its frequency response characteristics can be obtained
according to
G jω e jωθ
which can be written in rational form by substitution of the
Euler identity,
G jω e jωθ cosωθ j sin ωθ
Thus
AR G jω cos 2 ωθ sin 2 ωθ 1
sin ωθ
φ G jω tan 1
cos
ωθ
or
φ ωθ
Figure 13.6 Bode diagram for a time delay, e θs.
Figure 13.7 Phase angle plots for e θs and for the 1/1 and 2/2
Padé approximations (G1 is 1/1; G2 is 2/2).
Example
e s
G s
s 1
Let
1
G1 s e and G2 s
s 1
AR1 G1 j 1
s
AR2
1
1 2
AR AR1 AR2
1 2 tan 1
34
Nyquist Diagrams
Consider the transfer function
1
G s
2s 1
(13-76)
with
AR G jω
1
2ω 1
2
(13-77a)
and
φ G jω tan 1 2ω
(13-77b)
35
Figure 13.12 The Nyquist diagram for G(s) = 1/(2s + 1)
plotting Re G jω and Im G jω .
36
Figure 13.13 The Nyquist diagram for the transfer
function in Example 13.5:
5(8s 1)e6s
G( s)
(20s 1)(4s 1)
37
Frequency Response Characteristics of
Feedback Controllers
Proportional Controller. Consider a proportional controller with
positive gain
Gc s Kc
(13-57)
In this case Gc jω K c , which is independent of . Therefore,
AR c K c
(13-58)
and
φc 0
(13-59)
Proportional-Integral Controller. A proportional-integral (PI)
controller has the transfer function (cf. Eq. 8-9),
τI s 1
1
Gc s Kc 1
Kc
τ
s
τ
s
I
I
Substitute s=j:
(13-60)
jτ I 1
1
1
Gc j Kc 1
Kc
Kc 1
τ I j
jτ I
τI
j
Thus, the amplitude ratio and phase angle are:
AR c Gc jω K c 1
1
ωτ I
2
Kc
ωτ I
2
1
ωτ I
φc Gc jω tan 1 1/ ωτ I tan 1 ωτ I 90
(13-62)
(13-63)
10 s 1
G
s
2
Figure 13.9 Bode plot of a PI controller, c
10
s
Ideal Proportional-Derivative Controller. For the ideal
proportional-derivative (PD) controller (cf. Eq. 8-11)
Gc s Kc 1 τ D s
(13-64)
The frequency response characteristics are similar to those of a
LHP zero:
AR c K c
ωτ D
2
1
φ tan 1 ωτ D
(13-65)
(13-66)
Proportional-Derivative Controller with Filter. The PD controller
is most often realized by the transfer function
τDs 1
Gc s Kc
ατ
s
1
D
(13-67)
Figure 13.10 Bode
plots of an ideal PD
controller and a PD
controller with
derivative filter.
Idea: Gc s 2 4s 1
With Derivative
Filter:
4s 1
Gc s 2
0.4 s 1
PID Controller Forms
• Parallel PID Controller. The simplest form in Ch. 8 is
1
Gc s Kc 1
τDs
τ1s
Series PID Controller. The simplest version of the series PID
controller is
τ1s 1
Gc s Kc
(13-73)
τ D s 1
τ1s
Series PID Controller with a Derivative Filter.
τ1s 1 τ D s 1
Gc s Kc
τ1s τ D s 1
Figure 13.11 Bode
plots of ideal parallel
PID controller and
series PID controller
with derivative filter
(α = 1).
Idea parallel:
1
Gc s 2 1
4s
10 s
Series with
Derivative Filter:
10 s 1 4 s 1
Gc s 2
10
s
0.4
s
1