control chap10
TRANSCRIPT
![Page 1: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/1.jpg)
CONTROL SYSTEMS THEORY
Sinusoidal Tools
CHAPTER 10STB 35103
![Page 2: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/2.jpg)
Objectives To learn the definition of frequency
response To plot frequency response
![Page 3: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/3.jpg)
Introduction In previous chapters we learn to analyze
and design a control system using root locus method.
Another method that can be used is frequency response.
![Page 4: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/4.jpg)
Introduction What is frequency response?
The frequency response is a representation of the system's open loop response to sinusoidal inputs at varying frequencies.
The output of a linear system to a sinusoidal input is a sinusoid of the same frequency but with a different amplitude and phase.
The frequency response is defined as the amplitude and phase differences between the input and output sinusoids.
![Page 5: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/5.jpg)
Introduction Why do we use frequency response?
The open-loop frequency response of a system can be used to predict the behaviour of the closed-loop system .
we directly model a transfer function using physical data.
![Page 6: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/6.jpg)
Introduction Frequency response is consists of:
Magnitude frequency response, M(ω) Phase frequency response, ø(ω)
( )( )
( )
( ) ( ) ( )
o
i
o i
MM
M
![Page 7: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/7.jpg)
Introduction A transfer function Laplace form can be
change into frequency response using the following expression:
We can plot the frequency response in two ways: Function of frequency with separate magnitude and
phase plot. As a polar plot.
( ) ( )s j
G j G s
![Page 8: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/8.jpg)
Introduction Magnitude and phase plot
Magnitude curve can be plotted in decibels (dB) vs. log ω, where dB = 20 log M.
The phase curve is plotted as phase angle vs. log ω.
Data for the plots can be obtained using vectors on the s-plane drawn from the poles and zeros of G(s) to the imaginary axis.
![Page 9: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/9.jpg)
Introduction Magnitude response at a particular frequency
is the product of the vector length from the zeros of G(s) divided by the product of the vector lengths from the poles of G(s) drawn to points on the imaginary axis.
XXO
jω
jω1
σ
ACB
OD
1( )M j A •B
C•D
![Page 10: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/10.jpg)
Introduction The phase response is the sum of the
angles from the zeros of G(s) minus the sum of angles from the poles of G(s) drawn to points on imaginary axis.
XXO
jω
jω1
σO
1 1 2 3 4( ) [ ] [ ]j
13
1 32
![Page 11: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/11.jpg)
Introduction Example 10.1
Find the analytical expression for the magnitude frequency response and the phase frequency response for a system G(s) = 1/(s+2). Also, plot both the separate magnitude and phase diagrams and the polar plot.
![Page 12: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/12.jpg)
Introduction Solution:
First substitute s=jω in the system function and obtain
We always put complex number as numerator so we will multiply the above transfer function with the complex conjugate.
1( )
2G j
j
2
1( )
2
1 2
2 2
(2 )
( 4)
G jj
j
j j
j
![Page 13: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/13.jpg)
Introduction In order for us to plot the magnitude frequency
response we must find the magnitude of the transfer function.
Magnitude G(jω), M(ω)
Where G(jω)* is the conjugate of G(jω), so the magnitude for transfer function in the question is
( ) ( ) ( )G j G j G j
2 2
2
(2 ) (2 )( )
( 4) ( 4)
1
( 4)
j jG j
![Page 14: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/14.jpg)
Introduction The phase angle of G(jω), ø(ω)
2
2
(2 )( )
( 4)
1(2 )
( 4)
jG j
j
A B
1( ) tanB
A
1( ) tan2
![Page 15: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/15.jpg)
Introduction We can plot the magnitude frequency response and
phase frequency response
220log ( ) 1 4 vs. log M
1( ) tan 2 vs. log
![Page 16: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/16.jpg)
Introduction We can also plot the polar plot
2 1( ) ( ) 1 4 tan ( 2)M
![Page 17: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/17.jpg)
Introduction Exercise 10.1
Convert the following transfer function to frequency response. Find the magnitude frequency response and phase frequency response.
Solution
1( )
( 2)( 4)G s
s s
2 2
2
1( )
( 2)( 4)
1
4 2 8
1
8 6
G jj j
j j j
j
![Page 18: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/18.jpg)
Introduction
2
2
2 2
2
2 2 4 3 3 2 2
2
4 2
1
8 6
1 8 6
8 6 8 6
8 6
64 8 48 8 6 48 6 36
8 6
20 64
j
j
j j
j
j j j j j
j
![Page 19: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/19.jpg)
Introduction
![Page 20: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/20.jpg)
![Page 21: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/21.jpg)
Introduction Nyquist criterion
Nyquist criterion relates the stability of a closed-loop system to the open-loop frequency response and open-loop pole location.
This concept is similar to the root locus.
The most important concept that we need to understand when learning Nyquist criterion is mapping contours.
![Page 22: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/22.jpg)
Introduction Mapping contours
Mapping contours means we take a point on one contours and put it into a function, F(s), thus creating a new contours.
![Page 23: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/23.jpg)
![Page 24: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/24.jpg)
![Page 25: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/25.jpg)
Introduction When checking the stability of a system,
the shape of contour that we will use is a counter that encircles the entire right half-plane.
![Page 26: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/26.jpg)
Introduction The number of closed-loop poles in the right half
plane (also equals zeros of 1+ G(s)H(s)), Z The number of open-loop poles in the right half
plane , P The number of counterclockwise rotations about
(-1,0), N
N = P - Z The above relationship is called the Nyquist
Criterion; and the mapping through G(s)H(s) is called the Nyquist Diagram of G(s)H(s)
![Page 27: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/27.jpg)
Introduction Examples to determine the stability of a
system0, 0,
0, the system is stable
P N
Z P N
0, 2, ( ' ')
0 ( 2) 2,system unstable
P N clockwise ve
Z
![Page 28: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/28.jpg)
Sketching the Nyquist Diagram The contour that encloses the right half-plane
can be mapped through the function G(s)H(s) by substituting points along the contour into G(s)H(s).
The points along the positive extension of the imaginary axis yield the polar frequency response of G(s)H(s).
Approximation can be made to G(s)H(s) for points around the infinite semicircle by assuming that the vectors originate at the origin.
![Page 29: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/29.jpg)
Sketching the Nyquist Diagram Example 10.4
Sketch a nyquist diagram based on the block diagram below.
![Page 30: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/30.jpg)
Sketching the Nyquist Diagram
Solution The open loop transfer function G(s),
Replacing s with jω yields the frequency response of G(s)H(s), i.e.
500( )
( 1)( 10)( 3)G s
s s s
2 3
2 2 3 2
500( )
( 1)( 10)( 3)
( 14 30) (43 )500
( 14 30) (43 )
G jj j j
j
![Page 31: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/31.jpg)
Sketching the Nyquist Diagram
Magnitude frequency response
Phase frequency response
2 2 3 2
500( )
( 14 30) (43 )G j
31 1
2
(43 )( ) tan tan
14 30
BG j
A
![Page 32: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/32.jpg)
Sketching the Nyquist Diagram
Using the phase frequency response and magnitude frequency response we can calculate the key points on the Nyquist diagram. The key points that we will calculate are:
Frequency when it crosses the imaginary and real axis.
The magnitude and polar values during the frequency that crosses the imaginary and real axis.
The magnitude and polar values when frequency is 0 and ∞.
![Page 33: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/33.jpg)
Sketching the Nyquist Diagram
When a contour crosses the real axis, the imaginary value is zero. So, the frequency during this is,
2 3
2 2 3 2
2 3
2 2 3 2 2 2 3 2
( 14 30) (43 )500
( 14 30) (43 )
( 14 30) (43 )500
( 14 30) (43 ) ( 14 30) (43 )
j
j
real imaginary
3
2 2 3 2
(43 )0
( 14 30) (43 )
![Page 34: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/34.jpg)
Sketching the Nyquist Diagram
We need to find the frequency when imaginary is zero by finding the value of ω that could produce zero imaginary value.
There are actually two conditions that could produce zero imaginary.
First
Second
2 2 3 2
00
( 14 30) (43 )
3(43 )0
![Page 35: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/35.jpg)
Sketching the Nyquist Diagram
For the first condition, in order to get the numerator equals to zero we must find the root value of the numerator polynomial.
3
2 2 3 2 2 2 3 2
(43 ) 0
( 14 30) (43 ) ( 14 30) (43 )
3
1
2
3
(43 ) 0
0
6.5574
6.5574
There are three frequencies where the contour crosses the real axis.
![Page 36: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/36.jpg)
Sketching the Nyquist Diagram
For the second condition, the frequency values in the denominator that could produce zero imaginary value is infinity, ∞.
3 3
2 2 3 2
(43 ) (43 )
( 14 30) (43 )
2 2 3 2( 14 30) (43 )
![Page 37: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/37.jpg)
Sketching the Nyquist Diagram
When a contour crosses the imaginary axis, the real value is zero.
2 3
2 2 3 2
2 3
2 2 3 2 2 2 3 2
( 14 30) (43 )500
( 14 30) (43 )
( 14 30) (43 )500
( 14 30) (43 ) ( 14 30) (43 )
j
j
real imaginary
2
2 2 3 2
( 14 30)0
( 14 30) (43 )
![Page 38: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/38.jpg)
Sketching the Nyquist Diagram
There are two conditions that could produce zero real value.
First
Second
2 2 3 2
00
( 14 30) (43 )
2( 14 30)0
![Page 39: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/39.jpg)
Sketching the Nyquist Diagram
Calculate the frequency values for the first condition.
Calculate the frequency values for the second condition
2
2 2 3 2 2 2 3 2
2
( 14 30) 0
( 14 30) (43 ) ( 14 30) (43 )
( 14 30) 0
1.4639
2 2
2 2 3 2
( 14 30) ( 14 30)
( 14 30) (43 )
![Page 40: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/40.jpg)
Sketching the Nyquist Diagram
Now that we know the frequencies of the key points in our polar plot we will now calculate the magnitudes and phase for each key points frequency.
Cross real
Cross imaginary
![Page 41: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/41.jpg)
The new contour can be plot based on the key points in the previous table.
Sketching the Nyquist Diagram
0
1.4639
6.5574 AC
![Page 42: Control chap10](https://reader033.vdocuments.mx/reader033/viewer/2022050706/558c0ad6d8b42a52568b470a/html5/thumbnails/42.jpg)
Sketching the Nyquist Diagram Note that the semicircle with a infinite
radius, i.e., C-D, is mapped to the origin if the order the denominator of G(s) is greater than the order the numerator of G(s).