《自动化仪表与过程控制》课程学习资料：Frequency-Domain Analysis of Control System

Frequency-Domain Analysis of Control System 2022-2-11

2022-2-11 1 Frequency-Domain Analysis of Control System

Frequency Response Frequency response is the anal ysis of the response of systemswhen subjected to a sinusoidal change in input When a linear system is subjected to a sinusoidal input its ultimate response is also a sustained sinusoidal wave, with the same frequency The figure below compares the output response of a system (solid line)with a sinusoidal input ( dashed line )disturbing the system 2022-2-11

2022-2-11 2 Frequency Response Frequency response is the analysis of the response of systemswhen subjected to a sinusoidal change in input. When a linear system is subjected to a sinusoidal input, its ultimate response is also a sustained sinusoidal wave, with the same frequency. The figure below compares the output response of a system (solid line) with a sinusoidal input (dashed line) disturbing the system

Phase shift 0.5 10 20 Tim A m plitude r atio =b/a This particular graph shows the response of the system: G(S)=3 5s+i to the sinusoidal input of: u= sin(t). The figure shows that the response of the system lags the input changes slightly this lag is known as the phase shift. The ratio of the amplitudes of the input and output sinusoids is known as the amplitude ratio Both the magnitude and the phase shift of a system will change with the frequency of the input into the system 2022-2-11 3

2022-2-11 3 - 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5 0 1 0 2 0 3 0 T im e O u t p u t P h a s e S h ift A B A m p l i t u d e R a t i o = B / A This particular graph shows the response of the system: to the sinusoidal input of: u = sin (t). The figure shows that the response of the system lags the input changes slightly, this lag is known as the phase shift. The ratio of the amplitudes of the input and output sinusoids is known as the amplitude ratio. Both the magnitude and the phase shift of a system will change with the frequency of the input into the system. G s s ( )   3 5 1

If the general system represented by the following transfer function: G(3)=P(s)(where @(s)and P(s)are polynomials is terms of s)is subjected to a sinusoidal input of frequency a then the amplitude ratio of the resulting response will be given by: AR= modulus of@)=G(o), where GGo)is found of replacing s in g(byo. Similarly the phase shift will be given by: phase shift=zG(o). For a given sinusoidal input of a sin(ot+9), the output will be a sinusoid of: aG(jo)sin(ot+9+2G(o) Example l: Determine the phase angle and ar of the system S G(s)=-,Replace s by jo G()= JO 0-j Jo 0Jo 2022-2-11 4

2022-2-11 4 If the general system represented by the following transfer function: (where Q(s) and P(s) are polynomials is terms of s) is subjected to a sinusoidal input of frequency w, then the amplitude ratio of the resulting response will be given by: , where G(jw) is found of replacing s in G(s) by jw. Similarly the phase shift will be given by: . For a given sinusoidal input of , the output will be a sinusoid of: . G s Q s P s ( ) ( ) ( )  AR  modulus of G(jw )= G(jw)   phase shift  G(jw) a sinwt   aG(jw)sinwt   G(jw) Example 1: Determine the phase angle and AR of the system 1/s. G s s s j G j j j j j j ( ) , ( )     1 1 1 0 1 Replace by w, w w w w w w

Given a complex number of the form: y =a+jb, the modulus and argument of it are as follows √a2+b2,∠y=tani Therefore AR=G(jO) tan ∞=-90°orp=-tan- Example 2: Determine the phase angle and ar of the system 4 G(S)= when subjected to a +1 sinusoidal input r=2sin (3t+600) 2022-2-11 5

2022-2-11 5 Given a complex number of the form: y = a+jb, the modulus and argument of it are as follows: y a b y b a           2 2 1 , tan Therefore, AR G j o             ( ) , tan tan w w   1 90 1 1 or Example 2: Determine the phase angle and AR of the system when subjected to a sinusoidal input R=2sin(3t+60 0). G s s ( )   4 1

J+1 maltiply top and bottom by (a+l) 40+44 40 G(O 2+1m2+1 4 4 ox又 ((o) 0+ O2+1)Vo2+1 o=tan@ Y=aGlo)sin(ot+9+ Y=2.6sinl 3t-1 2022-2-11 6

2022-2-11 6     G j j G j j j G j Y aG j t Y t ( ) ( ) ( ) tan ( ) sin . sin w w w w w w w w w w w w w w  w w w                                      4 1 4 4 1 4 1 4 1 4 1 4 1 4 1 2 6 3 12 2 2 2 2 2 2 2 2 1 0 multiply top and bottom by (-j +1)

Bode Diagrams o a Bode diagram is a graphical method for illustrating how the phase angle and amplitude ratio of a particular system vary with frequency of the input sinusoid. For example consider the integrating system: G(s)=, the amplitude ratio and phase angle were determined above to be: AR= 90 The table below shows how these values vary with frequency AR 201og(ar) o 0.01 40dB 9 0.1 20dB 90° 1.0 OdB 90 0.1 -20dB 90° 100 0.01 40dB 2022-2-11 7

2022-2-11 7 Bode Diagrams ¨ A Bode diagram is a graphical method for illustrating how the phase angle and amplitude ratio of a particular system vary with frequency of the input sinusoid. For example consider the integrating system: , the amplitude ratio and phase angle were determined above to be: . The table below shows how these values vary with frequency: G s s ( )  1 AR    1 90 0 w ,  AR 20log(AR) 0.01 100 40dB 0.1 10 20dB 1.0 1 0dB 10 0.1 -20dB 100 0.01 -40dB w 90  90  90  90  90

These values can now be plotted. a Bode diagram has two plots, one for the ar (in db) and the other for the phase angle Each plot has a log-linear scale. The Bode plot for this example Is shown below 111 I ;} 11 50 10 101 10 10 Frequency (rad/sec) """ a-90 1 144 "" t重1 10 10 10 10 10 Frequency〔rad/set 2022-2-11

2022-2-11 8 These values can now be plotted. A Bode diagram has two plots, one for the AR (in dB) and the other for the phase angle. Each plot has a log-linear scale. The Bode plot for this example is shown below:

Bode diagram of a First order System For a first order system the ar and phase angle can be determined as follows G(S) GGo 1+)0 GaindB=20 log 1+7ol 1-7)1-7c 1+⑦)o1+7o1-7o1+T 1+ta jo 1(1+702) VI+T Gain,p=2010g 11+10 Gains 20log√1+7 ∠=tan-oT 2022-2-11 9

2022-2-11 9 Bode Diagram of a First Order System ¨ For a first order system the AR and phase angle can be determined as follows:   G s Ts G j Tj Gain Tj Tj Tj Tj Tj Tj T Tj T T T Gain T Gain T T dB dB dB ( ) ( ) log * log log tan                                    1 1 1 1 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 20 1 1 20 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 w w w w w w w w w w w w w w w w

For low frequencies, aT>1, therefore Gaind B=-20logoT Therefore, the gain plot can be represented by 2 asymptotes one line at odb and another line with -20dB/decade 0=1/T 20g1+7≈-3B O=2/T 20logV1+027≈6B O=10/7-20lg√1+02m2≈-20B For the phase angle tan ot 1+joT 0,p=00=1/7,=-45 →>∞φ>-90 2022-2-11 10

2022-2-11 10 For low frequencies, wT>1, therefore GaindB=-20logwT Therefore, the gain plot can be represented by 2 asymptotes: one line at 0dB and another line with -20dB/decade. w w w w w w             1 20 1 3 2 20 1 6 10 20 1 20 2 2 2 2 2 2 / log / log / log T T dB T T dB T T dB For the phase angle  w w w  w  w                     1 1 0 0 1 45 90 1 j T T T o o o tan ; / ; ;