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### Chapter 3: Dynamic Response

Part C: Transient-response analysis with MATLAB.

Introduction

- The practical procedure for plotting time response curves of systems higher than second-order is through computer simulation.
- In this part, computational approach to the transient-response analysis with MATLAB is presented through various examples.

Representation of a Linear System

A linear system can be representedeither:

- In state-variable form:

with the values of the matrices F, G, H and the constant J.

Or

- By its transfer function:

Either in numerator-denominator polynomial form,

Or in pole-zero form

Or in partial expansion form

Example 1: Standard State-Variable Form

- Consider a linear system described by:

F = [0 1;0 -0.05];

G = [0;0.001];

H = [0 1];

J = 0;

step(F,G,H,J)

% defines state variable matrices

% generates plot of unit-step response (with Time (sec)and Amplitudelabels on x- and y-axis respectively, andStep responsetitle )

Note: time vector is automatically determined when t is not explicitly included in the step command.

Unit-Step Response for a Control System defined in State-Variable formF = [0 1;0 -0.05];

G = [0;0.001];

H = [0 1];

J = 0;

sys = ss(F, 50*G, H, J);

step(sys)

% defines state variable matrices

% defines system by its state-space

matrices

% generates plot of 50-step response vs t

Note:State-variable form is also called state-space form

50-Step Response for a system defined in State-Variable formF = [0 1;0 -0.05];

G = [0;0.001];

H = [0 1];

J = 0;

sys = ss(F, G, H, J);

t = 0:0.2:100;

y=step(sys,t);

plot(t,y)

% defines state variable matrices

% defines system by its state-space

matrices

% setup time vector ( dt = 0.2 sec)

% plots unit step response versus time ranging from 0 to 100 sec (with x- and y-labels)

Unit-Step Response on specific time interval for a system defined in State-Variable formF = [0 1;0 -0.05];

G = [0;0.001];

H = [0 1];

J = 0;

sys = ss(F, G, H, J);

impulse(sys)

% defines state variable matrices

% defines system by its state-space

matrices

% generates plot of impulse response

(with labels & title)

Note: an alternative use of impulse command is:

impulse(F,G,H,J)

Impulse Response for a system defined in State-Variable formExample 2: Initial Conditions

- Consider a linear system such as:
- In state-variable form, it is described by:

F = [0 1;-10 -5];

G = [0;0];

H = [1 0];

J = 0;

t = 0:0.5:3;

y=initial(F,G,H,J,[2;1],t);

plot(t,y)

% defines state variable matrices

% set up time vector

% computes initial condition response

% generates plot of response

Note: Initial conditions are defined between [ ].

Initial Condition Response for a system defined in State-Variable formExample 3: Transfer function in numerator-denominator form

- Consider a linear system whose the transfer function is:

num = [0 0 25];

den = [1 4 25];

step(num,den)

% defines numerator

% defines denominator

% generates plot of unit-step response (with labels and title)

Unit-Step Response for a system Transfer Function defined in num/den polynomial formnum = [0 0 50];

den = [1 0.2 1];

step(num,den)

% defines numerator

% defines denominator

% generates plot of 50-step response (with labels and title)

50-Step Response for a system Transfer Function defined in num/den polynomial formt = 0:0.2:10;

zeta = [0 0.2 0.4 0.6 0.8 1];

for n = 1:6;

num = [0 0 1];

den = [1 2*zeta(n) 1];

[y(1:51,n),x, t] = step(num,den,t);

end

plot(t,y)

% setup time vector

% defines zeta,

numerator and

denominator

% generates 2-D plot of the n unit-step responses (on same graph)

Unit-Step Responses for system Transfer Functions defined byt = 0:0.2:10;

zeta = [0 0.2 0.4 0.6 0.8 1];

for n = 1:6;

num = [0 0 1];

den = [1 2*zeta(n) 1];

[y(1:51,n),x, t] = step(num,den,t);

end

mesh(t,zeta,y’)

% setup time vector

% defines zeta,

numerator and

denominator

% generates 3-D plot of the n unit-step responses (on same graph)

Unit-Step Responses for system Transfer Functions defined bynum = [0 0 0 1];

den = [1 1 1 0];

step(num,den)

% defines numerator

% defines denominator

% generates plot of unit-step response (with x- and y-labels)

Unit Step Response for a 3rd order system defined by its Transfer Functionnum = [0 0 1];

den = [1 0.2 1];

sys=tf(num,den);

impulse(sys)

% defines numerator

% defines denominator

% defines system by its transfer function

% generates plot of impulse response

Note: an alternative use of impulse command is:

impulse(num,den)

Impulse Response for a system Transfer Function defined in num/den polynomial formnum = [0 1 0];

den = [1 0.2 1];

step(num,den)

% defines numerator of sG(s)

% defines denominator

% generates plot of impulse response (with x- and y-labels)

Alternative approach to obtain Impulse ResponseExample 4: Transfer function in standard 2nd order system

- Consider a standard second order system:

natural

undamped

frequency

damping

ratio

w0 = 5;

damping_ratio = 0.4;

[num0,den] = ord2(w0,damping_ratio);

num = 5^2*num0;

printsys(num,den,’s’)

% defines natural undamped frequency

% defines damping ratio

% defines numerator

% prints num/den as a ratio of s-polynomials

num/den =

MATLAB Description of Standard Second Order SystemExample 5: Transfer function in pole-zero form

- Consider a linear system whose the transfer function is:

num = conv([1 2],[1 4]);

den = conv([1 1 0],[1 3]);

step(num,den)

% defines zero ratios

% defines pole ratios

% plots unit-step response

Unit-Step Response for a system Transfer Function defined in pole-zero formExample 6: Transfer function in Partial Expansion Form

- Consider a linear system whose the transfer function is:

r = [8/3 -3/2 -1/6];

p = [0 -1 -3];

K = [] ;

[num,den] = residue(r,p,K)

step(num,den)

% defines residues

% defines poles

% define additive constant

% convert partial expansion form to polynomial form

% plots unit-step response

Note: to see ratio use

printsys(num,den,’s’)

Unit-Step Response for a system Transfer Function defined in partial expansion formState-variable form

Transfer function:

In num-den polynomial form

In zero-pole form

In partial expansion form

Convertion[num,den] = ss2tf(F,G,H,J)

[z,p,k]=tf2zp(num,den)

[r,p,K]=residue(num,den)

[z,p,k] = ss2zp(F,G,H,J)

State-variable form

Transfer function:

In num-den polynomial form

In zero-pole form

In partial expansion form

Convertion[F,G,H,J] = tf2ss(num,den)

[num,den]=zp2tf(z,p,k)

[num,den]=residue(r,p,K)

[F,G,H,J] = zp2ss(z,p,k)

title (‘Step-response’);

grid;

sys = …;

t = 0:0.2:100;

y = step(sys,t);

plot (t,y);

xlabel(‘t (sec)’);

ylabel(‘response’)

% writes the title Step-response

% draws a grid between ticks

% defines system by …

% setup time vector ( dt = 0.2 sec)

% computes step response

% plots step response

% writes label t (sec) on x-axis.

% writes label response on y-axis.

Title, Grid & Labels on the graphical screentext(3.4, -0.06, ‘Y111’);

text(4.1,1.86,’\zeta’);

gtext(‘blabla’)

% writes Y111 beginning at the coordinates x=3.4, y=-0.06.

% writes at x=4.1, y=1.86

% waits until the cursor is positioned (using the mouse) at the desired position in the screen and then writes on the plot at the cursor’s location the text enclosed in simple quotes.

Note: any number of gtext command can be used in a plot.

Writing Text on the Graphical Screennum = [0 0 25];

den = [1 6 25];

t = 0:0.5:5;

y = step(num,den,t);

plot(t,y,’o’,t,1,’-’);

% defines numerator

% defines denominator

% defines time vector

% computes unit-step response

% plot of unit step response y and unit step input 1 using oooo and ---- symbols respectively.

Use of Symbols in graphnum = [0 0 25];

den = [1 6 25];

t = 0:0.5:5;

y = step(num,den,t);

plot(t,y,’x’,t,y,’-’);

% defines numerator

% defines denominator

% defines time vector

% computes step response

% plot of unit step response y using -x-x-x-x- symbols

Use of Symbols in graph (cont’d)Computing roots using MATLAB

Plotting pole(s) and zero(s) in the s-plane using MATLAB

Plotting Step-response versus a parameter range

Obtaining rise time, peak time, maximum overshoot and settling time using MATLAB

Additional Convenient MATLAB Commandspol= [1 4 3 2 1 4 4];

roots(pol)

ans =

-3.2644

-0.6046 + 0.9935i

-0.6046 - 0.9935i

0.6797 + 0.7488i

0.6797 - 0.7488i

-0.8858

Computing Rootsden= [1 5 11 23 28 12];

roots(den)

ans =

-3.0000

0.0000 + 2.0000i

0.0000 - 2.0000i

-1.0000 + 0.0000i

-1.0000 - 0.0000i

Stability Analysis by Computing Roots2 poles are in the RHP

num=[0 2 1];

den= [2 3 2];

zmap(num,den)

Plotting Poles and Zero in the s-domainpoles as crosses

zero as circle

xlabel(\'Time (sec)\');

ylabel(\'Amplitude\');

Title(\'Step-Response

versus K parameter\');

grid;

text(7.1,3.8,\'K=6.5\');

text(7.5,3.15,\'7\');

text(7.15,2.65,\'7.5\');

text(7.1,2.3,\'8\');

text(6.65,1.37,\'10\');

text(6.4,0.75,\'12.5\');

t=0:0.1:10;

K=[6.5 7 7.5 8 10 12.5];

for n=1:6

num=[K(n) K(n)];

den=[1 5 K(n)-6 K(n)];

[y(1:101,n),x,t]=step(num,den,t);

end

plot(t,y);

Step-Response versus a Parameter RangeReminder: Rise Time

- The rise time is the time requiredfor the responseto rise from 0% to 100% of its final value.

1

t

d

0.5

0

t

r

Note: for overdamped systems, the 10%

to 90% rise time is commonly used.

num= [0 0 25];

den=[1 6 25];

t=0:0.001:5;

[y,x,t]=step(num,den,t);

r=1;

while y(r) <1.0001;

r=r+1;

end;

rise_time=(r-1)*0.001

rise_time =

0.5540

Computing Rise Time using MATLABNo ;

Reminder: Peak Time

- The peak time is the time requiredfor the responseto reach the first peak of the overshoot.

t

p

1

t

d

0.5

0

t

r

num= [0 0 25];

den=[1 6 25];

t=0:0.001:5;

[y,x,t]=step(num,den,t);

[ymax,tp]=max(y);

peak_time=(tp-1)*0.001

peak_time =

0.7850

Computing Peak Time using MATLABNo ;

Reminder: Maximum Overshoot

- The maximum overshoot is the relative maximum peak value of the response curve measured from the final value.

t

p

M

p

1

t

d

0.5

0

t

r

Note: the maximum overshoot directly indicates

the relative stability of the system.

num= [0 0 25];

den=[1 6 25];

t=0:0.001:5;

[y,x,t]=step(num,den,t);

[ymax,tp]=max(y);

peak_time=(tp-1)*0.001

max_overshoot=ymax-1

peak_time =

0.7850

max_overshoot =

0.0948

Computing Maximum Overshoot using MATLABReminder: Settling Time

- The settling time is the time required for the response curve to reach and stay within a range about1% or 2% of the final steady-state value.

t

p

M

p

±1%

1

t

d

0.5

0

t

t

r

s

Note:t is the time it takes the

system transients to decay.

s

num= [0 0 25];

den=[1 6 25];

t=0:0.001:5;

[y,x,t]=step(num,den,t);

s=5001;

while y(s)>0.98 & y(s)<1.02;

s=s-1;

end;

settling_time=(s-1)*0.001

settling_time =

1.1880

Computing Settling Time using MATLAB(based on +/-2%)Results given by MATLAB are:

rise_time = 0.5540; peak_time = 0.7850,

max_overshoot = 0.0948, settling_time = 1.1880

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