Question

1) Use Simulink to plot the unit step response of the following block diagram for K-1, 2, 5 and find Mp, tp, ts from the figure. (116s2 +1187s+8260) K(s) K controller plant R(s) K(s) G(s) Y(s) 2) Find the state variable representation of closed loop system of (1) by using Simulink.

Answer 1

Since K(s) is an improper fraction..we use K(s)*G(s) in Simulink.

The above image is for K=1.

Block Parameters: Transfer Fcn Transfer Fcn The numerator coefficient can be a vector or matrix expression. The denominator coefficient must be a vector. The output width equals the number of rows in the numerator coefficient. You should specify the coefficients in descending order of powers of s. Parameters Numerator coefficients: [116 1187 8260] Denominator coefficients: [1 69 0] Absolute tolerance: auto State Name: (e.g., 'position') num(s) den(s) OK Cancel Help Apply

For K=2,just multiply each coefficient in numerator with 2. Simpilarly for K=5,multiply each coefficient with 5.

The unit-step response for K=1:

For K=2:

0.8 0.4 0.2 05 15 2.5 Ofiset0

For K=5:

0.6 0.4 05 15 2.5 Ofiset0

Now to find Mp,Tp,Ts we write the code as:

0.8 2.5 15 05 Ofiset0

clc;clear all;close all;

k=116*1;

z=[-5.1164 + 6.7104i -5.1164 - 6.7104i];

p=[0 -3 -3];

sys=zpk(z,p,k)

cl_tf=feedback(sys,1)

step(cl_tf)

grid on

stepinfo(cl_tf)

To find for K=2,replace K=116*1 with K=116*2 and K=5,put K=116*5 in the code.

The output of the codes are:

For K=1,

RiseTime: 0.0171

SettlingTime: 0.2255

SettlingMin: 0.9013

SettlingMax: 1.0410

Overshoot: 4.0996

Undershoot: 0

Peak: 1.0410

PeakTime: 0.0744

For K=2,

RiseTime: 0.0090

SettlingTime: 0.0782

SettlingMin: 0.9018

SettlingMax: 1.0204

Overshoot: 2.0365

Undershoot: 0

Peak: 1.0204

PeakTime: 0.0554

For K=5,

RiseTime: 0.0037

SettlingTime: 0.0063

SettlingMin: 0.9019

SettlingMax: 1.0065

Overshoot: 0.6543

Undershoot: 0

Peak: 1.0065

PeakTime: 0.0119.

For State Space Representation,the code will be,

clc;clear all;close all;

k=116*1;

z=[-5.1164 + 6.7104i -5.1164 - 6.7104i];

p=[0 -3 -3];

sys=zpk(z,p,k)

cl_tf=feedback(sys,1)

ss(cl_tf)

Now for K=2,replace k=116*1 with k=116*2 and for K=5,replace with k=116*5

The output will be like:

For K=1,

ans =

A =

x1 x2 x3

x1 -5.011 6.975 0

x2 -6.975 -5.011 0.7479

x3 0 0 -112

B =

u1

x1 0

x2 0

x3 16

C =

x1 x2 x3

y1 -5.022 2.044 7.25

D =

u1

y1 0

For K=2,

A =

x1 x2 x3

x1 -5.07 6.841 0

x2 -6.841 -5.07 0.5235

x3 0 0 -227.9

B =

u1

x1 0

x2 0

x3 16

C =

x1 x2 x3

y1 -7.143 2.569 14.5

D =

u1

y1 0

For K=5;

A =

x1 x2 x3

x1 -5.099 6.762 0

x2 -6.762 -5.099 0.329

x3 0 0 -575.8

B =

u1

x1 0

x2 0

x3 32

C =

x1 x2 x3

y1 -5.656 1.891 18.13

D =

u1

y1 0