Question

PROBLEM 1 Consider the transfer function T(S) =s5 +2s4 + 2s3 + 4s2 + s + 2 a) Using the Routh-Hurwitz method, determine whether the system is stable. If it is not stable, how many poles are in the right-half plane? b) Using MATLAB, compute the poles of T(s) and verify the result in part a) c) Plot the unit step response and discuss the results. (Report should include: Code, Figure 1.Unit step response, answers and conclusion)

Answer 1

The transfer function is

T(s)

a) The denominator polynomial of the transfer function is

$s^5+2s^4+2s^3+4s^2+s+2$

The polynomial is considered as

Routh-Hurwitz table is constructed as follow,

3 1 02 S C1 C2 1 di S 0

where,
12 dod3 0l1
,
0l1

where,
b1a3 - ajb2 C1 b1
,

where,
C162 b1C2 C1
and

Routh-Hurwitz method requires the denominator polynomial. The table is constructed as shown below,

$\begin{matrix} s^5 & 1 & 2 & 1\\ s^4 & 2 & 4 & 2 \end{matrix}$

2*2-1*4
,
2*2-1*4

2 240 543

All the values in the row are zero, therefore we need to take differential of the polynomial in the second row

ds

$\begin{matrix} s^5 & 1 & 2 & 1\\ s^4 & 2 & 4 & 2\\ s^3&8&8 \end{matrix}$

$c_1 = \frac{8*4-8*2}{8} = 2$,

$\begin{matrix} s^5 & 1 & 2 & 1\\ s^4 & 2 & 4 & 2\\ s^3&8&8\\s^2&2&2 \end{matrix}$

$d_1 = \frac{8*2-8*2}{2} = 0$

$\begin{matrix} s^5 & 1 & 2 & 1\\ s^4 & 2 & 4 & 2\\ s^3&8&8\\s^2&2&2 \\s^1&0\end{matrix}$

All the values in the row are zero, therefore we need to take differential of the polynomial in the second row

$\frac{dP(s)}{ds} = 4s$

$\begin{matrix} s^5 & 1 & 2 & 1\\ s^4 & 2 & 4 & 2\\ s^3&8&8\\s^2&2&2 \\s^1&4\end{matrix}$

$\begin{matrix} s^5 & 1 & 2 & 1\\ s^4 & 2 & 4 & 2\\ s^3&8&8\\s^2&2&2 \\s^1&4\\s^0&2\end{matrix}$

We can deduce the following points after constructing the Routh-Hurwitz table

1) The column array do not have any sign change. Thus, all the poles lie on the left-half plane.

2) While constructing the table, we got two times zero value in the first column. Thus, we have two pairs of imaginary poles. There are four poles on the imaginary axis.

3) Since, the column array didn't had positive value. Therefore, the system is not stable.

Four poles lie on the imaginary axis of right-half plane. Those four poles are obtained by solving the polynomial.

$s = \pm 1i$

Repeating poles at $s = \pm 1i$

The four poles are +1i, +1i, -1i, -1i. The remaining pole is calculated as follows

$\frac{s^5+2s^4+2s^3+4s^2+s+2}{s^4+2s^2+1} = s+2$

The fifth pole is at -2.

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b) Construct a transfer function using numerator and denominator polynomial and calculate the poles in MATLAB. The code for MATLAB is below,

------ START OF MATLAB CODE --------

sys = tf(1,[1,2,2,4,1,2]); % Transfer function
[p,z] = pzmap(sys); % Finding poles and zeros
disp(p)

--------- END OF MATLAB CODE ------

Run the above command in the MATLAB command window. The poles are displayed. The values of the poles are

-2.00000 + 0.00000i
-0.00000 + 1.00000i
-0.00000 - 1.00000i
0.00000 + 1.00000i
0.00000 - 1.00000i

Note: The values displayed are same as the values calculated from Routh-Hurwitz method.

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c) Plotting the step response for the transfer function using MATLAB.

----------- START OF MATLAB CODE --------

sys = tf(1,[1,2,2,4,1,2]); % Transfer function
t = 0:0.1:100; % Time length for step response
[y,t] = step(sys, t); % Step response of the transfer function
figure,plot(t,y) % Plotting the step response
grid on;
title('Step Response')

---------- END OF MATLAB CODE --------

The above code can be run in the MATLAB command window. The step response plot is shown below,

From the result, we can observe that when we provide constant input to the system. The system becomes unstable as the output value increases as time increases. Thus, the system is unstable is also proved by plotting the step response.

Step Response 30 20 10 -10 -20 100 -30 80 60 40 20

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