Determine the transfer function and the poles and zeros of the systems. Evaluate zero-state response to x(t) = u(t).

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10M

1 Answer

The given system equation is

$\frac {d^2y(t)}{dt^2}+ 5\frac{dy(t)}{dt} + 6 y(t) = x(t)$

Taking Laplace transform of the above equation

$[s^2Y(s) – s y(0-) – y’(0-)] + 5[s Y(s) – y(0-)] + 6Y(s) = X(s)$ ------------ 1

Considering all the initial values zero

Therefore equation 1 becomes

$s^2 Y(s) + 5s Y(s) + 6Y(s) = X(s)$

$[s^2 + 5s + 6]Y(s) = X(s)$


$\therefore H(s)=\frac{1}{s^2+5(s)+6}$......2

This is the required transfer function of the given system.

Equation 2 can be written as


From the above equation, the value of H(s) will never be zero.

Therefore, the system has no zeros.

From the above equation, the value of H(s) will be infinite for s = -2 and s = -3. Therefore, these are the values of poles.

$\therefore$ Poles of the system are at s = -2 and s = -3

Now, x(t) = u(t)

Taking Laplace transform of x(t)

$X(s) =\frac{1}{s}$ ------------ 3

By convolution theorem,

Y(s) = H(s).X(s)

Substituting the values of H(s) and X(s) from equations 2 and 3



Expanding above equation in partial fractions


Here$ k_O = s Y(s) |s = 0$

$=s\frac{1}{s(s+2)(s+3)} |s=0$

$\therefore k_0=\frac{1}{6}$

$k_1=(s+2)Y(s) |s=-2$

$=(s+2)\frac{1}{s(s+2)(s+3)} |s=-2$


$\therefore k_1=\frac{-1}{2}$

$k_2=(s+3)\frac{1}{s(s+2)(s+3)} |s=-3$


$\therefore k_2=\frac{1}{3}$

Substituting the values of $ k_O, k_1 , k_2$ in equation 4, we get


Taking inverse Laplace transform of above equation

$\therefore y(t)=\frac{1}{6}u(t)-\frac{1}{2}e^{-2t}u(t)+\frac{1}{3}e^{-3t} u(t)$

This is the response of the system

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