Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10M

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

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

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

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

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

This is the required transfer function of the given system.

Equation 2 can be written as

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

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

$Y(s)=\frac{1}{s^2+5(s)+6(s)}\frac{1}{s}$

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

Expanding above equation in partial fractions

$Y(s)=\frac{k_o}{s}+\frac{k_1}{(s+2)}+\frac{k_2}{(s+3)}$---------4

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

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

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

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

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

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

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

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

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

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

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

$Y(s)=\frac{1}{6}\frac{1}{s}-\frac{1}{2}\frac{1}{(s+2)}+\frac{1}{3}\frac{1}{(s+3)}$--------5

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