Question

For a system with a transfer function 1/(s^2 + 6s + 9) when subject to a unit step input, the output as a function of time is​

Answer 1

Step-by-step explanation:

Given:For a system with a transfer function 1/(s² + 6s + 9) when subject to a unit step input,

To find: The output as a function of time

Solution:

We know that input is expressed as R(s) in s-domain

R(s) is unit step function so,

$R(s) = \frac{1}{s}$

The transfer function G(s) is given

$G(s) = \frac{1}{ {s}^{2} + 6s + 9 } \\ \\ G(s) = \frac{1}{ {(s + 3)}^{2} }$

It is clear here that it has repeated poles at -3.

Output response

$C(s)= \frac{1}{s} . \frac{1}{ {(s + 3)}^{2} }$

Now, partial fraction the output response,so that it can be easily written in time-domain

$C(s)= \frac{1}{s {(s + 3)}^{2} } \\ \\ \frac{1}{s {(s + 3)}^{2} } = \frac{A}{s} + \frac{B}{(s + 3)} + \frac{C}{( {s + 3)}^{2} } \\ \\ \text{find \: coefficients \: } \\ \\ \frac{1}{s {(s + 3)}^{2} } = \frac{A( {s + 3)}^{2} + Bs(s + 3) + Cs}{s( {s + 3)}^{2} } \\ \\ \frac{1}{s {(s + 3)}^{2} } = \frac{A {s}^{2} + 6As + 9A + B {s}^{2} + 3sB + Cs }{s( {s + 3)}^{2} } \\ \\ \frac{1}{s {(s + 3)}^{2} } = \frac{(A + B) {s}^{2} + (6A + 3B + C)s + 9A }{s( {s + 3)}^{2} }$

Compare the coefficient

$A+ B = 0 \\ \\ 6A + 3B + C = 0 \\ \\ 9A = 1$

So,

$A= \frac{1}{9} \\ \\ B = \frac{ - 1}{9} \\ \\ C= - 6A - 3B \\ \\ C = \frac{ - 2}{3} + \frac{1}{3} \\ \\ C= \frac{ - 1}{3}$

Put the values of coefficients

Thus

$C(s) = \frac{1}{9s} - \frac{1}{9(s + 3)} - \frac{1}{3( {s + 3)}^{2} }$

Now,This can be easily transformed into time domain by taking inverse Laplace.

Convert C(s) this into c(t) :find the inverse Laplace transform

$c(t) = \frac{1}{9} - \frac{1}{9} {e}^{ - 3t} - \frac{1}{3} t {e}^{ - 3t}$

as these are the direct terms.

Final Answer:

$\boxed{\bold{\blue{c(t) = \frac{1}{9} - \frac{1}{9} {e}^{ - 3t} - \frac{1}{3} t {e}^{ - 3t} }}}$

Hope it helps you.

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