Find Impulse Response of LTI system given transfer function

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Homework Statement



Find the impulse response of a system with transfer function H(S) = (s+3)/(s^2+2s+1)

or H(S)=(s+3)/[(s+1)^2]

Homework Equations



Poles are s1=s2=-1

y = Ae^st + Be^st

The Attempt at a Solution



In my notes I do not have an answer for the case when there is only one pole (root) to the denominator of the transfer function.

I know the cover-up method and it doesn't look like it will work here.

Also tried solving for the step response and differentiating to get the impulse response as shown here: http://tinyurl.com/c5uhzcv

But I cannot find the step response that way either. Thanks!
 

Answers and Replies

  • #2
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In my notes I do not have an answer for the case when there is only one pole (root) to the denominator of the transfer function.
I'm sure you do -- look for mention of repeated roots.

A partial fraction expansion of a root repeated to the nth power involves sums of all fractions with the root raised to powers 1 through n. This is because the common denominator of these terms is (s+1)n and the numerator must be able to achieve a power in s of (n-1) to be completely general.

In this case,

[itex]H(s)=\frac{s+3}{(s+1)^2}=\frac{A}{s+1}+\frac{B}{(s+1)^2}[/itex]
 
  • #3
296
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In this case,

[itex]H(s)=\frac{s+3}{(s+1)^2}=\frac{A}{s+1}+\frac{B}{(s+1)^2}[/itex]
If you add the fractions back together, the A will add an As term and the B will supply a constant term in the numerator.

You already know the inverse transform of the first fraction. Whenever factors in the denominator are taken to a power, in the time domain they are multiplied by t to that power-1. So the second fraction will invert as t^1 times the inverse of 1/(s+1)
 

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