Find Impulse Response of LTI system given transfer function

AI Thread Summary
To find the impulse response of the LTI system with transfer function H(S) = (s+3)/[(s+1)^2], one must consider the repeated root in the denominator. The approach involves partial fraction expansion, where the transfer function can be expressed as H(s) = A/(s+1) + B/(s+1)^2. The inverse Laplace transform of the first term is known, while the second term requires multiplication by t to account for the repeated root. The solution emphasizes the need to manage the powers of the denominator effectively to derive the impulse response accurately. Understanding these concepts is crucial for solving similar problems in control systems.
f00lishroy
Messages
6
Reaction score
0

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!
 
Physics news on Phys.org
f00lishroy said:
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,

H(s)=\frac{s+3}{(s+1)^2}=\frac{A}{s+1}+\frac{B}{(s+1)^2}
 
aralbrec said:
In this case,

H(s)=\frac{s+3}{(s+1)^2}=\frac{A}{s+1}+\frac{B}{(s+1)^2}

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)
 
Back
Top