Find Impulse Response of LTI system given transfer function

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SUMMARY

The impulse response of the LTI system with the transfer function H(S) = (s+3)/(s^2+2s+1) or H(S) = (s+3)/[(s+1)^2] can be determined using partial fraction expansion due to the repeated root at s = -1. The correct approach involves expressing H(S) as H(s) = A/(s+1) + B/(s+1)^2, where A and B are constants to be determined. The inverse Laplace transform of each term yields the impulse response, with the second term requiring multiplication by t to the power of the repeated root minus one.

PREREQUISITES
  • Understanding of Laplace transforms and their properties
  • Familiarity with partial fraction decomposition
  • Knowledge of impulse and step responses in LTI systems
  • Ability to perform inverse Laplace transforms
NEXT STEPS
  • Study the method of partial fraction decomposition in detail
  • Learn about the inverse Laplace transform of repeated roots
  • Explore the relationship between step responses and impulse responses
  • Review examples of LTI systems with repeated poles in their transfer functions
USEFUL FOR

Students and engineers in control systems, signal processing, or electrical engineering who are working with LTI systems and need to understand impulse response derivation techniques.

f00lishroy
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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!
 
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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)
 

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