Impulse Response

  • Thread starter ryukyu
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  • #1
20
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1. Find the impulse response h(t) given: z'(t) + 4z(t) = 4x(t)




2. The attempt at a solution

I first decided to divide through by 4

(1/4)z'(t) + z(t) = x(t)

since we are looking for impulse response I made the following substitutions:

let z(t) = h(t)

let x(t) = dirac(t)

which yields

(1/4) h'(t) + h(t) = dirac(t)

at which point I don't know how to handle the problem anymore.
 

Answers and Replies

  • #2
68
1
I'm assuming you want an explicit solution for h(t)

there are other ways to solve that differential equation, but in signal processing courses you're usually taught to use laplace transforms.

assuming initial conditions are 0
you would get

(1/4)[s*H(s)-h(0)] + H(s) = 1/s
(1/4)[s*H(s)] + H(s) = 1/s
H(s)[s/4 + 1] = 1/s
H(s) = 4/(s*(s+4))
from here, you can use partial fraction expansion, and then take the inverse laplace transform of the fractions separately.
 
  • #3
20
0
Great! This gets me a bit further. Now I am just a bit confused about expressing my answer.

I did the partial fraction expansion and got a=(1/4) and b=(-1/4)

This gave me:

H(s)=(1/4)(1/s) - (1/4)(1/(s+4))

taking the Laplace Inverse

h(t) = (1/4)u(t) - (1/4)e^(-4t)u(t)

While in DiffEq, we ignored the u(t), I'm assuming since it has relevance to signals it should be kept, but I'm not sure.

If so my final solution should be
h(t) = (1/4)u(t)*[1-e^(-4t)] ??
 
  • #4
CEL
656
0
Great! This gets me a bit further. Now I am just a bit confused about expressing my answer.

I did the partial fraction expansion and got a=(1/4) and b=(-1/4)

This gave me:

H(s)=(1/4)(1/s) - (1/4)(1/(s+4))

taking the Laplace Inverse

h(t) = (1/4)u(t) - (1/4)e^(-4t)u(t)

While in DiffEq, we ignored the u(t), I'm assuming since it has relevance to signals it should be kept, but I'm not sure.

If so my final solution should be
h(t) = (1/4)u(t)*[1-e^(-4t)] ??

You must keep u(t), because if it is omitted your response would have nonzero values before t = 0, when the excitation was applied.
 

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