Impulse Response


by ryukyu
Tags: impulse response
ryukyu
ryukyu is offline
#1
Oct6-09, 01:43 PM
P: 20
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.
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
smk037
smk037 is offline
#2
Oct7-09, 08:05 AM
P: 68
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.
ryukyu
ryukyu is offline
#3
Oct7-09, 08:38 AM
P: 20
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)] ??

CEL
CEL is offline
#4
Oct8-09, 02:27 PM
P: 639

Impulse Response


Quote Quote by ryukyu View Post
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.


Register to reply

Related Discussions
Impulse response of system. Introductory Physics Homework 0
Impulse response Precalculus Mathematics Homework 4
help with impulse response and convolution? Introductory Physics Homework 15
impulse response and convolution? any help from anyone? Introductory Physics Homework 0
impulse response? I need a bit of help with this if anyone can? Introductory Physics Homework 2