Finding Impulse and Convolution Integral

Thread Starter

IronMod

Joined Jun 14, 2011
15
The method I am using for the impulse, h(t), is finding my H(s), and taking the inverse laplace. So assuming I am correct on my work there, I get to the convolution part and I can not seem to get it to work correctly. I think my error is the step where I actually integrate the convolution. From what I understand, we only care about the function from 0 to t, because the function after that just goes to zero since there is no overlap in the responses. I think I am messing up from the multiplication of [δ(τ)-10,000e^(-10000τ)]δ(t-τ).

The last part of the problem I believe is correct, that is what i am using my convolution to check towards. If anyone could help it would be greatly appreciated. I believe it is something very elementary I am missing. Thank you.
 

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Vahe

Joined Mar 3, 2011
75
Don't mix time functions and s-domain functions -- you can't write \(v_o(t) = H(s) v_i(t)\).
Your H(s) can be found by voltage division,

\(
H(s) = \frac{V_o(s)}{V_i(s)}= \frac{s L_2}{R+s L_1+s L_2} = \frac{s L_2}{R+s (L_1+ L_2)}
\)

where \(L_1=4 \, \text{mH}\), \(L_2=16\, \text{mH}\) and \(R=200\Omega\).

\(
H(s) = \frac{s \frac{L_2}{L_1+L_2}}{s + \frac{R}{L_1+s L_2}} = \frac{0.8 s}{s + 10}
\)

Vahe
 
Last edited:

Vahe

Joined Mar 3, 2011
75
One way to proceed is by using partial fraction expansion,

\(
H(s) = \frac{0.8s}{s+10000} = 0.8 - \frac{8000}{s+10000}
\)

Therefore,

\(
h(t) = 0.8 \delta (t) - 8000 e^{-10000t} u(t)
\)
 
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Thread Starter

IronMod

Joined Jun 14, 2011
15
Thank you for your reply. Even after doing it that way, I still end up with:

.8δ(t)-8,0000e^(-10,000t) u(t)

I am assuming this was just because the 0's were left off on accident when you did it?

I had a friend help me, and he was able to figure out how to solve part b where I was going wrong.
 
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