Finding the impulse response

asmani

Suppose that [itex]y(t)=x(t)\ast h(t)[/itex].

([itex]\ast[/itex] denotes convolution)

Here are the signals:



How to find h(t) in time domain? Does there exist such h(t)?

Thanks in advance.

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asmani

Bump!

the_emi_guy

Using the doublet function:

h(t) =
\begin{cases}
\delta', & \text{for } -\infty < t <0 \\
-\delta', & \text{for } 0 < t < \infty\\
\end{cases}

asmani

Thanks.

Let's consider the delta function as the limiting case of the following function:



Now, isn't your function the same as [itex]-\delta'(t)[/itex]?

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the_emi_guy

The doublet is the derivative of the Dirac delta function.

rude man

Think Fourier integral!

vela

You need to show your attempt at solving the problem on your own before you receive help here.

asmani

A clarification: This thread is moved from electrical engineering forum. Maybe it is a homework or coursework-type question, but actually it's not a homework or a coursework question, and I believe cannot be.

Anyway, here is my attempt:
$$\mathcal{F}\left \{ x(t) \right \}=sinc^2(f)\; ;\: \mathcal{F}\left \{ y(t) \right \}=2sinc(2f)$$
$$\mathcal{F}\left \{ h(t) \right \}=\frac{\mathcal{F}\left \{ y(t) \right \}}{\mathcal{F}\left \{ x(t) \right \}}=\frac{2sinc(2f)}{sinc^2(f)}=4\pi f\cot(2\pi f)$$
I guess that this function has no inverse Fourier transform, and thus there is no LTI system with x(t) as input and y(t) as output. Is this correct? If yes, how to prove?

I know that, but for t<0 we have δ_Δ(t)=0, which implies that for t<0, δ_Δ'(t)=0. That's why I think your function is the same as -δ'(t).

the_emi_guy

Think of it this way...

1 - What kind of circuit would convert your x(t) to your y(t)?

2 - Next, what would be the impulse response of that circuit? This would be your h(t).

rude man

What reason do you have for assuming your H(f) has no inverse? Have you tried doing the integration?

Ref: G.A. Campbell and R. M. Foster, "Fourier Integrals for Practical Applications", D. Van Nostrand 1958.

PS - I did not check to see that you did X(f) and Y(f) correctly ...

asmani

When the plot is as follows, I don't know how to do the integration, if the integral exists.



I guess there is no such circuit/LTI system/h(t). Let's consider two cases:

1. The input to the system S is y(2t), and the output is x(t). We can easily find that the impulse response of S is h(t)=y(2t).

2. The input to the system S' is y(t), and the output is x(t). We can observe that S' is the same S series with another system S'' which gives the output z(2t) for the input z(t). Obviously S'' is not a LTI system, so S' isn't.

That's why I think it's possible that the original system cannot be LTI either.

Thanks.

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rude man

You're not integrating the function you sketched, which I presume is your H(f).
You're trying to integrate ∫H(f)e^jωtdf over f = -∞ to f = +∞ with ω = 2πf.

asmani

Actually this argument was not correct!
Of course, that's h(0). Can you help me on this integration?

Thanks.

rude man

I'm wondering about your initial approach, now that I've looked at it some more.

Looking at your x(t) and y(t) graphs, and taking the Fourier integrals of both,

X(w) = (1/2π)∫(t+1)exp(-jwt)dt from t= -1 to 0 + (1/2π)∫(1-t)exp(-jwt)dt from t= 0 to +1.

Similarly,
Y(w) = (1/2π)∫exp(-jwt)dt from t = -1 to +1.

Then H(w) = Y(w)/X(w) and finally

h(t) = ∫H(w)exp(jwt)dw.

I tried to do the X(w) integration and found it pretty messy, which means subject to making mistakes. I used ∫e^axdx = e^ax/a and ∫xe^axdx = (e^ax/a²)(ax-1).

The Y(w) integration is of course much simpler. The h(t) inverse integral promises to be messy also.

So, best I can do for you is to suggest either you try to muddle through the integrations, or get a table of Fourier integrals like the one I recommended previously. I wish I had that table but I don't.

Of course, it may be that the best approach is graphic convolution, but I'm not inclined to try that myself.

asmani

I thought these are known results:
http://en.wikipedia.org/wiki/Rectang...gular_function
http://en.wikipedia.org/wiki/Triangular_function

rude man

Thank you! We can now both work this. I'll keep you apprised of any progress.