# How do you find the Impulse Response, h(t)

1. Apr 13, 2012

### martnll2

1. The problem statement, all variables and given/known data
How do i find the impulse response, h(t), given the following?

f(t) = 8t u(t)

y(t) = (-2e^-2t + 8e^-t + 4t - 6) u(t)

2. Relevant equations

f(t) convolved with y(t) = h(t)

3. The attempt at a solution

I thought that you divide out y(t) by 8t to make f(t) = u(t) then take the derivative of the modified y(t) expression, call it g(t) in order to get h(t) but the derivatives look nasty and I don't think it is right.

2. Apr 13, 2012

### RoshanBBQ

Do you know about the Laplace transform? What is the relationship between input, output, and the impulse response in that domain?

3. Apr 13, 2012

### martnll2

We didn't cover Laplace yet, that comes next week. Is there an alternate way to solve this?

4. Apr 13, 2012

### RoshanBBQ

I don't think so. And I think your relevant equation is wrong:

"f(t) convolved with y(t) = h(t)"

It should be f(t) * h(t) = y(t) where * is the convolution operator

5. Apr 13, 2012

### martnll2

You are correct. I did mess that one up.

6. Apr 14, 2012

### rude man

I realize somewhat belatedly that the problem is poorly described. What is f(t)? Is it the system input? It's customary to call the input x(t).

Last edited: Apr 14, 2012
7. Apr 14, 2012

### martnll2

We know δ(t) = du/dt ... I already tried dividing both sides by 8t, and taking the derivative, but it didn't seem to work. Here is my work.

f(t) = 8t u(t)
y(t) = [-2e^(-2t) + 8e^(-t) + 4t -6] u(t)

Is it right to change f(t) = u(t) and y(t) = g (t) = {[-2e^(-2t) + 8e^(-t) + 4t -6] u(t)} / 8t
then h(t) = dg/dt?

I tried this but got unreal coefficients for the impulse term.

8. Apr 14, 2012

### RoshanBBQ

The reason I'm showing you this is in an intro course to systems, this should be a plain fact stated in the book (It was in mine, and rude man's comment got my memory jogging). I doubt the problem expected you to derive this, and the real culprit is you most likely not reading your book.

We have (leaving off 8)
$$y(t) = \int\limits_{-\infty}^{\infty} h(\tau)f(t-\tau)d\tau$$
$$y(t) = \int\limits_{-\infty}^{\infty} h(\tau)(t-\tau)u(t-\tau)d\tau$$
$$y(t) = t\int\limits_{-\infty}^{\infty} h(\tau)u(t-\tau)d\tau-\int\limits_{-\infty}^{\infty} h(\tau)\tau u(t-\tau)d\tau$$
$$y(t) = t\int\limits_{-\infty}^{t} h(\tau)d\tau-\int\limits_{-\infty}^{t} h(\tau)\tau d\tau$$
tau is just a dummy variable so these are straight integrations.We know h(t) is zero when t < 0:
$$y(t) = t\int\limits_{0}^{t} h(\tau)d\tau-\int\limits_{0}^{t} h(\tau)\tau d\tau$$
But that looks a whole lot like integration by parts of
$$\int\limits_{0}^{t}\int\limits_{0}^{t}h(\tau)d \tau d \tau$$

In general, if an excitation is changed to its integral, the response also changes to the integral of the previous response. Since you will be encountering this, I'll go ahead and remind you that you have to use product rule with u(t) being one function and the other stuff being another to find the right answer.

Last edited: Apr 14, 2012
9. Apr 14, 2012

### rude man

Theorem:
If y(t) is the reponse to a ramp, then
y'(t) is the response to a step input, and
y''(t) is the response to an impulse input (Dirac delta function).

I can't give you the proof of this since you haven't had the Laplace transform.

The problem is easy after that.

10. Apr 14, 2012

### RoshanBBQ

You can do it in the time domain.

$$y_1(t) = \int\limits_{-\infty}^{\infty}h(\tau)f(t-\tau)d\tau$$
$$y_2(t) = \int\limits_{-\infty}^{\infty}h(\tau)f'(t-\tau)d\tau$$
Just by looking at this, it is clear that y_2(t) is the derivative with respect to time of y_1(t) since f is the only part that depends on time.

11. Apr 14, 2012

### rude man

I know what you say is right - has to be - but the math still eludes me. Never was too comfortable with the t's and tau's in the convolution integral, even when I had to know it.
So thanks for your contribution & the OP should have no more trouble with getting the solution.

12. Apr 15, 2012

### martnll2

I'm still confused. I don't know how you could use these integrals not knowing h(t) since the problem is asking to find h(t).

f(t) is the input signal, y(t) is the output signal. These are both given, and i know that y(t) = $\int h(\tau)f(t-\tau) d\tau$ from -∞ to ∞. I don't know how to use these facts to solve h(t)

13. Apr 15, 2012

### martnll2

Do I just take the derivative with respect to tau on both sides and solve h(t) that way?

14. Apr 15, 2012

### rude man

What RoshanBBQ said in his last post is equivalent to what I said in my last post.

I suggest you look carefully at my last post .... then, if you're not happy with my lack of proof, study RoshanBBQ's last post which essentially proves in the time domain what I would have demonstrated in the frequency domain, except apparently you haven't been exposed to the frequency domain yet.

As I said, I myself get a bit nervous with these convolution integrals. But it's the only way if you insist on a proof and are restricted to the time domain.

15. Apr 15, 2012

### RoshanBBQ

The ramp is the double integral of the impulse. So the ramp response is the double integral of the impulse response (as I showed in post 8). Differentiate your output twice to find the impulse response (with respect to time). Keep in mind to use product rule with u(t) being one function and (-2e^-2t + 8e^-t + 4t - 6) being another function for the first derivative. You will have to use product rule for the second derivative too. Then divide what you get by eight for the final answer. Finally, you can check yourself by doing the convolution integral with the input to recompute the output.