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

[martnll2]

Homework Statement

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)

Homework Equations

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

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.

 

[RoshanBBQ]

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

 

[martnll2]

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

 

[RoshanBBQ]

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

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

 

[martnll2]

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

You are correct. I did mess that one up.

 

[rude man]

Homework Statement

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)

Homework Equations

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

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.

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).

 

[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.

 

[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.

 

[rude man]

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.

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.

 

[RoshanBBQ]

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.

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.

 

[rude man]

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.

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.

 

[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)

 

[martnll2]

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

 

[rude man]

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

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.

 

[RoshanBBQ]

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

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.

Are you not reading your textbook?

 

[martnll2]

Yeah... I do read my textbook. I guess this is one of the main reasons I am changing my major; I can't comprehend anything! I need to see the professor tomorrow.