Convolution Problem -- Triangular and Rectangular pulses

[kosmocomet]

Homework Statement

Homework Equations

• y(t)=x(t)*h(t)=∫x(λ)⋅h(t-λ)⋅dλ

The Attempt at a Solution

[/B]
Is what I have the correct interpretation or or am I wrong?

Thanks

 

[dvscrobe]

kosmocomet, Aaah, another FE exam question. I don’t have the answer for you but this one, I’m watching. Dan.

 

[MathematicalPhysicist]

What's the question here?

 

[scottdave]

Wow... Convolution. Its been awhile. I can go look it up, but the way I remember is one "waveform" slides past the other and are multiplied. The easy way, if you've learned this would be to do Laplace Transforms then it's just straight multiplication in the s domain, then inverse Laplace

 

[scottdave]

I'm on my phone, so it is a little difficult reading your notes. But from what I can make out, it appears you are in the ballpark. Here's a video on convolution
Note that your problem is piecewise, so you need to consider that.

 

[dRic2]

This is what I would do. I don't think there is the need to bring up Laplace or other transforms.

Take the integral of the convolution:
$$\int_{-\infty}^{+\infty} x(t-\tau)h(\tau)d\tau$$
Let's see what happens if $\tau < 0$, well, of course, $h(\tau < 0) = 0$ by definition. So the lower limit of the integral becomes 0. Now what about the upper limit?
Looking again at at $h(\tau)$ if $\tau > 1$ again $h(\tau > 1 ) = 0$ by definition. So here we would be tempted to set the upper limit to $1$. But wait a sec: what about the other function? If $\tau > t$ then $x((t-\tau) < 0) = 0$ and the integral is zero. From this we conclude that we must break the integral in two parts and in the end you will have a case defined function like this:

$$(x * h)(t) = \begin{cases}
\text{if } t < 1 : \int_0^{t} x(t-\tau)h(\tau)d\tau \\
\text{if } t > 1: \int_0^{1} x(t-\tau)h(\tau)d\tau
\end{cases}
$$

 

[kosmocomet]

This is what I would do. I don't think there is the need to bring up Laplace or other transforms.

Take the integral of the convolution:
$$\int_{-\infty}^{+\infty} x(t-\tau)h(\tau)d\tau$$
Let's see what happens if $\tau < 0$, well, of course, $h(\tau < 0) = 0$ by definition. So the lower limit of the integral becomes 0. Now what about the upper limit?
Looking again at at $h(\tau)$ if $\tau > 1$ again $h(\tau > 1 ) = 0$ by definition. So here we would be tempted to set the upper limit to $1$. But wait a sec: what about the other function? If $\tau > t$ then $x((t-\tau) < 0) = 0$ and the integral is zero. From this we conclude that we must break the integral in two parts and in the end you will have a case defined function like this:

$$(x * h)(t) = \begin{cases}
\text{if } t < 1 : \int_0^{t} x(t-\tau)h(\tau)d\tau \\
\text{if } t > 1: \int_0^{1} x(t-\tau)h(\tau)d\tau
\end{cases}
$$

Thanks for the help. I think I understand the set up. The issue is when evaluating the integral, I get this:

Using the t>1 case, the integral becomes ∫(t-λ)*(1)⋅dλ , where x(t-λ)=(t-λ) and h(λ)=1. This is because the x graph is only valid when λ is between 1 and 0. Since t is between [1,2], then it has to use the decreasing line from the original graph which has the points (t-1,1) and (t,0). Use slope intercept to find equation. Is this correct?

 

[scottdave]

Thanks for the help. I think I understand the set up. The issue is when evaluating the integral, I get this:

Using the t>1 case, the integral becomes ∫(t-λ)*(1)⋅dλ , where x(t-λ)=(t-λ) and h(λ)=1. This is because the x graph is only valid when λ is between 1 and 0. Since t is between [1,2], then it has to use the decreasing line from the original graph which has the points (t-1,1) and (t,0). Use slope intercept to find equation. Is this correct?

Yes, you need an expression for the decreasing line. Use point-slope or some method to get a valid expression. You will only integrate it in the limits which are shown.

 

[dRic2]

Yes, I should also point that my previous post is missing something... No big deal because you figured it out on your own, but for $t>3$ the integral is always zero because $\tau \in [0, 1]$. The correct solution should be:
$$(f*g)(t) = \begin{cases} \text{if } t < 1 : \int_0^t x(t-\tau)d\tau \\
\text{if } 1 < t < 3 : \int_0^1 x(t-\tau)d\tau
\end{cases}
$$

 

[rude man]

I think the problem requires more sectioning. I came up with a total of four non-zero integrals for three different ranges of t.

Giving just the limits of integration they are $\int_0^t , \int_{-1+t}^1 , \int_1^t , and \int_{-1+t}^2$ spanning t=0 to t=3.
The first two integrals associate with the first slope, the last two with the second slope. Of course g(t-τ)h(τ) is different for the two slopes.

 

[Dr Transport]

If you use Fourier transforms the problem is fairly straightforward.

 

[rude man]

If you use Fourier transforms the problem is fairly straightforward.

I looked into that & did not find it so "straightforward". Integrating all those complex jt e^jwt terms gave me a headache!

With graphical technique the procedure is so much more elegant, visualizable, and - yes - straightforward

 

[dvscrobe]

Did we ever get the answer to this? What is y(t)?

 

[rude man]

Did we ever get the answer to this? What is y(t)?

Using graphical techniqe you should be able to come up with the four integrals of post 10. I hope they taught you that technique.
The challenge is to come up with the limits of integration which is why graphical technique is preferred.
Here's a hint that might help in this regard:

f(t)*g(t) = $\int_{-\infty}^{\infty} f(τ)g(t-τ) \, dτ$ so g(t) is chosen as the "sliding" function.
Say you are given g(t) = 1, 1 < t < 2, then

g(τ) = 1, 1 < τ < 2
g(-τ) = 1, -1 > -τ > -2
g(t-τ) = 1, -1+t > t-τ > -2+t

which gives -2+t and -1+t the limits of the "sliding" function g(t-τ) for all -∞ < τ < ∞ as it intersects with h(τ). The resulting limits of integration of ∫ h(τ)g(t-τ) dτ will be combinations of t and numbers. Note carefully the < vs. > symbols and number signs!

Applying this method to your problem should give you the four integrals alluded to in post 10. One thing I strongly suggest is splitting x(t) into two separate problems. Then combine the results. cf. attachment.