Calculating Convolution Integrals with Unit Step Functions

[dashkin111]

Homework Statement

Compute the following

$$y(t)=e^{-3t}u(t)\ast u(t+3)$$

Homework Equations

u(t) is the unit step function.

The Attempt at a Solution

I get confused with these for some reason...

$$y(t)= \int^{+\infty}_{-\infty}e^{-3 \tau}u(\tau)u(t-\tau+3)d\tau$$

This is where I have my first problem, trying to eliminate the step functions. I tried

$$y(t)= \int^{t+3}_{0}e^{-3 \tau}d\tau$$

Does that look right?

Finally integrating that with my limits gave:

$$[1/3 - e^{-3(t+3)}]u(t+3)$$

Now the step function I added at the end, and I'm not quite sure why... My main problem is dealing with the step functions, any suggestions/ confirmations I did it wrong or right?

 

[rbj]

$$y(t)= \int^{t+3}_{0}e^{-3 \tau}d\tau$$

Does that look right?

Finally integrating that with my limits gave:

$$[1/3 - e^{-3(t+3)}]u(t+3)$$

Now the step function I added at the end, and I'm not quite sure why... My main problem is dealing with the step functions, any suggestions/ confirmations I did it wrong or right?

the fact that the symbol you use for the (heaviside) unit step function is u(t) not "h(t)", which is what is used by a lot of folks in the mathematics discipline (and you didn't use the term "heaviside function") makes me think that this is an EE course. likely the first Linear System Theory course (or whatever your school calls it). no?

now, i think i see one error. let's break it up a little. let's say it's:

$$y(t) = h(t) \ * \ x(t)$$

or, maybe a better notation is:

$$y(t) = (h \ * \ x)(t)$$where

$$h(t) = e^{-3 t}u(t)$$
and
$$x(t) = u(t+3)$$

Convolution of h(t) against x(t) is

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} h(\tau) x(t-tau) \ d\tau = \int_{-\infty}^{+\infty} x(\tau) h(t-tau) \ d\tau$$

you can pick either integral. but be careful. it looks to me that you intended to pick

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} h(\tau) x(t-tau) \ d\tau$$

or

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} e^{-3 \tau}u(\tau) u((t-\tau)+3) \ d\tau$$

now remember that it is $\tau$ that is your independent variable inside the integral, not $t$. so you might want to express it as

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} e^{-3 \tau}u(\tau) u( \ -(\tau-(t+3))\ ) \ d\tau$$

you are correct that the bottom limit is always zero because of $u(\tau)$. but what is the top limit? $u(a)$ is zero whenever the argument, a<0 . likewise $u(-a)$ is zero whenever the argument, a>0. so $u(\ -(\tau-(t+3))\)$ is zero whenever $\tau-(t+3)>0$ or $\tau>t+3$. hmmm. so it looks like you did the integral right, at least for the case when the upper limit t+3 is greater than the lower limit of 0.

in the case that the upper limit t+3 is less than the lower limit of 0, then you have to recognize that for all $\tau$, at least one of those unit step functions are zero, so you are integrating something that is zero for all $\tau$, so your integral is zero. this happens whenever t+3 < 0 or t<-3. but your integral

$$y(t)= \int^{t+3}_{0}e^{-3 \tau}d\tau$$

still evaluates to something that is not necessarily zero when t<-3, so you have to fix the result so it is

$$y(t) = \frac{1}{3} \left(1 - e^{-3(t+3)} \right)$$

when t>-3 and

$$y(t) = 0$$

when t<-3 . how're you going to do that?

 

[dashkin111]

the fact that the symbol you use for the (heaviside) unit step function is u(t) not "h(t)", which is what is used by a lot of folks in the mathematics discipline (and you didn't use the term "heaviside function") makes me think that this is an EE course. likely the first Linear System Theory course (or whatever your school calls it). no?

now, i think i see one error. let's break it up a little. let's say it's:

$$y(t) = h(t) \ * \ x(t)$$

or, maybe a better notation is:

$$y(t) = (h \ * \ x)(t)$$


where

$$h(t) = e^{-3 t}u(t)$$
and
$$x(t) = u(t+3)$$

Convolution of h(t) against x(t) is

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} h(\tau) x(t-tau) \ d\tau = \int_{-\infty}^{+\infty} x(\tau) h(t-tau) \ d\tau$$

you can pick either integral. but be careful. it looks to me that you intended to pick

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} h(\tau) x(t-tau) \ d\tau$$

or

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} e^{-3 \tau}u(\tau) u((t-\tau)+3) \ d\tau$$

now remember that it is $\tau$ that is your independent variable inside the integral, not $t$. so you might want to express it as

$$y(t) = (h \ * \ x)(t) = \int_{-\infty}^{+\infty} e^{-3 \tau}u(\tau) u( \ -(\tau-(t+3))\ ) \ d\tau$$

you are correct that the bottom limit is always zero because of $u(\tau)$. but what is the top limit? $u(a)$ is zero whenever the argument, a<0 . likewise $u(-a)$ is zero whenever the argument, a>0. so $u(\ -(\tau-(t+3))\)$ is zero whenever $\tau-(t+3)>0$ or $\tau>t+3$. hmmm. so it looks like you did the integral right, at least for the case when the upper limit t+3 is greater than the lower limit of 0.

in the case that the upper limit t+3 is less than the lower limit of 0, then you have to recognize that for all $\tau$, at least one of those unit step functions are zero, so you are integrating something that is zero for all $\tau$, so your integral is zero. this happens whenever t+3 < 0 or t<-3. but your integral

$$y(t)= \int^{t+3}_{0}e^{-3 \tau}d\tau$$

still evaluates to something that is not necessarily zero when t<-3, so you have to fix the result so it is

$$y(t) = \frac{1}{3} \left(1 - e^{-3(t+3)} \right)$$

when t>-3 and

$$y(t) = 0$$

when t<-3 . how're you going to do that?

Wouldn't the y(t) = 0 when t <-3 be taken care of by just multiplying that whole thing by a step function u(t+3)


$$y(t) = 1/3[1-e^{-3(t+3)}]u(t+3)$$
?

 

[rbj]

Wouldn't the y(t) = 0 when t <-3 be taken care of by just multiplying that whole thing by a step function u(t+3)


$$y(t) = 1/3[1-e^{-3(t+3)}]u(t+3)$$
?

yes.

thus ends the lesson.