# [EE - Signals] Verify my unit-step work

1. Aug 30, 2007

### user101

1. The problem statement, all variables and given/known data

For the top example: $$x(t) = u(t+1) - 2u(t-1) + u(t-3)$$

For the bottom example: $$x(t) = (t+1)*u(t-1) - t*u(t) - u(t-2)$$

The example asked for a sketch, so sorry if my drawings aren't to scale.

2. Relevant equations

Unit step: $$u(t-t_0)$$...

* if $$t_0$$ is positive, there will be a shift to the RIGHT, which indicates a delay.
* if $$t_0$$ is negative, there will be a shift to the LEFT, which indicates an advance of the signal.
* if $$t_0$$ is zero, therefore having just $$u(t)$$, there is no advance or delay.

3. The attempt at a solution

Here is my attached work. I drew a graph for each term (for both problems), then combined them if it was either addition or subtraction. I split the graphs into sections, where I thought it would be appropriate. Any comments would be greatly appreciated.

The image is kind of long, sorry :(

http://img256.imageshack.us/img256/4845/pg1aj0.png [Broken]

Last edited by a moderator: May 3, 2017
2. Aug 31, 2007

### Staff: Mentor

4a is correct.

In 4b, look at (t+1)*u(t-1). The sketch is t*u(t-1).

3. Aug 31, 2007

### user101

For the first one in 4b, I have the delay in there and then I start at y = 1, going up by 1/1 for the slope. Isn't that correct?

4. Aug 31, 2007

### learningphysics

It should start at y = 2...

5. Aug 31, 2007

### user101

Woops! My mistake guys, I got it :)

for my final answer, the -t slope right after the y axis, is that correct? 0 - t - 0 = -t slope.

6. Aug 31, 2007

### learningphysics

Yes, that's right. At t = 1, y jumps from -1 to 1... then from t = 1 to t= 2, y=1. Then from t=2 onwards y = 0... everything looks good.

7. Aug 31, 2007

### user101

Hi - just a quick question, is my explanation sufficient. My professor told me to keep it as simple as possible:

http://img251.imageshack.us/img251/1512/picture2nv2.jpg [Broken]

Last edited by a moderator: May 3, 2017
8. Aug 31, 2007

### Staff: Mentor

Keep in mind

$$\int_{-\infty}^{+\infty}\,\delta(t-T)f(t)\,dt\,=\,f(T)$$ or f(-T) if the argument of the delta function is t+T.

9. Aug 31, 2007

### user101

Right, I tried to go by that rule. Did I do something wrong when applying it to my examples?

10. Aug 31, 2007

### Staff: Mentor

The answers are correct, and you're logic is correct.

11. Sep 2, 2007

### user101

Ok, great thanks!

12. Sep 3, 2007

### user101

Ook, if I have a discrete time signal, I'm assuming I do the same thing? Here is the question:

http://img164.imageshack.us/img164/493/picture2ot4.jpg [Broken]

n = -5, so the summation is equal to 0 since it is not within the proper bounds of 4 <-> 10.

The reason I'm asking for verification is because I haven't learnt this yet, so I'm learning on my own accord. The feedback I get tells me whether or not I learned it properly! =D

Last edited by a moderator: May 3, 2017
13. Sep 3, 2007

### learningphysics

Yes, that's right. The summation is 0.

delta[n+5] = 0 for all numbers except n = -5, when it equals 1.

14. Sep 4, 2007

### user101

Hi, I have a new one that I'm having trouble with. My friend got something totally different than what I did. Is this even remotely close? My final answer is the very bottom one.

Basically what I did was for the impulse functions, they are only true at a certain location and the last two ones, the unit step ones, are like any other unit steps graphs, only with discrete values. I'm pretty sure the unit step discrete ones are correct, but is my overall methodology correct?

Also, is there any other way of doing this quicker, instead of drawing each one out?

http://img68.imageshack.us/img68/6868/picture1xc8.jpg [Broken]

Last edited by a moderator: May 3, 2017
15. Sep 5, 2007

### DefaultName

Nvm, I figured it out.