[EE - Signals] Verify my unit-step work

[user101]

Homework Statement

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.

Homework 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.

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

 

[Astronuc]

4a is correct.

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

 

[user101]

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

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?

 

[learningphysics]

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?

It should start at y = 2...

But your final answer looks right.

 

[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.

 

[learningphysics]

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.

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.

 

[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

 

[Astronuc]

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.

 

[user101]

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.

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

 

[Astronuc]

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

 

[user101]

Ok, great thanks!

 

[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

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 learned this yet, so I'm learning on my own accord. The feedback I get tells me whether or not I learned it properly! =D

 

[learningphysics]

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

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

 

[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

 

[DefaultName]

Nvm, I figured it out.