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[EE - Signals] Verify my unit-step work

  1. Aug 30, 2007 #1
    1. The problem statement, all variables and given/known data

    For the top example: [tex]x(t) = u(t+1) - 2u(t-1) + u(t-3)[/tex]

    For the bottom example: [tex]x(t) = (t+1)*u(t-1) - t*u(t) - u(t-2)[/tex]

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

    2. Relevant equations

    Unit step: [tex]u(t-t_0)[/tex]...

    * if [tex]t_0[/tex] is positive, there will be a shift to the RIGHT, which indicates a delay.
    * if [tex]t_0[/tex] is negative, there will be a shift to the LEFT, which indicates an advance of the signal.
    * if [tex]t_0[/tex] is zero, therefore having just [tex]u(t)[/tex], 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 :(

    [​IMG]
     
    Last edited: Aug 31, 2007
  2. jcsd
  3. Aug 31, 2007 #2

    Astronuc

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    Staff: Mentor

    4a is correct.

    In 4b, look at (t+1)*u(t-1). The sketch is t*u(t-1).
     
  4. Aug 31, 2007 #3
    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?
     
  5. Aug 31, 2007 #4

    learningphysics

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    Homework Helper

    It should start at y = 2...

    But your final answer looks right.
     
  6. Aug 31, 2007 #5
    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.
     
  7. Aug 31, 2007 #6

    learningphysics

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    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.
     
  8. Aug 31, 2007 #7
    Hi - just a quick question, is my explanation sufficient. My professor told me to keep it as simple as possible:

    [​IMG]
     
  9. Aug 31, 2007 #8

    Astronuc

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    Keep in mind

    [tex]\int_{-\infty}^{+\infty}\,\delta(t-T)f(t)\,dt\,=\,f(T)[/tex] or f(-T) if the argument of the delta function is t+T.
     
  10. Aug 31, 2007 #9
    Right, I tried to go by that rule. Did I do something wrong when applying it to my examples?
     
  11. Aug 31, 2007 #10

    Astronuc

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    Staff: Mentor

    The answers are correct, and you're logic is correct.
     
  12. Sep 2, 2007 #11
    Ok, great thanks!
     
  13. Sep 3, 2007 #12
    Ook, if I have a discrete time signal, I'm assuming I do the same thing? Here is the question:

    [​IMG]

    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
     
  14. Sep 3, 2007 #13

    learningphysics

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

    delta[n+5] = 0 for all numbers except n = -5, when it equals 1.
     
  15. Sep 4, 2007 #14
    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?

    [​IMG]
     
  16. Sep 5, 2007 #15
    Nvm, I figured it out.
     
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