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Convolution exercise with block diagram and cases

  1. Mar 28, 2013 #1
    1. First of all, a block diagram is given where x(t) is the input and y(t) the output.The y(t) is asked when given the h(t) and x(t).We have in parallel h(t) and δ(t-1)*h(t) and those 2 go through a summer (the second one with minus and the first one with a plus) and give y(t).




    2. I know from a previous question that x(t) is a rectangular signal with length from -1 to 2.
    And also, h(t)=e^{2-t}u(t-2).




    3. i find that y(t)=x(t)*h(t)-x(t)*h(t-1) , cause δ(t-1)*h(t)=h(t-1) (time shift).
    When i have to find first the x(t)*h(t) i get 3 cases and get :
    1.0 for t<1
    2.1-e^{1-t} for t>=1 and t<4
    3.e^{4-t}-e^{1-t} for t>4.

    Of course i could do the same for x(t)*h(t-1) and get another 3 cases.
    Here's my main problem : the limits of the 2 convolutions are totally diferrent : should i get a lot of cases for y(t) with all the limits from the 2 convolutions or is a smarter-more accurate way of calculating y(t) ?
    I've been working this for hours and tomorrow i must have it solved with another 2 exercises :(
    Any help would be much appreciated.Thanks in advance !
     
  2. jcsd
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