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Convolution help.

  1. Sep 2, 2010 #1
    Hi there, I'm having trouble convolving two signals, according to this site "http://cnx.org/content/m11541/latest/" [Broken]" and its example, for the time period 0≤t<1 they've used the integral

    yt=∫dτ between 0≤t<1.

    My problem is, how did they get this integral?. I get that the height of the two signals is one, hence 1*dτ but I'm still hazy on this.

    I'm working on a question where the graph looks like this.
    [PLAIN]http://img705.imageshack.us/img705/6834/myconv.png [Broken]

    but i don't know how to get the integral at the position?.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 2, 2010 #2

    Zryn

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    Gold Member

    Its not the height that they're looking at but rather the equation of the line for that section of time.

    Since both of those signals appear as the straight line with equation y = 1, multiplied together (1*1) they give you the 'integral of 1'.

    Your lines will have different equations, in different time periods.

    Looking at Figure 4 shows an example of the different equations for different lines.
     
    Last edited: Sep 3, 2010
  4. Sep 2, 2010 #3
    I think that was the resulting y(t) graph. Not another input x(t) or h(t). I'm still a little confused of what to do.
     
  5. Sep 3, 2010 #4

    Zryn

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    Gold Member

    My apologies, you're right there about Figure 4 being the answer and not a 2nd example.

    So just to clarify, you had (red) f(t) and made it f(tau) and then flipped (blue) g(t) and made it g(t-tau), correct?

    Now you're looking for the integral for the period of time of -2 <= t <= 1 correct?

    Whats the equation of the line for red, from -2 to -1, using the dummy variable 'tau' instead of 't'? Do you remember the gradient 'm' being rise/run and the equation of the line being y = m*x + b (where 'b' is the y-intercept when x = 0) ?

    Similarly, whats the equation of the line initially for blue? What happens to the equation of the line when you flip and shift it?

    Now you should have two equations you can multiply together to get the equation for integration. Do you know the integration bounds for this section by the way?
     
  6. Sep 3, 2010 #5
    do u have soln. to this question?

    A quick tip before starting convolution questions.

    1) Know the distance of your blue signal. i.e. blue line has a horizontal distance of 1.

    this will make it easy for you know how to whats going on when u slide the curve into the other

    2) Identify or try and visualise the regions of integration before starting. But how many ROA are here? -1<t<-2, then -1<t-1<-2, then 0<t<1, then 0<t-1<1, then 1<t-1<2, then after these cases the blue line lies outside the red so it is 0

    NOTE:I PUT T-1 BUT FOR EASE OF VISUALISATION, HOWEVER WHEN CALCULATING YOU HAVE TO PUT ?<T<?. Don't Leave it as t-1

    then proceed with your calculations.
     
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