Convolution Help: Understanding Integrals for 0≤t<1

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The discussion focuses on understanding the convolution of two signals over the interval 0≤t<1, specifically how to derive the integral used in the convolution process. The integral is derived from the product of the two signals, which are both equal to one, resulting in the integral of one over the specified bounds. Participants clarify that the convolution involves flipping and shifting one of the signals, leading to different equations for each time period. They emphasize the importance of identifying the equations of the lines for each signal and visualizing the regions of integration to facilitate calculations. The conversation concludes with tips on determining the integration bounds and understanding the regions of overlap between the signals.
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Hi there, I'm having trouble convolving two signals, according to this site "http://cnx.org/content/m11541/latest/" " 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

but i don't know how to get the integral at the position?.
 
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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.
 
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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.
 
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, what's 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?
 
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 what's 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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