Convolution specific question

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ovikanobi

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My textbook has an example I believe has an error.
The question is to calculate the output y(t)=x(t)*h(t) where x(t)=(t-1)[u(t-1)-u(t-3)] and h(t)=u(t+1)-2u(t-2). (u(t) is the step function defined by u(t>=0)=1, 0 elsewhere.
The answer has 5 parts and I question only the 5th part for the range of t>5. The textbook says the answer is -2. My understanding is that it should be 0.
This may be the wrong forum to get such help, if it is, my apologies - maybe you'll be kind to suggest an alternative forum. Thank you for comments.
 
From what you have there, if u(t>=0)=1, it seems you end up with y(t>5)=0*0=0, as you mentioned. Could you share a picture of the question to confirm we're not missing anything?
 
I think that the "*" in y(t)=x(t)*h(t) is the convolution operator(click)

I have calculated y numerically and I get y(6) = y(7) = -2 which agrees with your textbook. NOTE: I used continuous convolution, not discrete in my calculation.

Please post your work so that we can try to determine where the mistake is
 
i don't know how to enter math equations here. I will generate a PDF and attach it sometime today. Thank you for helping out.
 
[MATH]y(t)=\int_{5}^{\infty}x(\tau)h(t-\tau)d\tau[/MATH] is the expression for calculating the convolution for that range. [MATH]x(\tau)[/MATH] for this range is 0. The product in the integral calculates to 0, which, if true, means [MATH]y(t)=0[/MATH]
 
here's a plot showing the 2 functions.
 

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Cubist said:
I think that the "*" in y(t)=x(t)*h(t) is the convolution operator(click)

I have calculated y numerically and I get y(6) = y(7) = -2 which agrees with your textbook. NOTE: I used continuous convolution, not discrete in my calculation.

Please post your work so that we can try to determine where the mistake is
Click to expand...
Did you use a tool/program to calculate this convolution numerically?
 
ovikanobi said:
[MATH]y(t)=\int_{5}^{\infty}x(\tau)h(t-\tau)d\tau[/MATH] is the expression for calculating the convolution for that range.
Click to expand...

The lower limit of the integral should remain as -∞. I think you're being asked to consider the case when \(t > 5\), not \(\tau > 5\)

But you can reduce the range of the integral because x(t)=0 for t<1 and t>3, therefore it's safe to say:-

[MATH]y(t)=\int_{1}^{3}x(\tau)h(t-\tau)d\tau[/MATH]
if t>5 then what's the lowest value, \( x_{min} \), that can be passed into the function \( h(x) \)? And what is the value of \(h(x)\) for all \(x \geq x_{min} \) ?
 
the procedure to calculate a convolution is about identifying ranges of [MATH]\tau[/MATH] where the product x()h() has a constant form - an expression that forms a function that the area underneath is the result [MATH]y(t)[/MATH] for that range. [MATH]\tau[/MATH] is the dummy integration variable and the range of integration is a range of [MATH]\tau[/MATH] in terms of [MATH]t[/MATH]. It is true that the question asks about the [MATH]y(t>5)[/MATH], but the integration range for the dummy variable for that interval is [MATH]5<\tau<\infty[/MATH]. The product of [MATH]x()h()[/MATH] for that range is 0 and integrating 0 yields 0.
The textbook's discussion for this interval just postulates that the area under the curve is -2 which is only true for [MATH]h()[/MATH] but not for the product [MATH]x()h()[/MATH] - that product is zero because [MATH]x(\tau)=0[/MATH] for that range.
Pictorially, once there is no overlap between the two functions, the result of the convolution is 0. The meaning of the 'overlap' (my understanding) is the product of these two functions is not equal to 0.
I still don't see what is wrong with my reasoning.
 
ovikanobi said:
the range of integration is a range of [MATH]\tau[/MATH] in terms of [MATH]t[/MATH]. It is true that the question asks about the [MATH]y(t>5)[/MATH], but the integration range for the dummy variable for that interval is [MATH]5<\tau<\infty[/MATH].
Click to expand...

Where did you get your limits \(5<\tau<\infty\) from?

The definition of convolution is shown on this page where the limits of the \(\tau\) integration run from -∞ to +∞. This integration can then (often) be simplified by using a variety of techniques, like ignoring any ranges of \(\tau\) for which one of the functions returns the value zero. However you are ignoring values of \(\tau\) where both functions are non-zero (specifically \(1<\tau<3\) )
 
As I mentioned before, the answer has 5 parts and [MATH]1<\tau<3[/MATH] is the third. The one I am having trouble with is [MATH]t\ge5[/MATH]. I really appreciate you taking the time to discuss this with me. I tried to get in touch with Mr. Van Veen but this book was published in 2005 and I doubt he still has a copy... :). Mr Haykin who's the primary author is very senior...
The formal definition of the convolution is as you point out, however, in solving problems, this range is broken up into ranges over which the product x()h() remains in one form (one math expression).
 
ovikanobi said:
As I mentioned before, the answer has 5 parts and [MATH]1<\tau<3[/MATH] is the third. The one I am having trouble with is [MATH]t\ge5[/MATH].
Click to expand...

This doesn't tell me why you think the integration limits must change between the different parts of this question. Even if the value of t changes, this doesn't necessarily mean that the optimised integration range (for \(\tau\)) must also change.
 
I am just following the textbook's example. The integration is performed for the middle 3 ranges with different limits (and x()h() product) each tme.
The first range and the last range - there is no integration in the example. The first range is clear to me as x() and h() have no overlap (both functions are 0 ) and consequently, y(t)=0, the faith range the text just declares that the area under the curve is -2. This last postulate is problematic to me for 2 reasons:
1) x()=0
2) this range is infinite with h(t)=-1 and x()=0, so how could the area underneath be a finite -2 ?

I have taken enough of your time, it would be ok if we stop this now, up to you...
 
o, I just figured out what my problem is. while x()=0 is 0 for t>5, there is still a non-zero overlap between the 2 functions in 1<t<3. I will add a picture later on... I am sure if I go through the integration of that overlapped region, the result would be -2. So, the integrated is (-1)(t-1) and limits of that integration need to be [MATH]1<\tau<3[/MATH][MATH]\int_1^3(-1)(\tau-1)d\tau=(-1)[3-1 - (1-1)]=-2[/MATH]once again math is right and i am wrong :)
 

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