Can someone explain the flip?

[TecHNooB]

I know it has to do with the fact that Tau is changed to t - Tau but I don't see how that changes the area of integration from -infinity to t --> infinity to zero or how that causes the area of integration to become fixed and deltas to move around. Having trouble equating

Edit: I guess the reason is obvious. You want the beginning of one function to be scaled by the beginning of the other function, hence 'flip' one of them. I just don't understand where the integration from infinity to zero comes from in the derivation/explanation.

Mind the formatting, here's what I'm confused about spelled out:

Sigma = t - Tau

u(t) = integral ( delta(Tau), -infinity to t ) with respect to Tau
= integral ( delta(t - sigma) infinity to zero ) with respect to negative sigma
= integral ( delta(t - sigma) zero to infinity ) with respect to sigma

Integrate from infinity to zero? With respect to negative sigma? Not making the connection.

 

[esun]

Think of it this way. You're converting tau to -sigma (in the first integral you're integrating with respect to tau, in the second one you're integrating with respect to -sigma).

So, in the first integral, you are integrating from tau = -infinity to tau = t. If we have sigma(tau) = t - tau (i.e., sigma as a function of tau is t minus tau), then we have:

sigma(tau = -infinity) = t - (-infinity) = t + infinity = +infinity
sigma(tau = t) = t - (t) = 0

Thus, the integral is now from +infinity to 0.

Make sense?

 

[TecHNooB]

So why is the integral from infinity to zero with respect to negative sigma? I don't see where the negative comes in

Actually, I think it might be because sigma is based on negative Tau. What's the significance of using T - Tau as opposed to T + Tau?

 

[esun]

It's just an arbitrary choice. You want to replace tau with t - sigma, so you define sigma = t - tau. This is just an arbitrary choice for convenience (because you want the end result to look a certain way).

Now, you can let the integral be with respect to whatever variable you want. It just so happens that we want the integral to be a certain form, so we pick the integration variable to be -sigma (if it helps, let k=-sigma and do the integral with respect to k). Then, we just have to adjust the limits based on the relationships between tau and sigma (and k, if you choose to use k).

There is no significance to choosing t-tau instead of t+tau, since again, the original substitution was arbitrary.

 

[TecHNooB]

Originally posted by: esun
Think of it this way. You're converting tau to -sigma (in the first integral you're integrating with respect to tau, in the second one you're integrating with respect to -sigma).

So, in the first integral, you are integrating from tau = -infinity to tau = t. If we have sigma(tau) = t - tau (i.e., sigma as a function of tau is t minus tau), then we have:

sigma(tau = -infinity) = t - (-infinity) = t + infinity = +infinity
sigma(tau = t) = t - (t) = 0

Thus, the integral is now from +infinity to 0.

Make sense?

So when you integrate with respect to negative sigma, your limits should be from +infinity to zero so that when you flip the sign, the limits become 0 to +infinity. But negative sigma is -t + Tau. So wouldn't that make the limits negative infinity to zero?

 

[TecHNooB]

Can someone who knows their limits of integration well explain the last part? (the post above)

 

[bobsmith1492]

Oh, I know my limits in integration and they are limited to pictures. Draw a picture... it's the only way I got through Signals & Systems.

 

[RedArmy]

Maybe some of these lecture notes can help. It's from the professor that taught my Signals and Systems class this past semester.

Text

 

[esun]

Sorry, I see the issue. I converted tau to sigma, not to -sigma in my first reply. I wrote up a little PDF that will hopefully clear up everything: http://eudean.com/misc/u.pdf

I think the point of confusion is when switching from d(-sigma) to d(sigma), we have to do two things: we have to swap the limits of integration AND we swap the signs of the limits of integration. So an integral from a to b becomes an integral from -b to -a. The last couple lines of the PDF go over this pretty explicitly.

 

[TecHNooB]

Originally posted by: esun
Sorry, I see the issue. I converted tau to sigma, not to -sigma in my first reply. I wrote up a little PDF that will hopefully clear up everything: http://eudean.com/misc/u.pdf

I think the point of confusion is when switching from d(-sigma) to d(sigma), we have to do two things: we have to swap the limits of integration AND we swap the signs of the limits of integration. So an integral from a to b becomes an integral from -b to -a. The last couple lines of the PDF go over this pretty explicitly.

ok this helps a lot. thanks nice pdf

 

[blahblah99]

Originally posted by: TecHNooB
I know it has to do with the fact that Tau is changed to t - Tau but I don't see how that changes the area of integration from -infinity to t --> infinity to zero or how that causes the area of integration to become fixed and deltas to move around. Having trouble equating

Edit: I guess the reason is obvious. You want the beginning of one function to be scaled by the beginning of the other function, hence 'flip' one of them. I just don't understand where the integration from infinity to zero comes from in the derivation/explanation.

Mind the formatting, here's what I'm confused about spelled out:

Sigma = t - Tau

u(t) = integral ( delta(Tau), -infinity to t ) with respect to Tau
= integral ( delta(t - sigma) infinity to zero ) with respect to negative sigma
= integral ( delta(t - sigma) zero to infinity ) with respect to sigma

Integrate from infinity to zero? With respect to negative sigma? Not making the connection.

The notes show a substitution, where sigma = t - tau. Performing the substitution results in a negative sigma.