Solve Convolution Shortcut for x(t)

[Color_of_Cyan]

Look again at hint #1 of post 35. How did I get from δ(x) to δ(-x)?

You multiplied the whole expression inside the Delta function by -1. And just to also make sure, it has nothing to do with the "x=T" in the "δ(x-a)" part of the other formula right? And it was necessary in order to solve for a.

What do I do now that I have the a? I have to plug it back into f so that it becomes f(a) now? What happens to T?

 

[rude man]

You multiplied the whole expression inside the Delta function by -1. And just to also make sure, it has nothing to do with the "x=T" in the "δ(x-a)" part of the other formula right? And it was necessary in order to solve for a.

What do I do now that I have the a? I have to plug it back into f so that it becomes f(a) now?

Yes.

What happens to T?

It goes away.

 

[Color_of_Cyan]

Okay, so would the final answer be this then?:
(1/3)[-exp(-4(t-2)) + exp(2(t-2))](t - 2)²

 

[rude man]

Okay, so would the final answer be this then?:
(1/3)[-exp(-4(t-2)) + exp(2(t-2))](t - 2)²

You're not dealing with the t² term correctly in moving it into the convolution integral.

 

[Color_of_Cyan]

So the T there in the convolution integral doesn't just go away then?

 

[rude man]

So the T there in the convolution integral doesn't just go away then?

OK, from the beginning:
f(t)*g(t) = ∫_-∞^+∞ f(T)g(t-T)dT = h(t).
So T appears in the integrand but, as in all cases of integrating with respect to a variable, the variable disappears after the integration limits are invoked. Freshman calculus.

 

[Color_of_Cyan]

Afraid I can't see what went wrong this time either then, are you able to explain post #54 a little more?

Sure the 't²' term is next to the 'δ,' but it's not in the convolution integral? I was thinking just replace the t with 'a = t-2' in everything. I'm just trying to go by the formula you gave, but even if I solved for T there in the Delta function (per the other 'tricks' I know) I think it would've been T = t-2 as well.

Thinking about it more carefully now, does the t² just completely disappear too? (At the very least I think there was something 'special' about the terms / variables next to the δ)

 

[rude man]

Afraid I can't see what went wrong this time either then, are you able to explain post #54 a little more?

Sure the 't²' term is next to the 'δ,' but it's not in the convolution integral? I was thinking just replace the t with 'a = t-2' in everything. I'm just trying to go by the formula you gave, but even if I solved for T there in the Delta function (per the other 'tricks' I know) I think it would've been T = t-2 as well.

Thinking about it more carefully now, does the t² just completely disappear too? (At the very least I think there was something 'special' about the terms / variables next to the δ)

post 56: g(t) = t²δ(t-2).

 

[Color_of_Cyan]

Oh that's right... does it just become 4 when you integrate it then? (Because now it's t = 2 for it next to the 'δ'?)

 

[rude man]

Oh that's right... does it just become 4 when you integrate it then? (Because now it's t = 2 for it next to the 'δ'?)

You mean t-2, not t=2, right?

Things are looking up! So what's your final answer?

 

[Color_of_Cyan]

(4/3)[-exp(-4(t-2)) + exp(2(t-2))]

 

[rude man]

(4/3)[-exp(-4(t-2)) + exp(2(t-2))]

May the gods be praised!

BTW realize I may have made a mistake myself at some point. Let me know if our answer agrees with your prof please?

 

[Color_of_Cyan]

Thank you so much for being patient with me, hah.

Are you able to check your "inbox" here?
Sure thing...

 

[rude man]

Thank you so much for being patient with me, hah.

Are you able to check your "inbox" here?
Sure thing...

Sure. You can let me know either on this forum or by private message.

 

[Color_of_Cyan]

Sorry, I keep saying -U when I mean U(-).
So you want to graph U{-(T - t)}.
Change (T-t) to t on the horizontal (T) axis and you'd be there.
Now, multiply the two graphs into one new graph.
It's real easy if you understand
0 x 0 = 0
0 x 1 = 0
1 x 0 = 0
1 x 1 = 1

For this, what if it was instead a ramp function or parabolic function instead of a step? Would I 'solve' for the integration limits the same way?

 

[rude man]

For this, what if it was instead a ramp function or parabolic function instead of a step? Would I 'solve' for the integration limits the same way?

I'll look at this also tomorrow. There may not be a finite answer if it's a ramp for example.

 

[rude man]

I meant to post also that I think I know what's with the missing U(t-2) term. Will try to send answer tomorrow.

 

[rude man]

I meant to post also that I think I know what's with the missing U(t-2) term. Will try to send answer tomorrow.

Well, it's tomorrow! (12:32 a.m.)

OK, here's the deal:
Mathematically, the U(t-2) does not belong.

Let
h1= 2exp(-4t)
h2 = exp(2t)
x = t²δ(t-2)
y = x*(h1*h2)
By the conventional formula for convolution, the answer is y = (4/3){[exp2(t-2)] - exp[-4(t-2)]} as we have derived. Per wikipedia:

but

so

Notice: No U(t-T) term.

On the other hand, assume h1 and h2 are the impulse responses for two separate networks connected in cascade (series). Then the impulse response for the combined network is h1*h2 = 2[exp(2t) - exp(-4t)].
Now we apply an input x = t²δ(t-2). Since the input x is zero until t = 2, so must be the output. So when we evaluate the convolution integral y = x*(h1*h2) = (4/3){[exp2(t-2)] - exp[-4(t-2)]} we need to multiply that expression by U(t-2) for the network since otherwise we don't get zero output for t < 2.

In other words: it depends on how the convolution integral is interpreted. If you omit the U(t-2) term you get a finite output for t < 2 which is wrong for the network, but U(t-2) does not appear in the mathematically derived convolution integral. So, bottom line, I'd say our answer is at least as valid as theirs.

This is similar to many problems in algebra. For example, area of a square of side x = x². What are the sides? You can't have negative sides even though (-x)(-x) = x² also.