Solve Convolution Shortcut for x(t)

[rude man]

Can I ask why?

I just can't see what I did wrong now. I pulled down the -6 and as far as I know don't have to worry about the +2t (because it is a constant here and it was in an exp)? Doesn't look like I did anything wrong with the exp(a+b) = exp(a)*exp(b) either. And I think the + c will always cancel due to it being a definite integral (Fundamental Theorem of Calculus).

Going back to post 29:
2 0∫t [exp(-6T + 2t)]dT
[/quote]
right

= 2(exp(-6T + 2t))/-6 and then +c |0t (the c will cancel itself out later)

wrong. What's with the "c"?? You're supposed to evaluate a definite integral, between 0 and t.
NM about exponentiation. That wasn't the problem. The problem is you didn't evaluate the definite integral.

Again, I recommend removing the exp(2t) term outside the integral sign before integrating:

2 0∫t [exp(-6T + 2t)]dT = 2exp(2t) ∫₀^t exp(-6T)dT

 

[Color_of_Cyan]

2 0∫t [exp(-6T + 2t)]dT = 2exp(2t) ∫₀^t exp(-6T)dT

Ahhh I didn't really see this :) Hopefully this is forgivable.

ok so (I think) the integral is actually this now:

2₀∫^t[exp(-6T)exp(2t)]dT

= 2exp(2t)₀∫^t[exp(-6T)dT

= [2exp(2t)exp(-6T)]/-6 | ₀^t

= -(1/3)exp(2t)exp(-6T) | ₀^t

= -(1/3)exp(2t)exp(-6t) + (1/3)exp(2t)

= -(1/3)exp(-4t) + (1/3)exp(2t)

= (1/3)[-exp(-4t) + exp(2t)]

And the u(t) would be permanently gone from this convolution result then? Do I take this result and just do convolution with the last term now?

 

[rude man]

Ahhh I didn't really see this :) Hopefully this is forgivable.

ok so (I think) the integral is actually this now:

2₀∫^t[exp(-6T)exp(2t)]dT

= 2exp(2t)₀∫^t[exp(-6T)dT

= [2exp(2t)exp(-6T)]/-6 | ₀^t

= -(1/3)exp(2t)exp(-6T) | ₀^t

= -(1/3)exp(2t)exp(-6t) + (1/3)exp(2t)

= -(1/3)exp(-4t) + (1/3)exp(2t)

= (1/3)[-exp(-4t) + exp(2t)]

Right!

And the u(t) would be permanently gone from this convolution result then? Do I take this result and just do convolution with the last term now?

Yes. Can you see how you combined the removal of the U(T) and U(t-T) functions and changing the limits of integration from + & - infinity to 0 to t? You have to really understand this.

Then, ready for the second convolution? It's easier than the first, thanks to the miracle of integrating the product of a function and a delta function. No integration knowledge necessary!

 

[Color_of_Cyan]

So it's this now:

(1/3)[-exp(-4t) + exp(2t)] * t²δ(t - 2)

To refresh:
(f * g)(t) = _-∞∫^∞f(t - τ)g(τ)dτ

f has to become f(t-T)
g has to become f(t)

If I make the delta function part be f(t-T) then it would be this?:

_-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - 2)]dT

Then, ready for the second convolution? It's easier than the first, thanks to the miracle of integrating the product of a function and a delta function. No integration knowledge necessary!

I think I still need work with this too. I am not sure if I understand the sampling property.

 

[rude man]

So it's this now:

(1/3)[-exp(-4t) + exp(2t)] * t²δ(t - 2)

To refresh:
(f * g)(t) = _-∞∫^∞f(t - τ)g(τ)dτ

If I make the delta function part be f(t-T) then it would be this?:

_-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - 2)]dT

check δ(t - 2).

I think I still need work with this too. I am not sure if I understand the sampling property.

Fact: δ(x) is an even function: δ(-x) = δ(x).
Sampling function: ∫f(x)δ(x-a)dx = f(a). The integration limits can be anything so long as they include x=a. So they can for example be + and - infinity. Notice you don't need to know how to integrate f(x)!

 

[Color_of_Cyan]

check δ(t - 2).

Ok so it was actually this then?: (t - T - 2):

_-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - T - 2)]dT ??

Fact: δ(x) is an even function: δ(-x) = δ(x).
Sampling function: ∫f(x)δ(x-a)dx = f(a). The integration limits can be anything so long as they include x=a. So they can for example be + and - infinity. Notice you don't need to know how to integrate f(x)!

I can't seem to understand or see what is f(a) or f(x) in this case. But, if everything above is correct so far now, then according to only the 'tricks' I already know with this then:

t = T + 2 (because it is inside the 'δ')

Which is then substituted above into this:

(1/3)[-exp(-4t) + exp(2t)][(t - T)²

and so it becomes this:

(1/3)[-exp(-4(T+2)) + exp(2(T+2))][(2)²

= (4/3)[-exp(-4(T+2)) + exp(2(T+2))]

or was it supposed to be

T = t + 2 or something?

 

[Color_of_Cyan]

Bump, was this the final answer? I think I had the correct procedure but not completely sure (and I think I probably did something wrong).

 

[rude man]

Ok so it was actually this then?: (t - T - 2):

_-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - T - 2)]dT ??

I urge you to think hard about post 35. I gave you 2 hints and both are important. Take the first hint first, then the second.

Correct up to here.
One last hint: x = T.
t is again a constant in the integration over T.
Figure out what a must be.

I have to quit here lest I wind up in trouble with the Authorities, divulging too much.
If you post your final answer I'll tell you if it's right or wrong.

 

[Color_of_Cyan]

Ah, I think you mean that it was actually supposed to be T = t - 2 instead (still inside the 'δ'):

Then still substituted into this:

(1/3)[-exp(-4t) + exp(2t)][(t - T)²

and so it becomes this:

(1/3)[-exp(-4(t-2)) + exp(2(t-2))][(t - (t - 2))²

= (1/3)[-exp(-4(t-2)) + exp(2(t-2))][(t - t + 2))²

= (1/3)[-exp(-4(t-2)) + exp(2(t-2))(2)²= (4/3)[-exp(-4(t-2)) + exp(2(t-2))] for my final answer (?).

 

[rude man]

Nope. What is your a? What is the delta function inside the integral?

 

[Color_of_Cyan]

Nope. What is your a? What is the delta function inside the integral?

'δ(t - T - 2)'

Fact: δ(x) is an even function: δ(-x) = δ(x).
Sampling function: ∫f(x)δ(x-a)dx = f(a). The integration limits can be anything so long as they include x=a.

Still can't see it from the hints you gave but I thought you meant just solve for the inside (and you can since the limits are from -∞ to +∞ right?). And if x = T I think the whole thing means it becomes an integral with dT and you solve for T so that's what I thought you meant, but I guess I'm still missing something.

 

[rude man]

'δ(t - T - 2)'

Right. So if that's δ(x-a) and x = T, what is a?

I think the whole thing means it becomes an integral with dT and you solve for T so that's what I thought you meant, but I guess I'm still missing something.

If it's an integral with dT you don't solve for T. T disappears ...

 

[Color_of_Cyan]

Well the first hint was kind of something that didn't really make much sense to me and afraid to try the first time, that it was δ(x) = δ(-x) so it could be both -T and +T??

I'm going to say now (flipping the negative sign inside) that a = -t + 2

 

[rude man]

Nope. What does your δ function look like inside the final convolution integral?

 

[Color_of_Cyan]

Okay wait, let me check this again.

So if that's δ(x-a) and x = T, what is a?

t - T - 2 = x - a
at the same time
x = T

t - T - 2 = T - a

t - 2 = 2T - a

a = -t + 2 + 2T?

 

[rude man]

Okay wait, let me check this again.
t - T - 2 = x - a
at the same time
x = T

t - T - 2 = T - a

t - 2 = 2T - a

a = -t + 2 + 2T?

Go back to post 44. Better yet, show us your entire final convolution integral.

 

[Color_of_Cyan]

_-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - T - 2)]dT

∫f(x)δ(x-a)dx = f(a)

So I would still have at least the "[(1/3)[-exp(-4t) + exp(2t)]" part because I would still pull them from the integral, right? Would I have to replace any of the variables in the delta function?

I think I should also probably say I never really took statistics so I don't really know much about dummy variables. If this is just algebra now I'm also having a difficult time understanding it.

 

[rude man]

_-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - T - 2)]dT

That is right. But comparing that with ∫f(x)δ(x-a)dx = f(a), with x = T, what is a? You need to get the delta function looking like δ(T - a). Look again at my hint back in post 35.

So I would still have at least the "[(1/3)[-exp(-4t) + exp(2t)]" part because I would still pull them from the integral, right?

No, you will replace t with T and since you're integrating over T you can't pull that expression out of the integral sign.

But as I said before when you finally 'get it' you will realize that no actual integration needs to be done.

I think I should also probably say I never really took statistics so I don't really know much about dummy variables. If this is just algebra now I'm also having a difficult time understanding it.

Has nothing to do with statistics. Dummy variables are used everywhere in integral calculus. Sometimes people 'cheat' and don't use them but that is rigorously incorrect.

Example: s(t) = ∫₀^t v(t')dt' is position s given velocity v from t=0 to t. It is incorrect to write ∫₀^t v(t)dt. t' is a dummy variable & disappears with the integration.

 

[Color_of_Cyan]

Okay, so _-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - T - 2)]dT

δ(t - T - 2)

Then to make it like you say as this:

∫f(x)δ(x-a)dx = f(a); x = T, what is a?

multiply the inside by -1 so the Delta function is this:

δ-1(T - t + 2)

-a = -t + 2, from the above?

a = t - 2

No, you will replace t with T and since you're integrating over T you can't pull that expression out of the integral sign.

Does this mean t = T + 2 for the rest of the expression?

Has nothing to do with statistics. Dummy variables are used everywhere in integral calculus. Sometimes people 'cheat' and don't use them but that is rigorously incorrect.

Example: s(t) = ∫₀^t v(t')dt' is position s given velocity v from t=0 to t. It is incorrect to write ∫₀^t v(t)dt. t' is a dummy variable & disappears with the integration.

No but with there you change them to help you integrate (if you're doing u-sub or by parts) you switch to U or dv then substitute, integrate, and then switch them back. Here I don't see how it's supposed to help (even though you said you don't really have to integrate anything in this case).

 

[rude man]

Okay, so _-∞∫^∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)²δ(t - T - 2)]dT

δ(t - T - 2)
a = t - 2

Good up to here.

Then to make it like you say as this:

multiply the inside by -1 so the Delta function is this:

δ-1(T - t + 2)

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

Does this mean t = T + 2 for the rest of the expression?

No but with there you change them to help you integrate (if you're doing u-sub or by parts) you switch to U or dv then substitute, integrate, and then switch them back. Here I don't see how it's supposed to help (even though you said you don't really have to integrate anything in this case).

Look again at hint #2 of post 35. Did I have to integrate f(x) to get ∫f(x)δ(x-a)dx?

 

[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?