Solve Convolution Shortcut for x(t)

[Color_of_Cyan]

Homework Statement

Find (use shortcut):

x(t) = 2e^-4tu(t) * e^2tu(t) * t²σ(t - 2)

Homework Equations

Convolution properties:

# "shape of Y (output) is different from x1, x2"

# x1 * x2 = x2 * x1

# x1 * (x2 + x3) = (x1 * x2) + (x1 * x3)

# x1(t) = * x2(t) * x3(t) * ...

# step * ramp = parabolic function

# e^atu(t) * e^btu(t) = (1/a-b)[e^at - e^bt]u(t)

The Attempt at a Solution

2e^-4tu(t) * e^2tu(t) * t²σ(t - 2)

just doing

2e^-4tu(t) * e^2tu(t) for now, and using the last property listed:= 2(1/(2+4))[e^-4t - e^2t]u(t)

= (1/3)[e^-4t - e^2t]u(t)

I am unsure how to go about doing convolution with the * t²σ(t - 2) at this point, please help. Thanks!

 

[rude man]

What is σ(t)? I've never seen that function before.

What are you trying to find? You already have x(t). Simplifying?

 

[Color_of_Cyan]

It's a impulse function supposedly (and sadly I don't seem to have any property listed where it's involved). And yes, trying to simplify and do convolution with all of this.

 

[rude man]

It's a impulse function supposedly (and sadly I don't seem to have any property listed where it's involved). And yes, trying to simplify and do convolution with all of this.

OK, I guess σ(t) is what the rest of us call δ(t)?

I suppose you can convolve the first & 2nd terms, the convolve the result with the third.

So why not write the formal convolution integral for the first two terms for a starter.

I'm afraid I don't know any 'shortcut' here.

 

[Color_of_Cyan]

Hah, yes, δ(t). (Of course I forgot the 'δ' symbol was here and my poor eyesight never saw it.)

I suppose you can convolve the first & 2nd terms, the convolve the result with the third.

So why not write the formal convolution integral for the first two terms for a starter.

I'm afraid I don't know any 'shortcut' here.

Okay

(f * g)(t) = 0∫^tf(t - τ)g(τ)dτ then, is this it?So if f = 2e^-4tu(t) and g = e^2tu(t). So it's something like= 0∫^t[2e^{-4(t - τ)}u(t - τ)e^2τu(τ)]dτI had to change all the function variables (including inside the 'u(t)') to τ also right? Do I have to also end up integratring the 'u(τ)'? If so, how?

 

[rude man]

Hah, yes, δ(t). (Of course I forgot the 'δ' symbol was here and my poor eyesight never saw it.)


Okay

(f * g)(t) = 0∫^tf(t - τ)g(τ)dτ then, is this it?

Permit me to write T = tau, it's a lot easier typing ...

So, proceeding from that integral, with f(t-T) = exp{2(t - T)}U(t - T) and g(T) = 2exp{-4T}U(T), I would change t - T to -(T - t) everywhere in the integrand. Now, graph U(T) and -U(T - t) vs. T and multiply them. Then get rid of the two U functions in the integrand by suitable choice of the upper and lower limits of integration. Then, perform the integration.

The second convolution is done the same way. Remember the sampling property of the delta function: ∫f(x)δ(x - a)dx with upper limit +∞ and lower limit -∞ = f(a).

 

[donpacino]

use laplace!
convolution in the time domain becomes multiplication in the frequency domain.

 

[rude man]

use laplace!
convolution in the time domain becomes multiplication in the frequency domain.

Laplace is limited to t > 0. This is not specified here. But you could use Fourier. I have looked at that but found it harder to do the inversion integral back to the time domain than just plowing ahead solely in the time domain.

BTW there is also the double-sided Laplace transform which no one uses.

 

[Color_of_Cyan]

Sorry for getting back to this late

Permit me to write T = tau, it's a lot easier typing ...

So, proceeding from that integral, with f(t-T) = exp{2(t - T)}U(t - T) and g(T) = 2exp{-4T})U(T), I would change t - T to -(T - t) everywhere in the integrand.

So you have the convolution for the first two terms as:

= ₀∫^t[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT

Does changing it to -(T - t) help simply it? It would appear as this:

= ₀∫^t[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT

Now, graph U(T) and -U(T - t) vs. T and multiply them.

Can you explain this? Graph any unit step function?

Laplace is limited to t > 0. This is not specified here. But you could use Fourier. I have looked at that but found it harder to do the inversion integral back to the time domain than just plowing ahead solely in the time domain.

BTW there is also the double-sided Laplace transform which no one uses.

In the class this convolution problem actually came before we formally covered Laplace / Fourier (which I will also need help on later).

 

[rude man]

So you have the convolution for the first two terms as:

₀∫^t[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT

I just noticed you meant the lower limit of integration to be 0. It should be -∞. (BTW the upper limit should also be ∞. You can make the upper limit t in this case, it makes no difference here). Otherwise, good.

Does changing it to -(T - t) help simply it? It would appear as this:

= _-∞∫^t[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT

Yes, it clarifies the next step.

Can you explain this? Graph any unit step function?

Yes I can. I goofed!
I should have said "graph U(T) and U[-(T-t)] vs. T".

Can you graph U(x) vs. x?
U(x - a) vs. x?
U(-x) vs. x?
U[-(x-a)} vs. x?
You can do them all just by remembering
U(ζ) = 0, ζ < 0
U(ζ) = 1, ζ > 0.

 

[Color_of_Cyan]

I just noticed you meant the lower limit of integration to be 0. It should be -∞. (BTW the upper limit should also be ∞. You can make the upper limit t in this case, it makes no difference here). Otherwise, good.

Should it be -∞ to +∞ for the whole property or just for this problem? I don't think I quite understand that. I'll take your word for it though, so I think it would now be:

_-∞∫^∞[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT

I should have said "graph U(T) and U[-(T-t)] vs. T".

Can you graph U(x) vs. x?
U(x - a) vs. x?
U(-x) vs. x?
U[-(x-a)} vs. x?
You can do them all just by remembering
U(ζ) = 0, ζ < 0
U(ζ) = 1, ζ > 0.

Isn't this just the unit step graph property? Or what's the name or term for this if it isn't? What would be 'ζ' (if this is what you are basing the graph on)? I still can not see how this is supposed to help unless you are trying to hint at doing graphical convolution.

 

[rude man]

Should it be -∞ to +∞ for the whole property or just for this problem? I don't think I quite understand that. I'll take your word for it though, so I think it would now be:

_-∞∫^∞[exp(2(t - T))u(t - T)][2exp(-4T)u(T)]dT

That is the general definition for the convolution integral. Often, depending on the problem, -∞ can be replaced by 0 and +∞ by t, but it's always correct to use the infinite + and - limits. Just keep in mind that if the integrand is identically zero below and/or above certain values then the infinite limits can be replaced by those respective values. This is the key idea I'm trying to point out in doing the U graphs. U(T)U{-(T-t)} has zero value below and above certain values. What are those values?

Isn't this just the unit step graph property? Or what's the name or term for this if it isn't? What would be 'ζ' (if this is what you are basing the graph on)? I still can not see how this is supposed to help unless you are trying to hint at doing graphical convolution.

No, I don't want you to do any graphical convolution. I am suggesting that by graphing the two U functions and multiplying them graphically you will know what the upper & lower limits of integration are for evaluating the actual convolution integral.

Maybe you can figure out the limits of integration by looking at U(T)U{-(T-t)} some other way, but doing it graphically is real easy.

ζ was just a dummy variable.

 

[Color_of_Cyan]

I am suggesting that by graphing the two U functions and multiplying them graphically you will know what the upper & lower limits of integration are for evaluating the actual convolution integral.

Maybe you can figure out the limits of integration by looking at U(T)U{-(T-t)} some other way, but doing it graphically is real easy.

ζ was just a dummy variable.

This will sound real bad, but I'm afraid I haven't really heard of graphical multiplication before.

Are you saying you want me to graph exp(-2(T - t)) and 2exp(-4T)?


Also my bad the integral in my last post was supposed to now be this:

_-∞∫^∞[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT

 

[rude man]

This will sound real bad, but I'm afraid I haven't really heard of graphical multiplication before.

Are you saying you want me to graph exp(-2(T - t)) and 2exp(-4T)?

No, I want you to graph U(T) and -U{(T-t)}, then multiply them into one graph.
Please start with graphig these two functions separately, we can't go on until you do.

Also my bad the integral in my last post was supposed to now be this:

_-∞∫^∞[exp(-2(T - t))u(-1(T - t))][2exp(-4T)u(T)]dT

[/quote]
That's fine, that was correct also but this way the integration becomes more apparent, with t = constant in the integration.

 

[Color_of_Cyan]

No, I want you to graph U(T) and -U{(T-t)}, then multiply them into one graph.
Please start with graphig these two functions separately, we can't go on until you do.

Okay, I'm not sure if these are correct I don't really know what T is:

https://imagizer.imageshack.us/v2/547x297q90/909/MvuSyH.jpg

https://imagizer.imageshack.us/v2/547x297q90/673/owhtZD.jpg

 

[rude man]

T is your horizontal axis.

You got U(T) right.
You got -U{(T-t)} wrong. You graphed -U(T) instead.
When is -U(x) = 0? = 1? Now let x = T - t. t is a constant in this. T is your variable. Try again ...

 

[Color_of_Cyan]

So it's shifted then, is that what you are saying?:

https://imagizer.imageshack.us/v2/547x297q90/537/vHAo1X.jpg

 

[rude man]

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

 

[Color_of_Cyan]

Sorry for getting back to this super late again (site moved and I had some comp problems).

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

Does it mean the first graph (u(t)) is 0 and the second graph (U(-1(T - t))) is 1? I can not really see yet as to how these help with convolution integrals?

 

[Color_of_Cyan]

Bump, also forgot to post the graph, does multiplying the two graphs mean it will be the same as u(T) if u(T) is 0?

https://imagizer.imageshack.us/v2/547x297q90/909/MvuSyH.jpg

 

[rude man]

Sorry for getting back to this super late again (site moved and I had some comp problems).
Does it mean the first graph (u(t)) is 0 and the second graph (U(-1(T - t))) is 1?

No. Graph the product U(T)*U{-(T-t)} as you go along the T axis from T = - infinity to T = + infinity. (You can actually go from a small negative T to a bit above T=t).

I can not really see yet as to how these help with convolution integrals?

Those two U's appear in your convolution ingtegral! Look again at post 13.

 

[Color_of_Cyan]

No. Graph the product U(T)*U{-(T-t)} as you go along the T axis from T = - infinity to T = + infinity. (You can actually go from a small negative T to a bit above T=t).
Those two U's appear in your convolution ingtegral! Look again at post 13.

Okay, so like this then?:

https://imagizer.imageshack.us/v2/479x260q90/674/Smf0nJ.jpg

I'm guessing now you can say that the limit goes from 0 to t or something similar then, right?

 

[rude man]

Okay, so like this then?:

https://imagizer.imageshack.us/v2/479x260q90/674/Smf0nJ.jpg

I'm guessing now you can say that the limit goes from 0 to t or something similar then, right?

Congratulations! You have scaled the heights.

Now, last big step for this first convolution: remove U(T)U{-(T-t)} from your convolution integral by suitably setting the upper and lower limits of integration.

 

[Color_of_Cyan]

₀∫^t[exp(-2(T - t))][2exp(-4T)]dT

and now here I may need to review all the math down from here but I think it goes like this:

= 2₀∫^t[exp(-2(T - t))][exp(-4T)]dT

= 2₀∫^t[exp(-6T - 2t)]dT

= 2[exp(-6T - 2t)]/-6

= -3[exp(-6T - 2t)] | ^t₀

= -3[exp(-8t)] ?

 

[rude man]

₀∫^t[exp(-2(T - t))][2exp(-4T)]dT

Great step!

and now here I may need to review all the math down from here but I think it goes like this:

= 2₀∫^t[exp(-2(T - t))][exp(-4T)]dT

= 2₀∫^t[exp(-6T - 2t)]dT

Check this - small error made. I also suggest since t is a constant here taking the t term outside the integral sign.

Review your freshman calculus if necessary; you'll sink if you don't get this firmed up.
[/QUOTE]

 

[Color_of_Cyan]

Oh right so it's actually this now:

= -3[exp(-6T + 2t)] | ^t₀

= -3[exp(-4t)]. Also forgot the T = 0 part though so would it actually be this?:

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

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

 

[rude man]

Oh right so it's actually this now:

= -3[exp(-6T + 2t)] | ^t₀

Doesn't look like you integrated.

Fact: ∫exp(bx)dx = (1/b) exp(bx) + constant

 

[Color_of_Cyan]

I think I skipped a step there... I was integrating this:

2₀∫^t[exp(-6T + 2t)]dT

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

oh woops i swapped the 2 and -6 so it's really this now:

= (-1/3)(exp(-6T + 2t))|₀^t

(Assuming if you remember doing (d/dT)e^T don't you just bring the derivative of whatever is in T down?)

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

 

[rude man]

I think I skipped a step there... I was integrating this:

2₀∫^t[exp(-6T + 2t)]dT

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

oh woops i swapped the 2 and -6 so it's really this now:

= (-1/3)(exp(-6T + 2t))|₀^t

(Assuming if you remember doing (d/dT)e^T don't you just bring the derivative of whatever is in T down?)

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

Sorry, still not right. You need to review basic integral calculus & maybe even handling exponentials. Like exp(a+b) = exp(a)*exp(b).

 

[Color_of_Cyan]

Sorry, still not right. You need to review basic integral calculus & maybe even handling exponentials. Like exp(a+b) = exp(a)*exp(b).

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).

 

[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)!