Solve Convolution Shortcut for x(t)

[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.