Question

Question : 8 - Signals :

What is its Convolution??

\(y(t) = u(t) * h(t)\), where h(t) is defined as:

\(h(t) = \begin{cases} e^{2t} &, t < 0 \\ e^{-3t} &, t>0 \end{cases}\)

Answer 1

Can I write h(t) as:

\(h(t) = e^{2t} \cdot u(-t) + e^{-3t} \cdot u(t)\)

Answer 2

So - something about convolutions, huh?
let me pull something up on them - I know that you can make them into a laplace - what is your foal? o.O

Answer 3

goal*
not foal XD

Answer 4

Step functions. Ah! well, I can definitely help you, but we would have to work slowly as I would remember as we go...

Answer 5

I would venture to say "Yes". You can write H(t) that way with the step function
h(t)

Answer 6

No, I don't want to go by Laplace..

I want to use Convolution evaluation by Convolution Integral..

\[f(t) * h(t) = \int\limits_{-\infty} ^ {\infty} f(\tau) h(t - \tau) d \tau\]

Answer 7

Why you have used capital H there with h(t) ?? I will remain be h(t) only, I have just represented my signal in other form.. I have not transformed it into other form.

Answer 8

I know - I realized that after I typed it, but it required effort to change it...

Answer 9

Otherwise, I understand that you do NOT want to use laplace XD

Answer 10

No, I will use Laplace when I am done with basic Convolution, there is still long way to Laplace..

According to my Notes, Still in between, Fourier Series, Fourier Transform and then Laplace will come..

Answer 11

so, lets start the convolution process - we don't really have ANY idea what u(t) is, so we should probably make that the f(t)

and therefore, h(t) is h(t)...

Answer 12

u(t) is a Unit Step Function..

Answer 13

Alright - well lets start. Lets start plugging in and yes - writing h(t) that way is PERFECT.

Answer 14

Okay, I will not bother you much.. I just want to tell you where I have reached..

Just see my steps and check whether you are finding any mistakes or not..

Answer 15

what we have so far makes sense :)

Answer 16

\[y(t) = \int\limits_{- \infty}^{\infty} [e^{2t} \cdot u(-\tau) + e^{-3\tau} \cdot u( \tau)] * u(t - \tau) d \tau\]

Answer 17

What is c = 0 there??

Answer 18

Another question - are you okay with me adding in other information, as such in the blue on your chart you just made? Or are those contributions annoying?

Answer 19

oh :o it means that when t is 0 or more, the function begins.

Answer 20

The step function u(t-c) turns "on" at c.

Answer 21

Another question - are you okay with me adding in other information, as such in the blue on your chart you just made? Or are those contributions annoying?

?? What does it mean?? I am telling you in advance, if at any point of time, I am sounding rue to me, then just suppose that I am just kidding..

Answer 22

that's a good start for your equation what ^

Answer 23

You can add any information any number of time, it is going to add to my knowledge only..

Answer 24

I think we have a communication gap in between.. :P

Answer 25

probably

Answer 26

And yes, c = 0 there, meaning there is no shifting for u(t) signal, it is starting from t = 0 only.. :)

Answer 27

and no worries - I just had this perception that you wanted me to watch and only intervene at mistakes... I wanted to make sure you were cool with commentary XD

Answer 28

Coming back to my steps, I have reached to:

\[y(t) = \int\limits_{-\infty}^{t} [e^{2t} u( - \tau) + e^{-3 \tau} u( \tau)] d \tau\]

Answer 29

Note: notice the infinite integration? see how from negative infinity to infinity, you have a exponential equation that will actually give you an answer, and how it shuts off at 0, and then another equation that gives you an answer from 0 to infinity when integrated is turned on?

It means you should get a number answer ^.^

Answer 30

Sorry, if I or my words have hurt you anywhere..

I just meant that let us begin our discussion from that point where I have solved this question..

Answer 31

onward

Answer 32

Slowly explain this note to me..

Note: notice the infinite integration? see how from negative infinity to infinity, you have a exponential equation that will actually give you an answer, and how it shuts off at 0, and then another equation that gives you an answer from 0 to infinity when integrated is turned on?
\

Answer 33

lol ok

Answer 34

:)

Answer 35

and I wasn't hurt man- I just figured I would ask because I wasn't sure what you wanted XD

Answer 36

so, see the black integration answer?