Question

Question : 7 - Signals:

The continuous time Convolution Integrals \(y(t) = cos( \pi t)[u(t+1) - u(t-1)] * u(t)\) is ??

\(*\) denotes Convolution here..

Answer 1

Why I am getting 0, I don't know..

Answer 2

You are typing or not??

Or you have been typing since my birth.. :P

Answer 3

@amistre64 your one look may clear my doubt..

Answer 4

For \(t\leq 1\) the convolution become zero. Otherwise, It does not become zero.

Answer 5

I want the total answer from -infinity to +infinity..

Answer 6

convolutions are not in my repetiore ... and neither is spelling repetiore apparently

Answer 7

Oh, it is okay @amistre64 .. That's why I said "may"

Answer 8

By distribution property, firstly I have separated the two convolutions..

Answer 9

for \(t\leq -1\) , \(y(t) = 0\)

for \(-1<t\leq1\), \(y(t)=\int \limits_{-1}^{t} \cos (\pi \tau)\;d\tau\)

for \(1<t\), \(y(t)=\int \limits_{-1}^{1} \cos(\pi \tau)\; d\tau\)

Answer 10

How you are analyzing it, can you tell me in detail?

Answer 11

\[y(t) = \cos(\pi t) \cdot u(t+1) * u(t) - \cos( \pi t) \cdot u(t-1) * u(t)\]

Answer 12

u(t) can be replaced but It will convert the function in the form of integral from -infinity to t..

Answer 13

Sure,

take \(g(t)=\cos(\pi t)[u(t+1)-u(t-1)]\) and \[h(t)=u(t)\]

take \[y(t)=\int_{-\infty}^\infty g(\tau)h(t-\tau) d\tau\]

graph of g(t) is in the picture

Answer 14

\[y(t) = \int\limits_{- \infty}^t \cos( \pi t) \cdot u(t+1) \cdot dt - \int\limits_{- \infty}^t \cos( \pi t) \cdot u(t-1) \cdot dt \\y(t) = \int\limits_{- 1}^t \cos( \pi t)\cdot dt - \int\limits_{1}^t \cos( \pi t) \cdot dt\]

Answer 15

And then I do:

\[y(t) = \int\limits_{-1}^1 \cos( \pi t) \cdot dt + \int\limits_1^t \cos(\pi t) dt - \int\limits_1^t \cos(\pi t) dt \\y(t) = \int\limits_{-1}^1 \cos( \pi t) \cdot dt\]

Answer 16

Where am I doing wrong??

Answer 17

You have only cosidered \[\cos(\pi t)[u(t+1)-u(t-1)]\] which is just integration. Where is \(u(t)\)???

The final convolution should be

\[\int_{-\infty}^\infty (\cos(\pi \tau)[u(\tau+1)-u(\tau-1)] )\ast u(t-\tau) d\tau\]

For this kind of problem a graphical method is more suitable

Answer 18

u(t) becomes integration..

Answer 19

See, here if we convolve, u(t) with u(t), then:

\[\int\limits_{- \infty} ^{\infty}u(\tau) \cdot u(t - \tau) d \tau \implies \int\limits_{-\infty}^t u(\tau) d(\tau)\]

Answer 20

This means convolving with u(t) will introduce just an integral from -infinity to t..

Answer 21

Are you getting what I am try to say??

Answer 22

But there is a \(\cos(\pi t)\) attached to u(t+1) and u(t-1) with it. So it is not just convolving u(t) with u(t)...

Answer 23

Okay, I will clear it when I will get some time, you explain your method..

Answer 24

yes, I can see that graph, cos(pi t) multiplied with rectangular function will given that.. :)

Answer 25

You are assuming that \(g(t)\ast u(t) = \int_{-\infty}^tg(\tau) d\tau\) which is wrong

Answer 26

And u(t) with u(t) was just an example, in general you can take any function..

Why it is wrong??

Answer 27

Okay, explain me how that is wrong, I am listening..

Answer 28

@myininaya your eyes will be definitely need here.. (Not physically though).. :)

Answer 29

*needed..

Answer 30

@BAdhi why you have stopped??

Answer 31

Explain that..

Answer 32

Sorry, I was wrong about \(g(t)\ast u(t) = \int_{-\infty}^t g(\tau)d\tau\). It is a correct property.

You can use this straight forward to find the answer but do not expand the \(cos(\pi t)[u(t+1)-u(t-1)]\). This denotes a region of cos(pi t) from -1 to 1.

so the convolution \(y(t) = \int_{-\infty}^t g(\tau) \; d\tau \)

What you have made a mistake is taking \(\tau\) as \(t\). t here is like a constant in the integral.

So for example if t<-1, the y(t) integral will be zero since \(g(\tau)\) is from -1 to 1. But it will not be 0 if t>-1. Since it denotes the area under the g(t) which is shown in the picture below

Answer 33

That seems to be right.. I was thinking the same that I should work with \(\tau\) there.. :)

Answer 34

I got it now.. :)

A real thanks.. :)