Convolution of two signals

sniya · 2012-11-07T03:43:54+00:00

hello.. please tell me what convolution is ?? and while getting the convolution of two signals , one of them is delayed,right??? so why do we do that??(i.e delaying one signal) kindly help...

Convolution of two signals

sniya

Member

Updated: Oct 25, 2024

Views: 897

hello..
please tell me what convolution is ?? and while getting the convolution of two signals ,
one of them is delayed,right??? so why do we do that??(i.e delaying one signal)
kindly help...

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Jeffrey Arulraj

Member • Nov 7, 2012

We are not exactly delaying as we are tending that delay by 0 and infinity in the case of CT signals the delay term vanishes and it is of no importance in the final output

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sniya

Member • Nov 7, 2012

huh?? isnt x(t)*y(t)=(integration -infinity to infinity)x(T)(y(T-t)dT.....where are we tending it with 0???

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Jeffrey Arulraj

Member • Nov 7, 2012

I meant only the positive half of the signal where the X(t) is multiplied with u(t) and the entire signal is defined only in the positive half of the time domain

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sniya

Member • Nov 7, 2012

do we always define the signal in only positive half????i guess i have to read about it again..
(though it doesnt really go through my head...) but please tell me about what convolution basically is.....

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Jeffrey Arulraj

Member • Nov 7, 2012

It is a way of signal multiplication That is multiplication op of two signals

And some time only signal are defined in positive half some times in both the halves and some times in negative half

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sniya

Member • Nov 7, 2012

ohk thank u😀

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Sagar07

Member • Nov 13, 2012

convolution is basically used to find the response of linear time invariant systems. In simple language, if we the price of 1 banana is 5 rs, then to calculate the price of 10 bananas, 5*10=50. Similarly, in technical language, if the output of the system to unit delta(1) is h(n), then to calculate output to input x(n), y(n)=x(n)*h(n), here,simply multiplication is replaced by convolution, but idea remains the same.

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sniya

Member • Nov 14, 2012

Sagar07
convolution is basically used to find the response of linear time invariant systems. In simple language, if we the price of 1 banana is 5 rs, then to calculate the price of 10 bananas, 5*10=50. Similarly, in technical language, if the output of the system to unit delta(1) is h(n), then to calculate output to input x(n), y(n)=x(n)*h(n), here,simply multiplication is replaced by convolution, but idea remains the same.
...understood😀....so why is there the term h(k-n) in the formula??? why not just h(k)???

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