Convolution of two signals

Replies

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  Jeffrey Arulraj

  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

  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

  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

  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

  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

  ohk thank u😀
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  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.
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  sniya

  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)???