22nd January 2006, 16:07 #1
Newbie level 6
hi all ,
i'm new to dsp . i want to know the exact purpose for convolution . Why cant we just multiply the two signals ?
any one ready to help me ?
22nd January 2006, 19:51 #2
Member level 1
when give delta function signal to input of some system ,
you will see on output of system some response.
22nd January 2006, 19:51
22nd January 2006, 23:57 #3
Full Member level 4
When a signal go through a linear system, its frequency components are affected by the linear system, the frequencies results is a weight composition of the system frequency response. What happens is a multiplication in frequency not in time. You will learn that multiplication in frequency domaing is equivalent of convolution in time domain. Multipling signals in time domain is associated with displacement in frequency, for instance modulation process, wich for its turn is equivalent of convolution in frequency.
Go ahead in your studying and you will son realise the diferrences.
22nd January 2006, 23:57
23rd January 2006, 05:31 #4
Full Member level 2
In mathematics and in particular, functional analysis, convolution is a mathematical operator which takes two functions f and g and produces a third function that in a sense represents the amount of overlap between f and a reversed and translated version of g. A convolution is a kind of very general moving average, as one can see by taking one of the functions to be an indicator function of an interval.
to get more understanding on Convolution go through these links...
23rd January 2006, 05:31
26th January 2006, 10:43 #5
Newbie level 6
Convolution is a simple mathematical operation which is fundamental to many common image processing operators. Convolution provides a way of multiplying together two arrays of numbers, generally of different sizes, but of the same dimensionality, to produce a third array of numbers of the same dimensionality.
This can be used in image processing to implement operators whose output pixel values are simple linear combinations of certain input pixel values.
26th January 2006, 11:09 #6
Newbie level 1
"Why cant we just multiply the two signals "
we want to find response of a filter(system ,..) to one input
assume x[n] pass through a filter h[n].
generally speaking , x[n] is a sample of a phenoma that depends on time(..) , (and so we dont have all of samples at the same time.)
we consider the actual consequence of signal samples by inversing signal samples : x[-n]
by this work , the first generated samples passes through filter sooner than last generetad samples.
2nd February 2006, 15:20 #7
Newbie level 1
It is basically a multiplication analogy in time domain
6th February 2006, 15:28 #8
Newbie level 2
In LTI systems the output is the convolution between the input signal and system impulse responce.
In DSP LTI systems like filters could be designed in time-domain by designing impulse responce and signal processing will take action by convoluting this impulse signal with the input signal.
8th February 2006, 07:51 #9
Full Member level 1
any signal can be represented as scaled summation version of scaled impulses. If you consider an LTI system the out put can be viewed as summation of the corresponding scaled impulse responses. This is nothing but convolution. This is obviously different from multiplication which will not give the output of the system for the corresponding input.