Interpretation of convolution sum

Hi
I was just reading about convolution and I understood that input signal can be expressed as weighted sum of scaled and delayed impulse responses. In continuous domain, the definition is clear to me. But I am not clear about discrete domain definition.



In a sampled system when we have fixed number of samples , how can k vary from -infinity to + infinity?

gary36 said:

Hi
I was just reading about convolution and I understood that input signal can be expressed as weighted sum of scaled and delayed impulse responses. In continuous domain, the definition is clear to me. But I am not clear about discrete domain definition. **broken link removed**
In a sampled system when we have fixed number of samples , how can k vary from -infinity to + infinity?

Click to expand...

Easily. If the function is not defined everywhere from segment numbers minus infinity to plus infinity, then only the range where the signal function is defined gets multiplied by the unit impulse function. Suppose the function is only defined for the segment numbers k =10 to 20 . This means all the terms in summation ranges from minus infinity thru 9 and 21 thru plus infinity will be zero. So the summation limits can include the whole range of integers, but only the signal segments ( k =10 to 20 in this case) will have significance and be considered for values other than zero.

Ratch

Ok, this means that I can have N number of samples, but we have to consider only those bounded region decided by the range of K. This makes it clear.

Now coming to the convention. Delayed samples are always taken to the right axis x(n-k). But in the convolution sum evaluation we take the image and flip it such that it falls under positive axis. This is not clear because as per definition , convolution sum= x*h(n-k). h(n-k) should be in positive axis only. Why flip it and then multiply accumulate?

gary36 said:

Ok, this means that I can have N number of samples, but we have to consider only those bounded region decided by the range of K. This makes it clear.

Now coming to the convention. Delayed samples are always taken to the right axis x(n-k). But in the convolution sum evaluation we take the image and flip it such that it falls under positive axis. This is not clear because as per definition , convolution sum= x*h(n-k). h(n-k) should be in positive axis only. Why flip it and then multiply accumulate?

Click to expand...

I am not sure what you mean. However, there is much material on this subject in textbooks and the internet that can illustrate and explain it better than I can.

Ratch

Perhaps the most intuitive explanation for convolution is actually finance. Discrete time is even better in fact. Assume that saved money gains interest at a rate of 5% per year. This gives an impulse response of h[n] = 1, 1.05, 1.1025, (1.05)^3, etc...

assume money is deposited in year 0, 1, and that money is withdrawn in year 2. then the amount of money in the account at year 10 is the sum of the money deposited plus the interest minus the amount withdrawn and the interest that would have accumulated.

The effect of year 0's deposit includes 10 years of interest, which is about 1.63 -- this is h[10]. The effect of the year 1 deposit has 9 years of interest, which is about 1.55 -- this is h[9]. The withdraw in year 2 removes some money and 8 years of interest that would have been had. This is about -1.477 -- which is -h[8].

notice that the effect of year n on year 10 is h[10-n], not h[10+n] or h[n-10].

edit: in terms of indexing, x[n] and h[n] can both be defined as having positive or negative indicies. This could occur in imaging where a negative index might represent pixels to the left and positive pixels to the right.