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I am trying to make proper use of the circular convolution property of DFT. I was taught that the DFT of x[n]*CircularConv*y[n], would be equal the product of the individual DFT's X[k],Y[k].

On the problem im trying to solve, the signal x[n] is convolved (Circular convolution) with the discrete impulse response y[n] to produce the output signal z[n]. (x[n]*y[n]=z[n]) Having the signals z[n],y[n]+their DFT's, and using the property mentioned above, can I conclude that X[k]=Z[k]/Y[k] immediately? are there any limitations? or am I doing it totally wrong?

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  • $\begingroup$ When you are using DFTs to find the response of an actual system, you need to be sure that the result of $$\text{"take DFTs, multiply pointwise, take inverse DFT"}$$ which gives the circular or periodic convolution of $x[n]$ and $y[n]$ actually computes the linear or aperiodic convolution of $x[n]$ and $y[n]$ that the system will give you. For example, if the system is an IIR filter, your DFT method might not work. $\endgroup$ Commented Jun 21, 2013 at 12:54

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Yes, that conclusion would be valid. However, as you suspected, there are some limitations. Consider the case where the system's frequency response $Y[k]$ contains one or more zeros. In that case, corresponding frequency bin in your estimate of the input signal $X[k]$ would diverge to infinity (because of the division by zero).

What you're really trying to perform by doing this sort of calculation is deconvolution, which is a complex topic in itself, but suffice it to say that deconvolution is often difficult; you illustrated one of the reasons why. This naive approach of just dividing in the frequency domain is known as a zero-forcing type of algorithm.

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  • $\begingroup$ Thanks for the quick answer! is there any proper way to do it, without taking this risk? or, how does this deconvolution work practically? anyhow, i'm trying to use the division result (input of the convolution), and seperate it to 2 analog sampled audio signals (one of them multiplied by a cosine of 8Khz). the result of the division looks that way [IMG]i40.tinypic.com/2mwe4nt.jpg[/IMG] [IMG]i41.tinypic.com/30tt5y0.jpg[/IMG] first of all, i'm trying to filter the signal which is the domain 0-3 Khz. i'm afraid my division 'damaged' the signal. every filter i tried produced noise $\endgroup$ Commented Jun 21, 2013 at 7:53
  • $\begingroup$ {peroper links: i40.tinypic.com/2mwe4nt.jpg i41.tinypic.com/30tt5y0.jpg } $\endgroup$ Commented Jun 21, 2013 at 7:54

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