If we're given an x(n) signal say x(n) = sin(2pix) and a filter (say a two point moving average filter) would we be able to use the filter beforehand and then dft to end up with the same result?
In order to use the DFT to implement convolution with an infinite length (or very long) input signal, we have to do the processing in blocks, by breaking the input up into sections. L is the symbol for the length of the section. Ultimately the DFT length N will be related to the block length L. Choosing L smaller means smaller DFTs, but more of them for a given input signal duration. Choosing L larger means large DFTs, but fewer for a given input signal duration. Leaving L as a variable allows you to choose the block length that works best for your application constraints.
![Fast Convolution: FFT-based, Overlap-Add, Overlap-Save, Partitioned [DSP #09]](/img/1.gif)
thank you!
thank you very much ..
If we're given an x(n) signal say x(n) = sin(2pix) and a filter (say a two point moving average filter) would we be able to use the filter beforehand and then dft to end up with the same result?
thanks for your amazing video
but in slide 2 why is the length of xr[n] = L ??
thanks in Advance
In order to use the DFT to implement convolution with an infinite length (or very long) input signal, we have to do the processing in blocks, by breaking the input up into sections. L is the symbol for the length of the section. Ultimately the DFT length N will be related to the block length L. Choosing L smaller means smaller DFTs, but more of them for a given input signal duration. Choosing L larger means large DFTs, but fewer for a given input signal duration. Leaving L as a variable allows you to choose the block length that works best for your application constraints.