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Considering an example, where the model is \begin{align} x[n] = \mathbf{h}^\mathsf{T}\mathbf{u}[n] + w[n]. \label{Eq1} \end{align} $\mathbf{h} = [h_0,h_1,\ldots,h_{M-1}]^\mathsf{T}$ of length $M$ which represents the coefficients of the LTI SISO channel and $\mathbf{u}[n] = [u[n], u[n-1], u[n-2],\ldots,u[n-M+1]$ is the input to the channel. Let $\mathbf{v} = [v_0,v_1,\ldots,v_{M-1}]^\mathsf{T}$ of length $M$ be the coefficients of the equalizer.

If the length of the equalizer is different, say the channel length is $M$ and equalizer length is $Q=2M-1$ then how would the convolution happen since there will be a mismatch in the number of elements of $\mathbf{h}$ and $\mathbf{v}$

I can do conv(h,w) in Matlab where h is the coefficients of the channel and w denotes the coefficients of the equalizer. Assuming known values of h = [1,0.2,0.6] and the estimated coefficients of the equalizer as `w = [0.8,0.1,0.2,0.3,0.5]'. How would the convolution work? Can somebody please show first few steps. The first few sample will be zero based on my calculation. But i am not sure.

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    $\begingroup$ Your question is not clear. First what is the difference between $w[n]$ and your model and your equalizer w ? I suppose you meant $v$ as equalizer ? Second, what is the input and output of your equalizer ? $\endgroup$ – AlexTP Nov 19 '17 at 19:43
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The discrete convolution sum operation is not restricted to equal length vectors. You can, and most of the time you do, convolve two different signals of arbitary lengths. Your confusion is probably with something else. The equalizer length can be different than that of the channel model length. That should not pose a problem but it would of course effect the equalization performance.

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