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3

It will match for circular convolution modulo $N$, where $N$ is 4 here. For finite length sequences product of DFT of 2 sequences is equivalent to DFT of circular-convolution of the 2 sequences. >> cconv([0,1,2,1], [0,1,-1,1], 4) ans = 0 1 2 1 >>ifft(fft([0,1,2,1]).*fft([0,1,-1,1])) ans = 0 1 2 1

2

I am very thankful to Dilip-sarwate and Gilles, who took their precious time to understand my problem and guide me. So, Now I'm going to write the correct solution to my question. Which is as follows: $$ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}[x(t)g(t)]e^{-j\omega t}dt$$ As we know : $$x(t) = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha ... 2 If x[n] and y[n] are both causal and starting at index 0, then the result of convolution will also be causal and it will start at index 0. Just plug in n=-4 in the expression for z[n], you will find that it will be 0.$$z[n] = \sum^{\infty}_{k=-\infty}x[k]y[n-k]$$First non-zero term in above expression is at n=0. Because for k<0, x[k] ... 2 This symbol denotes a rectangular pulse of length N:$$\mathcal{R}_N(n)=\begin{cases}1,&0\le n\le N-1\\0,&\textrm{otherwise}\end{cases}$$I'm not sure where it is defined for the first time, but this definition is clear from the equation above Eq. (5.24) on page 130 of the 3rd edition of Digital Signal Processing Using Matlab by V.K. Ingle and ... 2 The variable NI in the book is just a typo, it should be N instead (not N*L as in your code). Apart from that, remember that book was written about 25 years ago, and the code was run on a 33 MHz 486 PC. So in order to see some effects on today's computers, you should crank up the value of L. I've modified the code a bit (see below). Now the figure ... 1 This problem is related to deconvolution and equalization. You are basically undoing the effect of a filter by another filter, such that the total system has a flat response, i.e., has a unit impulse as its impulse response. From$$(h\star g)[n]=\delta[n]\tag{1}$$it follows that$$H(z)G(z)=1\tag{2}$$must be satisfied. So the solution to the problem is$$...

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If your delayed input is $\hat{x}(t)=x(t-\alpha)$, then $\hat{x}(t-\tau)=x(t-\tau-\alpha)$, and not $x(t-\tau+\alpha)$. From here everything will work out as expected.

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Bilinear interpolation can be viewed as traditional sample rate conversion (insert 0s between existing samples, convolve with a triangular kernel in each dimension). Somewhat unusually, the size of the (continous-time) triangular filter kernel is not scaled relative to resampling ratio before it is sampled; each sample is always the result of 4 neighbouring ...

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