# Tag Info

3

You may solve it by 3 steps: Show yourself that a Cyclic Convolution with a vector $\boldsymbol{e}_{i}^{N}$ is a Cyclic Shift Operator ${T}_{i - 1} \left( \cdot \right)$. Where $\boldsymbol{e}_{i}^{N}$ is defined as a vector of length $N$ which all its elements is zero but the $i$- th element which is 1. Decompose $\phi$ into $3$ vectors of ...

2

I think what you are asking is: what happens of I implement the convolution of two sequences of length M & N, using spectral multiplication with an FFT of length K? We defined "correct" here as "matching the result of a linear convolution". For simplicity we define the correct length $L = M + N -1$ Here is what you get K > L: The ...

2

The choice of the normalization factor is just a matter of convention. Note that the specific correspondence between convolution in the time domain and multiplication in the frequency domain with a scaling of $\sqrt{2\pi}$, as shown in your question, applies only to the unitary definition of the Fourier transform with angular frequency as the independent ...

1

A matrix method is far simpler I suppose: h x x1 x2 x3 h1 ⌈x1*h1 x2*h1 x3*h1 ⌉ h2 |x1*h2 x2*h2 x3*h2| h3 ⌊x1*h3 x2*h3 x3*h3 ⌋ h x 3 2 1 1 ⌈3 2 1 ⌉ -2 |-6 -4 -2| 3 ⌊9 6 3 ⌋ Now just add up the diagonals: y[0] = 3 y[1] = -4 y[2] = 6 y[3] = 4 y[4] = 3 p.s I don't know how to type in a matrix here.

1

If it is difficult for you to remember or calculate the convolution of two sequences then you may try doing it as polynomial multiplication. Think of x[n] and h[n] as polynomial coefficients. So we have Px = 3x^2 + 2*x + 1 Ph = 1x^2 - 2*x + 3 Remember that linear convolution of two sequences is polynomial multiplication. Therefore Py = Px * Ph Py = (3x^2 ...

1

Your $h(t)$ is an ideal low pass filter with cutoff frequency of $75$ Hertz. At the output, everything above $75$ Hertz will be gone and equal to zero. If you're thinking about the inverse system, you need to think about getting back the original input. What can you say about the original input signal at frequencies greater than $75$ Hertz? Isn't it anybody'...

1

The operation aimed to restore a source signal from the result of the signal's convolution with a transfer function is called deconvolution. Your question concerns the rather innocuous kind of deconvolution problems, because the transfer function is given as an input. The transfer function being one of the unknowns, the problem is called a blind ...

1

For part 3, there is an old trick dating from the time where computing was expensive. If you have a smoothed version $I_\sigma$ of a picture $I$, you can get a crude yet simple approximation of a Laplacian from $I -I_\sigma$, illustrated here with the Gaussian: It is indeed generalized in many unsharp filters, with other weights, or in multirate/multiscale ...

1

Hint: the convolution is not circular in that case as we used DCT instead of DFT. Luckily, that's not true! At the receiver, you're doing a DFT, and you've prepared the transmit signal such that a convolution with the channel looks like a circular one by prepending the CP. The fact that the data inside has something to do with the DCT is irrelevant there. ...

Only top voted, non community-wiki answers of a minimum length are eligible