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3

If you use the Euler's formula, you can simplify like this: $$[d]_{k,n} = \frac{\sqrt{2}}{N}\left( \cos{\left[ \frac{(k-1)(2n-1)\pi}{2N} \right]} e^{j\frac{2 \pi nk}{N}} \right)$$ I think we can't simplify more. PS: If you use your expressions of $A$ and $B$ and again the Euler's formula, you will get the same result.

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The cosine and sine terms can be combined to a single exponential function using: $$\cos(\theta) + j\sin(\theta) = e^{j\theta}$$

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No. Your problem doesn't put any constraints on the columns, so they're totally independent. Then, transforming them to a different base doesn't change that, at all. If the entries in different columns had nothing to do with other columns, that won't change. The fact that you arranged your data in a matrix doesn't magically allow any operation to introduce ...

2

Observe: $$\cos( \alpha ) e^{j \beta} = \frac{e^{j \alpha}+e^{-j\alpha}}{2} \cdot e^{j \beta} =\frac{1}{2} \left( e^{j (\beta + \alpha)}+e^{j(\beta - \alpha)} \right)$$ Where: $$\alpha = (k-1)(2n-1) \cdot \frac{\pi}{2N} = \frac{ 2nk - 2n - k + 1}{4} \cdot \frac{2\pi}{N}$$ $$\beta = nk \cdot \frac{2\pi}{N}$$ Plug this in and you can transform ...

1

TL;DR: Time shift in doable with LTI systems. Subtracting $1$ is non linear. In discrete systems, integer shifts can be implemented with filters that are zero everywhere, except at the shift location. This can be implemented via circulant matrices, akin to convolution. This can be made compatible with Operation 1 (taking border effects into accounts). The ...

1

can we solve x2 and x1 if y2 and y2 are known? Yes, provided that you mean $y_1$ and $y_2$ are known. If you have two linear equations your can (in most cases) solve for two unknown variables. If not, can we get the value of y2 based on the equation (1), No. $y_2$ doesn't show up in equation (1) at all, so you can't get it from there. which is very ...

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You can use, from Euler's formula, the fact that $$\exp\left(j\theta\right) = \cos\theta + j\sin\theta \qquad \text{with}\qquad j = \sqrt{-1}$$ and combine the sine and cosine into exponential terms. You basically have the first exponential inside the square bracket equal to $$\exp\big[j\left(6\pi nk + \pi -k\pi - 2\pi n \right)\big] = \exp\bigg\{j\big[... 1 The basis vectors are orthonormal (very nice!, half the job done), Now project/take inner product of the basis vectors with \mathbf{y} to get the expansion coefficients. In general for an orthnormal basis any \mathbf{y} in \mathbb R^N can be written as$$ \mathbf{y} = \sum_{k=0}^{k=N-1}a_k\mathbf{v_k} where $a_k = \mathbf{y}^T\mathbf{v_k}$, in this ...

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You need to find projection of $\vec{y}$ along unit vectors in the direction of each of the basis vectors $\mathbf v^{(i)}$. For finding unit vector in the direction of the vector, you just divide the vector by its magnitude. And, for finding projection along a unit-vector, you just take the dot-product with the unit-vector. So, the combined steps boil ...

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Y and Z should show the same values clear; %Create random vector x x = randn(32,8); %take ifft along each column y = ifft(x); Y = reshape(y,[],1); z = reshape(x,[],1); % d=conj(dftmtx(32))/32; X = kron(eye(8),d); Z = X*z;

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What you mean might be circulant matrix instead of toeplitz matrix. See section 3.4.4 in https://web.stanford.edu/~dntse/Chapters_PDF/Fundamentals_Wireless_Communication_chapter3.pdf about how the circular convolution in OFDM is represented by matrix operations (eq 3.130 onwards). First, in almost all standard OFDM systems, you can assume $D \le L$. The ...

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The assertion "almost always seem to be expressed in integers?" does not seem to be true, in my opinion. However, such kernels are pretty frequent in codes. They are quantized both in support (limited discrete support) and amplitude (signed integer values). $3\times 3$ masks with integer coefficients, as you showed, are very familiar, albethey ...

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The Laplacian kernel with the 4 in the middle results from summing second derivatives along the two axes ([1,-2,1]). Those are the right values to use, you can show this by writing out the math for the second derivative and set the distance h to 1 (or search for discrete approximation to derivative). This kernel hasnâ€™t been rounded, the values just happen to ...

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Marcus' answer is perfect! For any wireless channel, y = Hx + n, where: y = received signal x = transmitted signal n = noise in the channel For an AWGN channel, output 'y' = input 'x' plus 'noise' (AWGN) i.e. y = x + n Therefore, for AWGN channel, channel matrix or H-matrix is the identity matrix.

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