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8

It's not sufficient to only consider causality, you also need to check whether the inverse system is stable, otherwise it can't be implemented. If $G(z)$ has zeros on the unit circle, it cannot be inverted. If $G(z)$ has no zeros on the unit circle, but if there are zeros outside the unit circle, then there is no causal and stable inverse, because the zeros ...

3

The only thing that went wrong is your last statement: They're only zero if $k\ne 0$. Thus, you get three Dirac delta distributions. Let's look at your last equation, and work with that: \begin{align} x(n) &= \frac{1}{2\pi}\int\limits_{0}^{2\pi} \frac{e^{j\omega(n+2)} + e^{j\omega(n-2)} + 2e^{jn\omega}}{4}\,d\omega \tag1\label{orig}\\ &=\... 3 The system is nonlinear (bilinear in x), with a nonlinear law (square) on indices. Odds are the system is not invertible. One can try to prove it in its full generally, or try to find a counterexample. Let us try the lazy way, using properties at hand. The system is bilinear. Hence, the output for k x[n], k\in \mathbb{R} , will be k^2 y[n]. ... 2 It seems that we have the same homework. You probably are Greek. After having the same question, I came to the conclusion that in order for this system to be invertible, you have to prove that for any givenx_1(t), \, x_2(t)$$the following sentence is correct:$$ x_1(t) \ne x_2(t) \,\, {\Rightarrow} \,\, y_1(t) \ne y_2(t) $$As like you proved when a ... 2 What you haven't understood is the fact that$$\frac{1}{2\pi}\int_0^{2\pi}e^{j\omega k}d\omega=\delta[k]\tag{1}$$where \delta[k]=1 for k=0 and zero otherwise. Note that in (1) with k=0 you get$$\frac{1}{2\pi}\int_0^{2\pi}1\cdot d\omega=1$$2 In this problem of finding the inverse system (if it exists) its intuitive to try differentiating the integral as the system input/output is given by:$$y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{3t} x(\tau) d\tau$$Before differentiating te integral however, I would like to make this little change which is quite clear I assume:$$y(t/3) = \mathcal{T}\{x(...

2

This system is not invertible and a single counter-example is sufficient to prove it. First express the signal $x[n^2]$ as $w[n]$ and then the output is $$y[n]= x[n] \star w[n]$$ Then let an example input be $x[n] = \delta[n-3]$. Then you can verify easily that $w[n] = 0$ for all $n$. Hence the output $y[n]$ for the input $x[n]=\delta[n-3]$ is $$y[n] = x[... 1 Here is an example using Matlab/Octave code. There is no need to conjugate. x = rand(1000,1); y = rand(1000,1); z = conv(x,y,'full'); y2 = ifft( fft(z,1999) ./ fft(x,1999) ); norm(y - y2(1:1000)) / norm(y) ans = 1.5654e-13 1 Two bugs in your code. pol2cart() returns the real and imaginary part, but you only use the real part and discard the imaginary part You apply ifftshift() on one method but not the other Try instead [a,b] = pol2cart(artPhase, artMag); artInv = ifft(a+1i*b); % Time-domain signal plot(artInv-invSig); 1 It's not possible to recover the original signal X based on h perfectly !! Filter or conv are usually performed in time domain, so it's similar "But not exactly" to y = conv(X,h) , So in that case after you know h or in other words after you estimate h , you need to use any equalizer like ZF or MMSE to get back X. Of course, ... 1 In general, what you want is impossible! Not every operation is reversible. In frequency domain, that becomes obvious: a system's frequency response H(f) might have actual zeros at some points, and thus, no inverse filter exists. Multiplication with zero can't be undone, since no matter what was at that frequency before filtering, it's 0 afterwards and we ... 1 As pointed out in the comments, the inverse system is not realizable. However, it is possible to write down an expression for the impulse response of the inverse system. We want a system that results in a Dirac impulse when convolved with the given rectangular pulse h(t). This can be done as follows. We delay the rectangular impulse by T/2, so it starts ... 1 Note that the DFT transforms a periodic sequence into a periodic sequence. Hence, the result of your IFFT is intrinsically periodic and you should view it as such. The parts you are seeing at the end of it are thus just as well behind the start as they are before it. Hence, this part may as well be some form of pre-ringing, stemming from a nonzero group ... 1 Regardless of causality and stability, if you count poles and zeros at the origin and at infinity, the total number of poles always equals the total number of zeros. I'll show a few examples to make this obvious. First, take a polynomial$$P(z)=(z-z_0)(z-z_1)\ldots (z-z_M)\tag{1}$$It has M+1 zeros and no (finite) poles. However, it must have M+1 ... 1 If a discrete-time LTI system is causal (that is that the impulse response doesn't respond before the driving impulse), then the number of zeros must not exceed the number of poles. Consider this general transfer function:$$\begin{align} H(z) &= A\frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_M)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \\ \\ &= A\frac{z^M + b_1z^{...

1

I don't know if I understand exactly what you mean, because your question is very unclear (you may need to edit it), but I will try to answer what you may need in your task. For DFT you will need this equation: $X[k] = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j 2 \pi nk/N}$ where $k = 0, 1, . . . , N − 1$ But I guess you will be asking for IDFT more, so ...

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