# Tag Info

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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 ...

5

What you are looking for are what we, in the audio space, call comb filters. Comb filters may or may not have a feedback path, just like FIR and IIR filters. In fact, there is a generalized theory about designing comb filters based simply on designing any digital filter. In fact, you can think of every LTI digital filter as being a comb filter where the ...

4

The Invertible FIR Filter A constraint based on the first coefficient alone is developed as follows: From Cauchy's argument principle any FIR filter that meets the following constraint will be invertible (including marginal invertibility, change $\le$ to $<$ otherwise): $$\max\left(\arg \left( H(e^{j\omega}) \right)\right)-\min\left(\arg \left( H(e^{j\... 4 Dan's answer -- to compute H(s) as normal, and then compute 1/H(s) -- is equivalent to your suggestion of swapping the x and y (or doing it in one step by solving for H_{yx}(s) = \frac{X(s)}{Y(s)}). In the Laplace domain it's justified by noting that$$\frac{H(s)}{H(s)} = H(s)H_{yx}(s) = 1$$In theory this means that a system followed by its ... 4 Doing direct division in Frequency Domain means you are assuming cyclic / periodic boundary conditions for your data which I don't think is the proper assumption for your data. The model I'd pursue would be as I described in my answer to Deconvolution of Synthetic 1D Signals - How To? As one can see in that question the marked answer by @Hilmar is wrong. ... 3 A general approach would be to take the Laplace Transform of the equation and put it in form of a transfer function:$$H(s) = \frac{Y(s)}{X(s)}$$And then invert that and solve for the unit step response from 1/H(s) Note that the inverse Laplace transform of 1/H(s) would give the impulse response, since that implies an impulse is at the input to the ... 3 The two methods differ, above all, by their applicability to matrix classes. col (cholesky) decomposes Hermitian, positive-definite rectangular matrices into the product of a lower triangular matrix and its conjugate transpose; svd (singular value decomposition) factorizes any m×n matrix into the form UΣV*, where U and V are square real or compex unitary ... 3 It is not. It suffices to find a counter example. Let me tell you that "I am flat". Could you derive my actual value? So, any constant function f(t) =c is differentiable, and have the same derivative, f'(t)=0. Only from knowing that f'(t)=0, you cannot recover the original (constant) function. Hence the system is not invertible. 3 If the OP is actually interested in selecting only one individual frequency from the even or odd harmonics, then a moving average filter (MAF) would be ideal since this can provide a null at every other frequency when the frequencies are harmonically related. The low-pass MAF with all the coefficients as 1 will pass DC at f=0 and provide nulls spaced by 1/... 3 The only thing that went wrong is your last statement: They're only zero if k\ne 0. Thus, you get three Dirac delta distributions. Let's look at your last equation, and work with that:$$ \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}\\ &=\...

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 $$y[n] = x[... 2 The system is nonlinear (bilinear in x), with a nonlinear law (square) on indices. Odds are the system is not invertible. One can try to prove it in its full generally, or try to find a counterexample. Let us try the lazy way, using properties at hand. The system is bilinear. Hence, the output for k x[n], k\in \mathbb{R} , will be k^2 y[n]. ... 2 It seems that we have the same homework. You probably are Greek. After having the same question, I came to the conclusion that in order for this system to be invertible, you have to prove that for any given$$x_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) $$As like you proved when a ... 2 In this problem of finding the inverse system (if it exists) its intuitive to try differentiating the integral as the system input/output is given by:$$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

The system is not invertible because you can always add an arbitrary constant $c$ to any function $x(t)$ and the system will map it to same differentiated function $y(t)$. So, the mapping is not unique or one-to-one and hence not invertible.

2

You can use a window as an FIR filter (convolve with window in time), such that the zero crossings of the window transform are at $\omega_0 = 2\pi * 2k*200$ to filter the even harmonics, and phase shift the window by transform $\frac{\pi}{2}$ to filter the odd harmonics. For example, a rectangular window of length $M$ will have zero crossings at $\omega = \... 2 You have the obvious relationship$\omega=2\pi f$. So going from a Fourier transform expressed in terms of$\omega$to a Fourier transform expressed in terms of$f$is just a matter of substitution. The problem you've encountered is very likely in the argument of the Dirac delta impulse. Clearly,$\delta(\omega)=\delta(2\pi f)$, but you need to know the ... 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 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 Yes, your understanding is correct. The differentiation is not an invertible system. And, the reason is precisely, because of the possible unknown constant which gets nulled by differentiation. Basically, the differentiation of$f(t)$and$(f(t) + c)$is indistinguishable$f^{'}(t), which makes the system, many-to-one, and hence non-invertible. 1 If a discrete-time LTI system is causal (that is that the impulse response doesn't respond before the driving impulse), then the number of zeros must not exceed the number of poles. Consider this general transfer function: \begin{align} H(z) &= A\frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_M)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \\ \\ &= A\frac{z^M + b_1z^{... 1 I don't know if I understand exactly what you mean, because your question is very unclear (you may need to edit it), but I will try to answer what you may need in your task. For DFT you will need this equation: X[k] = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j 2 \pi nk/N} where k = 0, 1, . . . , N − 1 But I guess you will be asking for IDFT more, so ... 1 What you have as the magnitude response is |Y(e^{j\omega})| = \left| \omega \right|, for \omega \in [-1,1] and phase response is \angle{Y(e^{j\omega})} = -3\omega. So, Y(e^{j\omega}) = |Y(e^{j\omega})|. e^{j\angle{Y(e^{j\omega})} }. 1 You can express a complex signal y as |y|.e^{i \phi_y} where |y| is the magnitude and \phi_y is the phase. You can use Inverse the Fourier transform formula:\displaystyle y(t) = \frac{1}{2\pi}\int_{-\infty}^\infty Y(\omega) e^{j\omega t} d\omega$You can express$Y(\omega)$in$|y|.e^{i \phi_y}$form and then perform the IFFT. But I am not sure ... 1 The FFT and IFFT are almost identical operation: other than scaling (depending on convention) and complex conjugation they are the same. Ignoring the scaling (or assuming$1/\sqrt N$). We have Apply one time: spectrum Apply second time: time inverse original Apply third time: conjugate spectrum Apply 4th time: original signal For inverse inverse FFT: ... 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 ...

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