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You need to define what you mean by "invertible". Do you mean invertible by a causal and stable system? If yes, then any system that is not minimum-phase is not invertible (because the inverse system can't be causal and stable). Example of a system that cannot be inverted by a causal and stable system: a simple delay $y(t)=x(t-T)$, $T>0$, could only be ...


7

[EDIT] A necessary condition for invertibility is that any output has only one possible input (or injectivity, as proposed in comments). Since we are looking at counterexamples, we can look at when this condition is not satisfied. The null system, that turns every signal into a zero flat line, is not invertible, but a bit trivial. A system that computes a ...


6

No. It's only LTI (Linear and Time-Invariant) systems that can be modeled with convolution through a unique single impulse response. For example the systems $$ y(t) = g(t) x(t) $$ or $$ y[n] = \sum_{k=0}^{k < n} x[n-k] $$ are both linear but not time-invariant and their output $y[n]$ cannot be computed with the convolution operation ( $\star$ denoting ...


5

It's possible, in mathematics, to complete real numbers with "infinite" values, with sound topological properties; for instance non-standard analysis or the Extended real number line (discussion at Math StackExchange). However, in this context, for any standard real number $t$, the rule is to set $t\pm \infty = \pm \infty$ (see Arithmetic operations). So, ...


5

If the derivative exists at the given point, then it doesn't matter if you look (infinitesimally) into the future or into the past, you can do both, because both will give the same result: $$x'(t)=\lim_{h\to 0}\frac{x(t+h)-x(t)}{h}=\lim_{h\to 0}\frac{x(t)-x(t-h)}{h}\tag{1}$$ So a differentiator can be (theoretically) implemented by a causal system. ...


5

the Fourier Series for a single periodic function, having period $P$, $$ x(t) = x(t+P) \qquad \forall \ -\infty < t < +\infty $$ is $$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{j(2\pi k/P)t} $$ each harmonic $c_k$ is labelled with its harmonic number, $k$, in the above. the actual frequency of each harmonic, in Hz or units of 1/time, is $\...


5

You cannot sum $X_1[k]$ and $X_2[k]$ (the discrete spectra of the two signals) element-wise because for them $k$ represents different angular frequencies $\frac{k\,2\pi}{P_1}$ and $\frac{k\,2\pi}{P_2}$ in radians. Time domain addition will only translate to frequency domain addition if your frequency variable is equally proportional to angular frequencies ...


4

For the case of input process $\{X(t)\}$ being white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$, the output process $\{Y(t)\}$ is a strictly stationary zero-mean Gaussian process in which all the random variables have the same variance $\frac{N_0}{2}\int_{-\infty}^\infty |H(f)|^2 \,\mathrm df$ almost as you say. But the key point ...


4

The IIR filter doesn't have to be unstable, but it has the potential of being so; unlike the FIR case which doesn't have even the potential. One reason for the (potential) unstability of an IIR (adaptive) filter is the numerical issues due to coefficient quantization. When the poles are closer to unit circle this will be critical. This is especially ...


4

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle. Even within the case of the FIR filters - which are unconditionally stable- we ...


4

No; the system given by $$ y[n] = x[n] + n $$ is time-varying, due to the added term $n$. Your mistake is in the line : $$\ y_2[n] = x_2[n] + n = x[n-k]+(n-k) $$ which should be instead $$\ y_2[n] = x_2[n] + n = x[n-k]+n$$ and therefore implies that $$y_2[n] \neq y[n-k] $$.


3

This is a good question and something that I remember asking myself when I first learned about impulse responses and convolution. To understand this, it is first necessary to understand the significance of impulses and impulse responses. Referring to the image below, you can see that an impulse is an instantaneous like input and the impulse response is the ...


3

To expand on Justme's reply: From the looks of your code you seem to have confused what constitutes an LTI system. In your code the LTI system that I see (y1=2*z) scales an undelayed input signal by 2. The additional 2*cos(pi/t) is NOT part of the system, it is in effect just another sinusoid that you have added to your signal - thus giving it an ...


3

Focus on the first equation for EY. Back in the day when color television was being developed, the color signal had to be compatible with black and white TVs and vice versa. So the compatible brightness signal (luma Y) has to be calculated from the three primary color signals (R, G B) for transmission. Human visual system does not perceive brightnesses of ...


3

In general LTI System is invertible if it has neither zeros nor poles in the Fourier Domain (Its spectrum). The way to prove it is to calculate the Fourier Transform of its Impulse Response. The intuition is simple, if it has no zeros in the frequency domain one could calculate its inverse (Element wise inverse) in the frequency domain. Few remarks for the ...


3

(Adapted from this answer on dsp.SE) The reason that the impulse response (also called the unit pulse response for discrete-time systems) determines the output for arbitrary input $x$ to an LTI system is that The output of a linear time-invariant system in response to input $x$ is the sum of scaled and time-delayed versions of the impulse response. ...


3

Yes your computation is correct. Indeed, you could have seen this even easier, had you considered that the output computation sum was $$ y[n] = T\{x[n]\} = x[n-1] + x[n] + x[n+1] + x[n+2] .$$ And this is clearly an LTI (linear time-invariant), FIR (finite impulse response) system with associated impulse response $$h[n] = \delta[n-1] + \delta[n] + \delta[...


3

Any LTI system can be completely characterized (among other things) by it's transfer function or it's impulse response. If your filter represents an LTI system, that you can calculate it's output by either convolving the input with the impulse response or multiplying the transfer function with the spectrum of the input signal. In theory these things are ...


3

Whether LTI or not all systems are invertible if unique (distinct) inputs produce unique (distinct) outputs Causality and stability are later concerns for making sense of the obtained inverse system. For example the inverse to the delay system $$y[n] = x[n-d] $$ is $$y[n] = x[n+d] $$ Which is clearly noncausal for $d > 0$, and is not ...


3

In addition to all the answers that are correct in a mathematical sense, in a practical sense, a system whose frequency response goes below some finite but small-enough value will not be usefully invertable, even if a simple mathematical analysis would suggest that it is. In frequency-domain terms, the frequency response of a system's inverse will have gain ...


3

For a parallel system, the individual subsystems just add, so you simply have $H_{||} = H_A + H_B$ . That goes for both the transfer function and the impulse response, so you can simply do $h_b = h_{||} - h_a$


2

The notation $ \hat{y} \left( k \mid k - 1 \right) $ usually means this is an estimated value of $ y \left( k \right) $ given all the available data up to time index $ k - 1 $. So generally speaking, this is a prediction of one step in time of the data. The case above also suggests linear estimation. Namely, $ \hat{y} \left( k \right) $ is built using ...


2

You can use the Convolution Theorem to show that $ \int_{-\infty}^{\infty} h \left( \tau \right) x \left( t - \tau \right) d \tau $ is equivalent to Element Wise multiplication in the Frequency Domain. Hence for the case $ x \left( t \right) = A {e}^{s t} $ where $ s \in \mathbb{C} $ the output is just a multiplication by scalar of the input (The value of ...


2

No the converse is not true in general. Take for example the discrete-time ideal lowpass filter with impulse response $$ h[n] = \frac{ \sin( \omega_c n) }{ \pi n } ~~~,~~~-\infty < n <\infty$$ which describes an LTI system but it does not correspond to a difference equation of any kind. Indeed $h[n]$ is derived based on the inverse discrete-time ...


2

Convolution is equivalent to calculating an output pixel $y[i,j]$ as a weighted sum of the nearby input pixels $x[i+k,j+l]$, with the weight being a function of the relative spatial location $(k,l)$. In bilateral filtering the formula for an output pixel has a part $f(|x[i+k,j+l] - x[i,j]|)$ which is a non-linear function of pixel values. This makes ...


2

Just write down the convolution sum to see what's going on: $$y[n]=\sum_{k=-\infty}^{\infty}x[k]p[n-k]\tag{1}$$ where we define the elements of the sequences $x[n]$ and $p[n]$ as equal to zero for the index $n$ outside the range of non-zero values, i.e., outside $n\in[0,2]$. Taking into account those zero values, we can rewrite $(1)$ with finite summation ...


2

Certainly possible. You can think of it in the way you described but I would say it just becomes an adaptive thresholding assuming a bimodal distribution: We admit that there are two main modes in the data and we try to identify those by grouping the values around the means of the clusters. These groups form clusters. The means are positioned such that each ...


2

You cannot solve this problem using the Laplace transform. The reason is that the Laplace transform of the input signal doesn't exist. You could use the Fourier transform, but in this case there's an even simpler way to determine the output signal. You need to know one important property of linear time-invariant (LTI) systems: their response to a sinusoidal ...


2

Also consider the somewhat simpler "identity system", given a continuous signal $x(t)$: $$\begin{align}y(t) &= x(t)\tag{1}\\ &= \lim_{\Delta t\to0^-}x(t + \Delta t)\tag{2}\\ &= \lim_{\Delta t\to0^+}x(t + \Delta t)\tag{3}\\ &= \lim_{\Delta t\to0}\frac{x(t - \Delta t) + x(t+\Delta t)}{2}.\tag{4}\end{align}$$ This might be interpreted as the ...


2

You have to be clear what you mean by "invertible". Commonly, you want the inverse system to be causal and stable, and that puts certain restrictions on the original system. In the case of systems with rational transfer functions, you just have to look at the zeros of the transfer function, because they become the poles of the inverse system. If all zeros ...


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