17 votes
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Why LTI system cannot generate new frequencies?

One of the definitive features of LTI systems is that they cannot generate any new frequencies which are not already present in their inputs. One way to see why this is so, comes by observing the ...
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13 votes
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What is the difference between natural response and zero input response?

First it's important to realize that many authors use the terms zero-input response and natural response as synonyms. This convention is used in the corresponding wikipedia article, and for instance ...
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12 votes

Are all LTI systems invertible? If not, what is a good counterexample?

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 ...
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12 votes

Are there any real world applications for complex-valued signals or impulse responses?

Absolutely! Conjugates are mentioned in textbooks because conjugation has no effect on real signals, but it does on complex ones. This way, formulations are more general and apply to both real and ...
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10 votes
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Proof of time-invariance of continuous-time system

Let $y_1(t)$ be the response to the signal $x_1(t)$: $$y_1(t)=x_1(-t)\tag{1}$$ Now let $x_2(t)$ be a shifted version of $x_1(t)$: $$x_2(t)=x_1(t-T)\tag{2}$$ The response to $x_2(t)$ is $$y_2(t)=...
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10 votes

Why LTI system cannot generate new frequencies?

You can make a simple algebraic argument, given the premise that you provided. If: $$ Y(\omega) = X(\omega) H(\omega) $$ where $X(\omega)$ is the spectrum of the input signal and $H(\omega$) is the ...
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Why $y[n] = x[-n]$ is not time-invariant?

A time-invariant system is one that, when you shift the input signal, the output is shifted by the same amount. A system that reverses the signal cannot be time-invariant because when you shift the ...
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9 votes

What is the relationship between poles and system stability?

I agree with Peter K.'s answer, but I would like to add one important point: the two statements in the question are only true for causal systems. The most general statement about stability for LTI ...
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9 votes

A system that perfoms Fourier Transform operation - is it an LTI system?

The Fourier transform operator $\mathscr{F}$ is a linear one; i.e., $$\mathscr{F}\{x(t)\}=X(f) ~,~ \mathscr{F}\{y(t)\}=Y(f) \implies \mathscr{F}\{\alpha x(t) + \beta y(t) \} = \alpha X(f) + \beta Y(...
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8 votes

Do Causal Discrete-time systems have proper transfer functions?

Note that a stable and causal continuous-time transfer function does not need to be strictly proper but only proper, i.e. the degree of the numerator does not exceed the degree of the denominator, but ...
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8 votes
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How to determine if the system is invertible

It depends on what exactly you mean by "invertible". In system theory, what is often meant is if there is a causal and stable system that can invert a given system, because otherwise there might be an ...
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8 votes
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Discrete State Space Model - Why Are We Calculating $ x \left[ k + 1 \right] $ Instead of $ \dot{\boldsymbol{x}} \left( t \right) $?

I will ask you something that will give you intuition. How would you calculate the Gradient of an image? Image is a discretization of reality, so how would you estimate the gradient of the "Reality" ...
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8 votes

Is the first derivative operation on a signal a causal system?

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)=\...
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Are all LTI systems invertible? If not, what is a good counterexample?

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 ...
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8 votes
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Estimating the Signal by Deconvolution with a Prior on the Filter Coefficients and the Signal Samples

I would take approach based on Blind Deconvolution. Since we're dealing with ill posed problem some assumptions should be made. The intuitive approach would be using the information as a prior for ...
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7 votes
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find impulse response from step response

You can find the impulse response Let's take the case of a discrete system. If $s[n]$ is the unit step response of the system, we can write $$s[n]= u[n]\ast h[n]$$ where $h[n]$ is the impulse ...
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7 votes
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How to obtain impulse response from the differential equation of a system?

It looks like your transfer function is correct, but there's a small mistake in your partial fraction expansion: $$H(s)=\frac{2}{(s+4)(s+2)}=\frac{1}{s+2}-\frac{1}{s+4}\tag{1}$$ The corresponding ...
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7 votes
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Why does reversing the order of these two transfer functions give me different outputs?

If the input is a unit step, then the output of the first block in system 1 is not zero, but it is a Dirac delta impulse $\delta(t)$. Intuitively, the derivative is infinite at $t=0$ because of the ...
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7 votes
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why exponential term neglected in equation?

The magnitude of that complex exponential is 1. Recall from complex algebra: any complex number can be expressed as $z = r e^{j \phi}$ where $|z|=r$ is its magnitude and $\arg z = \phi$ is the ...
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Why is the impulse response function of this system 0?

This system $$ y(t) = t^2 x(t) $$ is not LTI and therefore does not have an impulse response of the form $h(t) = \mathcal{T}\{\delta(t)\}$. So your statement $h(t) = t^2 \delta(t)$ is not correct......
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Negative group delay and envelope advance

Answer : No, any causal LTI system with frequency response $H(f)$ cannot produce the output $y(t)$ in advance. And, the answer lies in the causality of input signal $x(t)$ being applied to $h(t)$. Any ...
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Are the output functions of a continuous-time LTI system necessarily continuous (in the calculus sense) for any given input functions?

Consider the identity system $y(t) = x(t)$. This system is LTI. If the input $x(t)$ is discontinuous, then the output $y(t)$ will be discontinuous too.
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7 votes

Are the output functions of a continuous-time LTI system necessarily continuous (in the calculus sense) for any given input functions?

To add an even worse example to MBaz (best possible) counterexample: The derivative $\frac{\mathrm d}{\mathrm dt}$ is an LTI system. $f(t)=|t|$ is a continuous function. $\frac{\mathrm d}{\mathrm dt}(|...
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6 votes

What Are Some Examples of Applications of LTI Systems

The normal convolution. The only difference is that all the frequency domain analysis is made under the assumption of periodic signal. The Matched Filter used for ...
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6 votes
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Ideas on Matrix Factorization / Transformations for $ {L}_{1} $ Minimization

Since $ \epsilon $ is a parameter you need to set, why not trade it with another parameter you need to set to create an easily solvable problem (Relaxation of the Problem)? You can transform the ...
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Conceptual Question from Signal Processing - Impulse Response and AR Coefficients

Impulse Response is basically the FIR coefficients of the system. Namely, a system $ H $ with an impulse response given by $ f [n] $ and a Filter $ F $ with an FIR representation of $ {f[0], f[1], \...
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What's the Difference between Convolution Kernel and Point Spread Function in the Context of Image Convolution?

The PSF (Point Spread Function) is the system response to Impulse Signal (Point). If your system model is LSI (Linear Spatially Invariant) then the output image of the system is applying the PSF as a ...
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6 votes
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Why is the output of an LTI system expressed as the convolution of the input with its impulse response?

it takes more work to do this for continuous-time systems, so i'll just do it for discrete-time and we can all wave our arms to extend to continuous-time. a general discrete-time input signal can be ...
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6 votes

System response: LTI system for $x[n] = \sin(\frac{\pi n}{4})$

You're definitely on the right track. The way you're trying to solve the problem is the best and simplest. You just need to realize that you need to evaluate the magnitude and phase of the frequency ...
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6 votes
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Initial conditions for the LTI systems described as a difference equations

The problem is that non-zero initial conditions cause a term in the output signal that does not depend on the input signal. This explains why a system with non-zero initial conditions can neither be ...
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