17
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 output's Fourier transform $Y(\omega) = H(\omega)X(\omega)$
$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \longleftrightarrow Y(\omega)=X(\omega)H(\omega),$$ ...
12
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 ...
12
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 complex valued signals. Complex numbers don't exist themselves, they are a mathematical construct.
Having said that, their mathematical properties can be ...
10
All eigenfunctions of an LTI system can be described in terms of complex exponentials, and complex exponentials form a complete basis of the signal space. However, if you have a system that is degenerate, meaning you have eigensubspaces of dimension >1, then the eigenvectors to the corresponding eigenvalue are all linear combination of vectors from the ...
10
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
$$y_2(t)=x_2(-t)=x_1(-t-T)\tag{3}$$
If the system were time-invariant, its response to $x_2(t)$ should be a shifted version of its response to $x_1(t)$:
$$y_2(t)=y_1(t-...
10
You can make a simple algebraic argument, given the premise that you provided. If:
$$
Y(\omega) = X(\omega) H(\omega)
$$
where $X(\omega)$ is the spectrum of the input signal and $H(\omega$) is the frequency response of the system, then it's obvious that if there is some $\omega$ in the input signal for which $X(\omega) = 0$, then $Y(\omega) = 0$ as well; ...
10
A time-invariant system is one that, when you shift the input signal, the output is shifted by the same amount.
A system that reverses the signal cannot be time-invariant because when you shift the input, the output is shifted the other way. $k$ and $-k$ are not the same amount.
$$
y[n-k] = x[k-n] = x[-n \mathbf{+} k]
$$
9
I agree with Peter K.'s answer, but I would like to add one important point: the two statements in the question are only true for causal systems. The most general statement about stability for LTI systems described by rational transfer function is:
An LTI system with a rational transfer function is stable if the region of convergence (ROC) of its transfer ...
9
First it's important to realize that many authors use the terms zero-input response and natural response as synonyms. This convention is used in the corresponding wikipedia article, and for instance also in this book. Even Proakis and Manolakis are not entirely clear about it. In the book you quoted you can find the following sentence on page 97:
[...] the ...
9
The Fourier transform operator $\mathscr{F}$ is a linear one; i.e.,
$$\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(f)$$ And therefore the system that implements it will be linear.
However time invariance, which is tested by the method
$$\mathcal{T}\{x(t)\}=y(t) \implies \...
8
I think you are constructing the convolution matrix incorrectly. Let's do an easy example with some numbers to see how it is constructed
Let's imagine that youhave the following vectors of dimensions $M$ and $N$, And let's for the sake of it, say that $M = 5$ and $N = 4$
$\mathbf{a} = [a_{1},a_{2},a_{3},a_{4}]^{T}$
$\mathbf{b} = [b_{1},b_{2},b_{3},b_{4},...
8
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 numerator and denominator degree can be equal. E.g.
$$H(s)=\frac{as^2+bs+c}{s^2+ds+e}$$
can represent a causal and stable system, as long as its poles are in ...
8
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.
...
7
A simpler version of Phonon's answer is as follows.
Suppose that $y$ denotes the response of the system to the unit step function.
Then, as discussed in this answer,
in general, $y$ is the sum of scaled and time-delayed copies of the impulse
response, and in this particular case, no scaling is required; only time
delays. Thus,
$$\begin{align}
y[0] &= h[...
7
It looks like your transfer function is correct, but there's a small mistake in your partial fraction expansion:
$$H(s)=\frac{2}{(s+4)(s+2)}=\frac{1}{s+2}-\frac{1}{s+4}\tag{1}$$
The corresponding impulse response is
$$h(t)=(e^{-2t}-e^{-4t})u(t)\tag{2}$$
The response to $x(t)=te^{-2t}u(t)$ is indeed most easily computed by solving the convolution integral:...
7
It depends on what exactly you mean by "invertible". In system theory, what is often meant is if there is a causal and stable system that can invert a given system, because otherwise there might be an inverse system but you can't implement it.
For linear time-invariant systems there is a straightforward method, as mentioned in the comments by Robert Bristow-...
7
If the input is a unit step, then the output of the first block in system 1 is not zero, but it is a Dirac delta impulse $\delta(t)$. Intuitively, the derivative is infinite at $t=0$ because of the step going from zero to one. Integrating the Dirac delta impulse will give you a step at the output of the first system.
The output of the first block in system ...
7
The magnitude of that complex exponential is 1. Recall from complex algebra: any complex number can be expressed as $z = r e^{j \phi}$ where $|z|=r$ is its magnitude and $\arg z = \phi$ is the argument. Using this note that
$$
|e^{-j\Omega \lambda}| = 1
$$
which is why it "disappeared".
7
I will ask you something that will give you intuition.
How would you calculate the Gradient of an image?
Image is a discretization of reality, so how would you estimate the gradient of the "Reality" if you're given only the image?
In the case above we use Finite Differences to approximate the continuous derivative.
So what actually is approximating $ \dot{...
7
This system
$$ 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... Hope this solves your confusion.
7
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 ...
7
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 the signal. Another idea is to add LPF assumption of the Filter by setting the sum of its coefficients to be 1 and non negative. Yet since we have Discrete ...
7
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 causal input $x(t)$ which has an identifiable beginning cannot truly be Narrow-Band or Band-Limited. It will have non-zero frequency content at all frequencies....
6
If the Z-transform of the feedforward section is divisible by the Z-transform of the feedback section, the filter is FIR.
Consider your example: $y[n] = y[n-1] + x[n] - x[n-3]$.
The Z-transform is $\mathrm Y(z)- z^{-1}\mathrm Y(z) = \mathrm X(z) - z^{-3}\mathrm X(z)$, and the Z-transform of the response is $\mathrm H(z) = \mathrm Y(z)/\mathrm X(z) = (1 - z^...
6
A signal does not have the properties time-invariant, linear, stable, casual
that you wish to ascribe to it; these are properties of systems. Properties
of signals that might be of interest include
bounded amplitude versus unbounded amplitude
discrete amplitude versus continuous amplitude
discrete-time versus continuous-time
periodic versus nonperiodic
...
6
Yes, this is equally true in the discrete systems case. The differentiation operation in this case is replaced to by first order difference. It don't think it has a universal symbol, but let's call it $D(\cdot)$. This operation is equivalent to filtering your signal with $y[n] = x[n] - x[n-1]$. Let's call this filter $d[n]$. I'm going to denote convolution ...
6
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 expressed as
$$ x[n] = \sum\limits_{i=-\infty}^{\infty} x[i] \delta[n-i] $$
where $\delta[n]$ is the Kronecker delta or "unit impulse function"
$$ \delta[n] \...
answered Jun 9 '15 at 19:38
robert bristow-johnson
16.2k3 gold badges29 silver badges68 bronze badges
6
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 response just for one frequency, namely the frequency of the sinusoidal input signal:
$$y[n]=\left|H(e^{j\omega_0})\right|\sin\left(n\omega_0+\phi(\omega_0)\...
6
You can find the impulse response
Let's take the case of a discrete system.
If $s[n]$ is the unit step response of the system, we can write
$$s[n]= u[n]\ast h[n]$$
where $h[n]$ is the impulse response of the system and $u[n]$ is the unit step function.
Now using commutative property you can write $$s[n]=h[n]\ast u[n]$$
Expanding convolution we get $$s[...
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