14
First, to say linear predictive coding (LPC) is "more tolerant of transmission or encoding errors" isn't entirely true. The form in which the coefficients is transmitted makes a big difference. For example, if the linear prediction coefficients are solved for, they can be very sensitive to quantization, much like high order IIR filter coefficients (this is ...
12
Yes, in some sense the definitions are different. I'll give you two points of view, the first one will support your keen observation, the other will provide evidence for the contrary. These two do not conflict with each other, it's a matter of semantics. If the second one confuses you stick with the first one. Prelude concluded, here goes.
Problem ...
11
The "projection" that is referred to is a vector projection. To calculate the projection of vector $\mathbf{a}$ onto vector $\mathbf{b}$, you use the inner product of the two vectors:
$$
\mathbf{a_{proj}} = \langle \mathbf{a}, \mathbf{b} \rangle \mathbf{b}
$$
$\mathbf{a_{proj}}$ in this case is the vector component of $\mathbf{a}$ that lies in the same ...
11
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 ...
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
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 ...
8
A somewhat visual complement to the other answers
You are talking about systems that are linear and time invariant.
Exponential functions have one peculiar property (and can be actually defined by it): doing a time translation results in the same function multiplied by a constant. So
$$ e^{t-t_0}=e^{-t_0}e^t$$
The red exponential could as well be the ...
8
The elements building a lumped system are thought of being concentrated at singular points in space. The classical example is an electrical circuit with passive elements like resistor, inductance and capacitor. The physical quantities current and voltage are functions of time (only). E. g. the current at a capacitor with capacity $C$ is given by
$$
i(t) = C\...
8
I think you are constructing the convolution matrix incorrectly. Let's do an easy example with some numbers to see how it is constructed
Let's imagine that youhave the following vectors of dimensions $M$ and $N$, And let's for the sake of it, say that $M = 5$ and $N = 4$
$\mathbf{a} = [a_{1},a_{2},a_{3},a_{4}]^{T}$
$\mathbf{b} = [b_{1},b_{2},b_{3},b_{4},...
8
I agree with Peter K.'s answer, but I would like to add one important point: the two statements in the question are only true for causal systems. The most general statement about stability for LTI systems described by rational transfer function is:
An LTI system with a rational transfer function is stable if the region of convergence (ROC) of its transfer ...
8
The Fourier transform operator $\mathscr{F}$ is a linear one; i.e.,
$$\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 \...
7
What you have listed is a state equation. Usually a state-space representation has two equations:
(i) $\dot{x}(t)=Ax(t)+Bu(t)$
(ii) $y(t)=Cx(t)+Du(t)$
One takes the input $u(t)$ and generates the state $x(t)$ and another that takes the input $u(t)$ and state $x(t)$ and generates the output $y(t)$. You are missing the second equation.
That aside, the ...
7
This answer is a response to a comment by the OP on on yoda's answer.
Suppose that $h(t)$, the impulse response of a continuous-time
linear time-invariant system, has the property that
$$\int_{-\infty}^{\infty} |h(t)| \mathrm dt = M$$ for
some finite number $M$. Then, for each and every
bounded input $x(t)$, the output $y(t)$ is bounded also.
If $|x(t)| ...
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,
$$\begin{align}
y[0] &= h[...
7
The magnitude of that complex exponential is 1. Recall from complex algebra: any complex number can be expressed as $z = r e^{j \phi}$ where $|z|=r$ is its magnitude and $\arg z = \phi$ is the argument. Using this note that
$$
|e^{-j\Omega \lambda}| = 1
$$
which is why it "disappeared".
7
[EDIT] A necessary condition for invertibility is that any output has only one possible input (or injectivity, as proposed in comments). Since we are looking at counterexamples, we can look at when this condition is not satisfied.
The null system, that turns every signal into a zero flat line, is not invertible, but a bit trivial.
A system that computes a ...
6
The term $x(2n)$ is not time invariant: it gives the even samples of $x$. Shift the input by 1 sample and you'll get the odd terms, which are different from the even terms shifted 1 place.
Your reasoning above is not correct. You should compare $T(x(n-k))$ to $T(x(n))|_{n-k}$ (that is, the original output, shifted by $k$ samples).
Edit:
These type of ...
6
Shortly, we have two kind of basic responses: time responses and frequency responses. Time responses test how the system works with momentary disturbance while the frequency response test it with continuous disturbance. Time responses contain things such as step response, ramp response and impulse response. Frequency responses contain sinusoidal responses.
...
6
As Dilip pointed out in the comment above, you can get the impulse response using the inverse Fourier transform. However, a slightly easier method might be to use the Laplace domain instead; it's more amenable to easy inverse transforming via transform tables. First, recall that the frequency response is really just the $s$-plane transfer function evaluated ...
6
It is important to realize that due to the LTI property the response to an impulse $\delta(n)$ describes the system completely, because any input signal can be written as a weighted sum of shifted impulses:
$$x(n)=\sum_kx(k)\delta(n-k)\tag{1}$$
Let $\mathcal{T}\{\cdot\}$ denote the system operator, so the response to a shifted impulse is
$$\mathcal{T}\{\...
6
You have your definition of causality wrong. It's actually much simpler and much more intuitive. A causal system is a system in which the output does not depend on future values of the input. This property is not exclusive linear systems and can apply to systems in general. Here are a few examples to illustrate the point:
Causal linear time-invariant system:...
6
Both filters can be re-written in the form:
\begin{equation}
H(z) = \alpha ( 1 -2z^{-1} +z^{-2} )
\end{equation}
Clearly, in the first case, $\alpha=-\frac{1}{3}$ and in the second $\alpha=-\frac{1}{4}$.
In other words, the shape of the filter's frequency response is the same, the only difference is the scaling.
I can think of two reasons of why the second ...
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
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
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 ...
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:
$$y[n]=\left|H(e^{j\omega_0})\right|\sin\left(n\omega_0+\phi(\omega_0)\...
6
If the input is a unit step, then the output of the first block in system 1 is not zero, but it is a Dirac delta impulse $\delta(t)$. Intuitively, the derivative is infinite at $t=0$ because of the step going from zero to one. Integrating the Dirac delta impulse will give you a step at the output of the first system.
The output of the first block in system ...
6
No. It's only LTI (Linear and Time-Invariant) systems that can be modeled with convolution through a unique single impulse response.
For example the systems
$$ y(t) = g(t) x(t) $$ or
$$ y[n] = \sum_{k=0}^{k < n} x[n-k] $$
are both linear but not time-invariant and their output $y[n]$ cannot be computed with the convolution operation ( $\star$ denoting ...
5
The z domain transfer function of the system is the z transform of the system impulse response, so start by taking the Z transform of h[n] ...
$$H[z] = -z^1 + 1 + 2z^{-1} + 2z^{-2} + z^{-3} -z^{-4}$$
You may be able to message this into a nicer form, but that isn't necessary.
Next, to get the the frequency response, replace z with $e^{jw}$
So this ...
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