New answers tagged continuous-signals
1
The spectrum that you give in your question corresponds to the "train of impulses" sampling process $$x(k) = \sum_{k=-\infty}^\infty x(t) \delta(t - kT_s). \tag1$$
This model is a very good one for tying together the Laplace and $z$ transforms, and for expressing the four kinds of Fourier transform in a unified framework.
However, in this framework,...
0
The statement is:
eliminating all points where the difference between $x$ and $i$ is at
least $r$
so eliminate those points for which
$$ |x - i | \gt r \tag{1}$$
Isn't that correct? One could argue whether the words say
$$ |x - i | \ge r $$
instead, which might be problematic at the border, but using (1) the statement seems OK to me.
The text after the ...
2
A standard way of describing a continuous curve before rigorous and provable defintions was:
What you can draw on a sheet of paper without lifting the pen
This is quite visual, and applies to signals as well. Of course this applies well to curves of finite length. Fractals and pathological functions like $x\mapsto x \sin 1/x$ (below) are examples of a need ...
1
The common definition of as system with memory (also called dynamic system) is that its output at any time instant $t$ depends on the input signal at times that are generally different from $t$. In that sense, a system described by $y(t)=x(t_0)$ clearly has memory, because the system needs to observe the input signal and remember its value at $t=t_0$.
On the ...
2
$Y(t)$ is not a wide-sense-stationary (WSS) process even though $X(t)$ is, and so it does not have an autocorrelation function with a single argument $\tau$ as you have written it; there are two arguments $t$ and $\tau$, or $t+\tau$ if you prefer. So the concept of PSD as you have learned it does not apply in this instance. Note that there is nothing random ...
1
This looks like homework, so I will just give a few hints .
The power spectrum is simply the magnitude squared of the spectrum so $PSD_x(f) = |X(f)|^2$
A modulator is NOT an LTI system. There is no transfer function and there is no impulse response. You can't apply LTI methods to this problem.
Multiplication in the time domain is equivalent to convolution ...
3
Both results are correct. Hence, the equality
$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=0}^{\infty}e^{-nT(1+j\omega)}\tag{1}$$
holds true.
One way to prove this is to realize that the term on the left-hand side of Eq. $(1)$ is periodic with period $2\pi/T$. Consequently, we can express this term by ...
1
Frequency response is the eigenvalue of the system when the input is an eigenfunction of the form $e$$j\omega n$.
Solve it by yourself to get $\displaystyle \sum_{k=-\infty}^\infty h[k]$e$-j\omega k$, which is the Fourier transform of $h[n]$.
1
As you corrupt with Poisson noise, perhaps Richardson Lucy is relevant to you?
https://en.m.wikipedia.org/wiki/Richardson–Lucy_deconvolution
0
One piece you are missing is the idea of causality.
Take, for example, a system with impulse response
$$ h(t) = e^{-at}u(t)+e^{at}u(-t) $$
This has poles at $+a$ and $-a$, but has a region of convergence that includes the imaginary axis and so is BIBO stable.
BIBO stability does not imply causality. Causality does not imply BIBO stability.
The only condition ...
1
It's important to realize that a $T$-periodic function has a discrete frequency spectrum with contributions at integer multiples of $\omega_0=2\pi/T$. Consequently, the spectrum $X(\omega)$ has the form
$$X(\omega)=\sum_kc_k\delta\left(\omega-\frac{2\pi k}{T}\right)\tag{1}$$
with constants $c_k$, which are just scaled versions of the Fourier coefficients of $...
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