New answers tagged fourier
0
I worked it out by myself, after reading a bunch of posts mentioning different methods of FFT output scaling. I still find this aspect of FFT processing heavily unsdocumented everywhere. I have not yet found any reliable source that explains what is the use of these scalings, which fields of sciences or what processing methods use them.
I have yet found ...
1
The input and the integrator output are shown below:
So, you should consider two cases: $t<0$ and $t>0$.
In the first case ($t<0$), you have
$$\begin{align}
y(t) &= \int_{-\infty}^t x(\tau)d\tau = -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}u(-\tau)d\tau \\
&= -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}d\tau = -\frac{1}{T}Te^{\...
0
$\sigma$ is associated with attenuation in "Control Theory" and there Laplace is more suitable than Fourier. $\sigma=0$ the transformation becomes Fourier.
There is unilateral and bilateral Laplace transformation & periodic signals and Laplace transform. @Matt L have explained somewhere here the strong relationship between Laplace and Fourier.
1
It's nice to see alternative paths, like this one: First of all, properties:
Frequency shifting: $x(t)e^{j2\pi f_0 t} \longleftrightarrow X(f-f_0)$
Duality: if $x(t) \longleftrightarrow X(f)$ then $X(t) \longleftrightarrow x(-f)$
You have correctly expressed the depicted transform: $$X(f) = \frac{1}{2}\mathrm{tri}\Big(\frac{f+f_0}{B}\Big) - \frac{1}{2}\...
1
Hints:
X
It is quite likely that your book contains a formula (either as a solved example or as a theorem or property of Fourier transforms) that looks like
$$\mathcal F\{x(t)\sin 2\pi f_0 t\} = \frac{X(f-f_0)-X(f+f_0)}{2i}
\tag{1}$$
where $i$ is a square root of unity. So, copy the formula into your notebook, and then use the hint given in your book (...
0
I may consider them as L^2 then following theorem holds.
https://en.wikipedia.org/wiki/Plancherel_theorem
https://de.wikipedia.org/wiki/Satz_von_Plancherel
https://ms.mcmaster.ca/craig/craig-pde-chap5.pdf
Also keen to learn if there is any other theorems.
1
(iii) seems be to correct. Spectrum of two frequencies.
The other choices can be verified,
(i) Fourier transform of $f(t=0)=1$ is 1. So it cannot be graph (iii)
(ii) Signal (iii) is sine wave and there is no spectrum associated with that.
(iv) It's Fourier transform of sine wave (iii) which is
$$f(t)=0.5\cdot \sin(2\pi Ft)$$
and it's transform is
$$\...
2
If you already know the - quite well-known :) - FT pair $$e^{-at}u(t) \longleftrightarrow \frac{1}{a+j2\pi f}$$ then by setting $a = 1/T$, you get $$e^{-\frac{t}{T}}u(t) \longleftrightarrow \frac{1}{\frac{1}{T} + j2\pi f}$$
Time delay property $$x(t-t_0) \longleftrightarrow X(f)e^{-j2\pi ft_0}$$ can be used now by setting $t_0 = T$, and then $$x(t-T) = e^{-\...
1
Properties are great if you use them properly. In this case it was easier to just crank out the integral as it is not terrible:
$X(f)=\frac{e^1}{T}\int_T^{\infty} e^{-\big(\frac{t-T}{T}\big)}e^{-j2\pi ft}dt$
Combine the terms to get
$X(f)=\frac{e^1}{T}\int_T^{\infty}e^{-t(j2\pi f + \frac{1}{T})}dt$
Now do the integral
$X(f)=\frac{e^1}{T}\bigg[\frac{e^{-...
2
I doubt that Hilmar will agree (maybe he will) but I will offer this answer as a counter to the one that has already been (prematurely) accepted by the questioner.
Hilmar's answer is wrong. It is factually wrong and analytically wrong.
1) Does the DTFT take only infinite input sequences?
No. There is nothing that prevents you from applying ...
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