New answers tagged fourier
5
Well this goes to show that Fourier series is just approximation that gets more and more correct when you add more harmonics. Take a look at this:
$\dfrac{4\sin\theta}{\pi}$ is just first harmonic.
Harmonics are integer multiple of base frequency as you can see: $\sin3\theta$, $\sin5\theta$ etc. And the more of them you add to the first harmonic the more ...
1
The second term can be expanded to
$$\begin{align}
2|a_k|\left(\frac{ e^{i(2\pi f_k t+\angle a_k)} + e^{-i(2\pi f_k t+\angle a_k)}}{2}\right)
&= |a_k|e^{i\angle a_k}e^{i(2\pi f_k t)}+|a_k|e^{-i\angle a_k}e^{-i(2\pi f_k t)} \\
&= a_k e^{i(2\pi f_k t)}+a_k^*e^{-i(2\pi f_k t)} \\
\end{align}$$
Since
$a_k=|a_k|e^{i\angle a_k}$ and $a_k^*=|a_k|e^{-i\...
2
Simple:
Euler's formula:
$$ e^{i\theta} = \cos(\theta) + i \sin(\theta) $$
and
$$ e^{-i\theta} = \cos(\theta) - i \sin(\theta) $$
Add them together:
$$ e^{i\theta} + e^{-i\theta} = 2 \cos(\theta) $$
You should see that in your equation.
From there it is usually stated:
$$ \cos(\theta) = \frac{ e^{i\theta} + e^{-i\theta}}{ 2 } $$
See my article
The ...
1
The magnitude response of both the sine and the cosine waves are the same . The difference is in the phase response. The extra j term in the definition of the FT of sine introduces the $\pi/2$ in the phase response and since there is a minus sign between the two delta functions of the FT of the sine wave ($j*pi*(\delta(\omega+\omega0)-\delta(\omega-\omega0))$...
1
The assumption you made is that sum of 2 complex exponentials with opposite phase is always $sin$. That is not true.
$x(t) = 4 + 4 e^{j\omega_0 4t + \pi /2} + 4e^{-j\omega_0 4t -\pi /2} + 2e^{j\omega_0 3 t + \pi /2} + 2e^{-j\omega_0 3 t +\pi /2} = 4 + 8cos(4\omega_0 t+\pi /2) + 4cos(3\omega_0t -\pi/2)$.
1
The video could've stopped at the answer:
$x(t)=4+ 2e^{j(3\omega t + \pi /2)} + 4e^{j(4\omega t - \pi /2)} + 2e^{-j(3\omega t + \pi /2)} + 4e^{-j(4\omega t - \pi /2)}$
And this answer can be read directly from the plots. The idea is that the $n^{th}$ term in the sum is equal to $|C_n|e^{j(n\omega t+\angle C_n)}$. You get this from the Fourier series ...
2
Let me put a practical answer with the following Matlab / Octave Code :
L = 2*1000; % signal sample count
n = 0:L-1; % discrete-time index
Fs = 44100; % sampling frequency
am = [1, 0.5, 2, 0.5, 0.3, 0.6, 0.1, 0.2]; % magnitudes of 8 components
fm = [882, 2646, 4410, 6615, 8820, 10000, 13230, 15876]; % frequencies
% Time domain ...
0
If you have a periodic signal, that is a repeating waveform, you can most easily find the constituent component parameters by aligning the Discrete Fourier Transform (DFT) frame on a whole number of cycles. Suppose you chose four. Then bins 0, 4, 8, 12, etc. contain the relevant values.
For a signal:
$$ x[n] = A \cos( \omega n + \phi ) $$
If
$$ \omega =...
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