0
$\begingroup$

enter image description here

I know that we can get the phasor representation(frequency domain) of a signal through Fourier transform, but the picture above gives a method that can get the phasor representation of a signal in a different way, in this method according to the writer, it numerically "demodulate" the signal to create a phasor representation of the signal, however, i can't understand that, anyone can help me? Thanks in advance!

$\endgroup$
1
$\begingroup$

This is simply about representing a sinusoidal signal $x(t)$ in the following way:

$$x(t) \ = \ \cos(\omega_0t+\phi) \ = \ a \cos(\omega_0 t) \ - \ b \sin(\omega_0 t)$$

The 'method' described in your question is about identifying the constants $a$ and $b$. The phasor of $x(t)$ is then $X=a+jb$, because

$$x(t)=\text{Re}\{Xe^{j\omega_0t}\}$$

Using basic trigonometric identities you get

$$a=\cos\phi\quad\text{and}\quad b=\sin\phi$$

But you can also compute these values by correlation (as shown in your question), which is equivalent to computing the Fourier coefficients of $x(t)$ (apart from a sign difference for the coefficient $b$). Of course, the only non-zero coefficients are the coefficients corresponding to the fundamental:

$$a=\frac{2}{T}\int_0^Tx(t)\cos(\omega_0 t) \ dt,\quad T=\frac{2\pi}{\omega_0}\\ b=-\frac{2}{T}\int_0^Tx(t)\sin(\omega_0 t) \ dt$$

These last two equations correspond to the first two steps described in your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.