Im very confused regarding how to determine the angle on the reduced or harmonic form representation of the Fourier series. Some books state the following:
$$f(t)=F_0+\sum_{n=1}^\infty |F_n |\cos(n\omega t +\theta)$$
$$F_0=a_0$$
$$|F_n |=\sqrt{a_n^2+b_n^2}$$
$$\theta =\arctan\left(\frac{-b_n}{a_n}\right) $$
Now, say you want to plot the phase spectra of a square wave, with the following Fourier series:
$$\frac{4A}{\pi} \sum_{n=1}^\infty \frac{1}{2n-1}\sin((2n-1)\omega t) \,\,\,\,\,\,\,\,\mathbf(1)$$
$$\mathbf A \text{ is the amplitude of the signal}$$
then:
$$F_0=a_0=0$$
$$|F_n |=\sqrt{a_n^2+b_n^2}=\frac{4A}{\pi}\frac{1}{2n-1}$$
$$\theta =\arctan\left(\frac{-b_n}{a_n}\right)=\arctan\left(\frac{-\frac{4A}{\pi}\frac{1}{2n-1}}{0}\right)=-90^\circ \,\,\,\,\,\,\,\,\mathbf(2a)$$
Its phase spectrum is the following:
And the Fourier series (equation $\mathbf(1)$ ) of the square waveform represented in its harmonic form is:
$$\frac{4A}{\pi} \sum_{n=1}^\infty \frac{1}{2n-1}\cos\left((2n-1)\omega t -90^\circ\right) \,\,\,\,\,\,\,\,\mathbf(2b)$$
Now here is where I get confused: other books state that the reduced or harmonic form representation is as follows (notice the negative sign on the angle):
$$f(t)=F_0+\sum_{n=1}^\infty |F_n |\cos(n\omega t \color{red}{\mathbf-}\theta)$$
$$F_0=a_0$$
$$|F_n |=\sqrt{a_n^2+b_n^2}$$
$$\theta =\arctan\left(\frac{b_n}{a_n}\right) $$
So for the same square wave of equation $\mathbf(1)$ from the example above, the harmonic representation with the negative sign should be:
$$F_0=a_0=0$$
$$|F_n |=\sqrt{a_n^2+b_n^2}=\frac{4A}{\pi}\frac{1}{2n-1}$$
$$\theta =\arctan\left(\frac{-b_n}{a_n}\right)=\arctan\left(\frac{\frac{4A}{\pi}\frac{1}{2n-1}}{0}\right)=90^\circ \,\,\,\,\,\,\,\,\mathbf(3a)$$
Thus:
$$\frac{4A}{\pi} \sum_{n=1}^\infty \frac{1}{2n-1}\cos\left((2n-1)\omega t -90^\circ\right) \,\,\,\,\,\,\,\,\mathbf(3b)$$
Notice that equation $\mathbf(3b)$ is exactly the same as $\mathbf(2b)$ but the angle $\theta$ in $\mathbf(3a)$ has a positive sign while the angle $\theta$ in $\mathbf(2a)$ has a negative sign.
So the phase spectrum using the angle calculated in $\mathbf(3a)$ is:
Which is the mirror image of the phase spectrum found using $\mathbf(2a)$
So bottom line: What is the correct way or the standard for the harmonic representation (plus sign or negative angle sign) and how should one decide the angle sign to plot the phase spectra ?
Best Answer
The two spectra should be the same, as (2b) and (3b) indicate. These two equations should lead you to the same result.
You compare (2a) with (3a). That's wrong because you don't take into account the sign before $\theta$ of the initial equation.