# Complex exp. Fourier series, finding $x(t)$ when $X(j\omega)$ is given as magnitude and phase plot

I'm watching Neso Academy series on Signals and Systems, and in one of the videos the problem is to find $$x(t)$$ when magnitude and phase plot are given. The plot looks like this:

When he finishes calculation he gets:

$$4 + 4cos(3 \omega t+ \dfrac\pi2) + 8cos(4 \omega t- \dfrac\pi2)$$

I understand the steps that are required to get this, but I don't understand why is it cosine? When you look at magnitude plot you have just 3 components: DC offset = 4 and 2 harmonics.

As far as I know the values on magnitude plot represent sine value and phase plot represents just phase shift of that sine wave. Without doing anything but looking at the plot I would write following:

$$4 + 4sin (3 \omega t+ \dfrac\pi2) + 8 sin(4wt-\dfrac\pi2)$$

And it's completely wrong, why is it so?

BTW link to the video.

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 ($$j*pi*(\delta(\omega+\omega0)-\delta(\omega-\omega0))$$, the phase is antisymmetric. The answer that he gets after calculation is just $$sin\omega$$ but expressed in terms of $$cos$$ as $$cos(\omega + \pi/2)$$.

• So the pure cosine function has same magnitude spectrum as sine function, the only difference being that cosine phase plot is zero everywhere, and sine phase plot is $\omega_0=-\dfrac\pi2$ and $-\omega_0=\dfrac\pi2$. Is this correct? Commented Mar 31, 2020 at 11:27
• Yes...that's right Commented Mar 31, 2020 at 13:27

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 equation which is $$x(t)=\sum_n C_n e^{jn\omega t}$$, where $$C_n$$ may be a complex coefficient meaning it has a magnitude and a phase. Re-write $$C_n$$ as $$C_n = |C_n|e^{j\angle C_n}$$, plug into the Fourier series equation and you get what I wrote above.

The extra steps that the video does is just to convert the complex exponential expression into cos/sin functions using Euler's formula. One result of Euler's formula is that $$2cos(\theta)=e^{j\theta}+e^{-j\theta}$$, and that is why in the end it turns out to be cosine and not sine.

The assumption you made is that sum of 2 complex exponentials with opposite phase is always $$sin$$. That is not true. $$x(t) = 4 + 4 e^{j\omega_0 4t + \pi /2} + 4e^{-j\omega_0 4t -\pi /2} + 2e^{j\omega_0 3 t + \pi /2} + 2e^{-j\omega_0 3 t +\pi /2} = 4 + 8cos(4\omega_0 t+\pi /2) + 4cos(3\omega_0t -\pi/2)$$.

• I think you made a mistake on the last term on the LHS: +pi/2 should be -pi/2 ? Commented Mar 31, 2020 at 19:42