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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:

Magnitude and phase plot

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.

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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 $cos(\omega + \pi/2)$.

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    $\begingroup$ 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? $\endgroup$ – Healow Mar 31 at 11:27
  • $\begingroup$ Yes...that's right $\endgroup$ – DSP Novice Mar 31 at 13:27
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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.

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

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  • $\begingroup$ I think you made a mistake on the last term on the LHS: +pi/2 should be -pi/2 ? $\endgroup$ – rrogers Mar 31 at 19:42

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