Suppose we take a periodic signal and perform fourier analysis over it . Now we have two ways of representing the fourier series of this particular signal , one is trigonometric fourier series and other is exponential fourier series as far as my understanding. Now the exponential fourier series gives the component signals in terms of complex exponentials .Now as far as my understanding the amplitude spectrum of trigonometric fourier series is twice that of exponential fourier series of a signal . So in the same way , I want to know what is the relation of phase spectrum of trigonometric fourier series and exponential fourier series. Please Help me with this as I have been stuck with this for a long time . I am not able to find any good resources in the internet for this . also suggest me any books or articles which discuss about this.


1 Answer 1


Given the Trigonometric Fourier Series for real $x(t)$ as:

$$x(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(\omega_n t)+b_n\sin(\omega_n t)$$

And the Exponential Fourier Series as:

$$x(t) = \sum_{n=-\infty}^\infty c_n e^{j\omega_n t}$$

The two are related by:

$$e^{j\omega t} = \cos(\omega t) + j\sin(\omega t)$$


$$c_n = a_n - j b_n$$

For both the trigonometric series and exponential series the phase is given by $\tan^{-1}(-b_n/a_n)$. (They are the same).

It may help to see this intuitively by further understanding the relationship between sinusoids and exponentials as given by Euler' formula, where we see the magnitude is doubled and the phase is the same; for example for the cosine:

$$2\cos(\omega t) = e^{j\omega t} + e^{-j\omega t}$$

Adding in a phase term:

$$2\cos(\omega t + \phi) = e^{j(\omega t+\phi)} + e^{-j(\omega t+\phi)}$$

Why this is so is clearer by reviewing the geometric operations of the terms. The general exponential expression $Ae^{j/\theta}$ with real $A$ and real $\theta$ is a phasor with magnitude $A$ and angle $\theta$, which would appear as follows when plotted on a complex plane:


Thus the term $e^{j\omega t}$ is a phasor with magnitude equal to one and phase linearly increasing with time in the positive direction. Plotted on the complex plane it would rotate counter-clockwise at a constant rate (for constant $\omega$) and is referred to specifically as a positive frequency, and with that it would specifically be plotted on the positive frequency axis in the frequency domain. Similarly $e^{-j\omega t}$ rotates clockwise (and therefore is a negative frequency plotted on the negative frequency axis) and when we sum them together we get a real cosine at twice the magnitude. Real sinusoids do not have a notion of "positive" or "negative" frequency and thus are plotted on a positive frequency axis only.

sum of phasors

With the exponential Fourier Series in the frequency domain we see the coefficients for the individual spinning phasors, while in the trigonemtric Fourier Series we see the coefficients for sinusoids that the phasors would sum to. Thus the exponential Fourier Series is "two-sided" with the frequency axis extending to $\pm \infty$ while the frequency axis for the trigonometric Fourier Series is "one-sided" extending from $0$ to $\infty$. This is consistent with the index used for the time domain reconstruction given by the first two formulas above. For the simple case of a cosine as demonstrated here, we would have the following two results in the frequency domain:


Note in this plot we see a fundamental property of the Fourier Series as well as Fourier Transform for real waveforms: the spectrum will always be "complex conjugate symmetric" which means the negative frequencies will have the same magnitude and opposite phase as the positive frequencies. Why this is so is hopefully clear from the graphic I showed demonstrating how $e^{j\omega t}$ sums with $e^{-j\omega t}$ (same magnitude, opposite phase) to be the real $2\cos(\omega t)$. The two phasors sum such that the angles cancel and the result will always stay on the real axis!

  • $\begingroup$ Thanks for answering. If they are same , then why the phase spectrum of exponential fourier series appears on both sides of the y axis and phase spectrum of trigonometric fourier series only on the positive side of y axis. In case of amplitude spectrum , the amplitude spectrum of trigonometric fourier series which appears on the positive side of y axis , divides into two parts and appears on the either side of y axis. Does the same thing happen with phase spectrum as well ?If yes , Can you please show me how they are related mathematically to the one you mentioned above in the answer. . $\endgroup$
    – amoghfyi
    Jan 21 at 14:05
  • $\begingroup$ I think I know your confusion and adding some more background for you...stay tuned $\endgroup$ Jan 21 at 14:07
  • $\begingroup$ thank you sir , I am looking forward to your answer . $\endgroup$
    – amoghfyi
    Jan 21 at 14:15
  • $\begingroup$ Thank you sir for answering , isn't the phase spectrum of exponential fourier series odd symmetric ? $\endgroup$
    – amoghfyi
    Jan 21 at 14:36
  • $\begingroup$ Yes complex conjugate symmetric, the negative frequencies have the opposite phase as the positive frequencies. The trigonometric Fourier Series has the same phase as the positive frequencies for the exponential Fourier Series. Read through my answer as hopefully I am now making that clear. $\endgroup$ Jan 21 at 14:37

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