# Exponential Fourier Series

A compact way of expressing the Fourier series in Eq. (A) is to put it in exponential form.
$$f(t)=a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega_{0} t+b_{n} \sin n \omega_{0} t\right) \tag{A}$$
This requires that we represent the sine and cosine functions in the exponential form using Euler's identity:
\begin{aligned}\cos n \omega_{0} t &=\frac{1}{2}\left[e^{j n \omega_{0} t}+e^{-j n \omega_{0} t}\right] \\ \sin n \omega_{0} t &=\frac{1}{2 j}\left[e^{j n \omega_{0} t}-e^{-j n \omega_{0} t}\right]\end{aligned} \tag{1}
Substituting Eq. (1) into Eq. (A) and collecting terms, we obtain
$$f(t)=a_{0}+\frac{1}{2} \sum_{n=1}^{\infty}\left[\left(a_{n}-j b_{n}\right) e^{j n \omega_{0} t}+\left(a_{n}+j b_{n}\right) e^{-j n \omega_{0} t}\right] \tag{2}$$
If we define a new coefficient $c_{n}$ so that
$$c_{0}=a_{0}, \quad c_{n}=\frac{\left(a_{n}-j b_{n}\right)}{2}, \quad c_{-n}=c_{n}^{*}=\frac{\left(a_{n}+j b_{n}\right)}{2} \tag{3}$$
then $f(t)$ becomes
$$f(t)=c_{0}+\sum_{n=1}^{\infty}\left(c_{n} e^{j n \omega_{0} t}+c_{-n} e^{-j n \omega_{0} t}\right) \tag{4}$$
or
$$f(t)=\sum_{n=-\infty}^{\infty} c_{n} e^{j n \omega_{0} t} \tag{5}$$
This is the complex or exponential Fourier series representation of $f(t)$. Note that this exponential form is more compact than the sine-cosine form in Eq. (A). Although the exponential Fourier series coefficients $c_{n}$ can also be obtained from $a_{n}$ and $b_{n}$ using Eq. (3), they can also be obtained directly from $f(t)$ as
$$c_{n}=\frac{1}{T} \int_{0}^{T} f(t) e^{-j n \omega_{0} t} d t \tag{6}$$
where $\omega_{0}=2 \pi / T$, as usual. The plots of the magnitude and phase of $c_{n}$ versus $n \omega_{0}$ are called the complex amplitude spectrum and complex phase spectrum of $f(t)$, respectively. The two spectra form the complex frequency spectrum of $f(t)$.
The exponential Fourier series of a periodic function f (t) describes the spectrum of f (t) in terms of the amplitude and phase angle of ac components at positive and negative harmonic frequencies.
The coefficients of the three forms of Fourier series (sine-cosine form, amplitude-phase form, and exponential form) are related by
$$A_{n} \angle \phi_{n}=a_{n}-j b_{n}=2 c_{n}$$or$$c_{n}=\left|c_{n}\right| \angle \theta_{n}=\frac{\sqrt{a_{n}^{2}+b_{n}^{2}}}{2} \angle -\tan ^{-1} b_{n} / a_{n} \tag{7}$$
if only $a_{n}>0$. Note that the phase $\theta_{n}$ of $c_{n}$ is equal to $\phi_{n}$.In terms of the Fourier complex coefficients $c_{n}$, the rms value of a periodic signal $f(t)$ can be found as
\begin{aligned}F_{\mathrm{rms}}^{2} &=\frac{1}{T} \int_{0}^{T} f^{2}(t) d t=\frac{1}{T} \int_{0}^{T} f(t)\left[\sum_{n=-\infty}^{\infty} c_{n} e^{j n \omega_{0} t}\right] d t \\&=\sum_{n=-\infty}^{\infty} c_{n}\left[\frac{1}{T} \int_{0}^{T} f(t) e^{j n \omega_{0} t}\right] \\&=\sum_{n=-\infty}^{\infty} c_{n} c_{n}^{*}=\sum_{n=-\infty}^{\infty}\left|c_{n}\right|^{2}\end{aligned} \tag{8}
or
$$F_{\mathrm{ms}}=\sqrt{\sum_{n=-\infty}^{\infty}\left|c_{n}\right|^{2}} \tag{9}$$
Equation (8) can be written as
$$F_{\mathrm{ms}}^{2}=\left|c_{0}\right|^{2}+2 \sum_{n=1}^{\infty}\left|c_{n}\right|^{2} \tag{10}$$
Again, the power dissipated by a $1-\Omega$ resistance is
$$P_{1 \Omega}=F_{\text {rms }}^{2}=\sum_{n=-\infty}^{\infty}\left|c_{n}\right|^{2} \tag{11}$$
which is a restatement of Parseval's theorem. The power spectrum of the signal $f(t)$ is the plot of $\left|c_{n}\right|^{2}$ versus $n \omega_{0}$. If $f(t)$ is the voltage across a resistor $R$, the average power absorbed by the resistor is $F_{\text {rms }}^{2} / R$; if $f(t)$ is the current through $R$, the power is $F_{\mathrm{rms}}^{2} R$. As an illustration, consider the periodic pulse train of Fig. 1.
Fig. 1: The periodic pulse train.
Our goal is to obtain its amplitude and phase spectra. The period of the pulse train is $T=10$, so that $\omega_{0}=2 \pi / T=\pi / 5$. Using Eq. (6),
\begin{aligned}c_{n} &=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) e^{-j n \omega_{0} t} d t=\frac{1}{10} \int_{-1}^{1} 10 e^{-j n \omega_{0} t} d t \\&=\left.\frac{1}{-j n \omega_{0}} e^{-j n \omega_{0} t}\right|_{-1} ^{1}=\frac{1}{-j n \omega_{0}}\left(e^{-j n \omega_{0}}-e^{j n \omega_{0}}\right) \\&=\frac{2}{n \omega_{0}} \frac{e^{j n \omega_{0}}-e^{-j n \omega_{0}}}{2 j}=2 \frac{\sin n \omega_{0}}{n \omega_{0}}, \quad \omega_{0}=\frac{\pi}{5} \\&=2 \frac{\sin n \pi / 5}{n \pi / 5}\end{aligned} \tag{12}
and
$$f(t)=2 \sum_{n=-\infty}^{\infty} \frac{\sin n \pi / 5}{n \pi / 5} e^{j n \pi t / 5} \tag{13}$$
Notice from Eq. (12) that $c_{n}$ is the product of 2 and a function of the form $\sin x / x$. This function is known as the sinc function; we write it as
$$\operatorname{sinc}(x)=\frac{\sin x}{x} \tag{14}$$
Some properties of the sinc function are important here. For zero argument, the value of the sinc function is unity,
$$\operatorname{sinc}(0)=1 \tag{15}$$
This is obtained applying L'Hopital's rule to Eq. (14). For an integral multiple of $\pi$, the value of the sinc function is zero,
$$\operatorname{sinc}(n \pi)=0, \quad n=1,2,3, \ldots \tag{16}$$
Also, the sinc function shows even symmetry. With all this in mind, we can obtain the amplitude and phase spectra of $f(t)$. From Eq. (12), the magnitude is
$$\left|c_{n}\right|=2\left|\frac{\sin n \pi / 5}{n \pi / 5}\right| \tag{17}$$
while the phase is
$$\theta_{n}=\left\{\begin{array}{rc}0^{\circ}, & \sin \frac{n \pi}{5} > 0 \\180^{\circ}, & \sin \frac{n \pi}{5}< 0 \end{array}\right. \tag{18}$$
Figure $2$ shows the plot of $\left|c_{n}\right|$ versus $n$ for $n$ varying from $-10$ to 10 , where $n=\omega / \omega_{0}$ is the normalized frequency. Figure $3$ shows the plot of $\theta_{n}$ versus $n$.
Fig. 2: The amplitude of a periodic pulse train.
Fig. 3: The phase spectrum of a periodic pulse train.
Both the amplitude spectrum and phase spectrum are called line spectra, because the value of $\left|c_{n}\right|$ and $\theta_{n}$ occur only at discrete values of frequencies. The spacing between the lines is $\omega_{0}$. The power visualization of the effect of circuit on a periodic & spectrum, which is the plot of $\left|c_{n}\right|^{2}$ versus $n \omega_{0}$, can also be plotted. Notice signal. that the sinc function forms the envelope of the amplitude spectrum.
Example 1: Find the exponential Fourier series expansion of the periodic function $f(t)=e^{t}, 0 < t < 2 \pi$ with $f(t+2 \pi)=f(t)$.
Solution: Since $T=2 \pi, \omega_{0}=2 \pi / T=1$. Hence,
$$\begin{array}{l}c_{n}=\frac{1}{T} \int_{0}^{T} f(t) e^{-j n \omega \omega_{0} t} d t=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{t} e^{-j n t} d t \\=\left.\frac{1}{2 \pi} \frac{1}{1-j n} e^{(1-j n) t}\right|_{0} ^{2 \pi}=\frac{1}{2 \pi(1-j n)}\left[e^{2 \pi} e^{-j 2 \pi n}-1\right] \\\end{array}$$
But by Euler's identity,
$$e^{-j 2 \pi n}=\cos 2 \pi n-j \sin 2 \pi n=1-j 0=1$$
Thus,
$$c_{n}=\frac{1}{2 \pi(1-j n)}\left[e^{2 \pi}-1\right]=\frac{85}{1-j n}$$
The complex Fourier series is
$$f(t)=\sum_{n=-\infty}^{\infty} \frac{85}{1-j n} e^{j n t}$$
We may want to plot the complex frequency spectrum of $f(t)$. If we let $c_{n}=\left|c_{n}\right| \angle \theta_{n}$, then
$$\left|c_{n}\right|=\frac{85}{\sqrt{1+n^{2}}}, \quad \theta_{n}=\tan ^{-1} n$$
By inserting in negative and positive values of $n$, we obtain the amplitude and the phase plots of $c_{n}$ versus $n \omega_{0}=n$, as in Fig. 4.
Fig. 4: The complex frequency spectrum of the function in Example 1: (a) amplitude spectrum, (b) phase spectrum.
Example 2: Find the complex Fourier series of the sawtooth wave in Fig. 5. Plot the amplitude and the phase spectra.
Fig. 5: For Example 2.
Solution: From Fig. 5, $f(t)=t, 0 < t < 1, T=1$ so that $\omega_{0}=2 \pi / T=2 \pi$. Hence,
$$c_{n}=\frac{1}{T} \int_{0}^{T} f(t) e^{-j n \omega o s t} d t=\frac{1}{1} \int_{0}^{1} t e^{-j 2 n \pi t} d t \tag{2.1}$$
But
$$\int t e^{a t} d t=\frac{e^{a t}}{a^{2}}(a x-1)+C$$
Applying this to Eq. (2.1) gives
\begin{aligned}c_{n} &=\left.\frac{e^{-j 2 n \pi t}}{(-j 2 n \pi)^{2}}(-j 2 n \pi t-1)\right|_{0} ^{1} \\&=\frac{e^{-j 2 n \pi}(-j 2 n \pi-1)+1}{-4 n^{2} \pi^{2}}\end{aligned} \tag{2.2}
Again,
$$e^{-j 2 \pi n}=\cos 2 \pi n-j \sin 2 \pi n=1-j 0=1$$
so that Eq. (2.2) becomes
$$c_{n}=\frac{-j 2 n \pi}{-4 n^{2} \pi^{2}}=\frac{j}{2 n \pi}$$
This does not include the case when $n=0$. When $n=0$,
$$c_{0}=\frac{1}{T} \int_{0}^{T} f(t) d t=\frac{1}{1} \int_{0}^{1} t d t=\left.\frac{t^{2}}{2}\right|_{1} ^{0}=0.5$$
Hence,
$$f(t)=0.5+\sum_{\substack{n=-\infty \\ n \neq 0}}^{\infty} \frac{j}{2 n \pi} e^{j 2 n \pi t}$$
and
$$\left|c_{n}\right|=\left\{\begin{array}{ll}\frac{1}{2|n| \pi}, & n \neq 0 \\0.5, & n=0\end{array}, \quad \theta_{n}=90^{\circ}, \quad n \neq 0\right.$$
By plotting $\left|c_{n}\right|$ and $\theta_{n}$ for different $n$, we obtain the amplitude spectrum and the phase spectrum shown in Fig. 6.
Fig. 6: For Example 2: (a) amplitude spectrum, (b) phase spectrum.