Exponential Fourier Series
A compact way of expressing the Fourier series in Eq. (A) is to put it in exponential form.
This requires that we represent the sine and cosine functions in the exponential form using Euler's identity:
Substituting Eq. (1) into Eq. (A) and collecting terms, we obtain
If we define a new coefficient $ c_{n} $ so that
then $ f(t) $ becomes
or
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
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 coefficients of the three forms of Fourier series (sine-cosine form, amplitude-phase form, and exponential form) are related by
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
or
Equation (8) can be written as
Again, the power dissipated by a $ 1-\Omega $ resistance is
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.
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),
and
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
Some properties of the sinc function are important here. For zero argument, the value of the sinc function is unity,
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,
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
while the phase is
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 $.
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.
$$f(t)=a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega_{0} t+b_{n} \sin n \omega_{0} t\right) \tag{A}$$
$$\begin{aligned}\cos n \omega_{0} t &=\frac{1}{2}\left[e^{j n \omega_{0} t}+e^{-j n \omega_{0} t}\right] \\
\sin n \omega_{0} t &=\frac{1}{2 j}\left[e^{j n \omega_{0} t}-e^{-j n \omega_{0} t}\right]\end{aligned} \tag{1}$$
$$f(t)=a_{0}+\frac{1}{2} \sum_{n=1}^{\infty}\left[\left(a_{n}-j b_{n}\right) e^{j n \omega_{0} t}+\left(a_{n}+j b_{n}\right) e^{-j n \omega_{0} t}\right] \tag{2}$$
$$c_{0}=a_{0}, \quad c_{n}=\frac{\left(a_{n}-j b_{n}\right)}{2}, \quad c_{-n}=c_{n}^{*}=\frac{\left(a_{n}+j b_{n}\right)}{2} \tag{3}$$
$$f(t)=c_{0}+\sum_{n=1}^{\infty}\left(c_{n} e^{j n \omega_{0} t}+c_{-n} e^{-j n \omega_{0} t}\right) \tag{4}$$
$$f(t)=\sum_{n=-\infty}^{\infty} c_{n} e^{j n \omega_{0} t} \tag{5}$$
$$c_{n}=\frac{1}{T} \int_{0}^{T} f(t) e^{-j n \omega_{0} t} d t \tag{6}$$
$$A_{n} \angle \phi_{n}=a_{n}-j b_{n}=2 c_{n}$$or$$c_{n}=\left|c_{n}\right| \angle \theta_{n}=\frac{\sqrt{a_{n}^{2}+b_{n}^{2}}}{2} \angle -\tan ^{-1} b_{n} / a_{n} \tag{7}$$
$$\begin{aligned}F_{\mathrm{rms}}^{2} &=\frac{1}{T} \int_{0}^{T} f^{2}(t) d t=\frac{1}{T} \int_{0}^{T} f(t)\left[\sum_{n=-\infty}^{\infty} c_{n} e^{j n \omega_{0} t}\right] d t \\&=\sum_{n=-\infty}^{\infty} c_{n}\left[\frac{1}{T} \int_{0}^{T} f(t) e^{j n \omega_{0} t}\right] \\&=\sum_{n=-\infty}^{\infty} c_{n} c_{n}^{*}=\sum_{n=-\infty}^{\infty}\left|c_{n}\right|^{2}\end{aligned} \tag{8}$$
$$F_{\mathrm{ms}}=\sqrt{\sum_{n=-\infty}^{\infty}\left|c_{n}\right|^{2}} \tag{9}$$
$$F_{\mathrm{ms}}^{2}=\left|c_{0}\right|^{2}+2 \sum_{n=1}^{\infty}\left|c_{n}\right|^{2} \tag{10}$$
$$P_{1 \Omega}=F_{\text {rms }}^{2}=\sum_{n=-\infty}^{\infty}\left|c_{n}\right|^{2} \tag{11}$$
Fig. 1: The periodic pulse train.
$$\begin{aligned}c_{n} &=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) e^{-j n \omega_{0} t} d t=\frac{1}{10} \int_{-1}^{1} 10 e^{-j n \omega_{0} t} d t \\&=\left.\frac{1}{-j n \omega_{0}} e^{-j n \omega_{0} t}\right|_{-1} ^{1}=\frac{1}{-j n \omega_{0}}\left(e^{-j n \omega_{0}}-e^{j n \omega_{0}}\right) \\&=\frac{2}{n \omega_{0}} \frac{e^{j n \omega_{0}}-e^{-j n \omega_{0}}}{2 j}=2 \frac{\sin n \omega_{0}}{n \omega_{0}}, \quad \omega_{0}=\frac{\pi}{5} \\&=2 \frac{\sin n \pi / 5}{n \pi / 5}\end{aligned} \tag{12}$$
$$f(t)=2 \sum_{n=-\infty}^{\infty} \frac{\sin n \pi / 5}{n \pi / 5} e^{j n \pi t / 5} \tag{13}$$
$$\operatorname{sinc}(x)=\frac{\sin x}{x} \tag{14}$$
$$\operatorname{sinc}(0)=1 \tag{15}$$
$$\operatorname{sinc}(n \pi)=0, \quad n=1,2,3, \ldots \tag{16}$$
$$\left|c_{n}\right|=2\left|\frac{\sin n \pi / 5}{n \pi / 5}\right| \tag{17}$$
$$\theta_{n}=\left\{\begin{array}{rc}0^{\circ}, & \sin \frac{n \pi}{5} > 0 \\180^{\circ}, & \sin \frac{n \pi}{5}< 0 \end{array}\right. \tag{18}$$
Fig. 2: The amplitude of a periodic pulse train.
Fig. 3: The phase spectrum of a periodic pulse train.
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,
But by Euler's identity,
Thus,
The complex Fourier series is
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
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.

$$\begin{array}{l}c_{n}=\frac{1}{T} \int_{0}^{T} f(t) e^{-j n \omega \omega_{0} t} d t=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{t} e^{-j n t} d t \\=\left.\frac{1}{2 \pi} \frac{1}{1-j n} e^{(1-j n) t}\right|_{0} ^{2 \pi}=\frac{1}{2 \pi(1-j n)}\left[e^{2 \pi} e^{-j 2 \pi n}-1\right] \\\end{array}$$
$$e^{-j 2 \pi n}=\cos 2 \pi n-j \sin 2 \pi n=1-j 0=1$$
$$c_{n}=\frac{1}{2 \pi(1-j n)}\left[e^{2 \pi}-1\right]=\frac{85}{1-j n}$$
$$f(t)=\sum_{n=-\infty}^{\infty} \frac{85}{1-j n} e^{j n t}$$
$$\left|c_{n}\right|=\frac{85}{\sqrt{1+n^{2}}}, \quad \theta_{n}=\tan ^{-1} n$$

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.
Solution: From Fig. 5, $ f(t)=t, 0 < t < 1, T=1 $ so that $ \omega_{0}=2 \pi / T=2 \pi $. Hence,
But
Applying this to Eq. (2.1) gives
Again,
so that Eq. (2.2) becomes
This does not include the case when $ n=0 $. When $ n=0 $,
Hence,
and
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. 5: For Example 2.
$$c_{n}=\frac{1}{T} \int_{0}^{T} f(t) e^{-j n \omega o s t} d t=\frac{1}{1} \int_{0}^{1} t e^{-j 2 n \pi t} d t \tag{2.1}$$
$$\int t e^{a t} d t=\frac{e^{a t}}{a^{2}}(a x-1)+C$$
$$\begin{aligned}c_{n} &=\left.\frac{e^{-j 2 n \pi t}}{(-j 2 n \pi)^{2}}(-j 2 n \pi t-1)\right|_{0} ^{1} \\&=\frac{e^{-j 2 n \pi}(-j 2 n \pi-1)+1}{-4 n^{2} \pi^{2}}\end{aligned} \tag{2.2}$$
$$e^{-j 2 \pi n}=\cos 2 \pi n-j \sin 2 \pi n=1-j 0=1$$
$$c_{n}=\frac{-j 2 n \pi}{-4 n^{2} \pi^{2}}=\frac{j}{2 n \pi}$$
$$c_{0}=\frac{1}{T} \int_{0}^{T} f(t) d t=\frac{1}{1} \int_{0}^{1} t d t=\left.\frac{t^{2}}{2}\right|_{1} ^{0}=0.5$$
$$f(t)=0.5+\sum_{\substack{n=-\infty \\ n \neq 0}}^{\infty} \frac{j}{2 n \pi} e^{j 2 n \pi t}$$
$$\left|c_{n}\right|=\left\{\begin{array}{ll}\frac{1}{2|n| \pi}, & n \neq 0 \\0.5, & n=0\end{array}, \quad \theta_{n}=90^{\circ}, \quad n \neq 0\right.$$

Fig. 6: For Example 2: (a) amplitude spectrum, (b) phase spectrum.
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