17.6 Exponential Fourier Series

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17.6 Exponential Fourier Series

A compact way of expressing the Fourier series in Eq. (17.3) 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:

$$\cos n\omega_0 t = \frac{1}{2} \left[ e^{jn\omega_0 t} + e^{-jn\omega_0 t} \right]$$

(17.54a)

$$\sin n\omega_0 t = \frac{1}{2j} \left[ e^{jn\omega_0 t} - e^{-jn\omega_0 t} \right]$$

(17.54b)

Practice Problem 17.9

Practice Problem 17.8

Example 17.9

Substituting Eq. (17.54) into Eq. (17.3) and collecting terms, we obtain

$$f(t) = a_0 + \frac{1}{2} \sum_{n=1}^{\infty} \left[ (a_n - jb_n)e^{jn\omega_0 t} + (a_n + jb_n)e^{-jn\omega_0 t} \right]$$

(17.55)

If we define a new coefficient cn so that

$$c_0 = a_0$$

, $c_n = \frac{(a_n - jb_n)}{2}$ , $c_{-n} = c_n^* = \frac{(a_n + jb_n)}{2}$ (17.56)

then f(t) becomes

$$f(t) = c_0 + \sum_{n=1}^{\infty} (c_n e^{jn\omega_0 t} + c_{-n} e^{-jn\omega_0 t})$$

(17.57)

or

or

$$f(t) = \sum_{n = -\infty}^{\infty} c_n^{ejn\omega_0 t}$$

(17.58)

This is the complex or exponential Fourier series representation of f(t). Note that this e xponential form is more compact than the sine-cosine form in Eq. (17.3). Although the exponential Fourier series coefficients cn can also be obtained from an and bn using Eq. (17.56), they can also be obtained directly from f(t) as

$$c_n = \frac{1}{T} \int_0^T f(t) e^{-e j n \omega_0 t} dt$$

(17.59)

where ω0 = 2π∕T, as usual. The plots of the magnitude and phase of cn versus nω0 are called the complex amplitude spectrum and complex phase spectrum of f (t), respectively. The two spectra form the comple x 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

$$A_n / \underline{\phi_n} = a_n - jb_n = 2c_n \tag{17.60}$$

cn = ∣cn∣⧸θn = ^√ ______ a n 2 + b n 2 ________ 2 ^⧸ −tan−1 bn∕an (17.61)

if only an > 0. Note that the phase θⁿ of cn is equal to n.

In terms of the F ourier complex coefficients cn, the rms v alue of a periodic signal f(t) can be found as

$$F_{\text{rms}}^2 = \frac{1}{T} \int_0^T f^2(t) \, dt = \frac{1}{T} \int_0^T f(t) \left[ \sum_{n=-\infty}^{\infty} c_n e^{jn\omega_0 t} \right] \, dt$$

\n

$$= \sum_{n=-\infty}^{\infty} c_n \left[ \frac{1}{T} \int_0^T f(t) e^{jn\omega_0 t} \, dt \right]$$

\n

$$= \sum_{n=-\infty}^{\infty} c_n c_n^* = \sum_{n=-\infty}^{\infty} |c_n|^2$$

\nor

or

$$F_{\rm rms} = \sqrt{\sum_{n=-\infty}^{\infty} |c_n|^2}$$

(17.63)

Equation (17.62) can be written as

$$F_{\rm rms}^2 = |c_0|^2 + 2 \sum_{n=1}^{\infty} |c_n|^2$$

(17.64)

Again, the power dissipated by a 1-Ω resistance is

$$P_{1\Omega} = F_{\text{rms}}^2 = \sum_{n=-\infty}^{\infty} |c_n|^2$$

(17.65)

which is a restatement of P arseval’s theorem. The power spectrum of the signal f(t) is the plot of ∣cn∣ 2 versus nω0. If f(t) is the voltage across a resistor R, the average power absorbed by the resistor is Frms ² ∕R; if f(t) is the current through R, the power is Frms ² R.

As an illustration, consider the periodic pulse train of Fig. 17.27. Our goal is to obtain its amplitude and phase spectra. The period of the pulse train is T = 10, so that ω0 = 2π∕T = π∕5. Using Eq. (17.59),

$$c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t)e^{-jn\omega_0 t} dt = \frac{1}{10} \int_{-1}^{1} 10e^{-jn\omega_0 t} dt$$ $$= \frac{1}{-jn\omega_0} e^{-jn\omega_0 t} \Big|_{-1}^{1} = \frac{1}{-jn\omega_0} (e^{-jn\omega_0} - e^{jn\omega_0})$$ $$= \frac{2}{n\omega_0} \frac{e^{jn\omega_0} - e^{-jn\omega_0}}{2j} = 2 \frac{\sin n\omega_0}{n\omega_0}, \qquad \omega_0 = \frac{\pi}{5}$$ $$= 2 \frac{\sin n\pi/5}{n\pi/5}$$

(17.66)

‒11 ‒9 ‒1 1 0 9 11 t 10 f(t) Figure 17.27 The periodic pulse train.

and

$$f(t) = 2 \sum_{n = -\infty}^{\infty} \frac{\sin n\pi/5}{n\pi/5} e^{jn\pi t/5}$$

(17.67)

Notice from Eq. (17.66) that cn 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

$$\text{sinc}(x) = \frac{\sin x}{x} \tag{17.68}$$

Some properties of the sinc function are important here. F or zero argument, the value of the sinc function is unity,

$$\text{sinc}(0) = 1\tag{17.69}$$

The sinc function is called the sampling function in communication theory, where it is very useful.

This is obtained by applying L’Hopital’s rule to Eq. (17.68). For an integral multiple of π, the value of the sinc function is zero,

$$sinc(n\pi) = 0, \qquad n = 1, 2, 3, ... \tag{17.70}$$

Also, the sinc function shows even symmetry. With all this in mind, w e can obtain the amplitude and phase spectra of f(t). From Eq. (17.66), the magnitude is

$$|c_n| = 2 \left| \frac{\sin n\pi/5}{n\pi/5} \right| \tag{17.71}$$

while the phase is

$$\theta_n = \begin{cases}\n0^\circ, & \sin \frac{n\pi}{5} > 0 \\ 180^\circ, & \sin \frac{n\pi}{5} < 0\n\end{cases}$$

\n(17.72)

Figure 17.28 shows the plot of ∣cn∣ versus n for n varying from −10 to 10, where n = ω∕ω0 is the normalized frequenc y. Figure 17.29 shows the plot of θn versus n. Both the amplitude spectrum and phase spec trum are called line spectra, because the v alues of ∣cn∣ and θn occur only at discrete v alues of frequencies. The spacing between the lines is ω0. The power spectrum, which is the plot of ∣cn∣ 2 versus nω0, can also be plotted. Notice that the sinc function forms the envelope of the amplitude spectrum.

effect of a circuit on a periodic signal. 2 1.87 │cn│

Examining the input and output spectra allows visualization of the

The amplitude of a periodic pulse train.

Example 17.10

Figure 17.28

Find the exponential Fourier series expansion of the periodic function f(t) = et , 0 < t < 2π with f(t + 2π) = f(t).

Solution:

Because T = 2π, ω0 = 2π∕T = 1. Hence,

$$c_n = \frac{1}{T} \int_0^T f(t)e^{-jn\omega_0 t} dt = \frac{1}{2\pi} \int_0^{2\pi} e^t e^{-jnt} dt$$ $$= \frac{1}{2\pi} \frac{1}{1 - jn} e^{(1 - jn)t} \Big|_0^{2\pi} = \frac{1}{2\pi(1 - jn)} \left[ e^{2\pi} e^{-j2\pi n} - 1 \right]$$

But by Euler’s identity,

$$e^{-j2\pi n} = \cos 2\pi n - j \sin 2 \pi n = 1 - j0 = 1$$

Thus,

$$c_n = \frac{1}{2\pi(1 - jn)} \left[ e^{2\pi} - 1 \right] = \frac{85}{1 - jn}$$

The complex Fourier series is

$$f(t) = \sum_{n = -\infty}^{\infty} \frac{85}{1 - jn} e^{jnt}$$

We may want to plot the complex frequency spectrum of f(t). If we let cn = ∣cn∣ ^⧸θn, then

$$|c_n| = \frac{85}{\sqrt{1 + n^2}}, \qquad \theta_n = \tan^{-1} n$$

By inserting in negative and positive values of n, we obtain the amplitude and the phase plots of cn versus nω0 = n, as in Fig. 17.30.

Figure 17.30

The complex frequency spectrum of the function in Example 17.10: (a) amplitude spectrum, (b) phase spectrum.

Obtain the complex Fourier series of the function in Fig. 17.1.

Practice Problem 17.10

Answer:

$$f(t) = \frac{1}{2} - \sum_{\substack{n=-\infty\\n \neq 0\\n = \text{odd}}} \frac{j}{n\pi} e^{jn\pi t}.$$

Find the complex Fourier series of the sawtooth wave in Fig. 17.9. Plot the amplitude and the phase spectra.

Example 17.11

Solution:

From Fig. 17.9, f(t) = t, 0 < t < 1, T = 1 so that ω0 = 2π∕T = 2π. Hence,

$$c_n = \frac{1}{T} \int_0^T f(t)e^{-jn\omega_0 t} dt = \frac{1}{T} \int_0^1 t e^{-j2n\pi t} dt$$

(17.11.1)

But

$$\int t e^{at} dt = \frac{e^{at}}{a^2} (ax - 1) + C$$

Applying this to Eq. (17.11.1) gives

$$c_n = \frac{e^{-j2n\pi t}}{(-j2n\pi)^2} (-j2n\pi t - 1) \Big|_0^1$$

=

$$\frac{e^{-j2n\pi} (-j2n\pi - 1) + 1}{-4n^2 \pi^2}$$

(17.11.2)

Again,

$$e^{-j2\pi n} = \cos 2\pi n - j \sin 2\pi n = 1 - j0 = 1$$

so that Eq. (17.11.2) becomes

$$c_n = \frac{-j2n\pi}{-4n^2\pi^2} = \frac{j}{2n\pi}$$

(17.11.3)

This does not include the case when n = 0. When n = 0,

$$c_0 = \frac{1}{T} \int_0^T f(t)dt = \frac{1}{1} \int_0^1 t \, dt = \frac{t^2}{2} \Big|_1^0 = 0.5 \tag{17.11.4}$$

Hence,

$$f(t) = 0.5 + \sum_{\substack{n=-\infty\\n\neq 0}}^{\infty} \frac{j}{2n\pi} e^{j2n\pi t}$$

(17.11.5)

and

$$|c_n| = \begin{cases} \frac{1}{2|n|\pi}, & n \neq 0\\ 0.5, & n = 0 \end{cases}, \qquad \theta_n = 90^\circ, \qquad n \neq 0 \qquad (17.11.6)$$

By plotting ∣cn∣ and θn for different n, we obtain the amplitude spectrum and the phase spectrum shown in Fig. 17.31.

Obtain the complex Fourier series expansion of f(t) in Fig. 17.17. Show the amplitude and phase spectra. Practice Problem 17.11

Answer:

$$f(t) = \sum_{\substack{n=-\infty\\n\neq 0}}^{\infty} \frac{j(-1)^n}{n\pi} e^{jn\pi t}$$

. See Fig. 17.32 for the spectra.

Figure 17.32

For Practice Prob. 17.11: (a) amplitude spectrum, (b) phase spectrum.