17.3 Symmetry Considerations

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17.3 Symmetry Considerations

We noticed that the Fourier series of Example 17.1 consisted only of the sine terms. One may w onder if a method e xists whereby one can kno w in advance that some F ourier coefficients would be zero and a void the unnecessary work involved in the tedious process of calculating them. Such a method does e xist; it is based on recognizing the e xistence of symmetry. Here we discuss three types of symmetry: (1) even symmetry, (2) odd symmetry, (3) half-wave symmetry.

17.3.1 Even Symmetry

A function f(t) is even if its plot is symmetrical about the vertical axis; that is,

$$f(t) = f(-t) \tag{17.16}$$

(17.18)

Examples of e ven functions are t 2 , t 4 , and cos t. Figure 17.10 sho ws more e xamples of periodic e ven functions. Note that each of these examples satisfies Eq. (17.16). A main property of an e ven function fe(t) is that:

$$\int_{-T/2}^{T/2} f_e(t) \, dt = 2 \int_0^{T/2} f_e(t) \, dt \tag{17.17}$$

because integrating from −T∕2 to 0 is the same as inte grating from 0 to T∕2. Utilizing this property, the Fourier coefficients for an even function become

$$a_0 = \frac{2}{T} \int_0^{T/2} f(t)dt$$

\n

$$a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t dt$$

\n

$$b_n = 0$$

Because bn = 0, Eq. (17.3) becomes a Fourier cosine series. This makes sense because the cosine function is itself e ven. It also mak es intuitive sense that an e ven function contains no sine terms gi ven that the sine function is odd.

To confirm Eq. (17.18) quantitatively, we apply the property of an even function in Eq. (17.17) in evaluating the Fourier coefficients in Eqs. (17.6), (17.8), and (17.9). It is convenient in each case to integrate over the interval −T∕2 < t < T∕2, which is symmetrical about the origin. Thus,

$$a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) dt = \frac{1}{T} \left[ \int_{-T/2}^{0} f(t) dt + \int_{0}^{T/2} f(t) dt \right]$$

(17.19)

We change variables for the inte gral over the interval −T∕2 < t < 0 by letting t = −x, so that dt = −dx, f(t) = f(−t) = f(x), since f(t) is an even function, and when t = −T∕2, x = T∕2. Then,

$$a_0 = \frac{1}{T} \left[ \int_{T/2}^0 f(x)(-dx) + \int_0^{T/2} f(t) dt \right]$$

=

$$\frac{1}{T} \left[ \int_0^{T/2} f(x) dx + \int_0^{T/2} f(t) dt \right]$$

(17.20)

showing that the two integrals are identical. Hence,

$$a_0 = \frac{2}{T} \int_0^{T/2} f(t) dt$$

(17.21)

as expected. Similarly, from Eq. (17.8),

$$a_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \cos n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \cos n\omega_0 t \, dt \right]$$

(17.22)

Typical examples of even periodic functions.

We make the same change of v ariables that led to Eq. (17.20) and note that both f(t) and cos nω0t are even functions, implying that f(−t) = f(t) and cos(−nω0t) = cos nω0t. Equation (17.22) becomes

$$a_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \cos(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]$$

=

$$\frac{2}{T} \left[ \int_{T/2}^0 f(x) \cos(n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]$$

=

$$\frac{2}{T} \left[ \int_0^{T/2} f(x) \cos(n\omega_0 x) dx + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]$$

(17.23a)

or

t

$$a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \tag{17.23b}$$

as expected. For bn, we apply Eq. (17.9),

$$b_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \sin n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \sin n\omega_0 t \, dt \right] \tag{17.24}$$

We make the same change of variables but keep in mind that f(−t) = f(t) but sin(−nω0t) = −sin nω0t. Equation (17.24) yields

$$b_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \sin(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$$

= $\frac{2}{T} \left[ \int_{T/2}^0 f(x) \sin n\omega_0 x dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$
= $\frac{2}{T} \left[ - \int_0^{T/2} f(x) \sin (n\omega_0 x) dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$
= 0 (17.25)

confirming Eq. (17.18).

17.3.2 Odd Symmetry

A function f(t) is said to be odd if its plot is antisymmetrical about the vertical axis:

$$f(-t) = -f(t) \tag{17.26}$$

Examples of odd functions are t, t 3 , and sin t. Figure 17.11 sho ws more examples of periodic odd functions. All these e xamples satisfy Eq. (17.26). An odd function fo(t) has this major characteristic:

∫

$$f_{-T/2}^{T/2} f_o(t) dt = 0$$

\n(17.27)

f(t)

(c)

Figure 17.11 Typical examples of odd periodic functions.

because integration from −T∕2 to 0 is the negative of that from 0 to T∕2. With this property, the Fourier coefficients for an odd function become

$$a_0 = 0, \t a_n = 0$$ $$b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt$$

(17.28)

which give us a Fourier sine series. Again, this makes sense because the sine function is itself an odd function. Also, note that there is no dc term for the Fourier series expansion of an odd function.

The quantitati ve proof of Eq. (17.28) follo ws the same proce dure tak en to pro ve Eq. (17.18) e xcept that f(t) is no w odd, so that f(t) = −f(t). With this fundamental b ut simple difference, it is easy to see that a0 = 0 in Eq. (17.20), an = 0 in Eq. (17.23a), and bn in Eq. (17.24) becomes

$$b_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \sin(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$$ $$= \frac{2}{T} \left[ -\int_{T/2}^0 f(x) \sin n\omega_0 x \, dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$$ $$= \frac{2}{T} \left[ \int_0^{T/2} f(x) \sin(n\omega_0 x) \, dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$$ $$b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \qquad (17.29)$$

as expected.

It is interesting to note that an y periodic function f(t) with neither even nor odd symmetry may be decomposed into e ven and odd parts. Using the properties of even and odd functions from Eqs. (17.16) and (17.26), we can write

$$f(t) = \underbrace{\frac{1}{2} [f(t) + f(-t)]}_{\text{even}} + \underbrace{\frac{1}{2} [f(t) - f(-t)]}_{\text{odd}} = f_e(t) + f_o(t)$$

(17.30)

Notice that fe(t) = __1 ² [ f(t) + f(−t)] satisfies the property of an even function in Eq. (17.16), while fo(t) = __1 ² [ f(t) − f(−t)] satisfies the property of an odd function in Eq. (17.26). The fact that fe(t) contains only the dc term and the cosine terms, while fo(t) has only the sine terms, can be e xploited in grouping the F ourier series expansion of f(t) as

$$f(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + \sum_{n=1}^{\infty} b_n \sin n\omega_0 t = f_e(t) + f_o(t)$$

(17.31)
even
odd

It follows readily from Eq. (17.31) that when f(t) is e ven, bn = 0, and when f(t) is odd, a0 = 0 = an.

Also, note the following properties of odd and even functions:

• 1. The product of two even functions is also an even function.
• 2. The product of two odd functions is an even function.
• 3. The product of an e ven function and an odd function is an odd function.
• 4. The sum (or dif ference) of two even functions is also an even function.
• 5. The sum (or difference) of two odd functions is an odd function.
• 6. The sum (or difference) of an even function and an odd function is neither even nor odd.

Each of these properties can be proved using Eqs. (17.16) and (17.26).

17.3.3 Half-Wave Symmetry

A function is half-wave (odd) symmetric if

$$f\left(t - \frac{T}{2}\right) = -f(t) \tag{17.32}$$

which means that each half-c ycle is the mirror image of the ne xt halfcycle. Notice that functions cos nω0t and sin nω0t satisfy Eq. (17.32) for odd values of n and therefore possess half-wave symmetry when n is odd. Figure 17.12 sho ws other examples of half-wave symmetric func tions. The functions in Figs. 17.11(a) and 17.11(b) are also half-w ave symmetric. Notice that for each function, one half-c ycle is the inverted version of the adjacent half-cycle. The Fourier coefficients become

Figure 17.12 Typical examples of half-wave odd symmetric functions.

showing that the Fourier series of a half-wave symmetric function con tains only odd harmonics.

To deri ve Eq. (17.33), we apply the property of half-w ave sym metric functions in Eq. (17.32) in e valuating the Fourier coefficients in Eqs. (17.6), (17.8), and (17.9). Thus,

$$a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) dt = \frac{1}{T} \left[ \int_{-T/2}^{0} f(t) dt + \int_{0}^{T/2} f(t) dt \right]$$

(17.34)

We change v ariables for the inte gral o ver the interv al −T∕2 < t < 0 by letting x = t + T∕2, so that dx = dt; when t = −T∕2, x = 0; and when t = 0, x = T∕2. Also, we k eep Eq. (17.32) in mind; that is, f(x − T∕2) = −f (x). Then,

$$a_0 = \frac{1}{T} \left[ \int_0^{T/2} f\left(x - \frac{T}{2}\right) dx + \int_0^{T/2} f(t) dt \right]$$

= $\frac{1}{T} \left[ - \int_0^{T/2} f(x) dx + \int_0^{T/2} f(t) dt \right] = 0$ (17.35)

confirming the expression for a0 in Eq. (17.33). Similarly,

$$a_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \cos n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \cos n\omega_0 t \, dt \right] \quad (17.36)$$

We make the same change of variables that led to Eq. (17.35) so that Eq. (17.36) becomes

$$a_n = \frac{2}{T} \left[ \int_0^{T/2} f\left(x - \frac{T}{2}\right) \cos n\omega_0 \left(x - \frac{T}{2}\right) dx + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right]$$

(17.37)

Because f(x − T∕2) = −f(x) and

$$\cos n\omega_0 \left( x - \frac{T}{2} \right) = \cos(n\omega_0 t - n\pi)$$

= $\cos n\omega_0 t \cos n\pi + \sin n\omega_0 t \sin n\pi$ (17.38)
= $(-1)^n \cos n\omega_0 t$

substituting these in Eq. (17.37) leads to

$$a_n = \frac{2}{T} \left[ 1 - (-1)^n \right] \int_0^{T/2} f(t) \cos n\omega_0 t \, dt$$

=

$$\begin{cases} \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t \, dt, & \text{for } n \text{ odd} \\ 0, & \text{for } n \text{ even} \end{cases}$$

(17.39)

confirming Eq. (17.33). By following a similar procedure, we can derive bn as in Eq. (17.33).

Table 17.2 summarizes the ef fects of these symmetries on the Fourier coef ficients. Table 17.3 pro vides the F ourier series of some common periodic functions.

TABLE 17.2

Effects of symmetry on Fourier coefficients.

Symmetry	a0	an	bn	Remarks
Even	a0 ≠ 0	an ≠ 0	bn = 0	Integrate over T∕2 and multiply by 2 to get the coefficients.
Odd	a0 = 0	an = 0	bn ≠ 0	Integrate over T∕2 and multiply by 2 to get the coefficients.
Half-wave	a0 = 0	a2n = 0	b2n = 0	Integrate over T∕2 and multiply by 2 to get the coefficients.
		a2n+1 ≠ 0	b2n+1 ≠ 0

TABLE 17.3

Find the Fourier series expansion of f(t) given in Fig. 17.13.

Figure 17.13 For Example 17.3.

Solution:

The function f(t) is an odd function. Hence a0 = 0 = an. The period is T = 4, and ω0 = 2π∕T = π∕2, so that

$$b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt$$

= $\frac{4}{4} \left[ \int_0^1 1 \sin \frac{n\pi}{2} t \, dt + \int_1^2 0 \sin \frac{n\pi}{2} t \, dt \right]$
= $-\frac{2}{n\pi} \cos \frac{n\pi t}{2} \Big|_0^1 = \frac{2}{n\pi} \left( 1 - \cos \frac{n\pi}{2} \right)$

Hence,

$$f(t) = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left( 1 - \cos \frac{n\pi}{2} \right) \sin \frac{n\pi}{2} t$$

which is a Fourier sine series.

Find the Fourier series of the function f(t) in Fig. 17.14.

t f(t) ‒2𝜋 ‒𝜋 𝜋 0 2𝜋 3𝜋 20 ‒20

Figure 17.14 For Practice Prob. 17.3.

Answer:

$$f(t) = -\frac{80}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin nt$$

, $n = 2k - 1$ .

Practice Problem 17.3

Example 17.4

Determine the Fourier series for the half-w ave rectified cosine function shown in Fig. 17.15.

A half-wave rectified cosine function; for Example 17.4.

Solution:

This is an even function so that bn = 0. Also, T = 4, ω0 = 2π∕T = π∕2. Over a period,

$$f(t) = \begin{cases} 0, & -2 < t < -1 \\ \cos \frac{\pi}{2} t, & -1 < t < 1 \end{cases}$$ $$a_0 = \frac{2}{T} \int_0^{T/2} f(t) dt = \frac{2}{4} \left[ \int_0^1 \cos \frac{\pi}{2} t dt + \int_1^2 0 dt \right]$$ $$= \frac{1}{2} \frac{2}{\pi} \sin \frac{\pi}{2} t \Big|_0^1 = \frac{1}{\pi}$$ $$a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t dt = \frac{4}{4} \left[ \int_0^1 \cos \frac{\pi}{2} t \cos \frac{n\pi t}{2} dt + 0 \right]$$

Put $\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$ . Then

But cos A cos B = __1 2 [cos(A + B) + cos(A − B)]. Then an = __1 2 ∫ 0 1 [ cos __ π 2 (n + 1)t + cos __ π 2 (n − 1)t ] dt

For n = 1,

$$a_1 = \frac{1}{2} \int_0^1 \left[ \cos \pi t + 1 \right] dt = \frac{1}{2} \left[ \frac{\sin \pi t}{\pi} + t \right] \Big|_0^1 = \frac{1}{2}$$

For n > 1,

$$a_n = \frac{1}{\pi(n+1)} \sin \frac{\pi}{2} (n+1) + \frac{1}{\pi(n-1)} \sin \frac{\pi}{2} (n-1)$$

For n = odd (n = 1, 3, 5, …), (n + 1) and (n − 1) are both even, so

$$\sin \frac{\pi}{2}(n+1) = 0 = \sin \frac{\pi}{2}(n-1), \quad n = \text{odd}$$

For n = even (n = 2, 4, 6, …), (n + 1) and (n − 1) are both odd. Also,

$$\sin\frac{\pi}{2}(n+1) = -\sin\frac{\pi}{2}(n-1) = \cos\frac{n\pi}{2} = (-1)^{n/2}, \qquad n = \text{even}$$

Hence,

$$a_n = \frac{(-1)^{n/2}}{\pi(n+1)} + \frac{-(-1)^{n/2}}{\pi(n-1)} = \frac{-2(-1)^{n/2}}{\pi(n^2 - 1)}, \qquad n = \text{even}$$

Thus,

$$f(t) = \frac{1}{\pi} + \frac{1}{2}\cos\frac{\pi}{2}t - \frac{2}{\pi}\sum_{n=\text{even}}^{\infty}\frac{(-1)^{n/2}}{(n^2 - 1)}\cos\frac{n\pi}{2}t$$

To avoid using n = 2, 4, 6, … and also to ease computation, we can replace n by 2k, where k = 1, 2, 3, … and obtain

$$f(t) = \frac{1}{\pi} + \frac{1}{2}\cos\frac{\pi}{2}t - \frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^k}{(4k^2 - 1)}\cos k\pi t$$

which is a Fourier cosine series.

Find the Fourier series expansion of the function in Fig. 17.16.

Answer:

$$f(t) = 16 - \frac{128}{\pi^2} \sum_{k=1}^{\infty} \frac{1}{n^2} \cos nt, n = 2k - 1.$$

Figure 17.16 For Practice Prob. 17.4.

Calculate the Fourier series for the function in Fig. 17.17.

Solution:

The function in Fig. 17.17 is half-wave odd symmetric, so that a0 = 0 = an. It is described over half the period as

$$f(t) = t, \qquad -1 < t < 1$$

T = 4, ω0 = 2π∕T = π∕2. Hence,

$$b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt$$

Instead of integrating f(t) from 0 to 2, it is more convenient to integrate from −1 to 1. Applying Eq. (17.15d),

$$b_n = \frac{4}{4} \int_{-1}^{1} t \sin \frac{n\pi t}{2} dt = \left[ \frac{\sin n\pi t/2}{n^2 \pi^2/4} - \frac{t \cos n\pi t/2}{n\pi/2} \right] \Big|_{-1}^{1}$$

= $\frac{4}{n^2 \pi^2} \left[ \sin \frac{n\pi}{2} - \sin \left( -\frac{n\pi}{2} \right) \right] - \frac{2}{n\pi} \left[ \cos \frac{n\pi}{2} - \cos \left( -\frac{n\pi}{2} \right) \right]$
= $\frac{8}{n^2 \pi^2} \sin \frac{n\pi}{2}$

since sin(−x) = −sin x is an odd function, while cos( −x) = cos x is an even function. Using the identities for sin nπ∕2 in Table 17.1,

$$b_n = \frac{8}{n^2 \pi^2} (-1)^{(n-1)/2}, \quad n = \text{odd} = 1, 3, 5, ...$$

Figure 17.17 For Example 17.5.

Thus,

$$f(t) = \sum_{n=1,3,5}^{\infty} b_n \sin \frac{n\pi}{2} t.$$

Practice Problem 17.5

Determine the Fourier series of the function in Fig. 17.12(a). Take A = 8 and T = 2π.

Answer:

$$f(t) = \frac{16}{\pi} \sum_{k=1}^{\infty} \left( \frac{-2}{n^2 \pi} \cos nt + \frac{1}{n} \sin nt \right), n = 2k - 1.$$

17.4 Circuit Applications

We find that in practice, many circuits are driven by nonsinusoidal periodic functions. To find the steady-state response of a circuit to a nonsinusoidal periodic excitation requires the application of a Fourier series, ac phasor analysis, and the superposition principle. The procedure usually involves four steps.

Steps for Applying Fourier Series:

• 1. Express the excitation as a Fourier series.
• 2. Transform the circuit from the time domain to the frequenc y domain.
• 3. Find the response of the dc and ac components in the Fourier series.
• 4. Add the individual dc and ac responses using the superposition principle.

The first step is to determine the Fourier series e xpansion of the excitation. For the periodic v oltage source sho wn in Fig. 17.18(a), for example, the Fourier series is expressed as

$$v(t) = V_0 + \sum_{n=1}^{\infty} V_n \cos(n\omega_0 t + \theta_n)$$

(17.40)

(The same could be done for a periodic current source.) Equation (17.40) shows that v(t) consists of tw o parts: the dc component V0 and the ac component Vn = Vn⧸θn with several harmonics. This Fourier series representation may be re garded as a set of series-connected sinusoidal sources, with each source ha ving its own amplitude and frequency, as shown in Fig. 17.18(b).

The third step is finding the response to each term in the Fourier series. The response to the dc component can be determined in the

(a) Linear network excited by a periodic voltage source, (b) Fourier series representation (time-domain).

frequency domain by setting n = 0 or ω = 0 as in Fig. 17.19(a), or in the time domain by replacing all inductors with short circuits and all capacitors with open circuits. The response to the ac component is obtained by applying the phasor techniques co vered in Chapter 9, as shown in Fig. 17.19(b). The network is represented by its impedance Z(nω0) or admittance Y(nω0). Z(nω0) is the input impedance at the source when ω is everywhere replaced by nω0, and Y(nω0) is the reciprocal of Z(nω0).

Finally, following the principle of superposition, we add all the individual responses. For the case shown in Fig. 17.19,

$$i(t) = i_0(t) + i_1(t) + i_2(t) + \cdots$$

= $\mathbf{I}_0 + \sum_{n=1}^{\infty} |\mathbf{I}_n| \cos(n\omega_0 t + \psi_n)$ (17.41)

where each component In with frequenc y nω0 has been transformed to the time domain to get in(t), and ψn is the argument of In.

Figure 17.19

Steady-state responses: (a) dc component, (b) ac component (frequency domain).

Let the function f(t) in Example 17.1 be the v oltage source vs(t) in the circuit of Fig. 17.20. Find the response vo(t) of the circuit.

Solution: