The Fourier series of functions is used to find the steady-state response of a circuit. There are four different types of symmetry that can be used to simplify the process of evaluating the Fourier coefficients.

The Effect of Symmetry

  1.  Even-function symmetry
  2. Odd-function symmetry
  3. Half-wave symmetry
  4. Quarter-wave symmetry

Even-Function Symmetry

A function is defined to be even if and only if

f(t) = f(-t)     1.1

If a function satisfies Eq. 1.1, then it is said to be even because polynomial functions with only even exponents have this type of behavior. For any even periodic functions, the equations for the Fourier coefficients simplify to the following:


$$a_{v} = \frac{2}{T}\int_{0}^{T/2} f(t)dt.$$     (1.2)

$$a_{k} = \frac{4}{T}\int_{0}^{T/2} f(t)\cos k\omega _{0}tdt.$$     (1.3)

$$b_{k} = 0$$ for all k     (1.4)


Noting for Eq. 1.4, that all coefficients are zero if the function is even. Below, Fig. 1.1 depicts an even periodic function. The two derivatives below follow exactly from Eq. 1.2 - 1.4. Through each derivation, $$t_{0} = -T/2$$ is selected and then we break the interval of integration into the range from -T/2 to 0 and 0 to T/2, or as follows


$$a_{v} = \frac{1}{T}\int_{-T/2}^{T/2} f(t)dt$$

$$= \frac{1}{T}\int_{-T/2}^{0}f(t)dt + \int_{0}^{T/2}f(t)dt.$$     (1.5) 

Figure 1.1 Even function of f(t) = f(-t)



Now, the variable of integration must be changed in the first integral on the right-hand side of Eq. 1.5. Particularly, we can let t = - and observe that f(t) = f(-x) = f(x) due to the fact that the function is even. Also noting that = T/2 when t = - T/2 and dt -dx. Thus


$$\int_{-T/2}^{0}f(t)dt = \int_{T/2}^{0}f(x)(-dx) = \int_{0}^{T/2}f(x)dx.$$     (1.6)


which does show that integrating from -T/2 to 0 is the same as integrating from 0 to T/2. Thus Eq. 1.5 is the same as Eq. 1.2. Deriving Eq. 1.3 can be completed as follows:


$$a_{k} = \frac{2}{T}\int_{-T/2}^{0}f(t)\cos k\omega _{0}tdt + \frac{2}{T}\int_{0}^{T/2}\cos k\omega _{0}tdt$$     (1.7)


$$\int_{-T/2}^{0}f(t)\cos k\omega _{0}tdt = \int_{T/2}^{0}f(x)\cos (-k\omega _{0}x)(-dx)$$

$$=-\int_{0}^{T/2}f(x)\cos k\omega _{0}xdx.$$     (1.8)


Similarly, as before, integrating from -T/2 to 0 is identical as integrating from 0 to T/2. By combining Eq. 1.7 with Eq. 1.8, Eq. 1.3 is produced. After this, all the b coefficients are zero when f(t) is an even periodic function, because integrating from -T/2 to 0 is the exact negative of the integration from 0 to T/2. Thus,


$$\int_{-T/2}^{0}f(t)\sin k\omega _{0}tdt = \int_{T/2}^{0}f(x)\sin (-k\omega _{0}x)(-dx)$$ 

$$=-\int_{0}^{T/2}f(x)\sin k\omega _{0}xdx.$$     (1.9)


Now, if Eqs. 1.2 and 1.3 are used to find the Fourier coefficients, the integration interval must be between 0 and T/2.


Odd-Function Symmetry

A periodic function is defined to be odd if

f(t) = -f(t)     (1.10)

A Function that satisfies Eq. 1.10 is said to be odd due to the fact that polynomial functions with only odd exponents behave this way. The expressions for the Fourier coefficients are as follows

$$a_{v} = 0;$$     (1.11)

$$a_{k}= 0,$$ for all k;     (1.12)

$$b_{k}=\frac{4}{T}\int_{0}^{T/2}f(t)\sin k\omega _{0}dt.$$     (1.13)

Figure 1.2


Looking at Eqs. 1.11 - 1.13, all the coefficients are zero if the periodic function is odd. The figure shown above illustrates an odd periodic function. The same method of derivation is used on Eqs. 1.11 - 1.13 as was used in the derivation of Eqs. 1.2 - 1.4. 

The evenness (oddness) of a function can be dismantled by shifting the periodic function along the time axis. Essentially this means that the wise choice of where t = 0 might give a function either odd or even symmetry. For instance, the triangular function in Fig 1.3 (a) is not even or odd. Nevertheless, the function can be made even, as illustrated in Fig 1.3 (b), or odd, as shown in Fig 1.3 (c). 


Figure 1.3 


Half-Wave Symmetry

A function is said to have half-wave symmetry if it satisfies the following constraint:

f(t) = -f(t - T/2)      (1.14)

Equation 1.14 expresses that a periodic function has a half-wave symmetry if, after it has been shifted by one-half of a period and inverted, it is said to be identical to the original periodic function. For instance, the periodic functions illustrated in Figures 1.2 and 1.3 possess half-wave symmetry, whereas those functions in Figures 1.4 and 1.5 do not possess such symmetry. For = 0, the half-wave symmetry does not exist as a function. 

If a given function does possess a half-wave symmetry, both a and bare defined as zero for an even value of k. Similarly, ais also zero due to the fact that an average value of a periodic function with this symmetry is zero. Expressions for the Fourier coefficients are as follows:


$$a_{v}=0,$$     (1.15)

$$a_{k}=0,$$     for k even     (1.16)

$$a_{k}=\frac{4}{T}\int_{0}^{T/2}f(t)\cos k\omega_{0}tdt,$$     for k odd     (1.17)

$$b_{k}=0,$$     for k even     (1.18)

$$b_{k}=\frac{4}{T}\int_{0}^{T/2}f(t)\sin k\omega_{0}tdt,$$     for k odd     (1.19)


The equations are derived from starting with Eqs 1.2 - 1.4 from the previous article, Learn About Fourier Coefficients. An interval of integration from -T/2 to T/2 is chosen and then this range is divided into the intervals -T/2 to 0 and 0 to T/2. 


$$a_{k}= \frac{2}{T}\int_{t_{0}}^{t_{0}+T}f(t)\cos k\omega_{0}tdt$$

$$= \frac{2}{T}\int_{-T/2}^{T/2}f(t)\cos k\omega_{0}tdt$$

$$= \frac{2}{T}\int_{-T/2}^{0}f(t)\cos k\omega_{0}tdt$$

$$+ \frac{2}{T}\int_{0}^{T/2}f(t)\cos k\omega_{0}tdt$$      (1.20)


From here, the variable in the first integral on the right-hand side is changed. 

x - T/2


T/2, if = 0

 = 0, if t = -T/2;

dt = dx                

Rewriting the first integral, 

$$\int_{-T/2}^{0}f(t)\cos k \omega_{0}tdt = \int_{0}^{T/2}f(x - T/2)\cos k\omega_{0}(x - T/2)dx$$     (1.21)


Considering that 

$$\cos k\omega_{0}(x - T/2) = \cos (k\omega_{0}x - k\pi) = \cos k\pi\cos k\omega_{0}x$$


and, by postulating,

f(x - T/2) = -f(Ix)

Thus Eq. 1.21 can now be written as 

$$\int_{-T/2}^{0}f(t)\cos k\omega_{0}tdt = \int_{0}^{T/2}[-f(x)]\cos k\omega_{0}tdt$$     (1.22)


By including Eq. 1.22 into Eq 1.20, 

$$a_{k}= \frac{2}{T}(1 - \cos k\pi)\int_{0}^{T/2}f(t)\cos k\omega_{0}tdt$$     (1.23)


However, $$\cos k\pi$$ is equal to 1 if k is even and -1 if is odd. 

To summarize, the representation of the Fourier series of a periodic function with a half-wave symmetry zero average value and only contains odd harmonics. 


Quarter-Wave Symmetry

If a function has half-wave symmetry and symmetry about the midpoint of the positive and negative half-cycles, the periodic function is said to have quarter--wave symmetry. This function is illustrated in Figure 1.4; the function in Fig 1.4(a) is said to have quarter-wave symmetry about the midpoint of the positive and negative half-cycles. The function in Fig 1.4(b) does not have this symmetry, but it does have half-wave symmetry. 

Figure 1.4

A function that possesses quarter-wave symmetry always can be made even or odd by choosing where = 0. For instance, the periodic function in Fig. 1.4(a) is odd and can be turned into an even function by shifting over T/4 unites either left or right along the t-axis. However, because the periodic function in Fig. 1.4(b) only possesses half-wave symmetry, it cannot ever be made even or odd. 

If the periodic function were to be made even, then 


$$a_{v} = 0,$$ due to half-wave symmetry

$$a_{k} = 0,$$ for k even, due to half-wave symmetry

$$a_{k} = \frac{8}{T}\int_{0}^{T/4}f(t) \cos k\omega_{0}tdt,$$ for odd

$$b_{k} = 0,$$ for all k, because the periodic function is even     (1.24)


The above Eqs. 1.24 are the results from the periodic function's symmetry in addition to it being even. If the quarter-wave symmetry is super-imposed on half-wave symmetry, aand afor k even can therefore be eliminated. Taking a look at the expression for ak and k odd, Eq. 1.19 demonstrates that when combining a quarter-wave symmetry with evenness, the range of integration shortens from 0 to T/2 to 0 to T/4. 

If a quarter-wave symmetric periodic function is made odd, 


 $$a_{v}=0,$$ due to the function being odd

$$a_{k}=0,$$ for all k, due to the function being odd

$$b_{k}=0,$$ for k even, due to half-wae symmetry

$$b_{k}=\frac{8}{T}\int_{0}^{T/4}f(t)\sin k\omega_{0}tdt,$$ for odd     (1.25)


The above Eqs of 1.25 come consequently because of quarter-wave symmetry as well as oddness. Similarly to the evenness, the quarter-wave symmetry allows for the interval of integration from 0 to T/2 to 0 to T/4 to be shortened. 


Coming Up

As of now, you should have a better understanding of the Fourier coefficients and the different types of symmetry that can happen. These five types, even, odd, half-wave, quarter-wave half-wave even, and quarter-wave half-wave odd are all used to simplify the computation of the Fourier coefficients. A few topics that will be covered next will go in depth to find the steady-state response of a linear circuit from a Fourier series, calculation average power with periodic functions, as well as the rms value of such periodic functions.