01.05.2017 Views

563489578934

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

68<br />

Signals and Spectra Chap. 2<br />

Orthogonal Functions<br />

Before the orthogonal series is studied, we define orthogonal functions.<br />

DEFINITION. Functions w n 1t2 and w m 1t2 are said to be orthogonal with respect to each<br />

other over the interval a 6 t 6 b if they satisfy the condition<br />

w n (t)w m<br />

*(t) dt = 0, where n Z m<br />

(2–77)<br />

La<br />

Furthermore, if the functions in the set {w n 1t2} are orthogonal, then they also satisfy the<br />

relation<br />

where<br />

b<br />

b<br />

w n (t)w* m (t) dt = e 0,<br />

La<br />

K n ,<br />

n Z m<br />

n = m f = K nd nm<br />

(2–78)<br />

d nm ! e 0, n Z m<br />

(2–79)<br />

1, n = m f<br />

Here, d nm is called the Kronecker delta function. If the constants K n are all equal to 1,<br />

the w are said to be orthonormal functions. †<br />

n 1t2<br />

Equation (2–77) is used to test pairs of functions to see if they are orthogonal. Any such<br />

pair of functions is orthogonal over the interval (a, b) if the integral of the product of the functions<br />

is zero. The zero result implies that these functions are “independent” or in “disagreement.”<br />

If the result is not zero, they are not orthogonal, and consequently, the two functions<br />

have some “dependence” on, or “alikeness” to each other.<br />

Example 2–12 ORTHOGONAL COMPLEX EXPONENTIAL FUNCTIONS<br />

Show that the set of complex exponential functions {e jnv0t } are orthogonal over the interval<br />

a 6 t 6 b, where b = a + T 0 , T 0 = 1/f 0 , w 0 = 2pf 0 , and n is an integer.<br />

Solution. Substituting w and w m 1t2 = e jnv 0t<br />

n 1t2 = e jnv 0t<br />

into Eq. (2–77), we get<br />

b<br />

a+T 0<br />

a+T 0<br />

w n (t)w m *(t) dt = e jnv0t e -jnv 0t<br />

dt = e j(n-m)v 0t<br />

dt<br />

La<br />

L a<br />

L a<br />

For m Z n, Eq. (2–80) becomes<br />

a+T 0<br />

(2–80)<br />

e j(n-m)v 0t<br />

dt = ej(n-m)v0a Ce j(n-m)2p - 1D<br />

= 0<br />

(2–81)<br />

La<br />

j(n - m)v 0<br />

since e j(n-m)2p = cos [2p(n - m)] + j sin [2p(n - m)] = 1. Thus, Eq. (2–77) is satisfied, and<br />

consequently, the complex exponential functions are orthogonal to each other over the interval<br />

† To normalize a set of functions, we take each old w n 1t2 and divide it by 2K n to form the normalized w n 1t2.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!