| written 8.9 years ago by | modified 3.9 years ago by |
Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3
Marks : 06
Years : MAY 2016
| written 8.9 years ago by | modified 3.9 years ago by |
Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3
Marks : 06
Years : MAY 2016
| written 8.9 years ago by |
Let $fm(x) =\cos mx, fn(x) =\cos nx $
For $m\neq n \\ \int\limits_{-\pi}^{\pi}fm (x) fn(x) dx = \int\limits_{-\pi}^{\pi} \cos mx \cos nx dx=\dfrac 12\int\limits_{-\pi}^{\pi}[\cos(m+n)x+\cos(m-n)x]dx \\ =\dfrac 12[\dfrac {\sin (m+n)x}{(m+n)}+\dfrac {\sin(m-n)x}{(m-n)}]_{-\pi}^{\pi} =0 $
For $m=n\\ \int\limits_{-\pi}^{\pi}[fn(x)]^2 dx =\int\limits_{-\pi}^{\pi} \cos^2(nx) dx=2 \int\limits_{0}^{\pi} \dfrac {[1+\cos(2nx)]}2 dx \\ =[x+\dfrac {\sin (2nx)}{2n}]_0^\pi = \pi\neq 0 $
Hence $[\cos x,\cos 2x , -]$ is orthogonal set
For orthonormal set
$\int\limits_{-\pi}^{\pi} \dfrac 1{\sqrt\pi}fn(x). \dfrac 1{\sqrt{\pi}}fn(x) dx=1 \\ \Rightarrow [\dfrac 1{\sqrt\pi}\cos x,\dfrac 1{\sqrt\pi}\cos 2x -]$
is orthonormal set