Fourier and Laplace 325
R.4.16 For example, let us verify Parseval’s theorem by means of the function f(t) =
2 sin(100t), evaluating by hand the average power of f(t) in
a. The time domain by using integration
b. The frequency domain by using addition
ANALYTICAL Solution
- Time domain solution
w 0 = 100 rad/s T = 2 π/w 0 = [2π/100] s
P
T
ave f t dt t dt
(^22)
0
2 100
100
2
2 sin 100
1
0
() { ( )}
T
∫∫
Pavet dt t
400
2
1
2
1
2
200
400
4
1
0 2
2 100
0
210
cos( )
{}
∫
00
Pave
400
2
1
2
2
100
2
W
- Frequency domain solution
ft t
ee
j
jt jt
() 2 sin( 100 ) 2
2
^100 ^100
by Euler’s identity
2 100
sin( tje je) jt^100 jt^100
Then
FFF01 1 01 1,,
and all the coeffi cients
Fn = 0, for n = 2, 3, ..., ∞
Then
PFFFn
n
n
aveW
−
∑
1
1
11 11 2
R.4.17 Any periodic wave f(t) can be approximated by cosine terms only; then
ft
a
cnwtn
n
() cos( n)
0
1
2 0
∞
∑^