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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


  1. 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



  1. 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




∞
∑^
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