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Responses to Periodic Inputs notes
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  • November 10, 2024
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Chapter 9 Responses to Periodic Inputs


9.1 Fourier Series

 The defining property of a periodic function f (t ) of period T is that f(t) = f (t+nT ),
where n is any integer. That is, the function repeats every period T.
 A remarkable theorem on periodic functions is Fourier’s theorem:

Statement A periodic

function f (t ) can be expressed, in general, as an infinite series of cosine
and sine functions:
∞
f (t )=a0 + ∑ ( ak cos kω0 t +b k sin kω 0 t )
k =1 (9.1.1)

where k is a positive integer, and a0 , a k , and b k are constants, known as
Fourier coefficients, that depend on f(t).
 The component having k =1 is the fundamental, whereas the component having k
equal a particular integer n is the nth harmonic. The series expression of Equation
9.1.1 is the Fourier series expansion (FSE) of f(t).
 Combining the sine and cosine terms, the FSE becomes:
∞
f (t )=c 0 + ∑ c k cos ( kω0 t +θ k )
k=1 (9.1.2)
bk
θk =−tan -1
where: c 0 =a0 , c k =√ a2k +b2k , and ak (9.1.3)


9.2 Fourier Analysis

Summary Given the four functions cosm0t, sinm0t, cosn0t, and sinn0t,
where m and n are integers, the integral of the product of any two of these
functions over a period T = 2 /0 is zero, except the products cos2n0t,
and sin2n0t, having m = n, in which case the integral is T/ 2:
t 0 +T
∫t 0
cosnω 0 t sin mω0 t dt=0
for all n and m (9.2.1)
t 0 +T t +T
∫t 0
cosn ω 0 t cos mω0 t dt=0=∫t 0 sin n ω0 t sin mω0 t dt
0 for n≠ m (9.2.2)




9-1/22

, t 0 +T T t 0 +T
∫t 0
cos2 n ω0 t dt = =
2 ∫t 0
sin2 n ω 0 t dt
(9.2.3)

 To determine a0 in the FSE, we integrate both sides of Equation 9.1.1 over a period:
t0 +T t0 +T ∞ t0 +T ∞ t0 +T
∫t 0
f (t )dt=∫t
0
a0 dt+ ∑ ∫t
0
ak coskω0 t dt+ ∑ ∫t
0
bk sin kω0 t dt
k=1 k=1

=a0 T + 0+0
1 t0 +T
a0 = ∫ f (t )dt
T t0
or, (9.2.4)


a0 is the average of f (t) over a period. It is the dc component of f(t), whereas

the cosine and sine terms are the ac component.

 To determine an , we multiply both sides of Equation 9.1.1 by cos nω o t and integrate
over a whole period, invoking Equations 9.2.1 to 9.2.3:
t 0 +T t0 +T ∞ t0 +T
∫t 0
f (t )cosnω 0 t dt=∫t
0
a0 cos nω0 tdt +∑ ∫t
0
ak coskω0 t cosnω0 t dt+
k=1
∞ t 0 +T T
∑ ∫t 0
bk sin kω0 t cos nω0 t dt =0+ an +0
2
k=1

This gives:
2 t0 +T
an = ∫ f (t ) cos nω0 t dt
T t0 (9.2.5)

 To determine bn , we multiply both sides of Equation 9.1.1 by sin nω o t and integrate
over a whole period, invoking Equations 9.2.1 to 9.2.3:
t 0 +T t0 +T ∞ t 0+T
∫t 0
f (t )sin nω0 t dt=∫t
0
a0 sin nω 0 tdt+ ∑ ∫t
0
a k cos kω0 t sin nω0 t dt+
k=1
∞ t0 +T T
∑ ∫t 0
bk sin kω0 t sin nω 0 t dt =0+0+ bn
2
k=1

This gives:
2 t 0+ T
bn = ∫ f (t ) sin nω 0 t dt
T t0 (9.2.6)

Summary a0 is the average of f(t) over a period, an is twice the average of
f(t)cosn0t over a period, and bn is twice the average of f(t)sin0t over a
period.



9-2/22

, Example 9.2.1 FSE of Sawtooth Waveform
It is required to derive the Fourier coefficients of the sawtooth waveform of Figure
9.2.1.
A
f st (t )= t .
Solution: During the interval 0≤t<T , T


[]
T fst(t)
1 T A A t2 A
a0 = ∫0 tdt= 2 =
T T T 2 0 2 ; A
2 T A 2A T
an =
T
∫0 T
t cos nω0 dt= 2 ∫0 t cos nω0 tdt
T . t
-T T 2T

[ ]
2 π /ω0
2A 1 t Figure 9.2.1
an = 2 2 2 cos nω 0 t+ sin nω 0 t =0
T n ω0 nω 0 0
. The
FSE does not have any cosine terms for reasons that will be explained below.


[ ]
2 π /ω 0
2 T A 2A T 2A 1 t
bn = ∫0 t sin nω 0 dt = 2 ∫0 t sin nω 0 tdt = T 2 n2 ω 2 sin ω0 t− nω cosnω0 t
T T T 0 0 0


=
2A
T 2
[ ]
2π
− 2 =−
nω 0
A
πn
, where 0T = 2 . The trigonometric form of fst(t) is therefore:

A A ∞ sin nω 0 t A A
f st (t )= − ∑
2 π n=1 n
= − sin ω0 t+
2 π 2 [
sin 2 ω 0 t sin3 ω 0 t
+
3
+.. . ] (9.2.7)
At the points of discontinuity, t = kT, where k is an integer. All the sinusoidal
+ −
terms vanish and f(t) = A/2, the average of the values of f (kT ) and f (kT ) .


Exponential Form
 The FSE can also be expressed in exponential form. It is convenient for this purpose
to change the index k to n:


[( ) ( )]
∞ jnω 0 t − jnω0 t jn ω 0 t − jn ω0 t
e +e e −e
f ( t )=a0 + ∑ an + bn
n=1 2 2j


[( ) (
a n− jb n jnω0 t a n + jb n − jn ω0 t
) ]
∞
=a0 + ∑ e + e
2 2
n =1
(9.2.8)
1
2( n
C n= a − jb n ) . a b
 Let Substituting for n and n from Equations 9.2.5 and 9.2.6:


9-3/22

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