Continuous-time Fourier analysis consists of the Fourier series, or Fourier expansion, and the Fourier transform, or Fourier integral. The former is discussed in this section. Continuous-time Fourier analysis will not be presented in depth but rather as a review.
Trigonometric expansion
The famous French mathematician Jean Baptiste Joseph Fourier demonstrated that a periodic waveform, such as the one in
Figure 1, can be expanded into sinsoidal components having frequencies which are the multiples of the fundamental frequency of the waveform.
Let’s begin with the time signal
x(t)x(t) size 12{x \( t \) } {}, periodic at period T0 (sec) or angular frequency
Ω0=2π/T0Ω0=2π/T0 size 12{ %OMEGA rSub { size 8{0} } =2π/T rSub { size 8{0} } } {} (rad/sec) or frequency
F
0
=1/
T
0
F
0
=1/
T
0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadAeadaWgaaWcbaGaaGimaaqabaGccqGH9aqpdaWcgaqaaiaaigdaaeaacaWGubWaaSbaaSqaaiaaicdaaeqaaaaaaaa@3B2E@
(Hz)
Figure 1. The trigonometric expansion, or series, is
x(t)=
a
0
+
∑
n=1
∞
a
n
cosn
Ω
0
t
+
∑
n=1
∞
b
n
sin
Ω
0
t
x(t)=
a
0
+
∑
n=1
∞
a
n
cosn
Ω
0
t
+
∑
n=1
∞
b
n
sin
Ω
0
t
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiaacMcacqGH9aqpcaWGHbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaabCaeaacaWGHbWaaSbaaSqaaiaad6gaaeqaaOGaci4yaiaac+gacaGGZbGaamOBaiabfM6axnaaBaaaleaacaaIWaaabeaakiaadshaaSqaaiaad6gacqGH9aqpcaaIXaaabaGaeyOhIukaniabggHiLdGccqGHRaWkdaaeWbqaaiaadkgadaWgaaWcbaGaamOBaaqabaGcciGGZbGaaiyAaiaac6gacqqHPoWvdaWgaaWcbaGaaGimaaqabaGccaWG0baaleaacaWGUbGaeyypa0JaaGymaaqaaiabg6HiLcqdcqGHris5aaaa@5C4F@
(1)
Where the coefficients are given by
a
0
=
1
T
0
∫
−
T
0
/2
T
0
/2
x(t)dt
a
0
=
1
T
0
∫
−
T
0
/2
T
0
/2
x(t)dt
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaaGimaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaacaWGubWaaSbaaSqaaiaaicdaaeqaaaaakmaapedabaGaamiEaiaacIcacaWG0bGaaiykaiaadsgacaWG0baaleaacqGHsisldaWcgaqaaiaadsfadaWgaaadbaGaaGimaaqabaaaleaacaaIYaaaaaqaamaalyaabaGaamivamaaBaaameaacaaIWaaabeaaaSqaaiaaikdaaaaaniabgUIiYdaaaa@48CE@
(2)
a
n
=
2
T
0
∫
−
T
0
/2
T
0
/2
x(t)cosn
Ω
0
tdt
a
n
=
2
T
0
∫
−
T
0
/2
T
0
/2
x(t)cosn
Ω
0
tdt
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaamOBaaqabaGccqGH9aqpdaWcaaqaaiaaikdaaeaacaWGubWaaSbaaSqaaiaaicdaaeqaaaaakmaapedabaGaamiEaiaacIcacaWG0bGaaiykaiGacogacaGGVbGaai4Caiaad6gacqqHPoWvdaWgaaWcbaGaaGimaaqabaGccaWG0bGaamizaiaadshaaSqaaiabgkHiTmaalyaabaGaamivamaaBaaameaacaaIWaaabeaaaSqaaiaaikdaaaaabaWaaSGbaeaacaWGubWaaSbaaWqaaiaaicdaaeqaaaWcbaGaaGOmaaaaa0Gaey4kIipaaaa@5045@
(3)
b
n
=
2
T
0
∫
−
T
0
/2
T
0
/2
x(t)sinn
Ω
0
tdt
b
n
=
2
T
0
∫
−
T
0
/2
T
0
/2
x(t)sinn
Ω
0
tdt
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadkgadaWgaaWcbaGaamOBaaqabaGccqGH9aqpdaWcaaqaaiaaikdaaeaacaWGubWaaSbaaSqaaiaaicdaaeqaaaaakmaapedabaGaamiEaiaacIcacaWG0bGaaiykaiGacohacaGGPbGaaiOBaiaad6gacqqHPoWvdaWgaaWcbaGaaGimaaqabaGccaWG0bGaamizaiaadshaaSqaaiabgkHiTmaalyaabaGaamivamaaBaaameaacaaIWaaabeaaaSqaaiaaikdaaaaabaWaaSGbaeaacaWGubWaaSbaaWqaaiaaicdaaeqaaaWcbaGaaGOmaaaaa0Gaey4kIipaaaa@504B@
(4)
In above integrals the limits were put as
−
T
0
/2
−
T
0
/2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiabgkHiTmaalyaabaGaamivamaaBaaaleaacaaIWaaabeaaaOqaaiaaikdaaaaaaa@3965@
and
T
0
/2
T
0
/2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalyaabaGaamivamaaBaaaleaacaaIWaaabeaaaOqaaiaaikdaaaaaaa@3878@
, but other limits can be used so long as the distance between them is the period
T
0
T
0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadsfadaWgaaWcbaGaaGimaaqabaaaaa@379C@
, e.g. 0 and
T
0
T
0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadsfadaWgaaWcbaGaaGimaaqabaaaaa@379C@
.
The expansion components have following meaning :
-
a
0
a
0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaaGimaaqabaaaaa@37A9@
: The average of the signal (or DC component)
-
a
1
cos
Ω
0
+
b
1
sin
Ω
0
t
a
1
cos
Ω
0
+
b
1
sin
Ω
0
t
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaaGymaaqabaGcciGGJbGaai4BaiaacohacqqHPoWvdaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWGIbWaaSbaaSqaaiaaigdaaeqaaOGaci4CaiaacMgacaGGUbGaeuyQdC1aaSbaaSqaaiaaicdaaeqaaOGaamiDaaaa@460E@
: The fundamental component (remember the sum of two sinusoids of the same frequeny is a sinusoid at that frequency, see Equation 5), or the first harmonic.
-
a
2
cos
ω
0
t+
b
2
sin
ω
0
t
a
2
cos
ω
0
t+
b
2
sin
ω
0
t
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaaGOmaaqabaGcciGGJbGaai4BaiaacohacqaHjpWDdaWgaaWcbaGaaGimaaqabaGccaWG0bGaey4kaSIaamOyamaaBaaaleaacaaIYaaabeaakiGacohacaGGPbGaaiOBaiabeM8a3naaBaaaleaacaaIWaaabeaakiaadshaaaa@4787@
: The second harmonic
-
a
3
cos
ω
0
t+
b
3
sin
ω
0
t
a
3
cos
ω
0
t+
b
3
sin
ω
0
t
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaaG4maaqabaGcciGGJbGaai4BaiaacohacqaHjpWDdaWgaaWcbaGaaGimaaqabaGccaWG0bGaey4kaSIaamOyamaaBaaaleaacaaIZaaabeaakiGacohacaGGPbGaaiOBaiabeM8a3naaBaaaleaacaaIWaaabeaakiaadshaaaa@4789@
: The third harmonic
- ...
Example 1 Find the Fourier expansion for the symmetric square wave of
Figure 2.
Solution
We observe straightaway that the DC component is zero since the positive and regative parts of the signal are equal:
a
0
=0
a
0
=0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaaGimaaqabaGccqGH9aqpcaaIWaaaaa@3973@
Of course when using
Equation 2a we will get the same result. Next, since the waveform is odd-symmetric (antisymmetric) (symmetric with respect to the origin), the cosine components are also zero :
a
n
=0 all n
a
n
=0 all n
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggadaWgaaWcbaGaamOBaaqabaGccqGH9aqpcaaIWaGaaGzbVlaaywW7caWGHbGaamiBaiaadYgacaaMf8UaamOBaaaa@4211@
It is left with the sine components given by
b
n
=
4A
T
0
∫
0
T
0
/2
sinn
Ω
0
tdt
=
4A
T
0
−1
n
ω
0
[
cosn
Ω
0
t
]
0
T
0
/2
=
−4A
2π
1
n
[
1−1
]=0 , n even
=
−4A
2π
1
n
[
−1−1
]=
4A
2π
1
n
, n old
b
n
=
4A
T
0
∫
0
T
0
/2
sinn
Ω
0
tdt
=
4A
T
0
−1
n
ω
0
[
cosn
Ω
0
t
]
0
T
0
/2
=
−4A
2π
1
n
[
1−1
]=0 , n even
=
−4A
2π
1
n
[
−1−1
]=
4A
2π
1
n
, n old
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@A2CB@
These
bnbn size 12{b rSub { size 8{n} } } {}coefficients can be put in a more concise form :
Thus the Fourier expansion is
x(t)=
∑
n=1
∞
4A
π
1
(2n−1)
sin(2n−1)
Ω
0
t , n=1, 2, 3,...
=
4A
π
(sin
Ω
0
t+
1
3
sin3
Ω
0
t+
1
5
sin5
Ω
0
t+...)
x(t)=
∑
n=1
∞
4A
π
1
(2n−1)
sin(2n−1)
Ω
0
t , n=1, 2, 3,...
=
4A
π
(sin
Ω
0
t+
1
3
sin3
Ω
0
t+
1
5
sin5
Ω
0
t+...)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@8A3F@
Figure 3 is the plot of the normalized coefficients with respect to the normalized angular frequency.
We know that the sum of two sinusoids of the same frequency is another sinusoid at that frequency, specifically
acosΩt+bsinΩt=
a
2
+
b
2
cos(Ωt+
tan
−1
b
a
)
acosΩt+bsinΩt=
a
2
+
b
2
cos(Ωt+
tan
−1
b
a
)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadggaciGGJbGaai4BaiaacohacqqHPoWvcaWG0bGaey4kaSIaamOyaiGacohacaGGPbGaaiOBaiabfM6axjaadshacqGH9aqpdaGcaaqaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGIbWaaWbaaSqabeaacaaIYaaaaaqabaGcciGGJbGaai4BaiaacohacaGGOaGaeuyQdCLaamiDaiabgUcaRiGacshacaGGHbGaaiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaakmaalaaabaGaamOyaaqaaiaadggaaaGaaiykaaaa@5712@
(5)
Because of this, expansion
Equation 1 can be changed to the form of amplitude and phase:
x(t)=
c
0
+
∑
n=1
∞
c
n
cos(n
Ω
0
t+
Φ
0
)
x(t)=
c
0
+
∑
n=1
∞
c
n
cos(n
Ω
0
t+
Φ
0
)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiaacMcacqGH9aqpcaWGJbWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSYaaabCaeaacaWGJbWaaSbaaSqaaiaad6gaaeqaaOGaci4yaiaac+gacaGGZbGaaiikaiaad6gacqqHPoWvdaWgaaWcbaGaaGimaaqabaGccaWG0bGaey4kaSIaeuOPdy0aaSbaaSqaaiaaicdaaeqaaOGaaiykaaWcbaGaamOBaiabg2da9iaaigdaaeaacqGHEisPa0GaeyyeIuoaaaa@5146@
(6)
Where
c
0
=
a
0
c
0
=
a
0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadogadaWgaaWcbaGaaGimaaqabaGccqGH9aqpcaWGHbWaaSbaaSqaaiaaicdaaeqaaaaa@3A87@
(7)
c
n
=
a
n
2
+
b
n
2
n=1, 2, 3,...
c
n
=
a
n
2
+
b
n
2
n=1, 2, 3,...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadogadaWgaaWcbaGaamOBaaqabaGccqGH9aqpdaGcaaqaaiaadggadaqhaaWcbaGaamOBaaqaaiaaikdaaaGccqGHRaWkcaWGIbWaa0baaSqaaiaad6gaaeaacaaIYaaaaaqabaGccaaMf8UaaGzbVlaad6gacqGH9aqpcaaIXaGaaiilaiaaysW7caaIYaGaaiilaiaaysW7caaIZaGaaiilaiaac6cacaGGUaGaaiOlaaaa@4E08@
(8)
Φ
n
=
tan
−1
−
b
n
a
n
n=1, 2, 3,...
Φ
n
=
tan
−1
−
b
n
a
n
n=1, 2, 3,...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiabfA6agnaaBaaaleaacaWGUbaabeaakiabg2da9iGacshacaGGHbGaaiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaakmaalaaabaGaeyOeI0IaamOyamaaBaaaleaacaWGUbaabeaaaOqaaiaadggadaWgaaWcbaGaamOBaaqabaaaaOGaaGzbVlaaywW7caWGUbGaeyypa0JaaGymaiaacYcacaaMe8UaaGOmaiaacYcacaaMe8UaaG4maiaacYcacaGGUaGaaiOlaiaac6caaaa@51DB@
(9)
In this expansion we can recognize
c0c0 size 12{c rSub { size 8{0} } } {} as the average component,
c1cos(ω0t+Φ1)c1cos(ω0t+Φ1) size 12{c rSub { size 8{1} } "cos" \( ω rSub { size 8{0} } t+Φ rSub { size 8{1} } \) } {} the fundamental component, and
c2cos(2ω0t+Φ2)c2cos(2ω0t+Φ2) size 12{c rSub { size 8{2} } "cos" \( 2ω rSub { size 8{0} } t+Φ rSub { size 8{2} } \) } {} the second harmonic...
The plot of the coefficients versus frequency is the magnitude spectrum, and the plot of the phase
ΦnΦn size 12{Φ rSub { size 8{n} } } {}versus ferquency is the phase spectrum. Both spectra are discrete or line spectra.
Example 2 Find the Fourier expansion of the waveform in
Example 1.
Solution
The coefficients are
c
0
=
a
0
c
n
=
a
n
2
+
b
n
2
=
b
n
, n=1, 2, 3,...
Φ
n
=
tan
−1
−
b
n
a
n
=−
90
0
(=−π/2
) , n=1, 2, 3,...
c
0
=
a
0
c
n
=
a
n
2
+
b
n
2
=
b
n
, n=1, 2, 3,...
Φ
n
=
tan
−1
−
b
n
a
n
=−
90
0
(=−π/2
) , n=1, 2, 3,...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@8091@
The magnitude spectrum is as previously, the phase spectrum is given in
Figure 4.
The expansion expression is
x(t)=
∑
n=1
∞
4A
π
1
2n−1
cos[
(2n−1)
Ω
0
t+
90
0
]
=
∑
n=1
∞
4A
π
1
2n−1
cos(2n−1)
Ω
0
t
x(t)=
∑
n=1
∞
4A
π
1
2n−1
cos[
(2n−1)
Ω
0
t+
90
0
]
=
∑
n=1
∞
4A
π
1
2n−1
cos(2n−1)
Ω
0
t
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOabaeqabaGaamiEaiaacIcacaWG0bGaaiykaiabg2da9maaqahabaWaaSaaaeaacaaI0aGaamyqaaqaaiabec8aWbaaaSqaaiaad6gacqGH9aqpcaaIXaaabaGaeyOhIukaniabggHiLdGcdaWcaaqaaiaaigdaaeaacaaIYaGaamOBaiabgkHiTiaaigdaaaGaci4yaiaac+gacaGGZbWaamWaaeaacaGGOaGaaGOmaiaad6gacqGHsislcaaIXaGaaiykaiabfM6axnaaBaaaleaacaaIWaaabeaakiaadshacqGHRaWkcaaI5aGaaGimamaaCaaaleqabaGaaGimaaaaaOGaay5waiaaw2faaaqaaaqaaiaaywW7caaMf8Uaeyypa0ZaaabCaeaadaWcaaqaaiaaisdacaWGbbaabaGaeqiWdahaaaWcbaGaamOBaiabg2da9iaaigdaaeaacqGHEisPa0GaeyyeIuoakmaalaaabaGaaGymaaqaaiaaikdacaWGUbGaeyOeI0IaaGymaaaaciGGJbGaai4BaiaacohacaGGOaGaaGOmaiaad6gacqGHsislcaaIXaGaaiykaiabfM6axnaaBaaaleaacaaIWaaabeaakiaadshaaaaa@755A@
Complex exponential expansion
The Fourier expansion in the form of complex exponentials are more fundamental since it is more compact and is related directly to the Fourier transform. The expansion is
x(t)=
∑
n=−∞
∞
X
n
e
jn
Ω
0
t
x(t)=
∑
n=−∞
∞
X
n
e
jn
Ω
0
t
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiaacMcacqGH9aqpdaaeWbqaaiaadIfadaWgaaWcbaGaamOBaaqabaGccaWGLbWaaWbaaSqabeaacaWGQbGaamOBaiabfM6axnaaBaaameaacaaIWaaabeaaliaadshaaaaabaGaamOBaiabg2da9iabgkHiTiabg6HiLcqaaiabg6HiLcqdcqGHris5aaaa@4AA9@
(10)
The two symmetric components
XnXn size 12{X rSub { size 8{n} } } {} and
X−nX−n size 12{X rSub { size 8{ - n} } } {} always appear in pairs and the sum of each pair is a real signal. Relations between the complex exponential and trigonometric coefficients are
X
0
=
a
0
=
c
0
X
0
=
a
0
=
c
0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfadaWgaaWcbaGaaGimaaqabaGccqGH9aqpcaWGHbWaaSbaaSqaaiaaicdaaeqaaOGaeyypa0Jaam4yamaaBaaaleaacaaIWaaabeaaaaa@3D5A@
(11)
X
n
=
a
n
+j
b
n
2
=
c
n
2
e
j
Φ
n
X
n
=
a
n
+j
b
n
2
=
c
n
2
e
j
Φ
n
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfadaWgaaWcbaGaamOBaaqabaGccqGH9aqpdaWcaaqaaiaadggadaWgaaWcbaGaamOBaaqabaGccqGHRaWkcaWGQbGaamOyamaaBaaaleaacaWGUbaabeaaaOqaaiaaikdaaaGaeyypa0ZaaSaaaeaacaWGJbWaaSbaaSqaaiaad6gaaeqaaaGcbaGaaGOmaaaacaWGLbWaaWbaaSqabeaacaWGQbGaeuOPdy0aaSbaaWqaaiaad6gaaeqaaaaaaaa@4828@
(12)
X
n
=
a
n
−j
b
n
2
=
c
n
2
e
−j
Φ
n
X
n
=
a
n
−j
b
n
2
=
c
n
2
e
−j
Φ
n
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfadaWgaaWcbaGaamOBaaqabaGccqGH9aqpdaWcaaqaaiaadggadaWgaaWcbaGaamOBaaqabaGccqGHsislcaWGQbGaamOyamaaBaaaleaacaWGUbaabeaaaOqaaiaaikdaaaGaeyypa0ZaaSaaaeaacaWGJbWaaSbaaSqaaiaad6gaaeqaaaGcbaGaaGOmaaaacaWGLbWaaWbaaSqabeaacqGHsislcaWGQbGaeuOPdy0aaSbaaWqaaiaad6gaaeqaaaaaaaa@4920@
(13)
The coefficients
XnXn size 12{X rSub { size 8{n} } } {} can be computed directly from
X
n
=
1
T
0
∫
0
T
0
x(t)
e
−jn
Ω
0
t
dt
X
n
=
1
T
0
∫
0
T
0
x(t)
e
−jn
Ω
0
t
dt
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfadaWgaaWcbaGaamOBaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaacaWGubWaaSbaaSqaaiaaicdaaeqaaaaakmaapedabaGaamiEaiaacIcacaWG0bGaaiykaaWcbaGaaGimaaqaaiaadsfadaWgaaadbaGaaGimaaqabaaaniabgUIiYdGccaWGLbWaaWbaaSqabeaacqGHsislcaWGQbGaamOBaiabfM6axnaaBaaameaacaaIWaaabeaaliaadshaaaGccaWGKbGaamiDaaaa@4CC4@
(14)
The limits of the integral can be
-T0/2-T0/2 size 12{"-T" rSub { size 8{0} } /2 } {}and
T0/2T0/2 size 12{T rSub { size 8{0} } /2 } {}instead of 0 and T as above.
Since the coefficients
XnXn size 12{X rSub { size 8{n} } } {} are generally complex, we write
X
n
=|
X
n
|
e
j
Φ
n
X
n
=|
X
n
|
e
j
Φ
n
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIfadaWgaaWcbaGaamOBaaqabaGccqGH9aqpdaabdaqaaiaadIfadaWgaaWcbaGaamOBaaqabaaakiaawEa7caGLiWoacaWGLbWaaWbaaSqabeaacaWGQbGaeuOPdy0aaSbaaWqaaiaad6gaaeqaaaaaaaa@42B1@
(15)
The variation of
∣Xn∣∣Xn∣ size 12{ lline "Xn" rline } {} is the magnitude spectrum, the variation of
ΦnΦn size 12{Φ rSub { size 8{n} } } {} is the phase spectrum of the signal. For a real signal, the magnitude spectrum is even-symmetric (symmetric), and the phase spectrum is odd-symmetric (antisymmetric).
Example 3 Find the Fourier expansion of a uniform impulse sequence.
Solution
Let’s consider an uniform sequence of impulse
Aδ(t)Aδ(t) size 12{Aδ \( t \) } {} of interval (period)
T0T0 size 12{T rSub { size 8{0} } } {} Figure 5a:
x(t)=
∑
k=−∞
+∞
Aδ(t−k
T
0
)
x(t)=
∑
k=−∞
+∞
Aδ(t−k
T
0
)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiaacMcacqGH9aqpdaaeWbqaaiaadgeacqaH0oazcaGGOaGaamiDaiabgkHiTiaadUgacaWGubWaaSbaaSqaaiaaicdaaeqaaOGaaiykaaWcbaGaam4Aaiabg2da9iabgkHiTiabg6HiLcqaaiabgUcaRiabg6HiLcqdcqGHris5aaaa@4B7E@
The expansion coefficients are given by
X
n
=
1
T
0
∫
−
T
0
/2
T
0
/2
Aδ(t)
e
−jn
Ω
0
t
dt
=
1
T
0
∫
−ε
+ε
Aδ(t)
e
−jn
Ω
0
t
dt
=
A
T
0
X
n
=
1
T
0
∫
−
T
0
/2
T
0
/2
Aδ(t)
e
−jn
Ω
0
t
dt
=
1
T
0
∫
−ε
+ε
Aδ(t)
e
−jn
Ω
0
t
dt
=
A
T
0
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@7F8E@
Figure 6 is the magnitude spectrum. From these coefficients we can synthesize the signal as
x(t)=
A
T
0
∑
n=−∞
+∞
e
jn
Ω
0
t
x(t)=
A
T
0
∑
n=−∞
+∞
e
jn
Ω
0
t
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacaGGOaGaamiDaiaacMcacqGH9aqpdaWcaaqaaiaadgeaaeaacaWGubWaaSbaaSqaaiaaicdaaeqaaaaakmaaqahabaGaamyzamaaCaaaleqabaGaamOAaiaad6gacqqHPoWvdaWgaaadbaGaaGimaaqabaWccaWG0baaaaqaaiaad6gacqGH9aqpcqGHsislcqGHEisPaeaacqGHRaWkcqGHEisPa0GaeyyeIuoaaaa@4C24@
The sinx/x function
Concerning in the Fourier analysis there is a special function we need to know, that is the
sinx/xsinx/x size 12{"sin"x/x} {}function, also called function, or sampling function . The variation on
sinx/xsinx/x size 12{"sin"x/x} {} with respect to x is shown in
Figure 6. It is an even-symmetric function with unit area, having a maximum of 1 at origin, and zero crossings at regular interval of
π
π
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiabec8aWbaa@379A@
. The distance between the origin and the first zero crossing is also
π
π
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiabec8aWbaa@379A@
. The function is oscillating and progressively decaying. The first minimum peaks have the value of -0.2178, and the next maximum peaks have the value of 0.1284.
The zero crossing points are determined by
sinx
x
=0 ⇒ sinx=0 ⇒ x=±nπ , n=1, 2, 3,...
sinx
x
=0 ⇒ sinx=0 ⇒ x=±nπ , n=1, 2, 3,...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalaaabaGaci4CaiaacMgacaGGUbGaamiEaaqaaiaadIhaaaGaeyypa0JaaGimaiaaywW7cqGHshI3caaMf8Uaci4CaiaacMgacaGGUbGaamiEaiabg2da9iaaicdacaaMf8UaeyO0H4TaaGzbVlaadIhacqGH9aqpcqGHXcqScaWGUbGaeqiWdaNaaGjbVlaacYcacaaMf8UaaGzbVlaad6gacqGH9aqpcaaIXaGaaiilaiaaysW7caaIYaGaaiilaiaaysW7caaIZaGaaiilaiaac6cacaGGUaGaaiOlaaaa@646D@
And the extrema of the function occur at
sinx=±1 ⇒ x=±(2n+1)
π
2
, n=1, 2, 3,...
sinx=±1 ⇒ x=±(2n+1)
π
2
, n=1, 2, 3,...
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiGacohacaGGPbGaaiOBaiaadIhacqGH9aqpcqGHXcqScaaIXaGaaGzbVlabgkDiElaaywW7caWG4bGaeyypa0JaeyySaeRaaiikaiaaikdacaWGUbGaey4kaSIaaGymaiaacMcadaWcaaqaaiabec8aWbqaaiaaikdaaaGaaGjbVlaacYcacaaMf8UaaGzbVlaad6gacqGH9aqpcaaIXaGaaiilaiaaysW7caaIYaGaaiilaiaaysW7caaIZaGaaiilaiaac6cacaGGUaGaaiOlaaaa@5EBF@
The first few peak values are
x=±3π/2,x=±5π/2,...x=±3π/2,x=±5π/2,... size 12{x= +- 3π/2,x= +- 5π/2, "." "." "." } {} Notice these are in middle of the zero crossings. However, du