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Fourier Series and Transform
ENGIN 211, Engineering Math
Periodic Functions and Harmonics
2
a a+T t
f(t) Period: π
Frequency: π = 1π
Angular velocity (or angular frequency):
π = 2ππ = 2ππ
Such a periodic function can be expressed in terms of a series of cosine and sine functions, known as the Fourier series:
π π‘ =π02 + οΏ½ ππ cosπππ‘ + ππ sinπππ‘
β
π=1
=π02 + οΏ½ ππ cos
2πππ‘π + ππ sin
2πππ‘π
β
π=1
where cosπππ‘ and sinπππ‘ are the harmonics.
1st (or fundamental) Harmonic: cosππ‘ and sinππ‘, (π = 1)
2nd Harmonic: cos2ππ‘ and sin 2ππ‘, (π = 2)
3rd Harmonic: cos3ππ‘ and sin 3ππ‘, (π = 3)
β¦.
Significance of the Harmonics
3
π = 1
π = 1,2
π = 1,2,3
π = 1,2,3,4
Fourier series of a square wave containing various orders of harmonics
Orthogonality
4
π π‘ and π π‘ are orthogonal to each other over the interval π β€ π‘ β€ π if
οΏ½ π π‘π
ππ π‘ ππ‘ = 0
In fact, the harmonics cosπππ‘ and sinπππ‘ (π = 0,1,2,β―) form an infinite collection of periodic functions that are mutually orthogonal on the interval β π/2 β€ π‘ β€ π/2 because
οΏ½ cos πππ‘π/2
βπ/2cos πππ‘ ππ‘ = 0, for π β π
οΏ½ sin πππ‘π/2
βπ/2sin πππ‘ ππ‘ = 0, for π β π
οΏ½ cos πππ‘π/2
βπ/2sin πππ‘ ππ‘ = 0
Fourier Coefficients
5
ππ =2ποΏ½ π π‘π/2
βπ/2cos πππ‘ ππ‘, for π = 0,1,2,β―
ππ =2ποΏ½ π π‘ sin πππ‘ ππ‘
π2
βπ2
, for π = 1,2,3,β―
Please note π0 has been included in the first set of integrals.
π π‘ =π02 + οΏ½ ππ cosπππ‘ + ππ sinπππ‘
β
π=1
Example: Determine the Fourier series for the function
π π‘ = οΏ½2 1 + π‘ β1 < π‘ < 00 0 < π‘ < 1
π π‘ + 2 = π π‘ β1 1 β2 2
2
π π‘
π‘ 0
Example (Contβd)
6
π π‘ =π02 + οΏ½ ππ cos
2πππ‘π
+ ππ sin2πππ‘π
β
π=1
=π02 + οΏ½ ππ cosπππ‘ + ππ sinπππ‘
β
π=1
The coefficients: ππ = β« π π‘1β1 cos πππ‘ ππ‘ = β« 2 1 + π‘0
β1 cos πππ‘ ππ‘, π = 0,1,2,β―
For π = 0
π0 = οΏ½ 2 1 + π‘0
β1ππ‘ = 2π‘ + π‘2 β1
0 = 1
For π β 0, we can use integral by parts
ππ = οΏ½ 2 1 + π‘0
β1cos πππ‘ ππ‘ =
2ππ 1 + π‘ sin πππ‘ β1
0 β οΏ½ sin πππ‘ ππ‘0
β1
=2ππ 2 1 β cos ππ = οΏ½
0, π even4ππ 2 , π odd
Example (Contβd)
7
Similarly
ππ = οΏ½ π π‘1
β1sin πππ‘ ππ‘ = οΏ½ 2 1 + π‘
0
β1sin πππ‘ ππ‘ = β
2ππ
, π = 1,2,3,β―
The first few terms of the Fourier series:
π π‘ =12
+4π2
cosππ‘ +19
cos3ππ‘ +1
25cos5ππ‘ + β―
β2π
sinππ‘ +12
sin2ππ‘ +13
sin3ππ‘ +14
sin4ππ‘ + β―
Odd and Even Functions
8
Odd function: π βπ = βπ π , symmetrical about the origin.
Sine function is odd.
Fourier series of an odd function contains only sine terms (ππ = 0,π = 0,1,2, β¦), because
οΏ½ π π‘ cos πππ‘ ππ‘π/2
βπ/2
= 0 if π π‘ is odd.
Even function: π βπ = π π , symmetrical about the π¦-axis
Cosine function is even.
Fourier series of an even function contains only cosine terms (ππ = 0,π = 1,2,3, β¦),
οΏ½ π π‘ sin πππ‘ ππ‘π/2
βπ/2
= 0 if π π‘ is even.
Example (Even Function)
9
π₯
π¦
π/2 3π/2 βπ/2 β3π/2
4
The function is symmetrical about the π¦-axis
π π₯ =π02 + οΏ½ ππ cosππ₯
β
π=1
π0 =1ποΏ½ π π₯ ππ₯π
βπ=
2ποΏ½ π π₯ ππ₯π
0
=2ποΏ½ 4ππ₯
π/2
0= 4
ππ =1ποΏ½ π π₯ cosππ₯ ππ₯
π
βπ=
2ποΏ½ π π₯ cosππ₯ ππ₯
π
0=
2ποΏ½ 4 cosππ₯ ππ₯
π/2
0
=8ππ
sinππ2 =
0, π = 2π, (π = 2,4,6,β― )8ππ , π = 4π β 3, (n = 1,5,9,β― )
β8ππ , π = 4π β 1, (π = 3,7,11,β― )
π = 1,2,3,β―
Example (Odd Function)
10
π₯
π¦
π 2π 0 βπ
2
β2
A shift of π/2 in x-axis and shift of 2 in y-axis in the previous example change it into an odd function
π π₯ = οΏ½ ππ sinππ₯β
π=1
ππ =1ποΏ½ π π₯ sinππ₯ ππ₯
π
βπ=
2ποΏ½ π π₯ sinππ₯ ππ₯
π
0
=4ποΏ½ sinππ₯ ππ₯
π
0=
4ππ 1 β β1 π = οΏ½
0, π even8ππ
, π odd
π π₯ βπ2
β 2 =8ππ
cos π₯ βπ2
β13
cos 3 π₯ βπ2
+15
cos 5 π₯ βπ2
β17
cos 7 π₯ βπ2
+ β―
π π₯ =8ππ
sin π₯ +13
sin 3π₯ +15
sin 5π₯ β17
sin7π₯ + β―
Connection between the two examples
Complex Fourier Series
11
Eulerβs Identity πππ = cosπ + π sinπ and πβππ = cosπ β π sinπ
Thus, cosπ = πππ+πβππ
2 and sinπ = πππβπβππ
2π
Now the Fourier series
π π‘ =π02
+ οΏ½ ππ cosπππ‘ + ππ sinπππ‘β
π=1
=π02
+ οΏ½ππ2
πππππ + πβππππ +ππ2π
πππππ β πβππππβ
π=1
=π02
+ οΏ½ππ2
+ πππ2
πβππππ +ππ2β π
ππ2
πππππβ
π=1
= οΏ½ ππ
β
π=ββ
πππππ
where the complex coefficients
ππ =ππ2β π
ππ2
=1ποΏ½ π π‘ cos πππ‘ β π sin πππ‘ ππ‘π2
βπ2
=1ποΏ½ π π‘ πβππππππ‘π2
βπ2
Obviously, πβπ = ππβ = 1π β« π π‘ πππππππ‘
π2βπ2
, and π0 = 1π β« π π‘ ππ‘
π2βπ2
= π02
since π0 = 0
Example 1 (Complex Fourier)
12
π‘
π(π‘)
π/2
1
βπ/2 π/2 βπ/2
π π‘ = οΏ½0, βπ/2 < π‘ < βπ/21, βπ/2 < π‘ < π/20, π/2 < π‘ < π/2
,
and π π‘ + π = π(π‘)
π(π‘) = β ππβπ=ββ πππππ, π = 2π
π
ππ =1ποΏ½ π(π‘)πππππππ‘
π/2
βπ/2=
1ποΏ½ πππππππ‘
π/2
βπ/2=
1ππππ π
πππποΏ½βπ/2
π/2
=1
ππππ πππππ/2 β πβππππ/2
=2
πππ sinπππ
2 =ππ
sin πππ/2πππ/2 =
ππ
sin πππ/ππππ/π =
ππ sinc
ππππ , for π β 0
sinc π₯ = sin π₯π₯
is undefined at π₯ = 0, but limπ₯β0
sinc π₯ = limπ₯β0
sin π₯π₯
= 1,
If we define sinc π₯ = οΏ½sin π₯π₯
, for π₯ β 01, for π₯ = 0
and notice π0 = 1π β« π(π‘)ππ‘π/2
βπ/2 = 1π β« 1ππ‘π/2
βπ/2 = ππ
,
then ππ = ππ
sinc ππππ
, for all π, ββ < π < β
Example (Complex Fourier)
13
π‘
π(π‘)
π
1
0 π
π π‘ = οΏ½1, 0 < π‘ < π 0, π < π‘ < π ,
and π π‘ + π = π(π‘)
π(π‘) = β ππβπ=ββ πππππ, π = 2π
π
ππ =1ποΏ½ π(π‘)πππππππ‘
π
0=
1ποΏ½ πππππππ‘
π
0=
1ππππ π
πππποΏ½0
π
=1
ππππ πππππ β 1 =πππππ/2
ππππ πππππ/2 β πβππππ/2
=2πππππ/2
πππ sinπππ
2 =ππ
sin πππ/2πππ/2 πππππ/2 =
ππ sinc
ππππ πππΟπ/π , Complex coefficients
The same result can be obtained from Example 1 by shifting the time origin by half the pulse width,
π π‘ = π π‘ +π2
= οΏ½ππ
sincππππ
β
π=ββ
ππππ π+π2 = οΏ½ππ
sincππππ
πππΟπ/πβ
π=ββ
πππππ
Complex Spectra
14
The Fourier coefficients of square-pulse wave all have
sinc ππππ
, what does it look like?
In example 2, we have
ππ = ππ
sinc ππππ
πππππ/π = ππ ππππ
In both examples, their amplitudes
ππ = ππ
sinc ππππ
for π β 0, and π0 = ππ
ππ describes the spectrum of π π‘ in the frequency domain
π0 π1 πβ1
π2 πβ2
Power in Time and Frequency Domains
15
Power content of a periodic function
π(π‘) = οΏ½ ππ
β
π=ββ
πππππ
is defined as the mean square value in time domain, and is related to the spectrum in the frequency domain
π =1ποΏ½ π2 π‘ ππ‘π/2
βπ/2= οΏ½ ππ 2
β
π=ββ
Because,
π =1ποΏ½ οΏ½ ππ
β
π=ββ
ππππππ(π‘)ππ‘π/2
βπ/2= οΏ½
πππ
β
π=ββ
οΏ½ ππππππ(π‘)ππ‘π/2
βπ/2
= οΏ½πππ
β
π=ββ
πβππ = οΏ½ ππ
β
π=ββ
πβπ = οΏ½ ππππββ
π=ββ
= οΏ½ ππ 2β
π=ββ
Continuous Spectrum
16
In the example of the periodic square pulse function,
π π‘ = β ππβπ=ββ πππππ = β ππβ
π=ββ ππ2πππ/π, and ππ = ππ
sinc ππππ
,
the spacing between two neighboring harmonics is πΏπ₯ = πππ
. If we increase the period π, this spacing πΏπ₯ decreases, and when π β β, the function has a single pulse and no longer periodic, the spacing πΏπ₯ β 0, the spectrum becomes continuous.
π β β
Fourier Transform
17
We start from a periodic function π π‘ = β ππβπ=ββ πππ2ππ/π , where ππ = 1
π β« π π πβππ2ππ/ππππ2βπ2
We can plug in ππ into the series and use πΏπ = 2ππ
π π‘ = οΏ½1ποΏ½ π π πβππ2ππ/ππππ2
βπ2
β
π=ββ
πππ2ππ/π =12π
οΏ½ πΏποΏ½ π π πβππππππππ2
βπ2
β
π=ββ
ππππππ
Let π β β, then πΏπ β 0, thus πΏπ = ππ, and ππΏπ = π,
π π‘ = οΏ½12π
οΏ½ π π πβπππππβ
ββ
β
π=ββ
ππππππ =12π
οΏ½12π
οΏ½ π π πβπππππβ
ββ
β
ββππππππ
So if we introduce πΉ π = 12π β« π π‘ πβπππππ‘β
ββ - Fourier transform
Then π π‘ = 12π β« πΉ πβ
ββ ππππππ - inverse Fourier Transform
As π β β, the discrete harmonic values ππ0 = 2ππ/π become a continuous value π, and the discrete spectrum ππ = ππ ππππ becomes the continuous spectrum πΉ π = πΉ π πππ π .
Example
18
πΉ π =12π
οΏ½ π π‘ πβπππππ‘β
ββ=
12π
οΏ½ πβπππππ‘π
0= β
1π 2ππ
πβππποΏ½0
π
=1
π 2ππ1 β πβπππ =
πβπππ/2
π 2ππππππ/2 β πβπππ/2 =
πβπππ/2
π 2ππ2π sin
ππ2
=ππβπππ/2
2π
sinππ2ππ2
=ππβπππ/2
2πsinc
ππ2
πΉ π
π
This is a continuous spectrum!
π 0
1
π(π‘)
π‘
π π‘ = οΏ½1, 0 < π‘ < π0, otherwise
Properties of Fourier Transform
19
Linearity β± πΌ1π1 π‘ + πΌ2π2 π‘ = πΌ1πΉ1 π + πΌ2πΉ2 π
Time shifting β± π π‘ β π‘0 = ππππ0πΉ π Frequency shifting β± πππ0ππ π‘ = πΉ π β π0
Time scaling β± π ππ‘ = 1ππΉ π
π
Symmetry β± πΉ π‘ = π βπ
Proof: We start with the inverse Fourier transform and then employ two variable substitutions
π π‘ = 12π β« πΉ πβ
ββ ππππππ = 12π β« πΉ π’β
ββ ππππππ’
Let π‘ β βπ, then π βπ = 12π β« πΉ π’β
ββ πβπππππ’ = 12π β« πΉ π‘β
ββ πβπππππ‘ = β± πΉ π‘
Example: β± 1 =? (hard to solve without using the symmetry)
Since β± πΏ(π‘) = 12π
, then using symmetry: β± 12π
= πΏ βπ = πΏ π , thus β± 1 = 2ππΏ π .
Cosine and Sine Transforms
20
πΉ π =12π
οΏ½ π π‘ πβπππππ‘β
ββ=
12π
οΏ½ π π‘ cosππ‘ + π sinππ‘ ππ‘β
ββ
If π π‘ is even function, β« π π‘ sinππ‘ ππ‘βββ = 0, then the cosine transform:
πΉπ π =2ποΏ½ π π‘ cosππ‘ ππ‘
β
0
If π π‘ is odd function, β« π π‘ cosππ‘ ππ‘βββ = 0, then the sine transform:
πΉπ π =2ποΏ½ π π‘ sinππ‘ ππ‘
β
0
For those functions that are defined only for π‘ β₯ 0, and the extension into π‘ < 0 can make them either even or odd, where cosine and sine transforms can then be used, respectively.
Example
21
Example: π π‘ = οΏ½ 1, 0 < π‘ < π 0, π β€ π‘ < β and
undefined in ββ < π‘ < 0
π 0
1 π(π‘)
π‘
π 0
1
π‘ βπ
π 0
1
π‘ βπ
β1
Sine transform:
πΉπ π =2ποΏ½ π π‘ sinππ‘ ππ‘β
0=
2ποΏ½ sinππ‘ ππ‘π
0 =
2π
1 β cosπππ
=2π
2π
sin2ππ2
=2πππ2
2sin2 ππ2ππ2
2 =ππ2
2πsinc2
ππ2
πΉπ π =2ποΏ½ π π‘ cosππ‘ ππ‘β
0=
2ποΏ½ cosππ‘ ππ‘π
0
=2π
sinπππ
=2ππsinc ππ
Cosine transform:
Summary Key points:
Periodic functions
Fourier series and coefficients
Significance of harmonics
Sine and cosine for odd and even functions
Complex Fourier series and discrete spectrum
Fourier transform and continuous spectrum
Properties of Fourier transform
22