Fourier Series (FS)
For periodic signal.
Signals can be constructed by summing sinusoids of different frequencies, amplitudes and phases.
Signals = A0+A1cos(2πf1t+θ1)+A2cos(2πf2t+θ2)+A3cos(2πf3t+θ3)+...
Periodic signals can be constructed by summing sinusoids that are harmonically related (Fouries Series).
Signals = A0+A1cos(2πf1t+θ1)+A2cos(2πf2t+θ2)+A3cos(2πf3t+θ3)+...
Where f2=2f1,f3=3f1,f4=4f1,f5=5f1
What is Fourier Series?
definition
It is a way to find out the mathematical representation of the periodic signals.
Periodic waveform = sum of sinusoids that are harmonically related
Harmonically related sinusoids: Sinusoids with frequencies that are integer multiples of a basic (fundamental) frequency
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Fourier Series Expansion
definition
x(t)=k=−∞∑∞Ckej(2πk0ft)
or
x(t)=k=−∞∑∞CkejT2πk0t
where
Ck=T01∫T00x(t)e−j2πT0ktdt
Find Fourier Series Coefficient
how to do
Steps:
-
Find Frequency (Periodic signal, f) and Period (T) of x(t)
-
Ck=f∫2−T2Tx(t)e−j2πT0ktdt
-
Consider k on both terms
-
One should be ∫2−T2T1dt , while another is 0 because of sin(π) (Use Euler’s Relation!)
-
Draw the Magnitude and phase spectrum of x(t)
Note: ∫eatdt=a1eat
examples
Steps :
- Find Frequency (Periodic signal, f) and Period (T) of x(t)
sin3πt=sin(2π⋅23t)
Since the Period is f1 ,
T1=32
2cos(5π+0.25π)=2cos(2π⋅25t+0.25π)
Since the Period is f1 ,
T2=52
Period of sin3πt+2cos(5π+0.25π) :
LCM of T1 and T2 = 2
- Ck=f∫2−T2Tx(t)e−j2πT0ktdt
Ck=21∫−11x(t)e−jπktdt
=21∫−11([sin(3πt)+2cos(5πt+0.25π)]e−jπkt)dt
- Apply Euler’s Formula on both terms
Term 1:
sin(3πt)
=2j1(ej3πt−e−j3πt)
Term 2:
2cos(5πt+0.25π)
=21(ej(5πt+0.25π)+e−j(5πt+0.25π))
=21(ej5πt⋅ej0.25π+e−j5πt+e−j0.25π)
- If the term is a sin function, muiltply jj
sin(3πt)=2j1(ej3πt−e−j3πt)⋅jj
=2−j(ej3πt−e−j3πt)
- For sin term, draw a factor of −1 to make − become +
Since −1=ejπ
sin(3πt)=2j1(ej3πt−e−j3πt)⋅jj=2−j(ej3πt−e−j3πt)
=2−j(ej3πt+(−1)⋅e−j3πt)=2−j(ej3πt+ejπ⋅e−j3πt)
Since −j=e−j0.5π
2−j(ej3πt+ejπ⋅e−j3πt)
=21e−j0.5π(ej3πt+ejπ⋅e−j3πt)
= 21(e−j0.5π⋅ej3πt+e−j0.5π⋅ejπ⋅e−j3πt)
- Now you have the polar form of 2 terms. Find the magnitude and the phase.
2(a)
Period is 3.
2(b)
Note x(t)=∑k=−∞∞Ckej(2πk0t) where Ck=T01∫T00x(t)e−j2πT0ktdt
Therefore
Ck=T01[∫022e−j2πT0ktdt+∫23(−1)e−j2πT0ktdt]
A is 2, B is -1.
2(c)
since k=0
C0=T01[∫022e0dt+∫23(−1)e0dt]
=31[∫022dt+∫23(−1)dt]=31[(2⋅2−2⋅0)+((−1)⋅3−(−1)⋅2)]
=31(4+(−1))=1
2(d)
Ck=31[∫022e−j2πT0ktdt+∫23(−1)e−j2πT0ktdt]
=32(∫02e−j32πktdt)−31(∫23e−j32πktdt)
Note ∫eatdt=a1eat
On one side:
32∫02e−j32πktdt
=32(−j32πk1)[e−j32πkt]02
=−jπk1[e32πkt]02 Now you found D is −jπk1
=−jπk1[e32πk2−e32πk0]
=−jπk1[e34πk−1]
On the other side:
31(∫23e−j32πktdt)
=31(−j32πkt1)[e−j32πkt]23
=−j2πk1[e−j32πkt]23 Now you found E is −j2πk1
=−j2πk1[e−j32πk3−e−j32πk2]
=−j2πk1[e−j2πk−e−j34πk]
Therefore:
=−jπk1[e34πk−1]−−j2πk1[e−j2πk−e−j34πk]
2(e)
Sub a 2 to make a common factor.
=−j2πk12[e34πk−1]−−j2πk1[e−j2πk−e−j34πk]
=−j2πk1[2e34πk−2]+j2πk1[e−j2πk−e−j34πk]
=j2πk1(e−j2πk−e−j34πk−2e34πk+2)
=j2πk1(e−j2πk−3e−j34πk+2)
Note e−j2π=1
=j2πk3(1−1e−j34πk)
Make 1 to be e0
=j2πk3(e0−e−j34πk)
=j2πk3(ej32πke−j32πk−e−j322πk)
Draw e−j32πk
=j2πk3e−j32πk(ej32πk−e−j32πk)
=2πk3e−j32πkj(ej32πk−e−j32πk)
Use Euler’s Formula : sinθ=2jejθ−e−jθ
=πk3e−j32πk2j(ej32πk−e−j32πk)
=πk3e−j32πksin(32πk)
G is πk3 and F is 32πk
2(f)
Ck=πk3e−j32πksin(32πk)
C1=π3e−j32πsin(32π)
=π3sin(32π)e−j32π
H=π3sin(32π)=4π33
θ=32π
Magnitude and Phase Spectrum
Magnitude Spectrum
Note it must be even function. (Symmetric)
y axis :
∣Ck∣=21A
Note: In Polar form,
z=ρe−jθ
where ρ is the y axis of Magnitude spectrum.
In Rectangular form z=a+jb,
∣Ck∣=a2+b2
Phase Spectrum
Note it must be odd function. (Anti-symmetric)
y axis :
∠Ck=tan−1realimaginary
Note: In Polar form,
z=ρe−jθ
where θ is the y axis of phase spectrum.
In Rectangular form z=a+jb,
∠Ck=tan−1ab
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Fourier Transform (FT)
For non-periodic signal (No period)
Describe a continuous-time, aperiodic signal in in terms of frequency content.
The frequency components are defined for all real values of the frequency.
Definition
- Fourier transform: aperiodic signal, T→∞
When T is increased,
the density of the frequency components increases
the envelope of the magnitude of the scaled spectral components remains the same.
T→∞, converge into a continuous spectrum
ckT=kω02sin2kω0=2kω0sin2kω0=sinc2πkω0
T→∞limckT=sinc2πω
sincλ=πλsinπλ
sinc0=1
When carry out Fourier Transform, the Time domain changes from Time t to Frequency ω or changes from Frequency ω to Time t.
Forward transform, i.e., time to frequency:
X(ω)=∫−∞∞x(t)e−jωtdt
Inverse transform, i.e., frequency to time:
x(t)=2π1∫−∞∞X(ω)ejωtdω
The function is basically the same with Fourier Series but T become a variable.
The transform pair notation:
x(t)↔X(ω)
Look at the Fourier Transform Table. (bottom)
Note ω=2πf=2πT1
Functions that you need to know
Retangular pulse
Delta Pulse Function
Pulse Function / Rectangular Function
pτ(t)
Exponential function
Exponential signal:
x(t)=e−btu(t)
- Time-domain representation
- If b>0, exp(-bt) -> 0
What is u(t)? - Unit Step Function
The Unit Step Function - u(t)
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More about The Laplace Transform
FT Properties
FT duality
Difference between FS and FT
## How to use the FT table?
examples
(a)
Note the formula e−btu(t)↔jω+b1 from the table.
In this case, obviously, b=5.
x(t)=2e−5tu(t)↔X(ω)=2⋅jω+51=jω+52
(b)
Note the property time scaling. The formula is x(at)↔a1X(aω)
v1(t)=x(2t)=2e−5t⋅2u(2t)↔V1(ω)=21X(2ω)=21⋅j2ω+52=jω+102
(c)
Note the property of both time scaling and time shift,
The formula is x(at)↔a1X(aω) and x(t−c)↔X(ω)e−jωc
v2(t)=x(2t−1)=x(2(t−21))↔V2(ω)=21X(2ω)e2−jω=jω+102e2−jω
(d)
v3(t)=2e(−5+j)tu(t)=2e−5tu(t)⋅ejt=x(t)ejt
Note the property is mulitplication by a complex exponential. The formula is x(t)ejω0t↔X(ω−ω0)
Therefore obviously ω0=1.
v3(t)=x(t)ejt↔V3(ω)=X(ω−ω0)=j(ω−1)+52
Questions of p(t)
(a)
Period(T)=4−0=4
τ=4
x(t)=2⋅p4(t−2)
(b)
Note the formula : pτ(t)↔τsinc2πτω
x(t)=2⋅p4(t−2)↔X(ω)=2⋅4sinc2π4ω=8sincπ2ω
(c)
y(t)=x(t)+(−x(−t))=x(t)−x(−t)
(d)
y(t)=x(t)−x(−t)↔Y(ω)=X(ω)−X(−ω)
Some Tricky Question
Let’s say you have this question.
It is obvious, this function contains time shift and can be transformed into pτ(t) using the common form pairs.
Step 1: Check the available functions
Notice there is a e−3jω.
x(t−c)↔X(ω)ejωc
Notice there is a sin function.
sincλ=πλsinπλ
pτ(t)↔τsinc2πτω
Step 2: Divide the parts
X(ω)=ω4sin(3ω)e−3jω=4ωsin(3ω)⋅e−3jω
Step 3: Find the λ variable
If you look closely, the formula sincλ=πλsinπλ requires a π.
Therefore, plug a ππ inside.
X(ω)=4ωsin(3ω)⋅e−3jω=4ωsin(3ωππ)⋅e−3jω=4ωsin(ππ3ω)⋅e−3jω
the λ=π3ω.
Step 4: Turn it to sincλ
To make the bottom part match the formula sincλ=πλsinπλ, plug another ππ.
X(ω)=4ωsin(ππ3ω)⋅e−3jω=4ωsin(ππ3ω)⋅3ππ3ππ⋅e−3jω
X(ω)=4⋅3ππ⋅ω⋅3ππsin(ππ3ω)⋅e−3jω=4⋅3ππ⋅π⋅π3ωsin(ππ3ω)⋅e−3jω
Now you have the term πλsinπλ. Make it sincλ .
X(ω)=12⋅sincπ3ω⋅e−3jω
Step 5: Find the τ variable
To carry Fourier transform pτ(t)↔τsinc2πτω, you need to make π become 2π.
X(ω)=12⋅sinc2π6ω⋅e−3jω
Now you get the variable τ. τ=6.
X(ω)=2⋅6sinc2π6ω⋅e−3jω
Step 6: Carry Fourier transform
X(t)=2⋅p6(t−3)↔X(ω)=2⋅6sinc2π6ω⋅e−3jω