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03 fourierAnalysis

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EE150 Signals and Systems
– Part 3: Fourier Series Representation of Periodic
Signals
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Objective
Recall:
Previously, we use the weighted sum (integral) of shifted
impulses to represent an input and then derive the convolution
sum (integral).
F In this chapter:
We use different basic signal, the complex exponential, to
represent the input.
F Why we use complex exponential?
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Eigenfunction of LTI System
A signal for which the system output is just a constant (possibly
complex) times the input is referred to as an eigenfunction of the
system.
Objective
The output to an input x(t) can be found easily if x(t) can be
expressed as weighted sum of the eigenfunctions.
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Eigenfunction of CT LTI Systems
Consider an input x(t) = est and a CT LTI system with impulse
response h(t)
Z ∞
y (t) =
h(τ )es(t−τ ) dτ
−∞
Z ∞
= est
h(τ )e−sτ dτ
−∞
R∞
Let H(s) = −∞ h(τ )e−sτ dτ (eigenvalue), then y (t) = H(s)est .
Hence, complex exponentials are eigenfunctions of LTI systems:
est → H(s)est
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Eigenfunction of DT LTI Systems
Consider an input x[n] = z n and a DT LTI system with impulse
response h[n]
y [n] =
∞
X
h[k]z
n−k
=z
n
k=−∞
∞
X
h[k]z −k
k=−∞
P
−k (eigenvalue), then y [n] = H(z)z n .
Let H(z) = ∞
k=−∞ h[k]z
Hence, complex exponentials are eigenfunctions of LTI systems:
z n → H(z)z n
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Usefulness of Complex Exponentials
Ex: If input x(t) = a1 es1 t + a2 es2 t + a3 es3 t , based on eigenfunction
property and superposition property, the the response is
y (t) = a1 H(s1 )es1 t + a2 H(s2 )es2 t + a3 H(s3 )es3 t
Generally, for a CT LTI system, if the input is a linear combination
of complex exponentials, then
X
X
x(t) =
ak esk t → y (t) =
ak H(sk )esk t .
k
k
Similarly, for a DT LTI system,
X
X
x[n] =
ak zkn → y [n] =
ak H(zk )zkn .
k
k
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Fourier Analysis
For Fourier analysis, we consider
1
CT: purely imaginary s = jω: ejωt
2
DT: unit magnitude z = ejω : ejωn
ejωn → H(ejω )ejωn
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Periodic signals & Fourier Series Expansion
Using “trigonometric sum” to describe periodic signal
can be tracked back to Babylonians who predicted
astronomical events similarly.
L. Euler (in 1748) and Bernoulli (in 1753) used the
“normal mode concept to describe the motion of a
vibrating string; though JL Lagrange strongly
criticized this concept.
Fourier (in 1807) had found series of harmonically
related sinusoids to be useful to describe the
temperature distribution through body, and he
claimed any periodic signal can be represented by
such series.
Dirichlet (in 1829) provide a precise condition under
which a periodic signal can be represented by a
Fourier series.
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Aside: an orthonormal set
Consider the set, S, of x(t) satisfying x(t) = x(t + T0 )
Dot-product (inner-product) defined as
ω0
< x1 (t), x2 (t) >=
2π
Z
π
ω0
− ωπ
0
x1 (t)x2∗ (t)dt
Consider the set, B, of functions in S
φk (t) = e jkω0 t ; ω0 =
2π
,k ∈ Z
T0
Observe that they are orthonormal
ω0
2π
Z
π
ω0
− ωπ
0
e jk1 ω0 t e −jk2 ω0 t dt =
(
0 k1 6= k2
1 k1 = k2
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Fourier’s Idea
The span of the orthonormal functions, B, covers most of S.
i.e. span(B) ≈ S
More precisely, under mild assumptions: x(t) is sum of
sinusoids, i.e.
Z π
X
ω0 ω0
x(τ )e −jkω0 τ dτ
x(t) =
ak e jkω0 t , where ak =
2π − π
k
ω0
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Example
P
Consider a periodic signal x(t) = 3k=−3 ak ejk2πt , where a0 = 1,
a1 = a−1 = 0.25, a2 = a−2 = 0.5, a3 = a−3 = 1/3.
Collecting each of the harmonic components having the same
fundamental frequency
x(t) = 1+
1
1
1 j2πt
e
+ e−j2πt + ej4πt + e−j4πt + ej6πt + e−j6πt
4
2
3
Using Euler’s relation
x(t) = 1 +
1
2
cos(2πt) + cos(4πt) + cos(6πt)
2
3
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Example cont.
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Periodic signals & Fourier Series Expansion cont.
Theorem (for reasonable functions):
x(t) may be expressed as a Fourier series:
x(t) =
∞
X
ak · e jkω0 t
k=−∞
(sum of sinusoids whose frequencies are multiple of ω0 , the
“fundamental frequency”.)
Where ak can be obtained by
Z
1
ak =
x(τ )e −jkω0 τ dτ - Fourier series coefficient.
T 0 T0
Note: e jkω0 t , for k = −∞ to ∞, are orthonormal function.
(Normal basic signal)
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Example 3.4
Let x(t) = 1 + sin(ω0 t) + 2 cos(ω0 t) + cos(2ω0 t + π4 ) has
fundamental frequency ω0 .
First, expanding x(t) in terms of complex exponentials
π
1
1 π
1
1
ejω0 t + 1 −
e−jω0 t + ej 4 ej2ω0 t + e−j 4 e−j2ω0 t
x(t) = 1+ 1 +
2j
2j
2
2
Then, the Fourier series coefficients can be directly obtained.
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Example 3.5 Square Wave
For a periodic square wave, the definition over one period is
1, |t| < T1 ,
x(t) =
0, T1 < |t| < T2 .
Periodic with fundamental period T and fundamental frequency
ω0 = 2π
T
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Example 3.5 Square Wave cont.
For k = 0,
1
a0 =
T
Z
T1
dt =
−T1
2T1
.
T
For k 6= 0,
1
ak =
T
Z
T1
e−jkω0 t dt =
−T1
2 sin(kω0 T1 )
sin(kω0 T1 )
=
.
kω0 T
kπ
When T1 = 41 T → 50% duty-cycle square wave
ak =
sin( kπ
2 )
.
kπ
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Example 3.5 Square Wave cont.
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Fourier Series
x(t) =
=
∞
X
ak e jkω0 t
k=−∞
∞
X
(ak cos(kω0 t) + jak sin(kω0 t))
k=−∞
= a0 +
X
((ak + a−k )cos(kω0 t) + j(ak − a−k )sin(kω0 t))
k>0
In general, ak is complex. Therefore, this is not the real and
imaginary decomposition.
However, this is the even and odd decomposition
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Fourier Series cont.
1
Odd/Even periodic functions
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Fourier Series cont.
2 Approximation by Truncating Higher Harmonics
If
ak (for |k| > N) are small, x(t) ≈ x̂(t) ≡
N
X
ak e jkω0 t
−N
The approximation error is
e(t) = x(t) − x̂(t)
How good is the approximation?
Metric: relative energy in the error over a period
R
< e(t), e(t) >
err =
=R
< x(t), x(t) >
π
ω0
−π
ω0
π
ω0
−π
ω0
e(t)e ∗ (t)dt
x(t)x ∗ (t)dt
|k|>N
|ak |2
k=−∞
|ak |2
P
= P∞
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Convergence of Fourier Series
Conditions for convergence of Fourier series is a deep subject
- Finite energy over a single period
- Dirichlet conditions:
Over any period, x(t) must be absolutely integrable
Finite number of extrema during any single period
Finite number of discontinuities in any finite interval of time;
each of these discontinuities is finite.
Check: https://en.wikipedia.org/wiki/Dini_test
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Signals Violating Dirichlet Conditions
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Gibbs Phenomenon
Overshoot ≈ 9% as N goes to ∞
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Properties of Continuous-Time Fourier Series
Assume x(t) is periodic with period T and fundamental frequency
ω0 = 2π/T .
x(t) and its Fourier-series coefficients ak are denoted by
x(t)←−FS
−−→ak
Assume
x(t)←−FS
−−→ak ,
y (t)←−FS
−−→bk
(using same T )
1
Linearity: z(t) = αx(t) + βy (t)←−FS
−−→αak + βbk
2
Time-shift:
−jkω0 t0 a
x(t − t0 )←−FS
k
−−→e
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Properties of C-T Fourier Series cont.
3
Time-reverse: x(−t)←−FS
−−→a−k
4
Time-scaling:
∞
P
x(αt) =
ak e jk(αω0 )t
k=−∞
5
Multiplication:
x(t)y (t)←−FS
−−→hk =
∞
P
al bk−l
− Convolution!
l=−∞
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Properties of C-T Fourier Series cont.
6
conjugation & conjugate symmetry:
∗
x ∗ ←−FS
−−→a−k
→ x(t) = x ∗ (t)
→ ak∗ = a−k
If x(t) is real and even,
→ ak = a−k = ak∗
→ Fourier Series Coefficients are real & even
If x(t) is real and odd,
→ ak = −a−k = −ak∗
→ Fourier Series Coefficients are purely
imaginary & odd
If x(t) real
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Properties of C-T Fourier Series cont.
7
dx(t)
FS
dt ←−
−−→jkω0 ak
8
Parseval’s Identity:
Rt
ak
FS
−∞ x(τ )dτ ←−
−−→ jkω0
1
T
R
T
|x(t)|2 dt =
∞
P
|ak |2
k=−∞
Proof.
1
T
Z
T
Z X
1
ak1 ak∗2 e j(k1 −k2 )ω0 t dt
T T
k1 ,k2
X
=
ak1 ak∗2 δ[k1 − k2 ]
|x(t)|2 dt =
k1 ,k2
=
X
|ak |2
k
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Example:
a x(t) = cos ω0 t
→ x(t) =
1
2
e jω0 t + e −jω0 t = 12 e jω0 t + 12 e −jω0 t
∴ a0 = 0, a1 = 21 , a−1 = 12 , ak = 0 otherwise
b
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Example: cont.
Recall from the previous example, we have
By changing of variable, we can obtain
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Example: cont.
Assume: a(t)←−FS
−−→ak ,
⇒ ck =



ak e
b(t)←−FS
−−→bk ,
1
a0 − ,
2
T
−jkω0 40
∴ ck =
,


k=0
where ω0 = 2π
k 6= 0
0
π
 sin(k 2 )
kπ
c(t)←−FS
−−→ck
× e −j
kπ
2
,
k=0
,
k 6= 0
1
T0
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Example: cont.
0
Assume: g (t)←−FS
−−→dk
0
∵ g (t) = 2 · c(t) with T0 = 2
∴ dk = 2 · ck
Assume: g (t)←−FS
−−→ek
∴ ek =
For k 6= 0 ⇒ ek =
For k = 0 e0 =
dk
2ck
2ck
=
=
jkω0
jkω0
jkπ T20
2 sin( kπ
) −j kπ
2
e 2
j(kπ)2
(T0 = 2)
The area under g (t) in one period
period
=
1
2
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Example 3.9
Suppose we have the following facts about a signal x(t)
1
x(t) is a real signal
2
x(t) is periodic with period T = 4, and it has Fourier series
coefficient ak
3
ak = 0 for k > 1
4
5
The signal with Fourier coefficient ck = e−jπk/2 a−k is odd
R
1
1
2
4 4 |x(t)| dt = 2
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Fourier Series for Discrete-Time Periodic Signal
Difference:
1
2
continuous-time: infinite series
discrete-time: finite series
no convergence issue in discrete-time
F Recall: x[n] is periodic with period N if x[n] = x[n + N]
The fundamental period is the smallest positive N which
satisfies the above eq.; and ω0 = 2π/N is the fundamental
frequency.
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An Orthonormal Set
Consider the set, T , of x[n] satisfying x[n] = x[n + N]
Dot-product (inner-product) defined as
N−1
1 X
< x1 [n], x2 [n] >=
x1 [n]x2∗ [n]
N
n=0
Consider the set, C , of N functions in T
µk [n] = e jkω0 n ; ω0 =
2π
,0 ≤ k ≤ N − 1
N
Observe that they are orthonormal
N−1
1 X jk1 ω0 m −jk2 ω0 m
e
e
=
N
m=0
(
0
k1 6= k2
1
k1 = k2
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Fourier Series for Discrete-Time Periodic Signal
Theorem
x[n] =
N−1
X
ak e jkω0 n , where ak =
N−1
1 X
x[m]e −jkω0 m
N
m=0
k=0
Proof:
x[n] =
⇔
⇔
N−1
X
N−1
1 X
x[m]e −jkω0 m e jkω0 n ,
N
k=0
N−1
X
1
x[n] =
N
x[n] =
m=0
N−1
X
m=0
x[m]
N−1
X
e −jkω0 m e jkω0 n ,
k=0
x[m]δ[n − m].
m=0
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Example 3.12
Discrete-time periodic square wave
ak =
=
N1
1 X
e−jk(2π/N)n
N
n=−N1
(
1 sin[2πk(N1 +1/2)/N]
,k
N
sin(πk/N)
2N1 +1
k
N ,
6= 0, ±N, ±2N, . . .
= 0, ±N, ±2N, . . .
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Properties of Discrete-Time Fourier Series
x[n] and y [n] are periodic with period N, ak and bk are periodic
with period N
FS
1
Linearity: Ax[n] + By [n] ←→ Aak + Bbk
2
Time shifting: x[n − n0 ] ←→ ak e−jk(2π/N)n0
3
Frequency shifting: ejM(2π/N)n x[n] ←→ ak−M
4
∗
Conjugation: x ∗ [n] ←→ a−k
5
6
7
8
9
FS
FS
FS
FS
Time reversal: x[−n] ←→ a−k
P
FS
Periodic convolution:
r =<N> x[r ]y [n − r ] ←→ Nak bk
FS P
Multiplication: x[n]y [n] ←→ l=<N> al bk−l
FS
First difference: x[n] − x[n − 1] ←→ (1 − e−jk(2π/N) )ak
P
P
Parseval’s relation: N1 n=<N> |x[n]|2 = k=<N> |ak |2
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Example 3.13
As x[n] = x1 [n] + x2 [n], we have ak = bk + ck
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Fourier Series and LTI-System
Recall: eigenfunction
For discrete-time, similarly we have
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Fourier Series and LTI-System cont.
H(s) and H(z) are referred to as “system function” (transfer
functions).
(C − T ) continuous-time:
Re{s} = 0 → s = jω
(D − T ) discrete-time: |z| = 1 → z = e jω

R∞
H(jω) = −∞ h(t)e −jωt dt, C − T ,
∞
P
⇒
H(e jω ) =
h[n]e −jωn , D − T .
n=−∞
H(jω) and H(e jω ) are the “frequency response” of the continuous
time and discrete-time system, respectively.
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Remarks
The transfer function, H(jω), is then
H(jω) ≡ |H(jω)| e jφ
with φ ≡ ∠H(jω)
The output is →
H(jω)e jωt = |H(jω)| e j(ωt+φ)
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Fourier Series and LTI-System
Note: For C.T., periodic signal, then
∞
X
x(t) =
k=−∞
ak e jkω0 t → y (t) =
∞
X
ak H(jkω0 )e jkω0 t
k=−∞
Similarly,
X
X
x[n] =
ak ejk(2π/N)n → y [n] =
ak H(e j2πk/N )e jk(2π/N)n
k=<N>
k=<N>
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Remarks cont.
1
Sinusoids are eigenfunctions for any LTI system.
2
H(jω) characterizes the “frequency response” of a linear
system.
Z ∞
H(jω) ≡
h(τ )e −jωτ dτ
−∞
H(jω) is said to be the Fourier Transform of the time domain
function h(t).
Both h(t) and H(jω) can be used to find the output for a
particular input.
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Filter
Filtering: A process that changes the relative amplitude (or
phase) of some frequency components.
F e.g.
frequency-shaping filter
(like equalizer in a Hi-Fi system)
frequency-selective filter
(like low-pass, band-pass, high-pass filters)
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Frequency Shaping Filter
E.g. Differentiator (a HPF or LPF?)
y (t) =
dx(t)
→ H(jω) = jω
dt
→high frequency component is amplified & low frequency
component is suppressed.→ HPF (high pass filter)
(a.k.a. as edge-enhancement filter in image processing)
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Frequency-Selective Filter
Select some bands of frequencies and reject others.
1
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Frequency-Selective Filter cont.
2 D-T
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Frequency-Selective Filter cont.
For D-T: H(e jω ) is periodic with period 2π
low frequencies: at around ω = 0, ±2π, ±4π, . . .
high frequencies: at around ω ± π, ±3π, . . .
Note: ideal filters are not realizable;
Practical filters have transition band, and may have ripple in
stopband and passband
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Summary
Developed Fourier series representation for both C-T and D-T
systems.
Properties of Fourier Series
Eigenfunction and Eigenvalue of LTI systems
System function (transfer function) and frequency response
Filtering
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