Periodic signals - Fourier series
Valentina Hubeika, Honza Černocký
DCGM FIT BUT Brno, {ihubeika,cernocky}@fit.vutbr.cz
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Why do we like exponentials.
Fourier series.
Basis.
FS of square signals.
Example from real world.
Hints.
Mean power and Parseval theorem.
1
By means of FS, we can decompose a periodic signal into a series of complex exponentials.
⇒ Any cosine function with an arbitrary phase can be expressed as sum of
ejx + e−jx
:
exponentials cos(x) =
2
C1 cos(ω1 t + φ1 ) =
=
C1 j(ω1 t+φ1 )
2 e
+
C1 −j(ω1 t+φ1 )
2 e
C1 jφ1 jω1 t
e
2 e
+
C1 −jφ1 −jω1 t
e
,
2 e
=
where c1 =
C1 jφ1
,
2 e
c−1 =
C1 −jφ1
2 e
2
=
are complex constants.
⇒ Transfer of an exponential through an LTI system results in the original
exponential scaled (multiplied) by a constant: the input signal is defined as x(t) = est ,
where s is a complex number. Mainly, we will be interested in the case when s is purely
imaginary, s = jω. However, we will demonstrate the computation on a general complex s.
3
LTI system output is given by convolution of the input signal and the impulse response:
Z +∞
Z +∞
y(t) = h(t) ⋆ x(t) =
h(τ )x(t − τ )dτ =
h(τ )es(t−τ ) dτ
−∞
−∞
The expression es(t−τ ) can be decomposed (using equation ea+b = ea eb ) and the
component independent of τ is moved in front of the integral
Z +∞
Z +∞
y(t) =
h(τ )est e−sτ dτ = est
h(τ )e−sτ dτ.
−∞
−∞
The integral becomes now a function of the complex number s and the system’s impulse
response, but NOT a function of the input! Let us denote the integral as a complex
function H(s). Factor est is the input, thus we can write:
y(t) = est H(s),
Here we see that the output is the same exponential multiplied by the constant defined as:
Z +∞
H(s) =
h(τ )e−sτ dτ.
−∞
4
... what happens when we multiply complex exp. c1 ejω1 t by complex H(s)?
Short recap:
• Every complex number can be decomposed into a module and an agument (phase):
H(s) = |H(s)|ej arg H(s) .
• Multiplication rule: z1 z2 = r1 r2 ej(φ1 +φ2 ) .
How will it be? We start off with:
H(s)c1 ejω1 t = . . .
We do nothing with ejω1 t and define a new complex constant where c′1 = H(s)c1 .
As both H(s) and c1 are complex, during multiplication we multiply the modules and add
up the arguments. The output of LTI system is thus a complex exponential that is scaled
(wider or thinner) and has a different phase. The ”speed“ of the exponential does not
change relative the original input signal.
5
jω1 t
Example: complex exponential c1 e
−j π
4 j100πt
e
= 2.5e
π
π
jπ
4
is multiplied by H(s) = 2e
π
π
c′1 = c1 H(s) = 2.5e−j 4 2ej 4 = 2.5 × 2e−j 4 +j 4 = 5
6
.
FOURIER SERIES
A periodic signal x(t) = x(t + T1 ), with fundamental period T1 , is decomposed into a
series of exponentials:
x(t) =
+∞
X
ck ejkω1 t
k=−∞
ω1 is the fundamental period of the signal, ω1 =
is called harmonic complex exponential.
7
2π
T1 .
Function ejkω1 t for k = 0, ±1, ±2 . . .
Im
1
0
−1
2
1
Re
0
0
0.01
0.02
t
1
Im
Im
1
0
−1
1
0
Re
−1
0
0.01
−1
1
0.02
t
0
0
Re
−1
0
0.01
0.02
t
0
0.02
t
0
Re
−1
0
0.01
0.02
t
1
Im
Im
−1
0.01
0
−1
1
1
0
−1
1
0
Re
1
Im
Im
1
−1
1
0
0
Re
−1
0
0.01
t
0.02
0
−1
1
8
0
Re
−1
0
0.01
t
0.02
Basic properties of the coefficients:
• For real signals x(t): ck and c−k are always complex conjugate:
ck = c⋆−k
• c0 is complex conjugate with itself, thus is real: c0 ∈ ℜ. As ejkω1 t equals to 1 for
k = 0, c0 must be direct component.
• We can rewrite the series into the form:
s(t) = c0 +
∞
X
[ck ejkω1 t + c−k e−jkω1 t ]
|
{z
}
k=1
Ck cos(kω1 t + φk )
• for k > 0
– magnitude of the k-th harmonic component is Ck = 2|ck |.
– initial phase of the k-th harmonic is φk = arg ck .
9
Recapitulation - basis and projection into basis
Basis
• we express one ’thing’ by means of some other ’thing’.
• how similar is some input (vector, signal, function) to some reference.
 
2


x=
x2
3
 
 
0
1




b2 =
b1 =
1
0
 
 
2
2

=3


= 2 x2 = [0 1]
x1 = [1 0]
3
3
x
b2
b1
. . . scalar product
10
x1
Rotated axes
b1 = 
y1 =
y2 =
3

√1
2
√1
2
[ √12
[− √12
x




√1 ] 
2
b2 = 
2
3

√1 ] 
2

− √12
√1
2


y1

2
 = 3.53
2
3

 = 0.707
11
b1
x=

y2
b2

How can we plot a higher dimensional space ?
q
x = [3 2 1 0 1 2 3 4], b1 =
y1 = 5.65, similarity
y2 = 2.41
4
0.4
3
0.3
1
8 [1
1 1 1 1 1 1 1], b2n =
similarity.
1.5
1
2
0.2
1
0.1
0.5
0
0
5
10
0
0
5
10
0.5
0
0
5
10
0
5
10
1.5
1
0.5
0
0
−0.5
−0.5
0
5
10
−1
12
1
2
cos(2π/8n)
General expression in an arbitrary dimensional space...
vector representation yn = bTn x,
matrix representation y = Bx
13
Properties of basis
• One coefficient should not be dependent on the others - ortogonality:
bTi bj = 0
• All basis are of the same ’importance’ - basis magnitude equal to 1
|bi | = 1
• Ortogonality and unit magnitude of basis gives us an orthonormal system
14
Signal and basis can be functions!
• one function can be represented by means of some other function
• how two functions are similar.
Analogiously to the scalar product:
• multiplication with the basis
R
P
• adding up - integral for continuous signals or sum
for descrete
Example - what is cosine similar to ?
15
To a direct signal ?
b(t) = 1
R1
0
x(t)
s(t)b(t)dt = 0 ... is not similar
b(t)
x(t)b(t)
1
2
1
0.8
1.8
0.8
0.6
1.6
0.6
0.4
1.4
0.4
0.2
1.2
0.2
0
1
0
−0.2
0.8
−0.2
−0.4
0.6
−0.4
−0.6
0.4
−0.6
−0.8
0.2
−0.8
−1
0
0.5
1
0
0
0.5
1
−1
0
0.5
16
1
Maybe to a cosine function ?
b(t) = 2 cos(2πt)
x(t)
1
R1
0
s(t)b(t)dt = 1 ... is similar
b(t)
x(t)b(t)
2
0.8
2
1.8
1.5
0.6
1.6
1
0.4
1.4
0.5
0.2
0
1.2
0
−0.2
1
0.8
−0.5
−0.4
0.6
−1
−0.6
0.4
−1.5
−0.8
−1
0
0.5
1
−2
0.2
0
0.5
1
0
0
0.5
17
1
2x faster cosine ?
b(t) = 2 cos(4πt)
x(t)
1
0.8
R1
0
s(t)b(t)dt = 0 ... is not similar
b(t)
x(t)b(t)
2
2
1.5
1.5
1
1
0.5
0.5
0
0
−0.5
−0.5
−1
−1
−1.5
−1.5
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
0
0.5
1
−2
0
0.5
1
−2
0
0.5
18
1
Now lets play with sine function s(t) = sin(2πt),
... no
x(t)
b(t)
1
x(t)b(t)
2
0.8
b(t) = 2 cos(2πt)
1
0.8
1.5
0.6
0.6
1
0.4
0.4
0.5
0.2
0
0.2
0
−0.2
0
−0.2
−0.5
−0.4
−0.4
−1
−0.6
−0.6
−1.5
−0.8
−1
0
0.5
1
−2
−0.8
0
0.5
1
−1
0
0.5
⇒ we will need cosine and sine !!!
19
1
R1
0
s(t)b(t)dt = 0
Shifted sine ...
s(t) = sin(2πt − π/5)
R1
s(t)b1 (t)dt = −0.59
b1 (t) = 2 cos(2πt)
0
R1
s(t)b1 (t)dt = 0.81
b2 (t) = 2 sin(2πt)
0
x(t)
b(t)
1
2
0.5
1
0
0
x(t)b(t)
0.5
0
−0.5
−1
−0.5
−1
−1
0
0.5
1
−2
−1.5
0
0.5
1
−2
0
b(t)
0.5
1
x(t)b(t)
2
2
1.5
1
1
0
0.5
−1
−2
0
0
0.5
1
−0.5
0
0.5
1
20
We will be working with complex exponentials !
that conjugate cosine and sine in themselves !
b1 (t) = ejωt = cos ωt + j sin ωt
R
coefficient calculation: c1 = x(t)b⋆1 (t)dt
product of such coefficient and the base is complex... so we need to add a complex
cojugate base:
b−1 (t) = e−jωt = cos ωt − j sin ωt
R
coefficient computation: c−1 = x(t)b⋆−1 (t)dt
21
Derivation of coefficients of FS
We said that a periodic function can be decomposed to the series:
x(t) =
+∞
X
ck ejkω1 t
k=−∞
The coefficients ck are computed using the formula:
Z
1
x(t)e−jkω1 t dt,
ck =
T1 T1
R
where T1 denotes integral over one period (typically, from 0 to T1 or from − T21 to
also can be from 25T1 to 26T1 )
22
T1
2 ,
but
Plotting FS coefficients
The coefficients are indexed from k = −∞ to ∞. Remember, they are tied with angular
frequencies: c1 is associated with ω1 , c2 with 2ω1 Coefficients with negative indeces are
associated with corresponding negative frequencies: c−1 is associated with −ω1 , c−2 with
−2ω1 Negative frequency means that the exponential c−k e−jkω1 t is rotating the opposite
direction that ck ejkω1 t . Their composition will give us a normal cosine Ck cos(kω1 t + φk ).
23
As the coefficients are complex numbers, we will plot their modules and aguments in
separate graphs along frequency axes. Thus we are expressing a signal in frequency
domain. Coefficients’ position and value will form a SPECTRUM of a signal. For a
periodic signal the spectrum is descrete (composed of vertical lines).
Example 1.
Derive FS coefficients for signal: x(t) = 5 cos(100πt) . The basic angular frequency is
ω1 = 100π. In this example we don’t integrate because we can rewrite cosine function as:
x(t) =
5 j100πt 5 −j100πt
e
+ e
,
2
2
thus, the coefficients’ values are :
5
c1 = ,
2
c−1
The coefficients are real.
24
5
=
2
25
Example 2.
Derive FS coefficients of the signal: x(t) = 5 cos(100πt − π4 ) Again, we will manage with
no integration. Either we can rewrite cosine into the form:
π
π
2.5e−j 4 ej100πt + 2.5e+j 4 e−j100πt ,
thus, for c1 and c−1 we get:
−j π
4
c1 = 2.5e
+j π
4
,
c−1 = 2.5e
OR we look at how the coefficients ck are related to the magnitude and initial phase of the
harmonics (here, only one): C1 = 5 and φ1 = − π4 .
C1
= 2.5,
|c1 | =
2
π
arg c1 = φ1 = − ,
4
so
−j π
4
c1 = 2.5e
.
Coefficient c−1 is a complex conjugate to c1 (same module, reverse angle):
|c−1 | =
C1
= 2.5,
2
arg c−1 = −φ1 =
26
π
,
4
so
π
c−1 = 2.5ej 4 .
27
Example 3. . . . let’s integrate!! We have a periodic signal of square impulses:

 D for − ϑ ≤ t ≤ ϑ
2
2
x(t) =
 0 for − T1 ≤ t < − ϑ a ϑ < t ≤ T1
2
2
2
2
with period T1
s(t)
D
−υ/2
υ/2
−υ/2+Τ1 Τ 1
28
t
Preparation 1. — Function sinc(·)
1
0.8
sinc(x) =

sin(x)




x



 1
for x 6= 0
sinc(x)
0.6
0.4
0.2
for x = 0
0
−2π
−π
0
π
2π
−0.2
−0.4
−10
... Matlab! sinc is defined as
sin(πx)
πx
−5
!!!
29
0
x
5
10
Preparation 2. — Computation of integral ejxy
I(x) =
Z
b
e±jxy dy
−b
a) for x = 0 :
I(0) = 2b
b) for x 6= 0 :
I(x)
·
e±jxy
=
±jx
¸b
−b
ejxb − e−jxb
=
=
jx
2 ejxb − e−jxb
sin bx
=
= 2b
.
x
2j
bx
Z
b
e±jxy dy = 2b sinc(bx).
−b
30
Spectrum of a periodic signal of square impulses
The signal is defined by ϑ, D, T1 .
ck
T1
2
1
T1
Z
+
1
=
T1
Z
+ϑ
2
=
−
T1
2
−ϑ
2
x(t)e−jkω1 t dt
D
−jkω1 t
De
dt =
T1
Z
+ϑ
2
e−jkω1 t dt.
−ϑ
2
When we define: t = y, b = ϑ/2 and x = kω1 , we obtain:
¶
µ
¶
µ
D ϑ
ϑ
ϑ
ϑ
ck =
2 sinc
kω1 = D sinc
kω1 .
T1 2
2
T1
2
Plotting results is not trivial :-(
• Prepare graphs ω-|ck | and ω-arg(ck ).
• Realize that the resulting coefficients are real numbers- either positive or negative
depending on the sign of the funtion sinc.
31
• Prepare an auxiliary function (dotted)
for aµgeneral
¯
¶¯ ω (we will be placing the
¯
¯ ϑ
ϑ
¯
ω ¯¯. Note, that the principal (central)
coefficients under the aux func): ¯D sinc
T1
2
fold of the function is twise as wide as the side folds. Very important point is ωa ,
where the function touchs the axis for the first time. To estimate ωa , set the argument
of the sinc equal to π:
2π
ϑ
ωa = π → ω a =
.
2
ϑ
• indicate the coefficients on the aux function at every fold of ω1 .
• argument of a positive coefficients (sinc(·) > 0) is 0. Argument of a negative
coefficient (sinc(·) < 0) must be ±π. Convention: for positive ω it will be +π and for
negative ω it will be −π. (It will not be incorrect when its the opposite thought).
32
ω
|c |
=2π/T
1
k
1
Dϑ/T
1
ωa=2π/ϑ
|Dϑ/T1 sinc (ϑω/2)|
−5ω −4ω
1
1
−ω
a
−2ω −1ω
1
1
0
1ω
1
2ω
1
ω
a
4ω
1
5ω
1
6ω
1
7ω
1
8ω
1
−−−> ω
arg ck
π
−5ω −4ω −3ω −2ω −1ω
1
1
1
1
1
0
1ω1 2ω1 3ω1 4ω1 5ω1 6ω1 7ω1 8ω1
−−−> ω
−π
33
c0 and mean of a signal
¯
µ ¶¯
¯
¯ ϑ
ϑ
¯
The coefficient c0 lies in the maximum of the auxiliary function ¯D sinc
ω ¯¯ for
T1
2
ω = 0. Its value is thus:
ϑ
c0 = D .
T1
We also know that the value of c0 must be equal to the direct component of the signal,
which is its mean
Z + T21
Z + ϑ2
1
1
1
Dϑ
+ϑ
x̄ =
x(t)dt =
Ddt =
[Dt]− ϑ2 =
.
T
2
T1 − 21
T1 − ϑ2
T1
T1
Proved!
34
6
Example 1. D = 6,
T
=
1
µs,
ϑ
=
0.5
µs.
Solution:
f
=
1
MHz,
ω
=
2
×
10
π rad/s,
1
1
1
¯
¯
¡ ϑ ¢¯
¯ ϑ
auxiliary function: ¯D T1 sinc 2 ω ¯ = |3sinc(0.25 × 10−6 ω)|. It touches zero when
0.25 × 10−6 ω = kπ, so ω = k4 × 106 π.
35
6
Example 2. D = 6,
T
=
1
µs,
ϑ
=
0.25
µs.
Solution:
f
=
1
MHz,
ω
=
2
×
10
π rad/s,
1
1
1
¯
¯
¡ ϑ ¢¯
¯ ϑ
auxiliary function: ¯D T1 sinc 2 ω ¯ = |1.5sinc(0.125 × 10−6 ω)|. It touches zeros for
0.125 × 10−6 ω = kπ, so for ω = k8 × 106 π.
36
Example 3. ... same as example 1 but the signal is normalized (mean is 0)
Same result with the only difference: c0 is equal to 0.
37
Again example 1. – how it is composed: k = 0, ±1, ±3, ±5
4
4
3
3
2
−1
0
2
−1
10
1
−6
x 10
5
0
1
−6
x 10
2
5
0
−2
2
0
0
0
−2
−1
0
1
−6 −5
x 10
−1
5
0
1
−6
x 10
−5
−1
10
0
1
−6
x 10
2
5
0
−2
2
0
0
0
−2
−1
0
1
−6 −5
x 10
−1
5
0
1
−6
x 10
−5
−1
10
0
1
−6
x 10
2
5
0
−2
2
0
0
0
−2
−1
0
1
−6 −5
x 10
−1
0
1
−6
x 10
38
−5
−1
0
1
−6
x 10
k = ±7, ±9, ±11, ±13
5
10
2
5
0
−2
2
0
0
0
−2
−1
0
1
−6 −5
x 10
−1
5
0
1
−6
x 10
−5
−1
10
0
1
−6
x 10
2
5
0
−2
2
0
0
0
−2
−1
0
1
−6 −5
x 10
−1
5
0
1
−6
x 10
−5
−1
10
0
1
−6
x 10
2
5
0
−2
2
0
0
0
−2
−1
0
1
−6 −5
x 10
−1
5
0
1
−6
x 10
−5
−1
10
0
1
−6
x 10
2
5
0
−2
2
0
0
0
−2
−1
0
1
−6 −5
x 10
−1
0
1
−6
x 10
39
−5
−1
0
1
−6
x 10
What else can we decompose ?
Wovel ’a’, Fs = 16 kHz,
selected 1 period, repeted 50× ⇒ xx.wav.
Coefficients of FS:
yytill14.wav – k = 0, ±1, ±2, ±3, ±4.
yytill19.wav – k = 0, ±1, ±2, ±3, ±4, ±5, ±6, ±7, ±8, ±9. etc. . .
1
0.5
0
−0.5
0
50
100
150
0.08
0.06
0.04
0.02
0
0
5
10
15
20
0
5
10
15
20
25
30
35
40
45
50
25
30
35
40
45
50
4
2
0
−2
−4
40
0.05
0.2
0
0
−0.05
0.10
50
100
−0.2
10
150
0
100
150
50
100
150
50
100
150
50
100
150
50
100
150
50
100
150
50
100
150
50
100
150
0
−0.1
0.10
50
100
−1
10
150
0
0
−0.1
0.10
50
100
−1
10
150
0
0
−0.1
0.10
50
100
−1
10
150
0
0
−0.1
0.10
50
100
−1
10
150
0
0
−0.1
0.10
50
100
−1
10
150
0
0
−0.1
0.10
50
100
−1
10
150
0
−0.1
50
0
0
50
100
−1
150
41
0
Properties of spectra of periodic signals
x(t)
ck
linearity
a xa (t) + b xb (t)
a ca,k + b cb,k
time shift
x(t − τ )
ck e−jkω1 τ
scale of time axis
x(mt)
ck
42
A little bit more on time shif
Z
1
x(t)e−jkω1 t dt. FS coefficients of a shifted
FS coefficients are computed as ck =
T1 T1
signal x(t − τ ) are then:
¯
¯
¯ r =t−τ ¯
¯
¯
Z
Z
¯
¯
1
1
x(t − τ )e−jkω1 t dt = ¯¯ t = r + τ ¯¯ =
x(r)e−jkω1 (r+τ ) dr =
ck,τ =
T1 T1
¯
¯ T1 T1
¯ dr = dt ¯
=
1
T1
Z
x(r)e−jkω1 r e−jkω1 τ dr = e−jkω1 τ ck
T1
How have the coefficients changed? (Recall, complex numbers multiplication)
• magnitude - NO, as |e−jkω1 τ | = 1: |ck,τ | = ck .
• phase - YES - shift by factor −kω1 τ : arg ck,τ = arg ck − kω1 τ .
43
Time shift – example
signal x(t) from example 1.: D = 6, T1 = 1 µs, ϑ = 0.5 µs. Shifted signal
x(t − 0.25 × 10−6 ): coefficients are multiplied by |e−jkω1 τ | = 1
2π
π
−6
τ
=
−k
(0.25
×
10
)
=
−k
:−kω1 τ = −k 2π
−6
T1
1×10
2.
44
45
Mean power of periodic signals – Parseval theorem
1
Ps =
T1
Z
|x(t)|2 dt =
T1
+∞
X
|ck |2 .
k=−∞
The signal power thus can be calculated not only through integration of | · |2 in time, but
also by summing up squares of all FS coefficients. Summation is usually easier realized
than integration.
Why? What is the power of k-th complex exponential:
Z
Z
1
1
|ck ejkω1 t |2 dt =
|ck |2 dt = |ck |2
T1 T1
T1 T1
The power is then |ck |2 , when powers of all exponentials are summed, we get overall
mean power Ps of the signal.
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Convergency of FS
3 Dirichlet properties:
1. x(t) must be absolutelly integrable:
Z
|x(t)| < ∞
T1
not fullfilled for example for: x(t) = 1/t for 0 < t ≤ 1, periodic with T1 = 1.
2. x(t) must have finitly variable: a finit interval must contain a finit number of minima
and maxima.
not fullfilled for example for: x(t) = sin 2π
t for 0 < t ≤ 1, periodic with T1 = 1.
3. on finite interval the signal must have finite number of discontinuities.
We are going to work with signals that fully satisfy these restrictions.
47
Conclusions
•
Spectrum of periodic signal is discrete.
•
When signal is narrowed — spectrum is spread.
•
When signal is delayed phase is dropped downward,
when signal is advanced phase is tipped upward
48