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Linear Systems
lecture 5
Fourier series
UNIVERSITY OF TWENTE.
academic year : 18-19lecture : 5build : August 28, 2018slides : 39
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
1 intro
LS
Today
1 State equations and stability2 Inner products3 Fourier series
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
2 1.1
LS
State equations and stability Section 2.6.5
TheoremSuppose an LTI system is described by an n-th order lineardifferential equation. Assume that the state equations invector form are
v′ = Av + x(t)b.y(t) = c · v + d x(t),
where A is an n × n-matrix, b and c are vectors, and d is aconstant. Let λ1, λ2, . . . , λn be the eigenvalues of A.
If all Reλk < 0 for all k = 1, . . . ,n, then the system isstable.If there exists an index k such that Reλk > 0, then thesystem is unstable.
If Reλk = 0 for some index k, then no conclusion canbe drawn.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
2 1.1
LS
State equations and stability Section 2.6.5
TheoremSuppose an LTI system is described by an n-th order lineardifferential equation. Assume that the state equations invector form are
v′ = Av + x(t)b.y(t) = c · v + d x(t),
where A is an n × n-matrix, b and c are vectors, and d is aconstant. Let λ1, λ2, . . . , λn be the eigenvalues of A.
If all Reλk < 0 for all k = 1, . . . ,n, then the system isstable.If there exists an index k such that Reλk > 0, then thesystem is unstable.
If Reλk = 0 for some index k, then no conclusion canbe drawn.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
3 1.2
LS
Trace and determinant
Let A =[
a bc d
]be a real 2× 2-matrix.
- The trace of A is tr(A) = a + d.- The determinant of A is det(A) = ad − bc.
Denote T = tr(A) and D = det(A). Thecharacteristic equation of A is
s2 − Ts + D = (s − λ1)(s − λ2) = 0. (∗)
Let λ1 and λ2 be the eigenvalues of A, thenT = λ1 + λ2 and D = λ1λ2.
The discriminant of (∗) is T 2 − 4D.T2 > 4D Equation (∗) has two real roots λ1 6= λ2.
T2 < 4D Equation (∗) has two non-real roots λ1 = λ2.
T2 = 4D Equation (∗) has one real roots λ1 = λ2.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
3 1.2
LS
Trace and determinant
Let A =[
a bc d
]be a real 2× 2-matrix.
- The trace of A is tr(A) = a + d.- The determinant of A is det(A) = ad − bc.
Denote T = tr(A) and D = det(A). Thecharacteristic equation of A is
s2 − Ts + D = (s − λ1)(s − λ2) = 0. (∗)
Let λ1 and λ2 be the eigenvalues of A, thenT = λ1 + λ2 and D = λ1λ2.
The discriminant of (∗) is T 2 − 4D.T2 > 4D Equation (∗) has two real roots λ1 6= λ2.
T2 < 4D Equation (∗) has two non-real roots λ1 = λ2.
T2 = 4D Equation (∗) has one real roots λ1 = λ2.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
3 1.2
LS
Trace and determinant
Let A =[
a bc d
]be a real 2× 2-matrix.
- The trace of A is tr(A) = a + d.- The determinant of A is det(A) = ad − bc.
Denote T = tr(A) and D = det(A). Thecharacteristic equation of A is
s2 − Ts + D = (s − λ1)(s − λ2) = 0. (∗)
Let λ1 and λ2 be the eigenvalues of A, thenT = λ1 + λ2 and D = λ1λ2.
The discriminant of (∗) is T 2 − 4D.T2 > 4D Equation (∗) has two real roots λ1 6= λ2.
T2 < 4D Equation (∗) has two non-real roots λ1 = λ2.
T2 = 4D Equation (∗) has one real roots λ1 = λ2.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
3 1.2
LS
Trace and determinant
Let A =[
a bc d
]be a real 2× 2-matrix.
- The trace of A is tr(A) = a + d.- The determinant of A is det(A) = ad − bc.
Denote T = tr(A) and D = det(A). Thecharacteristic equation of A is
s2 − Ts + D = (s − λ1)(s − λ2) = 0. (∗)
Let λ1 and λ2 be the eigenvalues of A, thenT = λ1 + λ2 and D = λ1λ2.
The discriminant of (∗) is T 2 − 4D.T2 > 4D Equation (∗) has two real roots λ1 6= λ2.
T2 < 4D Equation (∗) has two non-real roots λ1 = λ2.
T2 = 4D Equation (∗) has one real roots λ1 = λ2.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
4 1.3
LS
Second order LTI systems
TheoremSuppose an LTI system is described by a second-order lineardifferential equation. Assume that the state equations invector form are
v′ = Av + x(t)b.
y(t) = c · v + d x(t),
where A is a 2× 2-matrix.
If tr(A) < 0 and det(A) > 0, then the system is stable.
If tr(A) > 0 or det(A) < 0, then the system is unstable.
This theorem can be proved by using the results fromthe previous slide.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
4 1.3
LS
Second order LTI systems
TheoremSuppose an LTI system is described by a second-order lineardifferential equation. Assume that the state equations invector form are
v′ = Av + x(t)b.
y(t) = c · v + d x(t),
where A is a 2× 2-matrix.
If tr(A) < 0 and det(A) > 0, then the system is stable.
If tr(A) > 0 or det(A) < 0, then the system is unstable.
This theorem can be proved by using the results fromthe previous slide.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
5 1.4
LS
Example
Example Example 2.6.13
Investigate the stability of the LTI system whose statematrix is
A =[
2 −14 −3
].
T = tr(A) = 2 + (−3) = −1 < 0, so we can notconclude anything yet.D = det(A) = 2 · (−3)− 4 · (−1) = −2 < 0, hence thesystem is unstable.Alternatively: note that the discriminant isT 2 − 4D = 9 > 0, hence A has two distinct realeigenvalues.The eigenvalues are λ1 = −2 and λ2 = 1. SinceReλ2 = 1 > 0, the system is unstable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
5 1.4
LS
Example
Example Example 2.6.13
Investigate the stability of the LTI system whose statematrix is
A =[
2 −14 −3
].
T = tr(A) = 2 + (−3) = −1 < 0, so we can notconclude anything yet.
D = det(A) = 2 · (−3)− 4 · (−1) = −2 < 0, hence thesystem is unstable.Alternatively: note that the discriminant isT 2 − 4D = 9 > 0, hence A has two distinct realeigenvalues.The eigenvalues are λ1 = −2 and λ2 = 1. SinceReλ2 = 1 > 0, the system is unstable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
5 1.4
LS
Example
Example Example 2.6.13
Investigate the stability of the LTI system whose statematrix is
A =[
2 −14 −3
].
T = tr(A) = 2 + (−3) = −1 < 0, so we can notconclude anything yet.D = det(A) = 2 · (−3)− 4 · (−1) = −2 < 0, hence thesystem is unstable.
Alternatively: note that the discriminant isT 2 − 4D = 9 > 0, hence A has two distinct realeigenvalues.The eigenvalues are λ1 = −2 and λ2 = 1. SinceReλ2 = 1 > 0, the system is unstable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
5 1.4
LS
Example
Example Example 2.6.13
Investigate the stability of the LTI system whose statematrix is
A =[
2 −14 −3
].
T = tr(A) = 2 + (−3) = −1 < 0, so we can notconclude anything yet.D = det(A) = 2 · (−3)− 4 · (−1) = −2 < 0, hence thesystem is unstable.Alternatively: note that the discriminant isT 2 − 4D = 9 > 0, hence A has two distinct realeigenvalues.
The eigenvalues are λ1 = −2 and λ2 = 1. SinceReλ2 = 1 > 0, the system is unstable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
5 1.4
LS
Example
Example Example 2.6.13
Investigate the stability of the LTI system whose statematrix is
A =[
2 −14 −3
].
T = tr(A) = 2 + (−3) = −1 < 0, so we can notconclude anything yet.D = det(A) = 2 · (−3)− 4 · (−1) = −2 < 0, hence thesystem is unstable.Alternatively: note that the discriminant isT 2 − 4D = 9 > 0, hence A has two distinct realeigenvalues.The eigenvalues are λ1 = −2 and λ2 = 1. SinceReλ2 = 1 > 0, the system is unstable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
6 1.5
LS
Example
Example
Investigate the stability of the LTI system whose statematrix is
A =[−3 −1
5 1
].
T = tr(A) = 2 + (−3) = −2 < 0 andD = det(A) = (−3) · 1− 5 · (−1) = 2 > 0, hence thesystem is stable.Alternatively: note that the discriminant isT 2 − 4D = −4 < 0, hence A has two conjugatenon-real eigenvalues.The eigenvalues are λ1 = −1 + i and λ2 = −1− i.Since Reλ1 = Reλ2 = −1 < 0, the system is stable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
6 1.5
LS
Example
Example
Investigate the stability of the LTI system whose statematrix is
A =[−3 −1
5 1
].
T = tr(A) = 2 + (−3) = −2 < 0 andD = det(A) = (−3) · 1− 5 · (−1) = 2 > 0, hence thesystem is stable.
Alternatively: note that the discriminant isT 2 − 4D = −4 < 0, hence A has two conjugatenon-real eigenvalues.The eigenvalues are λ1 = −1 + i and λ2 = −1− i.Since Reλ1 = Reλ2 = −1 < 0, the system is stable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
6 1.5
LS
Example
Example
Investigate the stability of the LTI system whose statematrix is
A =[−3 −1
5 1
].
T = tr(A) = 2 + (−3) = −2 < 0 andD = det(A) = (−3) · 1− 5 · (−1) = 2 > 0, hence thesystem is stable.Alternatively: note that the discriminant isT 2 − 4D = −4 < 0, hence A has two conjugatenon-real eigenvalues.
The eigenvalues are λ1 = −1 + i and λ2 = −1− i.Since Reλ1 = Reλ2 = −1 < 0, the system is stable.
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
Fourier series
Fourier series ofreal signals
Linear Systems
LS.18-19[5]28-8-2018
6 1.5
LS
Example
Example
Investigate the stability of the LTI system whose statematrix is
A =[−3 −1
5 1
].
T = tr(A) = 2 + (−3) = −2 < 0 andD = det(A) = (−3) · 1− 5 · (−1) = 2 > 0, hence thesystem is stable.Alternatively: note that the discriminant isT 2 − 4D = −4 < 0, hence A has two conjugatenon-real eigenvalues.The eigenvalues are λ1 = −1 + i and λ2 = −1− i.Since Reλ1 = Reλ2 = −1 < 0, the system is stable.
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LS.18-19[5]28-8-2018
7 1.6
LS
Intermediate test demarcation line
This slide marks the end of the first part of the course.
The material required for the intermediate test consistsof everything up to this point.
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8 2.1
LS
Inner products
DefinitionLet ϕ and ψ be two complex-valued functions defined on theinterval [a, b]. The inner product of ϕ and ψ is
〈ϕ,ψ〉 =∫ b
aϕ(t)ψ(t) dt.
The inner product of ϕ and ψ is a complex number.
Properties
(1) 〈ϕ,ψ〉 = 〈ψ,ϕ〉,
(2) 〈αϕ1 + βϕ2, ψ〉 = α〈ϕ1, ψ〉+ β〈ϕ2, ψ〉;〈ϕ, αψ1 + βψ2〉 = α〈ϕ,ψ1〉+ β〈ϕ,ψ2〉,
(3) 〈ϕ,ϕ〉 ∈ R and 〈ϕ,ϕ〉 ≥ 0,
(4) 〈ϕ,ϕ〉 = 0 if and only if ϕ = 0.
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8 2.1
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Inner products
DefinitionLet ϕ and ψ be two complex-valued functions defined on theinterval [a, b]. The inner product of ϕ and ψ is
〈ϕ,ψ〉 =∫ b
aϕ(t)ψ(t) dt.
The inner product of ϕ and ψ is a complex number.
Properties
(1) 〈ϕ,ψ〉 = 〈ψ,ϕ〉,
(2) 〈αϕ1 + βϕ2, ψ〉 = α〈ϕ1, ψ〉+ β〈ϕ2, ψ〉;〈ϕ, αψ1 + βψ2〉 = α〈ϕ,ψ1〉+ β〈ϕ,ψ2〉,
(3) 〈ϕ,ϕ〉 ∈ R and 〈ϕ,ϕ〉 ≥ 0,
(4) 〈ϕ,ϕ〉 = 0 if and only if ϕ = 0.
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9 2.2
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Norm and distance
DefinitionThe norm of a function ϕ : [a, b]→ C is defined as
‖ϕ‖ =√〈ϕ,ϕ〉.
The norm of a function is sometimes called the lengthof a function.In some books, the norm is denoted with single bars:
‖ϕ‖ = |ϕ|.
DefinitionLet ϕ,ψ : [a, b]→ C be two complex-valued functions. Thedistance beween ϕ and ψ is defined as
dist(ϕ,ψ) = ‖ϕ− ψ‖ .
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9 2.2
LS
Norm and distance
DefinitionThe norm of a function ϕ : [a, b]→ C is defined as
‖ϕ‖ =√〈ϕ,ϕ〉.
The norm of a function is sometimes called the lengthof a function.
In some books, the norm is denoted with single bars:‖ϕ‖ = |ϕ|.
DefinitionLet ϕ,ψ : [a, b]→ C be two complex-valued functions. Thedistance beween ϕ and ψ is defined as
dist(ϕ,ψ) = ‖ϕ− ψ‖ .
UNIVERSITYOF TWENTE.
State equationsand stability
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LS.18-19[5]28-8-2018
9 2.2
LS
Norm and distance
DefinitionThe norm of a function ϕ : [a, b]→ C is defined as
‖ϕ‖ =√〈ϕ,ϕ〉.
The norm of a function is sometimes called the lengthof a function.In some books, the norm is denoted with single bars:
‖ϕ‖ = |ϕ|.
DefinitionLet ϕ,ψ : [a, b]→ C be two complex-valued functions. Thedistance beween ϕ and ψ is defined as
dist(ϕ,ψ) = ‖ϕ− ψ‖ .
UNIVERSITYOF TWENTE.
State equationsand stability
Inner products
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LS.18-19[5]28-8-2018
9 2.2
LS
Norm and distance
DefinitionThe norm of a function ϕ : [a, b]→ C is defined as
‖ϕ‖ =√〈ϕ,ϕ〉.
The norm of a function is sometimes called the lengthof a function.In some books, the norm is denoted with single bars:
‖ϕ‖ = |ϕ|.
DefinitionLet ϕ,ψ : [a, b]→ C be two complex-valued functions. Thedistance beween ϕ and ψ is defined as
dist(ϕ,ψ) = ‖ϕ− ψ‖ .
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10 2.3
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Orthogonal and orthonormal sets
Definition
Two functions ϕ,ψ : [a, b]→ C are said to beorthogonal if 〈ϕ,ψ〉 = 0.A set of non-zero functions {ϕ1, ϕ2, . . .} is anorthogonal set if 〈ϕi , ϕj〉 = 0 for all i 6= j.
Note that if i = j then 〈ϕi , ϕj〉 = 〈ϕi , ϕi〉 > 0.
DefinitionThe Kronecker delta function is defined as
δij ={
1 if i = j,0 if i 6= j.
A set of functions {ϕ1, ϕ2, . . .} is an orthonormal setif 〈ϕi , ϕj〉 = δij .
In an orthonormal set, every function has length 1.
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Orthogonal and orthonormal sets
Definition
Two functions ϕ,ψ : [a, b]→ C are said to beorthogonal if 〈ϕ,ψ〉 = 0.A set of non-zero functions {ϕ1, ϕ2, . . .} is anorthogonal set if 〈ϕi , ϕj〉 = 0 for all i 6= j.
Note that if i = j then 〈ϕi , ϕj〉 = 〈ϕi , ϕi〉 > 0.
DefinitionThe Kronecker delta function is defined as
δij ={
1 if i = j,0 if i 6= j.
A set of functions {ϕ1, ϕ2, . . .} is an orthonormal setif 〈ϕi , ϕj〉 = δij .
In an orthonormal set, every function has length 1.
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Periodic signals
LemmaLet x(t) be a periodic signal with period T > 0, then∫ T
0x(t) dt =
∫ a+T
ax(t) dt for all a ∈ R.
0 a T a+T
x(t)
NotationThe integral over the interval [a, a + T ] is denoted as∫
〈T〉x(t) dt.
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Periodic signals
LemmaLet x(t) be a periodic signal with period T > 0, then∫ T
0x(t) dt =
∫ a+T
ax(t) dt for all a ∈ R.
0 a T a+T
x(t)
NotationThe integral over the interval [a, a + T ] is denoted as∫
〈T〉x(t) dt.
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Proof of the lemma Self-tuition
We prove the case 0 < a < T .∫ a+T
ax(t) dt
=∫ T
ax(t) dt +
∫ a+T
Tx(t) dt
τ = t − T
=∫ T
ax(t) dt +
∫ a
0x(τ + T ) dτ
x(t) isperiodic
=∫ T
ax(t) dt +
∫ a
0x(t) dt
swapintegrals
=∫ a
0x(t) dt +
∫ T
ax(t) dt
=∫ T
0x(t) dt.
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Time-harmonic signals Recap
DefinitionA time-harmonic signal is a continuous-time signalf : R→ C of the form f (t) = ceiωt with constants c ∈ Cand ω ∈ R.
The constant ω is called the (angular) frequency ofthe signal.
If c = Aeiϕ0 with A, ϕ0 ∈ R then f (t) = Aei(ωt+ϕ0).The constant A is called the amplitude of the signal,and ϕ0 is called the phase.
Let T = 2π/ |ω |, then f (t) = ceiωt is periodic withperiod T :
f (t) = f (t + T ) for all t ∈ R.
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Time-harmonic signals Recap
DefinitionA time-harmonic signal is a continuous-time signalf : R→ C of the form f (t) = ceiωt with constants c ∈ Cand ω ∈ R.
The constant ω is called the (angular) frequency ofthe signal.
If c = Aeiϕ0 with A, ϕ0 ∈ R then f (t) = Aei(ωt+ϕ0).The constant A is called the amplitude of the signal,and ϕ0 is called the phase.
Let T = 2π/ |ω |, then f (t) = ceiωt is periodic withperiod T :
f (t) = f (t + T ) for all t ∈ R.
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Time-harmonic signals Recap
DefinitionA time-harmonic signal is a continuous-time signalf : R→ C of the form f (t) = ceiωt with constants c ∈ Cand ω ∈ R.
The constant ω is called the (angular) frequency ofthe signal.
If c = Aeiϕ0 with A, ϕ0 ∈ R then f (t) = Aei(ωt+ϕ0).The constant A is called the amplitude of the signal,and ϕ0 is called the phase.
Let T = 2π/ |ω |, then f (t) = ceiωt is periodic withperiod T :
f (t) = f (t + T ) for all t ∈ R.
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Time-harmonic signals
TheoremLet ω0 > 0. The set {einω0t |n ∈ Z} is an orthogonal setconsisting of periodic time-harmonic signals withperiod T = 2π/ω0.
Note that ϕn(t) = einω0t is periodic with periodpn = 2π/(|n |ω0). Observe that T = |n | pn , thereforeT is also a period of ϕn .
For the inner product we integrate over the domain[0,T ] for all signals ϕn .
〈ϕm , ϕn〉 =∫ T
0eimω0t einω0t dt
=∫ T
0eimω0te−inω0t dt
=∫ T
0ei(m−n)ω0t dt.
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Time-harmonic signals
TheoremLet ω0 > 0. The set {einω0t |n ∈ Z} is an orthogonal setconsisting of periodic time-harmonic signals withperiod T = 2π/ω0.
Note that ϕn(t) = einω0t is periodic with periodpn = 2π/(|n |ω0). Observe that T = |n | pn , thereforeT is also a period of ϕn .For the inner product we integrate over the domain[0,T ] for all signals ϕn .
〈ϕm , ϕn〉 =∫ T
0eimω0t einω0t dt
=∫ T
0eimω0te−inω0t dt
=∫ T
0ei(m−n)ω0t dt.
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Time-harmonic signals
TheoremLet ω0 > 0. The set {einω0t |n ∈ Z} is an orthogonal setconsisting of periodic time-harmonic signals withperiod T = 2π/ω0.
Note that ϕn(t) = einω0t is periodic with periodpn = 2π/(|n |ω0). Observe that T = |n | pn , thereforeT is also a period of ϕn .For the inner product we integrate over the domain[0,T ] for all signals ϕn .
〈ϕm , ϕn〉 =∫ T
0eimω0t einω0t dt
=∫ T
0eimω0te−inω0t dt
=∫ T
0ei(m−n)ω0t dt.
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LS
Time-harmonic signals
If m 6= n then
〈ϕm , ϕn〉 =∫ T
0ei(m−n)ω0t dt
= 1(m − n)ω0i ei(m−n)ω0t
∣∣∣T0
= 1(m − n)ω0i
[ei(m−n)ω0T − 1
]= 1
(m − n)ω0i[e(m−n)2πi − 1
]= 0,
This shows that {eint |n ∈ Z} is an orthogonal set.
If m = n then
‖ϕm ‖2 = 〈ϕm , ϕm〉 =∫ T
0ei(m−m)t dt =
∫ T
01 dt = T .
Corollary
The set{
1√T einω0t
∣∣∣ n ∈ Z}is an orthonormal set.
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Time-harmonic signals
If m 6= n then
〈ϕm , ϕn〉 =∫ T
0ei(m−n)ω0t dt
= 1(m − n)ω0i ei(m−n)ω0t
∣∣∣T0
= 1(m − n)ω0i
[ei(m−n)ω0T − 1
]= 1
(m − n)ω0i[e(m−n)2πi − 1
]= 0,
This shows that {eint |n ∈ Z} is an orthogonal set.If m = n then
‖ϕm ‖2 = 〈ϕm , ϕm〉 =∫ T
0ei(m−m)t dt =
∫ T
01 dt = T .
Corollary
The set{
1√T einω0t
∣∣∣ n ∈ Z}is an orthonormal set.
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Time-harmonic signals
If m 6= n then
〈ϕm , ϕn〉 =∫ T
0ei(m−n)ω0t dt
= 1(m − n)ω0i ei(m−n)ω0t
∣∣∣T0
= 1(m − n)ω0i
[ei(m−n)ω0T − 1
]= 1
(m − n)ω0i[e(m−n)2πi − 1
]= 0,
This shows that {eint |n ∈ Z} is an orthogonal set.If m = n then
‖ϕm ‖2 = 〈ϕm , ϕm〉 =∫ T
0ei(m−m)t dt =
∫ T
01 dt = T .
Corollary
The set{
1√T einω0t
∣∣∣ n ∈ Z}is an orthonormal set.
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LS
Decomposition of periodic signals
Holy grail
Let x(t) be a periodic signal with period T > 0. Withω0 = 2π/T , find coefficients cn ∈ C (with n ∈ Z) such that
x(t) =∞∑
n=−∞cn einω0t .
Questions
For which t ∈ R does the series converge?
For which t ∈ R does the equation hold?
Are the coefficients cn unique?
How can you find the coefficients cn?
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Fourier coefficients
Theorem
Let x(t) be a periodic signal with period T > 0, and letω0 = 2π/T . Suppose that
x(t) =∞∑
n=−∞cn einω0t ,
then for all n ∈ Z:
cn = 1T
∫〈T〉
x(t) e−inω0t dt Eq. 3.3.4
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Proof of the theorem
Let ϕn(t) = einω0t , then
〈x, ϕn〉 =∫〈T〉
x(t) einω0t dt
=∫〈T〉
x(t) e−inω0t dt. (1)
Now use that {ϕn |n ∈ Z} is an orthogonal set:
〈x, ϕn〉 =⟨ ∞∑
k=−∞ck eikω0t , ϕn
⟩
=⟨ ∞∑
k=−∞ck ϕk , ϕn
⟩=
∞∑k=−∞
ck 〈ϕk , ϕn〉
= · · ·+ 0 + cn 〈ϕn , ϕn〉+ 0 + · · ·= cnT . (2)
Combining (1) and (2) proves the theorem.
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Proof of the theorem
Let ϕn(t) = einω0t , then
〈x, ϕn〉 =∫〈T〉
x(t) einω0t dt
=∫〈T〉
x(t) e−inω0t dt. (1)
Now use that {ϕn |n ∈ Z} is an orthogonal set:
〈x, ϕn〉 =⟨ ∞∑
k=−∞ck eikω0t , ϕn
⟩
=⟨ ∞∑
k=−∞ck ϕk , ϕn
⟩=
∞∑k=−∞
ck 〈ϕk , ϕn〉
= · · ·+ 0 + cn 〈ϕn , ϕn〉+ 0 + · · ·= cnT . (2)
Combining (1) and (2) proves the theorem.
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Proof of the theorem
Let ϕn(t) = einω0t , then
〈x, ϕn〉 =∫〈T〉
x(t) einω0t dt
=∫〈T〉
x(t) e−inω0t dt. (1)
Now use that {ϕn |n ∈ Z} is an orthogonal set:
〈x, ϕn〉 =⟨ ∞∑
k=−∞ck eikω0t , ϕn
⟩
=⟨ ∞∑
k=−∞ck ϕk , ϕn
⟩=
∞∑k=−∞
ck 〈ϕk , ϕn〉
= · · ·+ 0 + cn 〈ϕn , ϕn〉+ 0 + · · ·= cnT . (2)
Combining (1) and (2) proves the theorem.
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Fourier coefficients
DefinitionLet x(t) be a piecewise continuous periodic signal withperiod T , and let ω0 = 2π/T . The Fourier coefficientsof x(t) are defined as
cn = 1T
∫〈T〉
x(t) e−inω0t dt
If cn are the Fourier coefficients of x(t), then this isdenoted as x(t)↔ cn .If x(t) is not continuous at t0, the value of x at t0 doesnot affect the Fourier coefficients.
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Fourier coefficients
Example Example 3.3.1
Find the Fourier coefficients ofthe periodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0
with period 2.
t
K
−Kx(t)
0 1−1
Note that T = 2 and ω0 = π.
cn = 1T
∫ T/2
−T/2x(t) e−inω0t dt
= 12
[∫ 0
−1−K e−inπt dt +
∫ 1
0K e−inπt dt
]= 1
2K[−∫ 0
−1e−inπt dt +
∫ 1
0e−inπt dt
].
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Fourier coefficients
Example Example 3.3.1
Find the Fourier coefficients ofthe periodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0
with period 2.
t
K
−Kx(t)
0 1−1
Note that T = 2 and ω0 = π.
cn = 1T
∫ T/2
−T/2x(t) e−inω0t dt
= 12
[∫ 0
−1−K e−inπt dt +
∫ 1
0K e−inπt dt
]= 1
2K[−∫ 0
−1e−inπt dt +
∫ 1
0e−inπt dt
].
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Fourier coefficients
Example Example 3.3.1
Find the Fourier coefficients ofthe periodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0
with period 2.
t
K
−Kx(t)
0 1−1
Note that T = 2 and ω0 = π.
cn = 1T
∫ T/2
−T/2x(t) e−inω0t dt
= 12
[∫ 0
−1−K e−inπt dt +
∫ 1
0K e−inπt dt
]= 1
2K[−∫ 0
−1e−inπt dt +
∫ 1
0e−inπt dt
].
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Example (continued)
If n 6= 0:
cn = 12K
[−∫ 0
−1e−inπt dt +
∫ 1
0e−inπt dt
]= 1
2K[ 1
inπ e−inπt∣∣∣0−1− 1
inπ e−inπt∣∣∣10
]= K
2inπ[(
1− einπ)− (e−inπ − 1)]
= K2inπ
(2− 2(−1)n
)=
2Kinπ if n is odd,
0 if n is even.
If n = 0:c0 = −1
2K∫ 0
−1e−i 0πt dt + 1
2K∫ 1
0e−i 0πt dt
= −12K
∫ 0
−11 dt + 1
2K∫ 1
01 dt
= −12K · 1 + 1
2K · 1 = 0.
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Example (continued)
If n 6= 0:
cn = 12K
[−∫ 0
−1e−inπt dt +
∫ 1
0e−inπt dt
]= 1
2K[ 1
inπ e−inπt∣∣∣0−1− 1
inπ e−inπt∣∣∣10
]= K
2inπ[(
1− einπ)− (e−inπ − 1)]
= K2inπ
(2− 2(−1)n
)=
2Kinπ if n is odd,
0 if n is even.
If n = 0:c0 = −1
2K∫ 0
−1e−i 0πt dt + 1
2K∫ 1
0e−i 0πt dt
= −12K
∫ 0
−11 dt + 1
2K∫ 1
01 dt
= −12K · 1 + 1
2K · 1 = 0.
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Piecewise continuity and smoothness
Definition
A function f is called piecewise continuous on [a, b]if f is continuous in every point of [a, b], except possiblyin a finite number points t1, t2, . . . , tn ∈ [a, b].Moreover, f (a+), f (b−) and f (t+
i ), f (t−i ) should existfor i = 1, 2, . . . ,n.A function f is called piecewise continuous on R if fis piecewise continuous on every interval [a, b].
If f is continuous in t then f (t−) = f (t+) = f (t).
Definition
A function f is called piecewise smooth on [a, b]if f ′ is piecewise continuous on [a, b].A function f is called piecewise smooth on R if f ispiecewise smooth on every interval [a, b].
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Piecewise continuity and smoothness
Definition
A function f is called piecewise continuous on [a, b]if f is continuous in every point of [a, b], except possiblyin a finite number points t1, t2, . . . , tn ∈ [a, b].Moreover, f (a+), f (b−) and f (t+
i ), f (t−i ) should existfor i = 1, 2, . . . ,n.A function f is called piecewise continuous on R if fis piecewise continuous on every interval [a, b].
If f is continuous in t then f (t−) = f (t+) = f (t).
Definition
A function f is called piecewise smooth on [a, b]if f ′ is piecewise continuous on [a, b].A function f is called piecewise smooth on R if f ispiecewise smooth on every interval [a, b].
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The Dirichlet conditions Sec. 3.4
TheoremLet f be piecewise smooth, then for every a < b1. the function f is absolutely integrable over [a, b], i.e.∫ b
a|f (t)| dt <∞.
2. the function f has a finite number of extrema in [a, b],3. the function f has a finite number of discontinuities in
[a, b], none of the discontinuities are infinite,4. the function f is bounded.
The above four conditions are called the Dirichletconditions.
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The Fundamental Theorem of Fourier Series
TheoremLet x(t) be a periodic signal with Fourier coefficients cn ,that satisfies the Dirichlet conditions. Then for every t ∈ Rwe have
∞∑n=−∞
cn einω0t = x(t+) + x(t−)2 .
The series∞∑
n=−∞cn einω0t is called the Fourier series
of x(t). It converges for every t ∈ R.
The expression x(t+) + x(t−)2 is denoted as x(t).
If x is continuous in t, then x(t) = x(t).
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The Fundamental Theorem of Fourier Series
TheoremLet x(t) be a periodic signal with Fourier coefficients cn ,that satisfies the Dirichlet conditions. Then for every t ∈ Rwe have
∞∑n=−∞
cn einω0t = x(t+) + x(t−)2 .
The series∞∑
n=−∞cn einω0t is called the Fourier series
of x(t). It converges for every t ∈ R.
The expression x(t+) + x(t−)2 is denoted as x(t).
If x is continuous in t, then x(t) = x(t).
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The Fundamental Theorem of Fourier Series
TheoremLet x(t) be a periodic signal with Fourier coefficients cn ,that satisfies the Dirichlet conditions. Then for every t ∈ Rwe have
∞∑n=−∞
cn einω0t = x(t+) + x(t−)2 .
The series∞∑
n=−∞cn einω0t is called the Fourier series
of x(t). It converges for every t ∈ R.
The expression x(t+) + x(t−)2 is denoted as x(t).
If x is continuous in t, then x(t) = x(t).
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The Fundamental Theorem of Fourier Series
TheoremLet x(t) be a periodic signal with Fourier coefficients cn ,that satisfies the Dirichlet conditions. Then for every t ∈ Rwe have
∞∑n=−∞
cn einω0t = x(t+) + x(t−)2 .
The series∞∑
n=−∞cn einω0t is called the Fourier series
of x(t). It converges for every t ∈ R.
The expression x(t+) + x(t−)2 is denoted as x(t).
If x is continuous in t, then x(t) = x(t).
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The Fundamental Theorem of Fourier Series
Example Example 3.3.1
Find the Fourier series of theperiodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0.
t
K
−Kx(t)
0 1−1
The Fourier coefficients are cn =
2Kinπ if n is odd,
0 if n is even.The Fourier series of x(t) is:
2Kiπ
∞∑n=−∞
n odd
1n einπt = x(t) = x(t+)+x(t−)
2 =
−K −1<t<0,K 0<t<1,0 n∈Z.
1−1
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The Fundamental Theorem of Fourier Series
Example Example 3.3.1
Find the Fourier series of theperiodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0.
t
K
−Kx(t)
0 1−1
The Fourier coefficients are cn =
2Kinπ if n is odd,
0 if n is even.
The Fourier series of x(t) is:
2Kiπ
∞∑n=−∞
n odd
1n einπt = x(t) = x(t+)+x(t−)
2 =
−K −1<t<0,K 0<t<1,0 n∈Z.
1−1
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The Fundamental Theorem of Fourier Series
Example Example 3.3.1
Find the Fourier series of theperiodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0.
t
K
−Kx(t)
0 1−1
The Fourier coefficients are cn =
2Kinπ if n is odd,
0 if n is even.The Fourier series of x(t) is:
2Kiπ
∞∑n=−∞
n odd
1n einπt = x(t) = x(t+)+x(t−)
2 =
−K −1<t<0,K 0<t<1,0 n∈Z.
1−1
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Spectra
DefinitionLet cn be the Fourier coefficients of x(t).
The sequence |cn | is called the amplitude spectrum ormagnitude spectrum of x(t).The sequence Arg cn is called the phase spectrumof x(t).Amplitude spectrum and phase spectrum are linespectra of x(t).
If cn = 0 then the argument of cn is defined as 0.
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Spectra
DefinitionLet cn be the Fourier coefficients of x(t).
The sequence |cn | is called the amplitude spectrum ormagnitude spectrum of x(t).The sequence Arg cn is called the phase spectrumof x(t).Amplitude spectrum and phase spectrum are linespectra of x(t).
If cn = 0 then the argument of cn is defined as 0.
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Spectra
Example Example 3.3.1
Find the line spectra of the pe-riodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0.
t
K
−Kx(t)
0 1−1
n cn |cn | Arg cn
odd 2Kinπ
2K|n |π sgn(n)π/2
even 0 0 0
n
|cn |1 2 3 4−4 −3 −2 −1
2K|x |π
n
Arg cn
π2
−π2
1 2 3 4
−4 −3 −2 −1
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Spectra
Example Example 3.3.1
Find the line spectra of the pe-riodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0.
t
K
−Kx(t)
0 1−1
n cn |cn | Arg cn
odd 2Kinπ
2K|n |π sgn(n)π/2
even 0 0 0
n
|cn |1 2 3 4−4 −3 −2 −1
2K|x |π
n
Arg cn
π2
−π2
1 2 3 4
−4 −3 −2 −1
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Spectra
Example Example 3.3.1
Find the line spectra of the pe-riodic signal
x(t) ={
K if 0 < t < 1,−K if −1 < t < 0.
t
K
−Kx(t)
0 1−1
n cn |cn | Arg cn
odd 2Kinπ
2K|n |π sgn(n)π/2
even 0 0 0
n
|cn |1 2 3 4−4 −3 −2 −1
2K|x |π
n
Arg cn
π2
−π2
1 2 3 4
−4 −3 −2 −1
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Real signals
TheoremIf x(t)↔ cn , then x(t)↔ c−n .
Corollary
The signal x(t) is real if and only if cn = c−n .
cn
c−nC
We may assume that if x(t) is real, then x(t) is real:
Corollary (as applied in practice)
The signal x(t) is real if and only if cn = c−n .
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Real signals
TheoremIf x(t)↔ cn , then x(t)↔ c−n .
Corollary
The signal x(t) is real if and only if cn = c−n .
cn
c−nC
We may assume that if x(t) is real, then x(t) is real:
Corollary (as applied in practice)
The signal x(t) is real if and only if cn = c−n .
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Real signals
TheoremIf x(t)↔ cn , then x(t)↔ c−n .
Corollary
The signal x(t) is real if and only if cn = c−n .
cn
c−nC
We may assume that if x(t) is real, then x(t) is real:
Corollary (as applied in practice)
The signal x(t) is real if and only if cn = c−n .
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Real signals
TheoremIf x(t)↔ cn , then x(t)↔ c−n .
Corollary
The signal x(t) is real if and only if cn = c−n .
cn
c−nC
We may assume that if x(t) is real, then x(t) is real:
Corollary (as applied in practice)
The signal x(t) is real if and only if cn = c−n .
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Proofs Self tuition
Proof of the theorem:Let x(t) be periodic with period T , and let ω0 = 2π/T .The Fourier coefficient of x(t) is1T
∫〈T〉
x(t) e−inω0t dt = 1T
∫〈T〉
x(t) e−i(−n)ω0t dt = c−n .
Proof of the corollary:⇒ This follows directly from the theorem.⇐ If cn = c−n , then
12[x(t+) + x(t−)
]=
∞∑n=−∞
cn einω0t
=∞∑
n=−∞cn e−inω0t =
−∞∑−n=∞
c−n ei(−n)ω0t
=−∞∑
m=∞cm eimω0t = 1
2[x(t+) + x(t−)
].
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The trigonometric Fourier series
Rewrite the Fourier series with sines and cosines:∞∑
n=−∞cn einω0t = c0 +
∞∑n=1
cn einω0t +−1∑
n=−∞cn einω0t
= c0 +∞∑
n=1cn einω0t +
∞∑n=1
c−n einω0t
= c0 +∞∑
n=1cn(cos nω0t + i sin nω0t)
+ c−n(cos nω0t − i sin nω0t)
= c0 +∞∑
n=1(cn + c−n) cos nω0t + i(cn − c−n) sin nω0t.
Define
an ={
c0 if n = 0,cn + c−n if n ≥ 1, , bn = i(cn − c−n), n ≥ 1,
then∞∑
n=−∞cn einω0t =
∞∑n=0
an cos nω0t +∞∑
n=1bn sin nω0t.
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The trigonometric Fourier series
Rewrite the Fourier series with sines and cosines:∞∑
n=−∞cn einω0t = c0 +
∞∑n=1
cn einω0t +−1∑
n=−∞cn einω0t
= c0 +∞∑
n=1cn einω0t +
∞∑n=1
c−n einω0t
= c0 +∞∑
n=1cn(cos nω0t + i sin nω0t)
+ c−n(cos nω0t − i sin nω0t)
= c0 +∞∑
n=1(cn + c−n) cos nω0t + i(cn − c−n) sin nω0t.
Define
an ={
c0 if n = 0,cn + c−n if n ≥ 1, , bn = i(cn − c−n), n ≥ 1,
then∞∑
n=−∞cn einω0t =
∞∑n=0
an cos nω0t +∞∑
n=1bn sin nω0t.
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Trigonometric Fourier coefficients
From cn to an and bn :
an ={
c0 if n = 0,cn + c−n if n ≥ 1
bn = i(cn − c−n), n ≥ 1
From an and bn to cn :
cn =
an − i bn
2 if n ≥ 1,a0 if n = 0,a−n + i b−n
2 if n < 0
an and bn are real if and only if cn = c−n .
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Trigonometric Fourier coefficients
From cn to an and bn :
an ={
c0 if n = 0,cn + c−n if n ≥ 1
bn = i(cn − c−n), n ≥ 1
From an and bn to cn :
cn =
an − i bn
2 if n ≥ 1,a0 if n = 0,a−n + i b−n
2 if n < 0
an and bn are real if and only if cn = c−n .
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Trigonometric Fourier coefficients
From cn to an and bn :
an ={
c0 if n = 0,cn + c−n if n ≥ 1
bn = i(cn − c−n), n ≥ 1
From an and bn to cn :
cn =
an − i bn
2 if n ≥ 1,a0 if n = 0,a−n + i b−n
2 if n < 0
an and bn are real if and only if cn = c−n .
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Trigonometric Fourier coefficients of real signals
TheoremLet x(t) be a periodic signal.
The signal x(t) is real if and only if all trigonometricFourier coefficients an and bn are real.If x(t) is real then
an ={
c0 if n = 0,2 Re cn if n ≥ 1
andbn = −2 Im cn , n ≥ 1.
If x(t) is real then its Fourier series
a0 +∞∑
n=1
(an cos nω0t + bn sin nω0t
),
is called the real Fourier series of x(t).
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Trigonometric Fourier coefficients
The Fourier coefficients an and bn can be found directly byintegration:
an =
1T
∫〈T〉
x(t) dt, n = 0,
2T
∫〈T〉
x(t) cos(2nπt
T
)dt, n ≥ 1
eq. 3.3.9a
eq. 3.3.9b
bn = 2T
∫〈T〉
x(t) sin(2nπt
T
)dt, n ≥ 1 eq. 3.3.9c
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Example: the block wave
DefinitionLet T > 0 and 0 ≤ a ≤ T . The even block wave withpulse width a is the periodic signal ba,T with period Tdefined by
ba,T (t) =
1 if |t| ≤ a/2,
0 if a/2 < t ≤ T/2.
t−3T/2 −T −T/2 T/2 T 3T/2−a/2 a/2
1
0
ba,T
a
T
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The block wave
Compute the trigonometric Fourier coefficients an .If n = 0 then
a0 = 1T
∫〈T〉
ba,T (t) dt = 1T
∫ a/2
−a/21 dt = a
T .
If n ≥ 1 then
an = 2T
∫〈T〉
ba,T (t) cos(nω0t) dt
= 2T
∫ a/2
−a/2cos
(2nπtT
)dt
= 2T ·
T2nπ
[sin(2nπt
T
) ]a/2
−a/2
= 1nπ
[sin(nπa
T
)− sin
(−nπa
T
)]= 2
nπ sin(nπa
T
)
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The block wave
Compute the trigonometric Fourier coefficients an .If n = 0 then
a0 = 1T
∫〈T〉
ba,T (t) dt = 1T
∫ a/2
−a/21 dt = a
T .
If n ≥ 1 then
an = 2T
∫〈T〉
ba,T (t) cos(nω0t) dt
= 2T
∫ a/2
−a/2cos
(2nπtT
)dt
= 2T ·
T2nπ
[sin(2nπt
T
) ]a/2
−a/2
= 1nπ
[sin(nπa
T
)− sin
(−nπa
T
)]= 2
nπ sin(nπa
T
)
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The block wave
Compute the trigonometric Fourier coefficients bn .
For n ≥ 1 we have
bn = 2T
∫〈T〉
ba,T (t) sin(nω0t) dt
= 2T
∫ a/2
−a/2sin(2nπt
T
)dt
= − 2T ·
T2nπ
[cos
(2nπtT
) ]a/2
−a/2
= − 1nπ
[cos
(nπaT
)− cos
(−nπa
T
)]= 0.
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The block wave
For all n ∈ Z we have
cn =
an − i bn
2 if n > 0,a0 if n = 0,a−n + i b−n
2 if n < 0
=
a0 if n = 0,
12a|n | if n 6= 0
=
aT if n = 0,
1nπ sin
(nπaT
)if n 6= 0.
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The block wave
Note that if n 6= 0 then
cn = 1nπ sin
(nπaT
)= 1
nπ ·nπaT Sa
(nπaT
)= a
T sinc(n a
T
).
Also, for n = 0
c0 = aT = a
T sinc(
0 aT
),
by definition of sinc(0).
Conclusion:
cn = aT sinc
(n a
T
)for all n ∈ Z
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The block wave
Note that if n 6= 0 then
cn = 1nπ sin
(nπaT
)= 1
nπ ·nπaT Sa
(nπaT
)= a
T sinc(n a
T
).
Also, for n = 0
c0 = aT = a
T sinc(
0 aT
),
by definition of sinc(0).
Conclusion:
cn = aT sinc
(n a
T
)for all n ∈ Z
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The block wave
Note that if n 6= 0 then
cn = 1nπ sin
(nπaT
)= 1
nπ ·nπaT Sa
(nπaT
)= a
T sinc(n a
T
).
Also, for n = 0
c0 = aT = a
T sinc(
0 aT
),
by definition of sinc(0).
Conclusion:
cn = aT sinc
(n a
T
)for all n ∈ Z
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The block wave
Definition
Define the duty cycle of the block wave ba,T as ρ = aT .
TheoremFor the Fourier coefficients of the block wave ba,T we have
cn = ρ sinc(nρ)
−3ρ
−2ρ
−ρ 0 ρ
2ρ
3ρ
−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7
ρ
ρ sinc
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The block wave
Definition
Define the duty cycle of the block wave ba,T as ρ = aT .
TheoremFor the Fourier coefficients of the block wave ba,T we have
cn = ρ sinc(nρ)
−3ρ
−2ρ
−ρ 0 ρ
2ρ
3ρ
−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7
ρ
ρ sinc
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The block wave
Definition
Define the duty cycle of the block wave ba,T as ρ = aT .
TheoremFor the Fourier coefficients of the block wave ba,T we have
cn = ρ sinc(nρ)
−3ρ
−2ρ
−ρ 0 ρ
2ρ
3ρ
−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7
ρ
ρ sinc
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The block wave
Definition
Define the duty cycle of the block wave ba,T as ρ = aT .
TheoremFor the Fourier coefficients of the block wave ba,T we have
cn = ρ sinc(nρ)
−3ρ
−2ρ
−ρ 0 ρ
2ρ
3ρ
−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7
ρ
ρ sinc