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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

1 State equations and stability2 Inner products3 Fourier series

Inner products

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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.

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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.

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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.

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Let A =[

a bc d

s2 − Ts + D = (s − λ1)(s − λ2) = 0. (∗)

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Let A =[

a bc d

s2 − Ts + D = (s − λ1)(s − λ2) = 0. (∗)

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Let A =[

a bc d

s2 − Ts + D = (s − λ1)(s − λ2) = 0. (∗)

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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.

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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.

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Example

Example Example 2.6.13

Investigate the stability of the LTI system whose statematrix is

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.

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Example

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.

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Example

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.

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Example

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.

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Example

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.

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Example

A =[−3 −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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Example

A =[−3 −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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Example

A =[−3 −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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Example

A =[−3 −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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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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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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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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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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Norm and distance

‖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|.

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Norm and distance

‖ϕ‖ =√〈ϕ,ϕ〉.

‖ϕ‖ = |ϕ|.

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Norm and distance

‖ϕ‖ =√〈ϕ,ϕ〉.

‖ϕ‖ = |ϕ|.

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10 2.3

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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10 2.3

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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11 2.4

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

NotationThe integral over the interval [a, a + T ] is denoted as∫

〈T〉x(t) dt.

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11 2.4

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

NotationThe integral over the interval [a, a + T ] is denoted as∫

〈T〉x(t) dt.

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12 2.5

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 +

0x(τ + T ) dτ

x(t) isperiodic

=∫ T

ax(t) dt +

0x(t) dt

swapintegrals

=∫ a

0x(t) dt +

ax(t) dt

=∫ T

0x(t) dt.

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13 2.6

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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13 2.6

f (t) = f (t + T ) for all t ∈ R.

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13 2.6

f (t) = f (t + T ) for all t ∈ R.

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14 2.7

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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14 2.7

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

0ei(m−n)ω0t dt.

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14 2.7

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

0ei(m−n)ω0t dt.

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15 2.8

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

(m − n)ω0i[e(m−n)2πi − 1

This shows that {eint |n ∈ Z} is an orthogonal set.

If m = n then

‖ϕm ‖2 = 〈ϕm , ϕm〉 =∫ T

0ei(m−m)t dt =

01 dt = T .

Corollary

The set{

1√T einω0t

∣∣∣ n ∈ Z}is an orthonormal set.

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15 2.8

If m 6= n then

〈ϕm , ϕn〉 =∫ T

0ei(m−n)ω0t dt

∣∣∣T0

= 1(m − n)ω0i

(m − n)ω0i[e(m−n)2πi − 1

This shows that {eint |n ∈ Z} is an orthogonal set.If m = n then

‖ϕm ‖2 = 〈ϕm , ϕm〉 =∫ T

0ei(m−m)t dt =

01 dt = T .

Corollary

The set{

1√T einω0t

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15 2.8

If m 6= n then

〈ϕm , ϕn〉 =∫ T

0ei(m−n)ω0t dt

∣∣∣T0

= 1(m − n)ω0i

(m − n)ω0i[e(m−n)2πi − 1

This shows that {eint |n ∈ Z} is an orthogonal set.If m = n then

‖ϕm ‖2 = 〈ϕm , ϕm〉 =∫ T

0ei(m−m)t dt =

01 dt = T .

Corollary

The set{

1√T einω0t

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16 3.1

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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17 3.2

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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18 3.3

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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18 3.3

〈x, ϕn〉 =∫〈T〉

x(t) einω0t dt

=∫〈T〉

〈x, ϕn〉 =⟨ ∞∑

=⟨ ∞∑

∞∑k=−∞

ck 〈ϕk , ϕn〉

= · · ·+ 0 + cn 〈ϕn , ϕn〉+ 0 + · · ·= cnT . (2)

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18 3.3

〈x, ϕn〉 =∫〈T〉

x(t) einω0t dt

=∫〈T〉

〈x, ϕn〉 =⟨ ∞∑

=⟨ ∞∑

∞∑k=−∞

ck 〈ϕk , ϕn〉

= · · ·+ 0 + cn 〈ϕn , ϕn〉+ 0 + · · ·= cnT . (2)

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19 3.4

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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20 3.5

Find the Fourier coefficients ofthe periodic signal

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0

with period 2.

−Kx(t)

0 1−1

Note that T = 2 and ω0 = π.

cn = 1T

∫ T/2

−T/2x(t) e−inω0t dt

[∫ 0

−1−K e−inπt dt +

0K e−inπt dt

2K[−∫ 0

−1e−inπt dt +

0e−inπt dt

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20 3.5

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0

with period 2.

−Kx(t)

0 1−1

cn = 1T

∫ T/2

[∫ 0

0K e−inπt dt

2K[−∫ 0

−1e−inπt dt +

0e−inπt dt

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20 3.5

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0

with period 2.

−Kx(t)

0 1−1

cn = 1T

∫ T/2

[∫ 0

0K e−inπt dt

2K[−∫ 0

−1e−inπt dt +

0e−inπt dt

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21 3.6

Example (continued)

If n 6= 0:

cn = 12K

[−∫ 0

−1e−inπt dt +

0e−inπt dt

inπ e−inπt∣∣∣0−1− 1

inπ e−inπt∣∣∣10

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

−11 dt + 1

2K∫ 1

= −12K · 1 + 1

2K · 1 = 0.

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21 3.6

Example (continued)

If n 6= 0:

cn = 12K

[−∫ 0

−1e−inπt dt +

0e−inπt dt

inπ e−inπt∣∣∣0−1− 1

inπ e−inπt∣∣∣10

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

−11 dt + 1

2K∫ 1

= −12K · 1 + 1

2K · 1 = 0.

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22 3.7

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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22 3.7

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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23 3.8

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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24 3.9

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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24 3.9

∞∑n=−∞

cn einω0t = x(t+) + x(t−)2 .

The series∞∑

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24 3.9

∞∑n=−∞

cn einω0t = x(t+) + x(t−)2 .

The series∞∑

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24 3.9

∞∑n=−∞

cn einω0t = x(t+) + x(t−)2 .

The series∞∑

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25 3.10

Find the Fourier series of theperiodic signal

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0.

−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:

∞∑n=−∞

1n einπt = x(t) = x(t+)+x(t−)

−K −1<t<0,K 0<t<1,0 n∈Z.

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25 3.10

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0.

−Kx(t)

0 1−1

2Kinπ if n is odd,

0 if n is even.

The Fourier series of x(t) is:

∞∑n=−∞

−K −1<t<0,K 0<t<1,0 n∈Z.

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25 3.10

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0.

−Kx(t)

0 1−1

2Kinπ if n is odd,

0 if n is even.The Fourier series of x(t) is:

∞∑n=−∞

−K −1<t<0,K 0<t<1,0 n∈Z.

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26 3.11

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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26 3.11

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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27 3.12

Spectra

Find the line spectra of the pe-riodic signal

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0.

−Kx(t)

0 1−1

n cn |cn | Arg cn

odd 2Kinπ

2K|n |π sgn(n)π/2

even 0 0 0

|cn |1 2 3 4−4 −3 −2 −1

2K|x |π

Arg cn

−π2

1 2 3 4

−4 −3 −2 −1

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27 3.12

Spectra

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0.

−Kx(t)

0 1−1

n cn |cn | Arg cn

odd 2Kinπ

2K|n |π sgn(n)π/2

even 0 0 0

|cn |1 2 3 4−4 −3 −2 −1

2K|x |π

Arg cn

−π2

1 2 3 4

−4 −3 −2 −1

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27 3.12

Spectra

x(t) ={

K if 0 < t < 1,−K if −1 < t < 0.

−Kx(t)

0 1−1

n cn |cn | Arg cn

odd 2Kinπ

2K|n |π sgn(n)π/2

even 0 0 0

|cn |1 2 3 4−4 −3 −2 −1

2K|x |π

Arg cn

−π2

1 2 3 4

−4 −3 −2 −1

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28 4.1

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 .

c−nC

We may assume that if x(t) is real, then x(t) is real:

Corollary (as applied in practice)

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28 4.1

Real signals

Corollary

c−nC

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28 4.1

Real signals

Corollary

c−nC

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28 4.1

Real signals

Corollary

c−nC

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29 4.2

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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30 4.3

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

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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30 4.3

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

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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31 4.4

Trigonometric Fourier coefficients

From cn to an and bn :

c0 if n = 0,cn + c−n if n ≥ 1

bn = i(cn − c−n), n ≥ 1

From an and bn to 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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31 4.4

c0 if n = 0,cn + c−n if n ≥ 1

bn = i(cn − c−n), n ≥ 1

an − i bn

2 if n ≥ 1,a0 if n = 0,a−n + i b−n

2 if n < 0

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31 4.4

c0 if n = 0,cn + c−n if n ≥ 1

bn = i(cn − c−n), n ≥ 1

an − i bn

2 if n ≥ 1,a0 if n = 0,a−n + i b−n

2 if n < 0

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32 4.5

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

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 +∞∑

(an cos nω0t + bn sin nω0t

is called the real Fourier series of x(t).

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33 4.6

The Fourier coefficients an and bn can be found directly byintegration:

∫〈T〉

x(t) dt, n = 0,

∫〈T〉

x(t) cos(2nπt

)dt, n ≥ 1

eq. 3.3.9a

eq. 3.3.9b

bn = 2T

∫〈T〉

x(t) sin(2nπt

)dt, n ≥ 1 eq. 3.3.9c

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34 4.7

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

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35 4.8

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

If n ≥ 1 then

an = 2T

∫〈T〉

ba,T (t) cos(nω0t) dt

∫ a/2

−a/2cos

(2nπtT

= 2T ·

[sin(2nπt

) ]a/2

−a/2

= 1nπ

[sin(nπa

)− sin

(−nπa

nπ sin(nπa

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35 4.8

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

If n ≥ 1 then

an = 2T

∫〈T〉

ba,T (t) cos(nω0t) dt

∫ a/2

−a/2cos

(2nπtT

= 2T ·

[sin(2nπt

) ]a/2

−a/2

= 1nπ

[sin(nπa

)− sin

(−nπa

nπ sin(nπa

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36 4.9

The block wave

Compute the trigonometric Fourier coefficients bn .

For n ≥ 1 we have

bn = 2T

∫〈T〉

ba,T (t) sin(nω0t) dt

∫ a/2

−a/2sin(2nπt

= − 2T ·

(2nπtT

) ]a/2

−a/2

= − 1nπ

(nπaT

)− cos

(−nπa

)]= 0.

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37 4.10

The block wave

For all n ∈ Z we have

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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38 4.11

The block wave

Note that if n 6= 0 then

cn = 1nπ sin

(nπaT

nπ ·nπaT Sa

(nπaT

T sinc(n a

Also, for n = 0

c0 = aT = a

T sinc(

by definition of sinc(0).

Conclusion:

cn = aT sinc

)for all n ∈ Z

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38 4.11

The block wave

cn = 1nπ sin

(nπaT

nπ ·nπaT Sa

(nπaT

T sinc(n a

Also, for n = 0

c0 = aT = a

T sinc(

Conclusion:

cn = aT sinc

)for all n ∈ Z

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38 4.11

The block wave

cn = 1nπ sin

(nπaT

nπ ·nπaT Sa

(nπaT

T sinc(n a

Also, for n = 0

c0 = aT = a

T sinc(

Conclusion:

cn = aT sinc

)for all n ∈ Z

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39 4.12

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

−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7

ρ sinc

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39 4.12

The block wave

Definition

cn = ρ sinc(nρ)

−3ρ

−2ρ

−ρ 0 ρ

−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7

ρ sinc

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39 4.12

The block wave

Definition

cn = ρ sinc(nρ)

−3ρ

−2ρ

−ρ 0 ρ

−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7

ρ sinc

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39 4.12

The block wave

Definition

cn = ρ sinc(nρ)

−3ρ

−2ρ

−ρ 0 ρ

−7 −6 −5 −4 −3 −2 1 2 3 4 5 6 7

ρ sinc

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