EI331 Signals and Systems Lecture 9 Bo Jiang John Hopcroft Center - - PowerPoint PPT Presentation

EI331 Signals and Systems

Lecture 9 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

March 26, 2019

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Contents

• 1. Eigenvalues and Eigenfunctions
• 2. CT Fourier Series
• 3. Properties of CT Fourier Series

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Recap

Rn DT signals CZ CT signals basis {e1, . . . , en} {δk : k ∈ Z} {δτ : τ ∈ R}1 general x x =

n

• k=1

xkek x =

• k∈Z

x[k]δk x =

• R

x(τ)δτdτ Image under linear transformation basis {f1, . . . , fn} {hk : k ∈ Z} {hτ : τ ∈ R} general x Ax =

n

• k=1

xkfk T(x) =

• k∈Z

x[k]hk T(x) =

• R

x(τ)hτdτ LTI – T(x) = x ∗ h T(x) = x ∗ h

1Not standard usage of term “basis”; understood in terms of

integral representation on next line.

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Eigenvalues and Eigenvectors

In linear algebra, v ∈ Rn is eigenvector of matrix A with associated eigenvalue λ, if Av = λv Suppose A has n distinct eigenvalues λ1, . . . , λn associated to eigenvectors v1, . . . , vn, respectively.

• v1, . . . , vn form a basis of Rn, i.e. ∀x ∈ Rn has following

representation x =

n

• k=1

akvk

• image under A

Ax =

n

• k=1

akAvk =

n

• k=1

akλkvk

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Change of Basis

standard basis eigenbasis of A basis vectors {e1, . . . , en} {v1, . . . , vn} decomposition x =

n

• k=1

xkek x =

n

• k=1

akvk image under A Ax =

n

• k=1

xkfk Ax =

n

• k=1

akλkvk Expansions of x and Ax

• in standard basis: same coefficients {xk}, different

basis vectors {ek} → {fk}

• in eigenbasis: same basis vectors {vk}, different

coefficients {ak} → {λkak}

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Change of Basis

standard basis eigenbasis DT basis vectors {δk : k ∈ Z} decomposition x =

• k∈Z

x[k]δk image under T T(x) =

• k∈Z

x[k]hk “standard basis” eigenbasis CT basis vectors {δτ : τ ∈ R} decomposition x =

• R

x(τ)δτdτ image under T T(x) =

• R

x(τ)hτdτ

? ?

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Eigenvalues and Eigenfunctions of LTI Systems

CT Exponential est T(est) =

• R

h(τ)es(t−τ)dτ = est

• R

h(τ)e−sτdτ, est is eigenfunction of T with associated eigenvalue H(s) =

• R

h(τ)e−sτdτ (system function) DT Exponential zn T(zn) =

• k∈Z

h[k]zn−kdτ = zn

k∈Z

h[k]z−k, zn is eigenfunction of T with associated eigenvalue H(z) =

• k∈Z

h[k]z−k (system function)

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Response to Linear Combination of Eigenfunctions

CT T

• k

akeskt

• =
• k

akH(sk)eskt DT T

• k

akzn

k

• =
• k

akH(zk)zn

k

Input vs. output

• linear combinations of same exponentials
• different coefficients {ak} → {H(sk)ak}/{H(zk)ak}

Questions

• Which functions are linear combinations of exponentials?
• How to find coefficients ak?

Fourier analysis focuses on ejωt for CT and ejωn for DT

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Contents

• 1. Eigenvalues and Eigenfunctions
• 2. CT Fourier Series
• 3. Properties of CT Fourier Series

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CT Fourier Series

Recall CT signal is periodic with period T if x = τTx

• r

x(t) = x(t + T), ∀t ∈ R

• fundamental period T: smallest positive period
• fundamental frequency ω0 = 2π

T

sin(ω0t), cos(ω0t), ejω0t periodic with fundamental frequency ω0 Fourier series represent periodic signals in terms of harmonically related sinusoids or complex exponentials x(t) = a0 +

∞

• k=1

[ak cos(kω0t) + bk sin(kω0t)] x(t) =

∞

• k=−∞

ckejkω0t

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Joseph Fourier

Jean-Baptiste Joseph Fourier (from Wikipedia)

1807, M´ emoire sur la propagation de la chaleur dans les corps solides

(Memoir on the propagation of heat in solid bodies)

1822, Th´ eorie analytique de la chaleur

(The Analytical Theory of Heat) (https://gallica.bnf.fr)

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Harmonics

ω 2ω0 3ω0 4ω0 5ω0 6ω0 ω0 1 2 3 4 5 6 D C f u n d a m e n t a l / fi r s t h a r m

• n

i c s e c

• n

d h a r m

• n

i c t h i r d h a r m

• n

i c f

• u

r t h h a r m

• n

i c fi f t h h a r m

• n

i c s i x t h h a r m

• n

i c |k|: harmonic #

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Harmonics

ω 2ω0 3ω0 4ω0 5ω0 6ω0 ω0 t T = 2π

ω0

t T = 2π

ω0

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Orthonormality of Harmonics

Recall for k = 0, b

a

ejkω0tdt = ejkω0b − ejkω0a jkω0 Since ejkω0t has period T, over any period 1 T

• T

ejkω0tdt = 1 T t0+T

t0

ejkω0tdt = δ[k] Define inner product between two signals with period T by f, g = 1 T

• T

f(t)g(t)dt {ejkω0t : k ∈ Z} is orthonormal system of functions ejkω0t, ejmω0t = 1 T

• T

ej(k−m)ω0tdt = δkm = δ[k − m]

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Orthogonality of Harmonics

For sines and cosines, sin(kω0t), sin(mω0t) = 1 2δ[k − m] − 1 2δ[k + m] cos(kω0t), cos(mω0t) = 1 2δ[k − m] + 1 2δ[k + m] sin(kω0t), cos(mω0t) = 0 Proof. Use Euler’s formula and orthonormality of {ejkω0t : k ∈ Z}

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Orthogonality of Harmonics

sin(ω0t) sin(2ω0t) t O T = 2π

ω0

t O T = 2π

ω0

sin(ω0t) cos(ω0t)

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Fourier Coefficients

Suppose x has period T and Fourier series representation x(t) =

∞

• k=−∞

ckejkω0t, ω0 = 2π T Find Fourier coefficients using orthonormality of {ejkω0t} x, ejmω0t =

∞

• k=−∞

ckejkω0t, ejmω0t =

∞

• k=−∞

ckejkω0t, ejmω0t =

∞

• k=−∞

ckδ[m − k] = cm

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Complex Fourier Series

Synthesis equation x(t) =

∞

• k=−∞

ˆ x[k]ejkω0t =

∞

• k=−∞

ˆ x[k]ejk 2π

T t

N-th partial sum SN(x)(t) =

N

• k=−N

ˆ x[k]ejkω0t Analysis equation ˆ x[k] = x, ejkω0t = 1 T

• T

x(t)e−jkω0tdt = 1 T

• T

x(t)e−jk 2π

T tdt

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Trigonometric Fourier Series

For sines and cosines, sin(kω0t), sin(mω0t) = 1 2δ[k − m] − 1 2δ[k + m] cos(kω0t), cos(mω0t) = 1 2δ[k − m] + 1 2δ[k + m] sin(kω0t), cos(mω0t) = 0 Synthesis equation x(t) = a0 +

∞

• k=1

[ak cos(kω0t) + bk sin(kω0t)] Analysis equation ak = 2 − δ[k] T

• T

x(t) cos(kω0t)dt, bk = 2 T

• T

x(t) sin(kω0t)dt

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Equivalence of Two Forms

Complex form x(t) =

∞

• k=−∞

ˆ x[k]ejkω0t Trigonometric form x(t) = a0 +

∞

• k=1

[ak cos(kω0t) + bk sin(kω0t)] Conversion of coefficients (by Euler’s formula)      a0 = ˆ x[0] ak = ˆ x[k] + ˆ x[−k], k ≥ 1 bk = j(ˆ x[k] − ˆ x[−k]), k ≥ 1      ˆ x[0] = a0 ˆ x[k] = 1

2(ak − jbk),

k ≥ 1 ˆ x[k] = 1

2(ak + jb−k),

k ≤ −1 Negative frequencies in complex form introduced for mathematical convenience, no physical significance

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Examples

• Example. x(t) = ejω0t, ˆ

x[k] = δ[k − 1] Spectrum ˆ x[k] k 1 1 −3 −2 −1 0 2 3

• Example. x(t) = cos(ω0t + φ) = ejφ

2 ejω0t + e−jφ 2 e−jω0t,

ˆ x[1] = 1 2ejφ, ,ˆ x[−1] = 1 2e−jφ, ˆ x[k] = 0, k = ±1 Magnitude spectrum |ˆ x[k]| k

1 2

−1

1 2

1 −3 −2 2 3 Phase spectrum arg ˆ x[k] k 1 φ −1 −φ −3 −2 2 3

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Example: Triangle Wave

In one period, x(t) = 1 − 2|t| T , |t| ≤ T 2 t x(t) − T

2 T 2

Fourier coefficients k = 0, ˆ x[0] = 1 T T/2

−T/2

• 1 − 2|t|

T

• dt = 1

2 k = 0, k odd, ˆ x[k] = 1 T T/2

−T/2

• 1 − 2|t|

T

• e−jkω0tdt =

2 π2k2 k = 0, k even, ˆ x[k] = 0

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Example: Triangle Wave

ˆ x[k] =     

1 2,

k = 0

2 π2k2,

k = 0 odd 0, k = 0 even Spectrum for fixed T, frequency spacing ∆ω = 2π

T

ωk = k 2π

T

k

1 2

1 3 5 −1 −3 −5 2 4 6 −2 −4 −6

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Example: Triangle Wave

S0(x)(t) =

• k=−0

ˆ x[k]ejkω0t

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Triangle Wave

S1(x)(t) =

1

• k=−1

ˆ x[k]ejkω0t

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Triangle Wave

S3(x)(t) =

3

• k=−3

ˆ x[k]ejkω0t

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Triangle Wave

S5(x)(t) =

5

• k=−5

ˆ x[k]ejkω0t

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Triangle Wave

S7(x)(t) =

7

• k=−7

ˆ x[k]ejkω0t

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Triangle Wave

S9(x)(t) =

9

• k=−9

ˆ x[k]ejkω0t

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Triangle Wave

S19(x)(t) =

19

• k=−19

ˆ x[k]ejkω0t

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Periodic Square Wave

In one period, x(t) =

• 1,

|t| < T1 0, T1 < |t| < T/2 t x(t) − T

2 T 2

−T1 T1 −T T −2T 2T

Fourier coefficients k = 0, ˆ x[0] = 1 T T1

−T1

1dt = 2T1 T k = 0, ˆ x[k] = 1 T T1

−T1

e−jkω0tdt = 2 sin(kω0T1) kω0T = sin(kω0T1) kπ

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Example: Periodic Square Wave

ˆ x[k] = sin

• k 2πT1

T

• kπ

= 2 sin(ωkT1) ωkT Spectra for fixed T and different T1 ωk = k 2π

T T1 T = 1 4

k 2 −2

T1 T = 1 8

k 4 −4

T1 T = 1 16

k 8 −8

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Example: Periodic Square Wave

ˆ x[k] = sin

• k 2πT1

T

• kπ

= 2 sin(ωkT1) ωkT Spectra for fixed T1 and different T ωk = k 2π

T T1 T = 1 4

ω

π T1 T1 T = 1 8

ω

π T1 T1 T = 1 16

ω

π T1

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Example: Periodic Square Wave

S0(x)(t) =

• k=−0

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Periodic Square Wave

S1(x)(t) =

1

• k=−1

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

27/31

Example: Periodic Square Wave

S3(x)(t) =

3

• k=−3

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

27/31

Example: Periodic Square Wave

S5(x)(t) =

5

• k=−5

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

27/31

Example: Periodic Square Wave

S7(x)(t) =

7

• k=−7

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Periodic Square Wave

S9(x)(t) =

9

• k=−9

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Periodic Square Wave

S13(x)(t) =

13

• k=−13

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Periodic Square Wave

S19(x)(t) =

19

• k=−19

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Periodic Square Wave

S29(x)(t) =

29

• k=−29

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Example: Periodic Square Wave

S39(x)(t) =

39

• k=−39

ˆ x[k]ejkω0t

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Gibbs Phenomenon

Partial sums of Fourier series “ring” at jump discontinuity

• overshoot gets closer and closer to discontinuity
• overshoot approaches ≈ 9% of jump size

1 2 3 4 5 6 7 8 0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2

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Contents

• 1. Eigenvalues and Eigenfunctions
• 2. CT Fourier Series
• 3. Properties of CT Fourier Series

30/31

Properties of CT Fourier Series

Fourier series establishes correspondence between periodic functions and doubly infinite sequences x

FS

← − − → ˆ x

• r

x(t)

FS

← − − → ˆ x[k] Linearity If x, y have same period T, so does their linear combination ax + by, and

• ax + by = aˆ

x + bˆ y Proof. ( ax + by)[k] = ax + by, ejkω0t = ax, ejkω0t + by, ejkω0t = aˆ x[k] + bˆ y[k]

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Properties of CT Fourier Series

Time shifting If x has period T and ω0 = 2π

T ,

• τt0x = E−ω0t0ˆ

x

• r

x(t − t0)

FS

← − − → e−jkω0t0ˆ x[k] where (Eaˆ x)[k] = ejakˆ x[k] Time shift ⇐ ⇒ linear phase change in frequency Proof. ( τt0x)[k] = τt0x, ejkω0t = x, τ−t0ejkω0t = x, ejkω0(t+t0) = x, ejkω0t0ejkω0t = e−jkω0t0x, ejkω0t = e−jkω0t0ˆ x[k] = (E−ω0t0ˆ x)[k]

• Example. cos(t) = 1

2ejt + 1 2e−jt, sin(t) = cos(t − π 2) = −j 2 ejt + j 2e−jt