EI331 Signals and Systems Lecture 15 Bo Jiang John Hopcroft Center - - PowerPoint PPT Presentation

EI331 Signals and Systems

Lecture 15 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

April 16, 2019

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Contents

• 1. Recap
• 2. Properties of CT Fourier Transform

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Recap: Fourier Transform for L1(R) Functions

Fourier transform and its inverse as principal values F{x}(jω) = lim

T→∞

T

−T

x(t)e−jωtdt F−1(X)(t) = lim

W→∞

1 2π W

−W

X(jω)ejωtdω

• Theorem. (Pointwise inversion)
• 1. If x ∈ L1(R) satisfies Dirichlet conditions on all finite intervals,

(F−1 ◦ F{x})(t) = x(t+) + x(t−) 2 , ∀t

• 2. If X ∈ L1(R) satisfies Dirichlet conditions on all finite intervals,

(F ◦ F−1{X})(jω) = X(jω+) + X(jω−) 2 , ∀ω

• 3. If x ∈ C(R) ∩ L1(R) and F{x} ∈ L1(R), then F−1 ◦ F{x} = x

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Recap: Fourier Transform for L1(R) Functions

Fourier transform pairs case x(t) X(jω) 1 e−atu(t), Re a > 0 1 a + jω 3 e−a|t|, Re a > 0 2a a2 + ω2 3 e−at2, a > 0 π

ae− ω2

4a

1 u(t + T) − u(t − T) 2 sin(ωT) ω 2 sin(ωct) πt u(ω + ωc) − u(ω − ωc)

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Recap: Fourier Transform for Generalized Functions

Schwarz space S = S(R) of rapidly decreasing functions S = {φ ∈ C∞(R) : φℓ,k < ∞, ∀ℓ, k ∈ N} where φℓ,k = supt∈R |tℓφ(k)(t)|. Note F{S} = S. Fourier transform of tempered distributions (x, φ)

• R

x(ξ)φ(ξ)dξ

• If xn → x in distribution, then F{x} limn F{xn} in distribution

(x, φ) = lim

n (xn, φ) =

⇒ (F{x}, φ) lim

n (F{xn}, φ),

∀φ ∈ S

• Alternatively,

(F{x}, φ) (x, F{φ}), ∀φ ∈ S

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Recap: Fourier Transform for Generalized Functions

Fourier transform pairs x(t) X(jω) δ(t) 1 δ(t − t0) e−jωt0 1 2πδ(ω) ejω0t 2πδ(ω − ω0)

• R

δ(t −t0)e−jωtdt = e−jωt0,

• R

ej(ω1−ω2)tdt = 2πδ(ω1 −ω2)

• R

X(jω)ejωtdω =

• R
• R

x(τ)ejω(t−τ)dτdω =

• R

x(τ)2πδ(t−τ)dτ = 2πx(t)

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Recap: Fourier Transform for Periodic Functions

Fourier series x(t) =

∞

• k=−∞

ˆ x[k]ejkω0t Fourier transform X(jω) =

∞

• k=−∞

2πˆ x[k]δ(ω − kω0) Equivalent representations

• ˆ

x[k] is amplitude of component at ωk = kω0

• ˆ

x[k]δ(ω − kω0) is density of component at ωk = kω0

• Cf. discrete random variable x in probability

Ef(X) =

• n

f(xn)pn =

• R

f(x)p(x)dx with p(x) =

• n

pnδ(x−xn)

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Recap: Relation between FS and FT

• Aperiodic signal x with finite support of length < T
• Periodic extension xT of x

xT(t) =

∞

• k=−∞

x(t − kT)

• NB. In other words, x is one period of periodic signal xT

x

F

← − − → X xT

FS

← − − → ˆ xT xT

F

← − − → XT ω0 = 2π T ˆ xT[k] = 1 T X(jkω0) XT(jω) =

∞

• k=−∞

2πˆ xT[k]δ(ω − kω0) =

∞

• k=−∞

ω0X(jkω0)δ(ω − kω0)

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Contents

• 1. Recap
• 2. Properties of CT Fourier Transform

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Properties of CT Fourier Transform

Linearity F{ax + by} = aF{x} + bF{y} Time shifting F{τt0x} = E−t0F{x}

• r

x(t − t0)

F

← − − → e−jωt0X(jω) where (Eaf)(s) = ejasf(s) Proof.

• R

x(t − t0)e−jωtdt =

• R

x(s)e−jω(s+t0)ds = e−jωt0

• R

x(s)e−jωsds Frequency shifting F{Eω0x} = τω0ˆ x

• r

ejω0tx(t)

F

← − − → X(j(ω − ω0))

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Example: Multipath Effect

Transmitter: x(t) Receiver: y(t) = a0x(t) + a1x(t − τ) Y(jω) = a0X(jω) + a1e−jωτX(jω) = (a0 + a1e−jωτ)X(jω) Let x(t) = u(t + T) − u(t − T). X(jω) = 2 sin(ωT) ω Y(jω) = (a0 + a1e−jωτ)2 sin(ωT) ω For a0 = a1 = 1, Y(jω) = 4e−jωτ/2 cos(ωτ 2 )sin(ωT) ω x(t) y(t) ω |X(jω)|

2T

π T

ω |Y(jω)|

4T

π T π τ

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Example: Modulation

Baseband signal: x(t) Modulated signal y(t) = x(t) cos(ωct) = 1 2x(t)(ejωct + e−jωct) Y(jω) = 1 2X(j(ω − ωc)) + 1 2X(j(ω + ωc)) For x(t) = u(t + T) − u(t − T), Y(jω) = sin((ω − ωc)T) ω − ωc + sin((ω + ωc)T) ω + ωc t x(t)

1 −T T

ω X(jω)

2T

π T

t y(t)

1 −T T

ω Y(jω)

T

−ωc ωc

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Properties of CT Fourier Transform

Time reversal F{Rx} = RF{x}

• r

x(−t)

F

← − − → X(−jω) Conjugation F{x∗} = RF{x}

• r

x∗(t)

F

← − − → X∗(−jω)

• Proof. F{x∗}(jω) =
• R x∗(t)e−jωtdt =
• R x(t)ejωtdt

∗ = X∗(−jω) Symmetry

• x even ⇐

⇒ X even, x odd ⇐ ⇒ X odd

• x real ⇐

⇒ X(−jω) = X∗(jω)

• x real and even ⇐

⇒ X real and even

• x real and odd ⇐

⇒ X purely imaginary and odd

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Example

For a > 0, x(t) =

• e−at,

t > 0 −eat, t < 0 Let y(t) = e−atu(t), with Y(jω) = 1 a + jω Since x(t) = y(t) − y(−t). X(jω) = Y(jω) − Y(−jω) = 1 a + jω − 1 a − jω = −2jω a2 + ω2 x real & odd, X purely imaginary & odd t x(t)

O 1 −1

ω

O

|X(jω)|

1/a −a a

ω arg X(jω)

O − π

2 π 2

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Time and Frequency Scaling

For a = 0, F{Sax} = 1 |a|S 1

aF{x}

• r

x(at)

F

← − − → 1 |a|X jω a

• Proof. Change variables by τ = at.

For a > 0, ∞

−∞

x(at)e−jωtdt = ∞

−∞

x(τ)e−j ω

a τ dτ

a = 1 aX jω a

• For a < 0,

∞

−∞

x(at)e−jωtdt = −∞

∞

x(τ)e−j ω

a τ dτ

a = −1 aX jω a

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Time and Frequency Scaling

F{Sax} = 1 |a|S 1

aF{x}

• r

x(at)

F

← − − → 1 |a|X jω a

• compression in time ⇐

⇒ stretching in frequency stretching in time ⇐ ⇒ compression in frequency e.g. audio: faster playback = ⇒ higher pitch x(t) = e−(at)2

F

← − − → X(jω) = √π |a| e− 1

4( ω a )2

t x(t)

a = 1/2 a = 1 a = 2

ω X(jω)

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Differentiation

x′(t)

F

← − − → jωX(jω), x(k)(t)

F

← − − → (jω)kX(jω)

• Proof. Differentiate under integral sign

x(t) = 1 2π

• R

X(jω)ejωtdω = ⇒ x′(t) = 1 2π

• R

jωX(jω)ejωtdω y = x′ is output of differentiator with frequency response ω |H(jω)| H(jω) = F{δ′}(jω) =

• R

δ′(t)e−jωtdt = − d dte−jωt

• 0 = jω
• Y(jω) = H(jω)X(jω)
• amplifies high frequencies
• suppresses low frequencies
• DC completely eliminated

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Integration

y(t) = t

−∞

x(τ)dτ

F

← − − → Y(jω) = 1 jωX(jω) + πX(0)δ(ω) Since x(t) = y′(t), by differentiation property X(jω) = jωY(jω) = ⇒ Y(jω) = 1 jωX(jω) for ω = 0 Intuitively, y(t) has DC component ¯ y = lim

T→∞

1 2T T

−T

y(t)dt = lim

T→∞

1 2T T

−T

• R

x(τ)u(t − τ)dτdt =

• R

x(τ)

• lim

T→∞

1 2T T

−T

u(t − τ)dt

• dτ = 1

2X(0) So Y also has component 2π¯ yδ(ω) = πX(0)δ(ω)

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Integration

• Example. Unit step function

u(t) = t

−∞

δ(τ)dτ U(jω) = 1 jωF{δ}(jω) + πF{δ}(0)δ(ω) = 1 jω + πδ(ω)

• Example. Sign function

sgn(t) =

• 1,

t > 0 −1 t < 0 Since sgn = 2u − 1, F{sgn}(jω) = 2U(jω) − 2πδ(ω) = 2 jω t u(t) O 1

1 2

ω O |U(jω)| π t sgn(t) O 1 −1

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Integration

• Example. Unit ramp function u−2(t) = tu(t)

u−2(t) = t

−∞

u(τ)dτ Integration property suggests F{u−2}(jω) = 1 jωU(jω) + πU(0)δ(ω) = − 1 ω2 + π jωδ(ω) + πU(0)δ(ω) But what’s U(0) = 1

j0 + πδ(0)? What’s π jωδ(ω)? Not well-defined!

Integration property not applicable here! Will see F{u−2} = − 1

ω2 + jπδ′(ω)

Rule of thumb. Applicable when formula is well-defined

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Example

t x

O 1 − T

2 T 2

t x1 = x′

O − T

2 2 T

− 2

T T 2

t x2 = x′′

O − T

2 2 T T 2 2 T

− 4

T

x(t) =

• 1 − 2|t|

T u(t + T 2 ) − u(t − T 2 )

• X2(jω) = 2

T [ej ωT

2 −2+e−j ωT 2 ] = − 8

T sin2 ωT 4

• X1(jω) = X2(jω)

jω +πX2(0)δ(ω) = −8 sin2 ωT

4

• jωT

X(jω) = X1(jω) jω +πX1(0)δ(ω) = 8 sin2 ωT

4

• ω2T

ω X(jω)

T 2

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Example: Sign Function sgn

x(t) = sgn(t)

F

← − − → X(jω) = 2 jω

• Neither x nor X is in L1(R), above interpreted in the

sense of distribution, i.e. (X, φ) = (x, F{φ}), ∀φ ∈ S

• r for approximating sequence xn → x,

(X, φ) = lim

n (Xn, φ),

∀φ ∈ S

• Since

2 jω not integrable at ω = 0, X should be interpreted

as principal value, often denoted X(jω) = pv

• 2

jω

• , i.e.

(X, φ) =

• R

pv 2 jω

• φ(ω)dω lim

ǫ→0

• |ω|≥ǫ

2 jωφ(ω)dω

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Example: Sign Function sgn

• Claim. Principal value below is well-defined
• R

pv 1 ω

• φ(ω)dω lim

ǫ→0

• |ω|≥ǫ

1 ωφ(ω)dω Proof.

|ω|≥1 1 ωφ(ω)dω well-defined, since φ is rapidly decreasing

• Since
• 1≥|ω|≥ǫ

1 ωdω = 0,

• 1≥|ω|≥ǫ

1 ωφ(ω)dω =

• 1≥|ω|≥ǫ

φ(ω) − φ(0) ω dω

• By Intermediate Value Theorem,
• φ(ω)−φ(0)

ω

• = |φ′(ξ)| ≤ φ0,1
• 1≥|ω|≥ǫ

1 ωφ(ω)dω

• =
• 1≥|ω|≥ǫ
• φ(ω) − φ(0)

ω

• dω ≤ 2φ0,1

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Example: Sign Function sgn

Claim. F{sgn}(jω) = pv 2 jω

• Proof. Let xa(t) = sgn(t)e−a|t| for a > 0. xa → x pointwise and

in distribution as a → 0 (legal to push limit inside integral) lim

a→0

• R

xa(t)φ(t) =

• R

lim

a→0 xa(t)φ(t) =

• R

x(t)φ(t)dt Recall Xa(jω) = −2jω a2 + ω2 For ω = 0, Xa(jω) →

2 jω pointwise as a → 0. But need

• R

Xa(jω)φ(ω)dω →

• R

pv 2 jω

• φ(ω)dω

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Example: Sign Function sgn

Claim. lim

a→0

• R

ω a2 + ω2φ(ω)dω = lim

ǫ→0

• |ω|≥ǫ

1 ωφ(ω)dω

• Proof. Since

ω a2+ω2φ(ω) ∈ L1(R),

• R

ω a2 + ω2φ(ω)dω = lim

ǫ→0

• |ω|≥ǫ

ω a2 + ω2φ(ω)dω Need to show lim

a→0 lim ǫ→0 |J| = 0, where

J =

• |ω|≥ǫ
• ω

a2 + ω2 − 1 ω

• φ(ω)dω =
• |ω|≥ǫ

−a2 ω(a2 + ω2)φ(ω)dω

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Example: Sign Function sgn

Proof (cont’d). Using 0 =

• |ω|≥ǫ

−a2 ω(a2+ω2)dω, we obtain

J =

• |ω|≥ǫ

−a2 a2 + ω2 · φ(ω) − φ(0) ω dω. By Intermediate Value Theorem, | φ(ω)−φ(0)

ω

| = |φ′(ξ)| ≤ φ0,1 |J| ≤

• |ω|≥ǫ
• −a2

a2 + ω2 · φ(ω) − φ(0) ω

• dω ≤
• |ω|≥ǫ

a2 a2 + ω2φ0,1dω ≤

• R

a2 a2 + ω2φ0,1dω = aπφ0,1 Therefore, lim

a→0 lim ǫ→0 |J| = 0

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Example: Unit Step Function

u(t)

F

← − − → U(jω) = 1 jω + πδ(ω)

• NB. More precisely, U(jω) = pv
• 1

jω

• + πδ(ω)
• Proof. Note u(t) = 1

2 sgn(t) + 1

• 2. By linearity

U(jω) = 1 2F{sgn}(jω) + 1 2F{1}(jω) = 1 jω + πδ(ω)

• Remark. Can use approximation approach but with caution.
• Let xa(t) = e−atu(t) for a > 0. Note xa → u pointwise and

in distribution as a → 0.

• For ω = 0, Xa(jω) =

1 a+jω → 1 jω pointwise as a ↓ 0.

• Tempting to conclude U(jω) =

1 jω.

• But need convergence in distribution !

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Duality

F{f}(η) =

• R

f(ξ)e−jηξdξ F−1{f}(η) = 1 2π

• R

f(ξ)ejηξdξ Identical except for

• different signs in exponent of complex exponential
• constant factor

1 2π

F{f}(η) =

• R

f(ξ)e−jηξdξ =

• R

f(−ξ)ejηξdξ = F−1{2πRf}(η) Duality f

F

← − − → g ⇐ ⇒ g

F

← − − → 2πRf ⇐ ⇒ Rg

F

← − − → 2πf

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Duality

F{f}(η) =

• R

f(ξ)e−jηξdξ F−1{f}(η) = 1 2π

• R

f(ξ)ejηξdξ Identical except for

• different signs in exponent of complex exponential
• constant factor

1 2π

F{f} =

• R

f(ξ)e−jηξdξ ∗ =

• R

f ∗(ξ)ejηξdξ = F−1{2πf ∗} Duality f

F

← − − → g ⇐ ⇒ g∗

F

← − − → 2πf ∗

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Duality

u(t + a) − u(t − a)

F

← − − → 2 sin(aω) ω 2 sin(at) t

F

← − − → 2π[u(ω + a) − u(ω − a)] t x1(t)

1 −T T

ω X1(jω)

2T

π T

t x2(t)

W π

π W

ω X3(jω)

1 −W W

F F

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Duality

• Example. x(t) = 1

t (more precisely, pv

1

t

• )

Know sgn(t)

F

← − − → 1 jω By duality −1 jt

F

← − − → 2π sgn(ω)

• switch LHS and RHS, switch ω and t
• do either of following (but not both)

◮ take complex conjugate of both sides ◮ substitute ω → −ω or t → −t (but not both)

• multiply RHS by 2π

By linearity 1 t

F

← − − → −j2π sgn(ω)

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Duality

• Example. x(t) =

1 1+t2

Know e−|t|

F

← − − → 2 1 + ω2 By duality 2 1 + t2

F

← − − → 2πe−|ω|

• switch LHS and RHS, switch ω and t
• do either of following (but not both)

◮ take complex conjugate of both sides ◮ substitute ω → −ω or t → −t (but not both)

• multiply RHS by 2π

By linearity 1 1 + t2

F

← − − → πe−|ω|

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Duality

Differentiation in frequency −jtx(t)

F

← − − → dX(jω) dω

• Proof. Differentiate under integral sign

X(jω) =

• R

x(t)e−jωtdt = ⇒ dX(jω) dω =

• R

(−jt)x(t)e−jωtdt

• Example. Unit ramp function u−2 = tu(t)

F{u−2} = jF{−jtu} = jdU(jω) dω = − 1 ω2 + jπδ′(ω) Integration in frequency −1 jtx(t) + πx(0)δ(t)

F

← − − → ω

−∞

X(jλ)dλ