EI331 Signals and Systems Lecture 31 Bo Jiang John Hopcroft Center - - PowerPoint PPT Presentation

EI331 Signals and Systems

Lecture 31 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

June 13, 2019

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Contents

• 1. Unilateral Laplace Transform
• 2. Review

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Unilateral Laplace Transform

Recall the (bilateral) Laplace transform of a CT signal x X(s) = ∞

−∞

x(t)e−stdt The unilateral Laplace transform of x is X(s) = ∞

0−

x(t)e−stdt also denoted X = UL{x}, x(t)

UL

← − − → X(s) ROC of X is always a right half-plane

• NB. Information about x(t) at t < 0 is lost in X.
• NB. L{x} = UL{x} iff x(t) is causal, i.e. x(t) = 0 for t < 0

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Examples

The calculation of unilateral Laplace transforms is almost the same as for bilateral Laplace transforms.

• Example. x1(t) = e−atu(t),

X1(s) = X1(s) = 1 s + a, Re s > −Re a

• Example. x2(t) = e−a(t+1)u(t + 1) = x1(t + 1),

bilateral X2(s) = esX1(s) = es s + a, Re s > −Re a unilateral X2(s) = ∞ e−a(t+1)e−stdt = e−a s + a, Re s > −Re a

• NB. X2 = X2, since x2 is noncausal.
• NB. X2 = esX1, time-shift property fails for unilateral transform.

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Examples

The calculation of unilateral Laplace transforms is almost the same as for bilateral Laplace transforms.

• Example. x3(t) = e−a|t| with Re a > 0

bilateral X3(t) = −2a s2 − a2, −Re a < Re s < Re a unilateral X3(s) = 1 s + a, Re s > −Re a

• NB. X3 = X1, since x3 = x1
• NB. X3 = X1, since x3 and x1 differ only for t < 0
• Example. x(t) = δ(t) + 2δ′(t) + etu(t)

X(s) = X(s) = 1 + 2s + 1 s − 1, Re s > 1

• NB. The lower limit of integration is 0− so δ and δ′ contribute to X

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Examples

The calculation of inverse unilateral Laplace transforms is the same as for bilateral Laplace transforms, but we can only recover x(t) for t ≥ 0!

• Example. Given unilateral Laplace transform

X(s) = 1 (s + 1)(s + 2) = 1 s + 1 − 1 s + 2 the only possibility for ROC is Re s > −1. Thus x(t) = e−t − e−2t, t ≥ 0 X provides no information about x(t) for t < 0

• Example. X(s) = s2−3

s+2 = −2 + s + 1 s+2. The ROC must be

Re s > −2, and x(t) = −2δ(t) + δ′(t) + e−2t, t ≥ 0

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Properties of Unilateral Laplace Transform

Many properties are the same as for bilateral Laplace transform Property Signal ULT – x(t) X(s) – y(t) Y(s) Linearity ax(t) + by(t) aX(s) + bY(s) Shifting in s-domain es0tx(t) X(s − s0) Time scaling x(at), a > 0

1 aX( s a)

Conjugation x∗(t) X∗(s∗) Differentiation in s domain −tx(t)

d dsX(s)

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Convolution Property of Unilateral Laplace Transform

If x(t) = y(t) = 0 for t < 0, then (x ∗ y)(t)

UL

← − − → X(s)Y(s), ROAC ⊃ ROACX ∩ ROACY

• Proof. Note (x ∗ y)(t) = 0 for t < 0.

UL{x ∗ y} = L{x ∗ y} = L{x}L{y} = UL{x}UL{y}

• Caution. The assumption x(t) = y(t) = 0 for t < 0 is important!
• Example. x(t) = δ(t) − δ(t − 1), y(t) = δ(t + 1),

X(s) = 1 − e−s, Y(s) = 0, Note (x ∗ y)(t) = δ(t + 1) − δ(t), and UL{x ∗ y} = −1 = X(s)Y(s)

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Integration in Time Domain

If x(t)

UL

← − − → X(s), with ROAC = R then t

0−

x(τ)dτ

UL

← − − → 1 s X(s), with ROAC ⊃ R ∩ {Re s > 0}

• Proof. Apply the convolution property to ˜

x ∗ u, where ˜ x(t) =

• x(t),

t ≥ 0 0, t < 0

• Example. x(t) = δ(t), X(s) = 1.

t

0−

x(τ)dτ = u(t)

UL

← − − → 1 s , Re s > 0

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Differentiation in Time Domain

If x(t)

UL

← − − → X(s), with ROC = R and lim

t→+∞ x(t)e−st = 0 for s ∈ R0, then

d dtx(t)

UL

← − − → sX(s) − x(0−), with ROC ⊃ R ∩ R0

• NB. R0 ⊃ ROAC of X(s) (also true for bilateral Laplace transform)
• Proof. Use integration by parts

∞

0−

x′(t)e−stdt = x(t)e−st ∞

t=0− + s

∞

0−

x(t)e−stdt = −x(0−) + sX(s)

• NB. Had we used the definition X(s) =

∞

0+ x(t)e−stdt, we would

have d

dtx(t) UL

← − − → sX(s) − x(0+)

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Differentiation in Time Domain

• Example. x(t) = e−t, X(s) =

1 s+1 with ROAC Re s > −1. By the

differentiation property, x′(t)

UL

← − − → s s + 1 − 1 = − 1 s + 1, ROC ⊃ {Re s > −1} By direct calculation, x′(t) = −e−t

UL

← − − →= − 1 s + 1, ROC = ROAC = {Re s > −1}

• Example. x(t) = e−tu(t), X(s) =

1 s+1 with ROAC Re s > −1. By

the differentiation property, x′(t)

UL

← − − → s s + 1, ROC ⊃ {Re s > −1} By direct calculation, x′(t) = δ(t) − e−tu(t), and UL{x′} = − 1 s + 1 + 1 = s s + 1, ROC = ROAC = {Re s > −1}

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Differentiation in Time Domain

Under appropriate conditions, e.g. x(k)(t) = O(eγt), we can extend the differentiation property to higher derivatives, x(n)(t)

UL

← − − → snX(s) −

n−1

• k=0

sn−1−kx(k)(0−) Since solutions to constant coefficient ODEs is of the form x(t) =

n

• k=1

pi(t)eait where pi are polynomials, the above condition is satisfied. Unilateral Laplace transform is useful for solving ODEs with nonzero initial condtions.

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Initial Value Theorem

• Theorem. If the unilateral Laplace transform X(s) of x(t) is a

proper rational function, then x(0+) = lim

s→∞ sX(s)

• Proof. X(s) has partial fraction expansion,

X(s) =

r

• i=1

Ni

• ki=1

Ai,ki (s + ai)ki x(t) =

r

• i=1

Ni

• ki=1

Ai,kitki−1 (ki − 1)!e−ait, t ≥ 0 x(0+) =

r

• i=1

Ai,1 = lim

s→∞ r

• i=1

Ni

• ki=1

Ai,kis (s + ai)ki = lim

s→∞ sX(s)

• Example. x(t) = 2e−t + e−2t, X(s) =

3s+5 s2+3s+2, x(0+) = lim s→∞ sX(s) = 3.

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Initial Value Theorem

If X(s) is rational but not proper, then X(s) =

n

• k=0

aksk + X1(s) where X1(s) is a proper rational function. Taking inverse transform, x(t) =

n

• k=0

akδ(k)(t) + x1(t) so x(0+) = x1(0+) = lim

s→∞ sX1(s)

• Example. x(t) = δ(t) + 2e−t, X(s) = 1 +

2 s+1 = s+3 s+1,

x(0+) = 2 = lim

s→∞

2s s + 1 = lim

s→∞ sX(s)

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Example

Suppose a CT LTI system has the following properties.

• 1. The system is causal
• 2. The system function H(s) is a proper rational function and

has only two poles, at s = −2 and s = −4.

• 3. For input x(t) = 1, the output is y(t) = 0
• 4. The impulse response satisfies h(0+) = 4

Determine the system function H(s).

• Solution. By 1 and 2

H(s) = as + b (s + 2)(s + 4), Re s > −2 By 3, H(0) = 0, so b = 0. By 4, lim

s→∞ sH(s) = 4, so a = 4, and

H(s) = 4s (s + 2)(s + 4), Re s > −2

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Final Value Theorem

• Theorem. Suppose x(t) does not contain δ or its derivatives for

t ≥ 0. If X(s) converges for real s > 0 and the x(t) has a finite limit as t → +∞, then lim

t→+∞ x(t) = lim s→0+ sX(s)

• Proof. Since A lim

t→+∞ x(t) exists, x(t) is bounded on (0, +∞),

i.e. |x(t)| ≤ M. For any ǫ > 0, ∃T s.t. |x(t) − A| < ǫ for t ≥ T. For s > 0, sX(s) − A = s ∞ [x(t) − A]e−stdt |sX(s) − A| ≤ s T + ∞

T

• |x(t) − A|e−stdt = I1 + I2

where I1 ≤ sT(M + A) and I2 ≤ ǫs ∞

T e−stdt ≤ ǫs

∞

0 e−stdt = ǫ.

lim

s→0+ |sX(s) − A| ≤ ǫ =

⇒ lim

s→0+ |sX(s) − A| = 0 =

⇒ lim

s→0+ sX(s) = A

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Final Value Theorem

• Example. x(t) = 2 + 3e−t + e−2t,

X(s) = 2 s + 3 s + 1 + 1 s + 2, Re s > 0 Check lim

s→0+ sX(s) = 2 = lim t→+∞ x(t)

The Final Values Theorem can be extended to allow finitely many δ and its derivatives in x(t) on the positive real axis. We can rewrite x(t) as x(t) = x1(t) +

n

• i=1

aiδ(ki)(t − ti)

UL

← − − → X(s) = X1(s) +

n

• i=1

aiskie−sti lim

t→+∞ x(t) = lim t→+∞ x1(t) = lim s→0+ sX1(s) = lim s→0+ sX(s)

• Example. x(t) = δ(t) + 2, X(s) = 1

s + 1, x(+∞) = 2 = lim s→0+ sX(s).

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Linear Constant-coefficient ODE

Consider ODE

N

• k=0

ak dky dtk =

M

• k=0

bk dkx dtk with initial condition y(k)(0−), k = 0, 1, . . . , N − 1, and causal input, i.e. x(t) = 0 for t < 0. Take unilateral Laplace transform of both sides

N

• k=0

ak

• skY(s) −

k−1

• ℓ=0

sk−1−ℓy(ℓ)(0−)

• =

M

• k=0

bkskX(s) so Y(s) = M

k=0 bksk

N

k=0 aksk X(s)

• zero-state response

+ N

k=0 ak

k−1

ℓ=0 sk−1−ℓy(ℓ)(0−)

N

k=0 aksk

• zero-input response

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Example

Consider ODE y′′(t) + 3y′(t) + 2y(t) = 2x′(t) + 6x(t) with initial condition y(0−) = c0, y′(0−) = c1 and x(t) = e−tu(t). Take unilateral Laplace transform of both sides [s2Y(s) − sc0 − c1] + 3[sY(s) − c0] + 2Y(s) = 2sX(s) + 6X(s) = 2s + 6 s + 1 so Y(s) = 2s + 6 (s2 + 3s + 2)(s + 1) + c0s + (3c0 + c1) s2 + 3s + 2 = 2s + 6 (s + 1)2(s + 2)

• zero-state response

+ c0s + (3c0 + c1) (s + 1)(s + 2)

• zero-input response

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Example (cont’d)

Zero-state response Yzs(s) = 2s + 6 (s + 1)2(s + 2), Re s > −1 For t ≥ 0, yzs(t) = Res[Yziest, −2] + Res[Yziest, −1] = (2s + 6)est (s + 1)2

• s=−2

+ d ds (2s + 6)est s + 2

• s=−1

= 2e−2t + (4t − 2)e−t Zero-input response Yzi(s) = c0s + (3c0 + c1) (s + 1)(s + 2) , Re s > −1 yzi(t) = Res[Yzsest, −2] + Res[Yzsest, −1] = (2c0 + c1)e−t − (c0 + c1)e−2t, t ≥ 0

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Contents

• 1. Unilateral Laplace Transform
• 2. Review

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• Complex numbers and complex functions

◮ arithmetic and representations ◮ argument is multivalued, principal value (−π, π] ◮ simply and multiply connected domains ◮ continuity of functions

• Analytic functions

◮ analytic at z0 iff differentiable on some open disk B(z0, r) ◮ f(z) = u(x, y) + jv(x, y) is analytic iff u, v are continuously differentiable and satisfy Cauchy-Riemann equation ux = vy, uy = −vx ◮ rational function R(z) = N(z)

D(z) is analytic on C except for zeros

• f D(z) (aka poles of R(z))

◮ elementary analytic functions

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• Complex Integration

◮ Cauchy’s Theorem

• C

f(z)dz = 0 ◮ Cauchy’s Integral Formula f (k)(z) = n! j2π

• C

f(ζ) (ζ − z)n+1 dζ

• Series

◮ power series and disk of convergence ◮ Laurent series and annulus of convergence (cf. z-transform)

• Residue

◮ Res[f, z0] = 1

j2π

• C f(z)dz = c−1,

z0 ∈ C − 1

j2π

• C f(z)dz = −c−1,

z0 = ∞ ◮ Residue Theorem

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• Signal Representations

◮ time domain: linear combination of δ, convolution ◮ frequency domain: linear combination of sinusoids (CT/DT Fourier series/transforms), spectrum ◮ s-domain: bilateral/unilateral Laplace transforms ◮ z-domain: bilateral/unilateral z-transforms

• Sampling

◮ Nyquist rate ◮ spectra of sampled signals

• Systems

◮ properties of systems (causality, stability, linearity, time-invariance...) ◮ representations of LTI systems (differential/difference equations, block diagrams, impulse response, frequency response, system function) ◮ eigenfunction property of LTI systems ◮ filtering

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• Transforms (Fourier, Laplace, Z)

◮ forward and inverse transforms ◮ properties ◮ analysis of LTI systems

◮ given system, find response to input ◮ given input output pairs, identify system

• ODE

◮ classical method in time domain ◮ transform method

◮ initial rest/LTI: Fourier, bilateral Laplace ◮ nonzero initial condition: unilateral Laplace

• Difference equations

◮ classical method in time domain ◮ transform method

◮ initial rest/LTI: Fourier, bilateral z-transform ◮ nonzero initial condition: unilateral z-transform

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Good Luck with Finals!