EI331 Signals and Systems Lecture 28 Bo Jiang John Hopcroft Center - - PowerPoint PPT Presentation

EI331 Signals and Systems

Lecture 28 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

June 4, 2019

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Contents

• 1. Analysis of DT LTI Systems by Z-transform
• 2. Block Diagram Representations
• 3. Unilateral Z-transform

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DT System Function

Recall the response of a DT LTI system to the input x[n] is y[n] = (x ∗ h)[n] where h is the impulse response of the system. If x and h have z-transforms, the convolution property implies Y(z) = X(z)H(z) in their common ROC. If the ROC has a nonempty interior point, the system function (aka transfer function) H(z) uniquely determines h and hence system properties through the Laurent series expansion H(z) =

∞

• n=−∞

h[n]z−n

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Causality

Recall h is right-sided iff its ROC is the exterior of a circle, i.e. R1 < |z| < ∞ H(z) =

∞

• n=N1

h[n]z−n The following conditions are equivalent

• 1. N1 ≥ 0
• 2. H(z−1) =

∞

• n=N1

h[n]zn is a convergent power series on |z| <

1 R1

• 3. 0 is a removable singularity of H(z−1)
• 4. ∞ is removable singularity of H(z)
• 5. lim

z→0 H(z−1) exists and is finite

• 6. lim

z→∞ H(z) exists and is finite1

1This is the more precise statement of ∞ ∈ ROC.

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Causality

An LTI system with system function H(z) is causal iff

• 1. the ROC is the exterior of a circle
• 2. lim

z→∞ H(z) exists and is finite

causal ⇐ ⇒ ROC is the exterior of a circle including ∞ An LTI system with rational system function H(z) = N(z)

D(z) is causal

iff

• 1. the ROC is |z| > |p|, where p is the outermost pole
• 2. deg D ≥ deg N
• Example. H(z) = z3−2z2−z

z2+ 1

4z+ 1 8 cannot be the system function of a

causal system.

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Stability

Recall an LTI system is stable iff its impulse response h ∈ ℓ1, i.e.

∞

• n=−∞

|h[n]| < ∞ i.e. H(z) converges absolutely on the unit circle |z| = 1, so its ROC R1 < |z| < R2 must satisfy R1 < 1 < R2. stable ⇐ ⇒ ROC includes the unit circle |z| = 1 A causal LTI system with rational system function H(z) is stable iff all its poles are inside the unit circle.

• Example. A causal system with H(z) =

1 1−az−1 is stable iff |a| < 1

• Example. A system with H(z) =

1 1−az−1 where |a| > 1 and ROC

|z| < |a| is also stable, but it is noncausal.

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Example

Re Im

1

X

3 2

X

− 1

2

H(z) = 1 2

• z

z − 3

2

− z z + 1

2

• There are two poles p1 = − 1

2 and p2 = 3 2.

• 1. |z| > 3

2, causal, unstable

h1[n] = 1 2 3 2 n u[n] − 1 2

• −1

2 n u[n] 2.

1 2 < |z| < 3 2, noncausal, stable

h2[n] = −1 2 3 2 n u[−n−1]−1 2

• −1

2 n u[n]

• 3. |z| < 1

2, noncausal, unstable

h3[n] = −1 2 3 2 n u[−n−1]+1 2

• −1

2 n u[−n−1]

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Linear Constant-coefficient Difference Equations

LTI system with input and output related by

N

• k=0

aky[n − k] =

M

• k=0

bkx[n − k] Take z-transform of both sides

N

• k=0

akz−kY(z) =

M

• k=0

bkz−kX(z) so H(z) = Y(z) X(z) = M

k=0 bkz−k

N

k=0 akz−k

System function is always rational Difference equation does not specify ROC! Need additional conditions (e.g. stability, causality) to determine h[n].

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Example

Consider LTI system with input and output related by y[n] − 1 2y[n − 1] = x[n] + 1 3x[n − 1] System function H(z) = 1 + 1

3z−1

1 − 1

2z−1

Two possibilities for ROC: |z| > 1

2 and |z| < 1 2.

• 1. If |z| > 1

2

1 1 − 1

2z−1 = ∞

• n=0

1 2 n z−n

Z−1

← − − → 1 2 n u[n] By linearity and time-shifting property, h[n] = 1 2 n u[n] + 1 3 1 2 n−1 u[n − 1]

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Example (cont’d)

Consider LTI system with input and output related by y[n] − 1 2y[n − 1] = x[n] + 1 3x[n − 1] System function H(z) = 1 + 1

3z−1

1 − 1

2z−1

Two possibilities for ROC: |z| > 1

2 and |z| < 1 2.

• 2. If |z| < 1

2

1 1 − 1

2z−1 =

−2z 1 − 2z = −

∞

• n=1

2nzn

Z−1

← − − → − 1 2 n u[−n − 1] By linearity and time-shifting property, h[n] = − 1 2 n u[−n − 1] − 1 3 1 2 n−1 u[−n]

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Example (cont’d)

Consider LTI system with input and output related by y[n] − 1 2y[n − 1] = x[n] + 1 3x[n − 1] If we use Fourier transform, then frequency response is H(ejω) = 1 + 1

3e−jω

1 − 1

2e−jω

and h[n] = 1 2 n u[n] + 1 3 1 2 n−1 u[n − 1] Why only one possibility?

• Fourier transform method assumes stability, requiring that

ROC of H(z) contain the unit circle, i.e. |z| > 1

2

• In general, not applicable to unstable systems

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Example

Consider LTI system with input and output related by y[n] − 3 4y[n − 1] + 1 8y[n − 2] = 2x[n] System function H(z) = 2 1 − 3

4z−1 + 1 8z−2

By partial fraction expansion H(z) = 2 (1 − 1

2z−1)(1 − 1 4z−1) =

4 1 − 1

2z−1 −

2 1 − 1

4z−1

Take inverse z-transform to obtain impulse response      h1[n] = 4 1

2

n u[n] − 2 1

4

n u[n], ROC: |z| > 1

2

h2[n] = −4 1

2

n u[−n − 1] − 2 1

4

n u[n], ROC: 1

4 < |z| < 1 2

h3[n] = −4 1

2

n u[−n − 1] + 2 1

4

n u[−n − 1], ROC: |z| < 1

4

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Example (cont’d)

Find response to x[n] = 1

4

n u[n]. X(z) = 1 1 − 1

4z−1,

|z| > 1 4 z-transform of response Y(z) = H(z)X(z) = 2 (1 − 1

2z−1)(1 − 1 4z−1) ·

1 1 − 1

4z−1,

• NB. Output well-defined only for |z| > 1

2 and 1 4 < |z| < 1 2!

• Exercise. Verify that h3 ∗ x is not well-defined.

By partial fraction expansion (see slides 20-22 of Lecture 19) Y(z) = − 4 1 − 1

4z−1 −

2 (1 − 1

4z−1)2 +

8 1 − 1

2z−1

Take inverse z-transform to obtain response for first two cases.

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Inverse Z-transform of

1 (1−az−1)n

Recall 1 1 − aζ =

∞

• n=0

anζn, |aζ| < 1 and 1 1 − aζ = −

−1

• n=−∞

anζn, |aζ| > 1 Repeat the derivation on slide 23 of Lecture 19, and let ζ = z−1, n + m − 1 m − 1

• anu[n]

Z

← − − → 1 (1 − az−1)m, |z| > |a| and (−1)m −n − 1 m − 1

• anu[−n − m]

Z

← − − → 1 (1 − az−1)m, |z| < |a|

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Example

Re Im

1 − 2

3

X

1 2

Re Im

1

X

− 2

3

Re Im

1

X

1 2

Causal LTI system described by y[n] − 1 2y[n − 1] = x[n] + 2 3x[n − 1] Find response to x[n] = (− 2

3)nu[n]

H(z) = 1 + 2

3z−1

1 − 1

2z−1,

|z| > 1 2 X(z) = 1 1 + 2

3z−1,

|z| > 2 3 Y(z) = 1 1 − 1

2z−1, |z| > 1

2 = ⇒ y[n] = 1 2 n u[n] ROC of Y is larger than intersection of ROCs

• f X and H due to pole-zero cancellation at

z = − 2

3.

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Example

Suppose an LTI system satisfies the following,

• 1. output for input x1[n] = ( 1

6)nu[n] is y1[n] = [a( 1 2)n + 10( 1 3)n]u[n]

• 2. output for input x2[n] = (−1)n is y2[n] = 7

4(−1)n

Want to find H(z). X1(z) = 1 1 − 1

6z−1,

|z| > 1 6 Y1(z) = a 1 − 1

2z−1 +

10 1 − 1

3z−1,

|z| > 1 2 System function H(z) = Y1(z) X1(z) = [(a + 10) − (5 + a

3)z−1][1 − 1 6z−1]

(1 − 1

2z−1)(1 − 1 3z−1)

, By 2, H(−1) = 7

4 =

⇒ a = −9 = ⇒ H(z) = (1 − 2z−1)(1 − 1

6z−1)

(1 − 1

2z−1)(1 − 1 3z−1)

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Example (cont’d)

Three possibilities for ROC of H(z) = (1 − 2z−1)(1 − 1

6z−1)

(1 − 1

2z−1)(1 − 1 3z−1)

• |z| < 1

3

• 1

3 < |z| < 1 2

• |z| > 1

2

By the convolution property, ROC of Y1 contains the intersection

• f the ROCs of X1 and H =

⇒ ROC of H is |z| > 1

2 =

⇒ stable. Also H(∞) = 1 is finite = ⇒ causal H(z) = 1 − 13

6 z−1 + 1 3z−2

1 − 5

6z−1 + 1 6z−2

Difference equation y[n] − 5 6y[n − 1] + 1 6y[n − 2] = x[n] − 13 6 x[n − 1] + 1 3x[n − 2]

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Contents

• 1. Analysis of DT LTI Systems by Z-transform
• 2. Block Diagram Representations
• 3. Unilateral Z-transform

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System Interconnections

LTI systems in series connection Y = (XH1)H2 = X(H1H2) = (XH2)H1 H = H1H2 x H1(z) H2(z) y H(z) commutative H = H2H1 x H2(z) H1(z) y H(z)

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System Interconnections

LTI systems in systems in parallel connection h = h1 + h2 x h1 h2 + y h H = H1 + H2 x H1(z) H2(z) + y H(z)

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System Interconnections

LTI systems in systems in feedback connection y = h1 ∗ (x − h2 ∗ y) h =? x + + h1 h2 y

• h

Y = H1X − H1H2Y H = Y X = H1 1 + H1H2 x + + H1(z) H2(z) y

• H(z)

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Example

Causal LTI systems with system function H(z) = 1 1 − az−1, |z| > |a| Equivalent description by difference equation y[n] − ay[n − 1] = x[n] with initial rest condition. x[n] + y[n] z−1 a y[n − 1]

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Example

Causal LTI systems with system function H(z) = 1 − bz−1 1 − az−1 =

• 1

1 − az−1

• (1 − bz−1),

|z| > |a| x[n] +

w[n]

y[n] z−1 a

w[n − 1] w[n]

+ z−1

w[n − 1]

−b x[n] +

w[n]

y[n] z−1 a

w[n − 1]

+ −b

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Example

Causal LTI systems with system function H(z) = 1 (1 + 1

2z−1)(1 − 1 4z−1) =

1 1 + 1

4z−1 − 1 8z−2

Equivalent description by different equation y[n] + 1 4y[n − 1] − 1 8y[n − 2] = x[n] x[n] + y[n] z−1

− 1

4

y[n − 1]

+ z−1

1 8

y[n − 2]

Direct form

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Example (cont’d)

Causal LTI systems with system function H(z) = 1 1 + 1

2z−1 ·

1 1 − 1

4z−1

Cascade form x[n] + z−1

− 1

2

y[n] + z−1

1 4

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Example (cont’d)

Causal LTI systems with system function H(z) = 2/3 1 + 1

2z−1 +

1/3 1 − 1

4z−1

Parallel form x[n] + +

2 3

z−1

− 1

2 1 3

y[n] + z−1

1 4

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Contents

• 1. Analysis of DT LTI Systems by Z-transform
• 2. Block Diagram Representations
• 3. Unilateral Z-transform

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Unilateral Z-transform

The (bilateral) z-transform of a DT signal x X(z) =

∞

• n=−∞

x[n]z−n The unilateral z-transform of x is X(z) =

∞

• n=0

x[n]x−n also denoted X = UZ{x}, x[n]

UZ

← − − → X(z) Note UZ{x} = Z{xu}, where u is the unit step function.

• ROC of X is always the exterior of a circle, including ∞

◮ For rational X(z) = N(z)

D(z), we always have deg N ≤ deg D

• Z{x} = UZ{x} iff x[n] = 0 for n < 0

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Examples

The calculation of unilateral z-transforms is almost the same as for bilateral z-transforms. Remember to sum over only n ≥ 0.

• Example. x1[n] = anu[n],

X1(z) = X1(z) = 1 1 − az−1, |z| > |a|

• Example. x2[n] = an+1u[n + 1] = x1[n + 1],

X2(z) = zX1(z) = z 1 − az−1, |z| > |a| But X2(z) =

∞

• n=0

x[n]z−n =

∞

• n=0

an+1z−n = a 1 − az−1, |z| > |a|

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Examples

The calculation of inverse unilateral z-transforms is the same as for bilateral z-transforms, but we can only recover x[n] for n ≥ 0!

• Example. Given unilateral z-transform

X(z) = 3 − 5

6z−1

(1 − 1

4z−1)(1 − 1 3z−1) =

1 1 − 1

4z−1 +

2 1 − 1

3z−1

the only possibility for ROC is |z| ≥ 1

• 3. Thus

x[n] = 1 4 n + 2 2 3 n , n ≥ 0 X provides no information about x[n] for n < 0 Example. z2 z − a is not a valid unilateral z-transform, since deg D = 1 < 2 = deg N

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Properties of Unilateral Z-transform

Many properties are the same as for bilateral z-transform Property Signal Unilateral z-transform – x[n] X(z) – y[n] Y(z) Linearity ax[n] + by[n] aX(z) + bY(z) Scaling in z domain zn

0x[n]

X(z/z0) Time expansion x(k)[n] X(zk) Conjugation x∗[n] X∗(z∗) Differentiation in z domain nx[n] −z d

dzX(z)

Initial Value Theorem x[0] = lim

z→∞ X(z)

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Properties of Unilateral z-transform

Convolution Property. If x[n] = y[n] = 0 for n < 0, then (x ∗ y)[n]

UZ

← − − → X(z)Y(z)

• Proof. Note (x ∗ y)[n] = 0 for n < 0 and UZ{x} = Z{x} if x = xu.

UZ{x ∗ y} = Z{x ∗ y} = Z{x}Z{y} = UZ{x}UZ{y}

• Caution. The assumption x[n] = y[n] = 0 for n < 0 is important!
• Example. x[n] = δ[n] − δ[n − 1], y[n] = δ[n + 1],

X(z) = 1 − z−1, Y(z) = 0, Note (x ∗ y)[n] = δ[n + 1] − δ[n], and UZ{x ∗ y} = −1 = X(z)Y(z)

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Properties of Unilateral z-transform

Time delay. x[n − m]

UZ

← − − → x−m

• X(z) +

m

• k=1

x[−k]zk

• Proof.

UZ{x[n − m]} =

∞

• n=0

x[n − m]z−n = z−m

∞

• k=−m

x[k]z−k = z−m

• X(z) +

−1

• k=−m

x[k]z−k

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Properties of Unilateral z-transform

Time advance. x[n + m]

UZ

← − − → zm

• X(z) −

m−1

• k=0

x[k]z−k

• Proof.

UZ{x[n + m]} =

∞

• n=0

x[n + m]z−n = zm

∞

• k=m

x[k]z−k = zm

• X(z) −

m−1

• k=0

x[k]z−k

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Linear Constant-coefficient Difference Equations

Consider difference equation

N

• k=0

aky[n − k] =

M

• k=0

bkx[n − k] with initial condition y[−1], y[−2], . . . , y[−N] and causal input, i.e. x[n] = 0 for n < 0. Take unilateral z-transform of both sides

N

• k=0

akz−k

• Y(z) +

−1

• ℓ=−k

y[ℓ]z−ℓ

• =

M

• k=0

bkz−kX(z) so Y(z) = M

k=0 bkz−k

N

k=0 akz−k X(z)

• zero-state response

− N

k=0 akz−k −1 ℓ=−k y[ℓ]z−ℓ

N

k=0 akz−k

• zero-input response

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Example

Consider difference equation y[n] + 2y[n − 1] = x[n] with initial condition y[−1] = β and x[n] = αu[n]. Take unilateral z-transform of both sides Y(z) + 2z−1(Y(z) + y[−1]z) = X(z) = α 1 − z−1 so Y(z) = 1 1 + 2z−1 · α 1 − z−1

• zero-state response

+ −2β 1 + 2z−1

• zero-input response

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Example (cont’d)

Y(z) = 1 1 + 2z−1 · α 1 − z−1

• zero-state response

+ −2β 1 + 2z−1

• zero-input response

Zero-input response. α = 0 yzi[n] = −2β(−2)n, n ≥ 0 Zero-state response. β = 0, system at initial rest Yzs(z) = 1 1 + 2z−1 · α 1 − z−1 = 2α/3 1 + 2z−1 + α/3 1 − z−1 yzs[n] = 2α 3 (−2)n + α 3 , n ≥ 0

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Example

Consider difference equation y[n] − y[n − 1] − 2y[n − 2] = x[n] + 2x[n − 2] with initial condition y[−1] = 2, y[−2] = − 1

2 and x[n] = u[n].

Take unilateral z-transform of both sides Y(z)−z−1(Y(z)+y[−1]z)−2z−2(Y(z)+y[−1]z+y[−2]z2) = (1+2z−2)X(z) so Y(z) = 1 + 2z−2 1 − z−1 − 2z−2 · 1 1 − z−1

• zero-state response

+ 2z−1y[−1] + y[−1] + 2y[−2] 1 − z−1 − 2z−2

• zero-input response

= 1 + 2z−2 (1 + z−1)(1 − 2z−1)(1 − z−1)

• zero-state response

+ 4z−1 + 1 (1 + z−1)(1 − 2z−1)

• zero-input response

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Example (cont’d)

For the zero-input response, Yzi(z) = 4z−1 + 1 (1 + z−1)(1 − 2z−1) = −1 1 + z−1 + 2 1 − 2z−1 so yzi[n] = 2n+1 + (−1)n+1, n ≥ 0 For the zero-state response, Yzs(z) = 1 + 2z−2 (1 + z−1)(1 − 2z−1)(1 − z−1) = 1/2 1 + z−1 + 2 1 − 2z−1 + −3/2 1 − z−1 so yzs[n] = 1 2(−1)n + 2n+1 − 3 2, n ≥ 0