[PPT] - EI331 Signals and Systems Lecture 19 Bo Jiang John Hopcroft Center PowerPoint Presentation

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EI331 Signals and Systems

Lecture 19 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

April 30, 2019

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Convolution Property of DTFT

(x ∗ h)[n]

F

← − − → X(ejω)H(ejω)

NB. Dual of multiplication property of CTFS

x(t)y(t)

FS

← − − → (ˆ x ∗ ˆ y)[k]

NB. Similar to CTFT, applicable when formula is well-defined

Proof. F{x ∗ h}(ejω) =

n∈Z

(x ∗ h)[n]e−jωn =

n∈Z
m∈Z

x[m]h[n − m]e−jωn =

m∈Z

x[m]

n∈Z

h[n − m]e−jωn =

m∈Z

x[m]e−jωmH(ejω) = X(ejω)H(ejω)

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Frequency Response of LTI System

LTI system T

fully characterized by impulse response h

y = T(x) = h ∗ x

also fully characterized by frequency response H = F{h},

if H is well-defined

◮ BIBO stable system, h ∈ ℓ1 ◮ other systems: accumulator h = u, ...

Typically, convolution property implies Y = F{y} = HX = F{h}F{x} Instead of computing x ∗ h, can do y = F−1(F{h}F{x})

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Example

Response of LTI system with impulse response h[n] = anu[n] to input x[n] = bnu[n], where |a| < 1, |b| < 1 Method 1. Direct convolution y = x ∗ h Method 2. Solve difference equation with initial rest condition y[n] − ay[n − 1] = bnu[n] Method 3. Fourier transform. H = 1 1 − ae−jω , X = 1 1 − be−jω = ⇒ Y = 1 (1 − ae−jω)(1 − be−jω) If a = b, Y(ejω) = a/(a−b)

1−ae−jω − b/(a−b) 1−be−jω ,

y[n] =

1 a−b(an+1 − bn+1)u[n]

If a = b, Y(ejω) =

d da

a

1−ae−jω

, y[n] =

d daan+1u[n] = (n + 1)anu[n]

Or use Y(ejω) = j

aejω d dω

a

1−ae−jω

and differentiation property

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Example

Response of LTI system with impulse response h[n] = anu[n] to input x[n] = cos(ω0n), |a| < 1. Frequency response H(ejω) = 1 1 − ae−jω Method 1. Use eigenfunction property x[n] = 1 2ejω0n + 1 2e−jω0n y[n] = 1 2H(ejω0)ejω0n + 1 2H(e−jω0)e−jω0n = Re ejω0n 1 − ae−jω0 = 1 √1 − 2a cos ω0 + a2 cos

ω0n − arctan

a sin ω0 1 − a cos ω0

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Example

Method 2. Use Fourier transform of X X(ejω) =

∞

k=−∞

[πδ(ω − ω0 + 2kπ) + πδ(ω + ω0 + 2kπ)] Y(ejω) =

∞

k=−∞

πH(ej(ω0−2kπ))δ(ω − ω0 + 2kπ) +

∞

k=−∞

πH(e−j(ω0+2kπ))δ(ω + ω0 + 2kπ) = H(ejω0)

∞

k=−∞

πδ(ω − ω0 + 2kπ) + H(e−jω0)

∞

k=−∞

πδ(ω + ω0 + 2kπ) y[n] = 1 2H(ejω0)ejω0n + 1 2H(e−jω0)e−jω0n

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Example: Ideal Bandstop Filter

x × (−1)n Hlp(ejω) × (−1)n Hlp(ejω) + y Hhp(ejω) ω Hlp(ejω)

1 −W W π −π 2π −2π

0 < W < π/2 ω H(ejω)

1 −W W π − W π −π 2π −2π

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Multiplication Property of DTFT

x[n]y[n]

F

← − − → 1 2πX ⊛ Y = 1 2π

2π

X(ejθ)Y(ej(ω−θ))dθ

NB. Dual of periodic convolution property of CTFS

x ⊛ y

FS

← − − → Tˆ xˆ y Proof. F{xy}(ejω) =

n∈Z

x[n]y[n]e−jωn =

n∈Z

1 2π

2π

X(ejθ)ejθndθ

y[n]e−jωn

= 1 2π

2π

X(ejθ)

n∈Z

y[n]e−j(ω−θ)n

dθ

= 1 2π

2π

X(ejθ)Y(ej(ω−θ))dθ

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Example

x[n] = x1[n]x2[n], where x1[n] = sin(3πn/4) πn , x2[n] = sin(πn/2) πn X =

1 2πX1 ⊛ X2 = 1 2π ˜

X1 ∗ X2 periodic aperiodic ω X1(ejω)

1 − π

2 π 2

π −π 2π −2π

ω X2(ejω)

1 − 3π

4 3π 4

π −π 2π −2π

ω ˜ X1(ejω)

1 − π

2 π 2

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Example

x[n] = x1[n]x2[n], where x1[n] = sin(3πn/4) πn , x2[n] = sin(πn/2) πn X =

1 2πX1 ⊛ X2 = 1 2π ˜

X1 ∗ X2 =

1 2π ∞

k=−∞

τ2kπ(˜ X1 ∗ ˜ X2) periodic aperiodic aperiodic ω ˜ X1(ejω)

1 − π

2 π 2

ω ˜ X2(ejω)

1 − 3π

4 3π 4

ω Y =

1 2π(˜

X1 ∗ ˜ X2)

1/2 − π

2 π 2

−π π −2π 2π

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Example

x[n] = x1[n]x2[n], where x1[n] = sin(3πn/4) πn , x2[n] = sin(πn/2) πn X =

1 2πX1 ⊛ X2 = 1 2π ˜

X1 ∗ X2 =

1 2π ∞

k=−∞

τ2kπ(˜ X1 ∗ ˜ X2) periodic aperiodic aperiodic ω ˜ X1(ejω)

1 − π

2 π 2

ω ˜ X2(ejω)

1 − 3π

4 3π 4

ω X =

1 2π(X1 ⊛ X2)

1/2 − π

2 π 2

−π π −2π 2π

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Example

x[n] = x1[n]x2[n], where x1[n] = sin(3πn/4) πn , x2[n] = sin(πn/2) πn X =

1 2πX1 ⊛ X2 = 1 2π ˜

X1 ∗ X2 =

1 2π ∞

k=−∞

τ2kπ(˜ X1 ∗ ˜ X2) periodic aperiodic aperiodic ω ˜ X1(ejω)

1 − π

2 π 2

ω ˜ X2(ejω)

1 − 3π

4 3π 4

ω X =

1 2π(X1 ⊛ X2)

1/2 − π

2 π 2

−π π −2π 2π

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Example: DT Modulation

y[n] = x[n]c[n], where c[n] = cos(ωcn) X(ejω)

A ωM −ωM −2π 2π

C(ejω)

π π −2π 2π ωc −ωc 2π + ωc 2π − ωc −2π + ωc −2π − ωc

Y(ejω)

A 2

−2π −2π 2π −ωc ωc ωc + ωM 2π − ωc − ωM

No overlap between replicas:

ωc > ωM

ωc + ωM < π = ⇒ ωM < π

2

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Example: DT Demodulation

z[n] = y[n]c[n], where c[n] = cos(ωcn) Y(ejω)

A 2

−2π −2π 2π −ωc ωc ωc + ωM 2π − ωc − ωM

C(ejω)

π π −2π 2π ωc −ωc 2π + ωc 2π − ωc −2π + ωc −2π − ωc

Z(ejω)

A 2

ωM −ωM −2π 2π 2ωc − ωM −W W 2

Recover X by lowpass filtering Y with W ∈ (ωM, 2ωc − ωM)

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Linear Constant-coefficient Difference Equations

Frequency response of LTI system described by

N

k=0

aky[n − k] =

M

k=0

bkx[n − k] Method 1. Use eigenfunction property x[n] = ejωn = ⇒ y[n] = H(ejω)ejωn Substitution into difference equation yields

N

k=0

akH(ejω)ejω(n−k) =

M

k=0

bkejω(n−k) H(ejω) = M

k=0 bke−jωk

N

k=0 ake−jωk

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Linear Constant-coefficient Difference Equations

Method 2. Take Fourier transform of both sides F N

k=0

aky[n − k]

= F

M

k=0

bkx[n − k]

By linearity and time-shifting property

N

k=0

ake−jωkY(ejω) =

M

k=0

bke−jωkX(ejω) By convolution property H(ejω) = Y(ejω) X(ejω) = M

k=0 bke−jωk

N

k=0 ake−jωk

Frequency response is rational function of e−jω

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Example

Consider LTI system described by y[n] − 3 4y[n − 1] + 1 8y[n − 2] = 2x[n] Frequency response H(ejω) = 2 1 − 3

4e−jω + 1 8e−j2ω

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

2e−jω)(1 − 1 4e−jω) =

4 1 − 1

2e−jω −

2 1 − 1

4e−jω

Take inverse Fourier transform to obtain impulse response h[n] = 4 1 2 n u[n] − 2 1 4 n u[n]

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Example

Find response to x[n] = 1

4

n u[n]. Fourier transform of response Y(ejω) = H(ejω)X(ejω) = 2 (1 − 1

2e−jω)(1 − 1 4e−jω) ·

1 1 − 1

4e−jω

By partial fraction expansion Y(ejω) = − 4 1 − 1

4e−jω −

2 (1 − 1

4e−jω)2 +

8 1 − 1

2e−jω

Take inverse Fourier transform to obtain response y[n] =

−4

1 4 n − 2(n + 1) 1 4 n + 8 1 2 n u[n]

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Partial Fraction Expansion

H(ejω) = 2 1 − 3

4e−jω + 1 8e−j2ω

Replace e−jω by z, H(z−1) = 2 1 − 3

4z + 1 8z2 =

2 (1 − 1

2z)(1 − 1 4z) =

A 1 − 1

2z +

B 1 − 1

4z

A =

(1 − 1

2z)H(z−1)

z=2

= 2 1 − 1

4z

z=2

= 4 B =

(1 − 1

4z)H(z−1)

z=4

= 2 1 − 1

2z

z=4

= −2

NB. Terms take form

A (s − r)m for CT, but A (1 − rz)m for DT

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Partial Fraction Expansion

Y(ejω) = 2 (1 − 1

2e−jω)(1 − 1 4e−jω) ·

1 1 − 1

4e−jω

Replace e−jω by z, Y(z−1) = 2 (1 − 1

2z)(1 − 1 4z)2 =

A1,1 1 − 1

2z +

A2,1 1 − 1

4z +

A2,2 (1 − 1

4z)2

To find A1,1

1. Multiply both side by 1 − 1

2z,

(1 − 1 2z)Y(z−1) = 2 (1 − 1

4z)2 = A1,1 + A2,1(1 − 1 2z)

1 − 1

4z

+ A2,2(1 − 1

2z)

(1 − 1

4z)2

2. Evaluate at z = 2, i.e. root of 1 − 1

2z = 0

A1,1 =

(1 − 1

2z)Y(z−1)

z=2

= 2 (1 − 1

4z)2

z=2

= 8

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Partial Fraction Expansion

Y(z−1) = 2 (1 − 1

2z)(1 − 1 4z)2 =

A1,1 1 − 1

2z +

A2,1 1 − 1

4z +

A2,2 (1 − 1

4z)2

To find A2,2

1. Multiply both side by (1 − 1

4z)2,

(1 − 1 4z)2Y(z−1) = 2 1 − 1

2z = A1,1(1 − 1 4z)2

1 − 1

2z

+ A2,1(1 − 1 4z) + A2,2

2. Evaluate at z = 4, i.e. root of 1 − 1

4z = 0

A2,2 =

(1 − 1

4z)2Y(z−1)

z=4

= 2 1 − 1

2z

z=4

= −2

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Partial Fraction Expansion

Y(z−1) = 2 (1 − 1

2z)(1 − 1 4z)2 =

A1,1 1 − 1

2z +

A2,1 1 − 1

4z +

A2,2 (1 − 1

4z)2

To find A2,1

1. Multiply both side by (1 − 1

4z)2,

(1 − 1 4z)2Y(z−1) = 2 1 − 1

2v = A1,1(1 − 1 4z)2

1 − 1

2z

+ A2,1(1 − 1 4z) + A2,2

2. Differentiate both sides

d dz(1 − 1 4z)2Y(z−1) = 1 (1 − 1

2z)2 = d

dz A1,1(1 − 1

4z)2

1 − 1

2z

+

−1

4

A2,1
3. Evaluate at z = 4, i.e. root of 1 − 1

4z = 0

A2,1 = (−4) d dz(1 − 1 4z)2Y(z−1)

z=4

= −4 (1 − 1

2z)2

z=4

= −4

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Inverse Fourier Transform of

1 (1−ae−jω)n Recall anu[n]

F

← − − → 1 1 − ae−jω , |a| < 1 i.e. 1 1 − az =

∞

n=0

anzn, z = e−jω Differentiate m − 1 times w.r.t. z (m − 1)!am−1 (1 − az)m =

∞

n=m−1

an n! (n − m + 1)!zn−(m−1) 1 (1 − az)m =

∞

n=m−1

an−m+1

n

m − 1

zn−m+1 =

∞

n=0

an n + m − 1 m − 1

zn

Thus n + m − 1 n

anu[n]

F

← − − → 1 (1 − ae−jω)m

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Impulse-train Sampling

n x[n]

−8−7−6−5−4−3−2−1 0 1 2 3 4 5 6 7 8

n p[n]

−8−7−6−5−4−3−2−1 0 1 2 3 4 5 6 7 8

1 n xp[n]

−8−7−6−5−4−3−2−1 0 1 2 3 4 5 6 7 8

DT signal x[n] DT impulse train p[n] =

∞

k=−∞

δ[n − kN] Sampled signal xp[n] = x[n]p[n] =

∞

k=−∞

x[kN]δ[n − kN]

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Impulse-train Sampling

Spectrum of impulse train P(ejω) = ωs

∞

k=−∞

δ(ω − kωs), ωs = 2π N Spectrum of sampled signal Xp = 1 2πX ⊛ P = 1 2π ˜ X ∗ P = 1 2πX ∗ ˜ P where ˜ X, ˜ P are respective parts of X, P within one period of 2π Xp(ejω) = 1 N

∞

k=−∞

˜ X(ej(ω−kωs)) = 1 N

N−1

k=0

X(ej(ω−kωs))

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Impulse-train Sampling

First view: Replicating ˜ X at kωs for k ∈ Z Xp(ejω) = 1 N

∞

k=−∞

˜ X(ej(ω−kωs)) ω X(ejω)

A ωM −ωM π −π −2π 2π

ω P(ejω)

2π N

ωs −2π 2π

ω Xp(ejω)

A N

ωs 2ωs 2π −2π

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Impulse-train Sampling

Second view: Replicating X at kωs for k = 0, 1, . . . , N − 1 Xp(ejω) = 1 N

N−1

k=0

X(ej(ω−kωs)) ω X(ejω)

A ωM −ωM π −π −2π 2π

ω P(ejω)

2π N

ωs 2ωs −2π 2π

ω Xp(ejω)

A N

ωs 2ωs 2π −2π

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Impulse-train Sampling

ω X(ejω)

A ωM −ωM π −π −2π 2π

ω P1(ejω)

2π N1

ωs1 −2π 2π

ωs1 > 2ωM ω Xp1(ejω)

A N1

ωs1 2π −2π

recoverable ω P2(ejω)

2π N2

ωs2 −2π 2π

ωs2 < 2ωM ω Xp2(ejω)

A N2

ωs2 2π −2π

undersampling, aliasing

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DT Sampling Theorem

Band-limited DT signal x[n] whose spectrum X(ejω) = 0 for ωM < |ω| ≤ π is uniquely determined by its samples x[kN], k ∈ Z if ωs 2π N > 2ωM Given {x[kN] : k ∈ Z}, x[n] can be reconstructed as follows

1. construct xp[n] =

∞

k=−∞

x[kN]δ[n − kN]

2. send xp through lowpass filter with gain N and cutoff

frequency ωc ∈ (ωM, ωs − ωM), i.e. H(ejω) = N[u(ω + ωc) − u(ω − ωc)]

3. output xr[n] with Xr(ejω) = Xp(ejω)H(ejω) is same as x[n]

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Reconstruction in Time Domain

x[n] × p[n] =

∞

k=−∞

δ[n − kN] xp[n] H(ejω)

−ωc ωc N

xr[n] Impulse response of lowpass filter h[n] = N sin(ωcn) πn x recovered by band-limited interpolation using sinc function x[n] = xr[n] = (xp ∗ h)[n] =

∞

k=−∞

x[kN]N sin(ωc(n − kN)) π(n − kN)

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Decimation

n x[n]

−8−7−6−5−4−3−2−1 0 1 2 3 4 5 6 7 8

n xp[n]

−8−7−6−5−4−3−2−1 0 1 2 3 4 5 6 7 8

n xb[n]

−2−1 0 1 2

DT signal x[n] Sampled signal xp[n] =

∞

k=−∞

x[kN]δ[n−kN] Decimated signal xb[n] = x[nN] Reduces samples

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Decimation

Xb(ejω) =

n∈Z

xb[n]e−jωn =

n∈Z

x[nN]e−jωn =

k∈NZ

x[k]e−jωk/N (k = nN) =

k∈Z

xp[k]e−j ω

N k

(xp[n] = 0 for n ∈ Z \ NZ) = Xp(ejω/N) Recall Xp(ejω) has period 2π/N, so Xb(ejω) has period 2π. If no aliasing in sampling

Xb is stretched version of X in each period
Xb(ejω) = Xp(ejω/N), but Xb(ejω) = X(ejω/N)

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Decimation

ω X(ejω)

A ωM −ωM π −π −2π 2π

ω P(ejω)

2π N

ωs = 2π

N

−2π 2π

ω Xp(ejω)

A N

ωs 2ωs 2π −2π

ω Xb(ejω) = Xp(ejω/N)

A N

−NωM NωM 2π −2π −2π 2π

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DT Processing of CT Signals

xc(t) C/D conversion xd[n] DT system yd[n] D/C conversion yc(t) Benefits of DT processing

versatility of digital electronic devices
resilience against noise

In practice

C/D implemented by analog-to-digital converter
D/C implemented by digital-to-analog converter

Idealization

no quantization
C/D by impulse-train sampling
D/C by ideal reconstruction

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C/D Conversion

xc(t) × p(t) xp(t) conversion of CT impulse train to DT sequence xd[n] = xc(nT) C/D conversion Assumption

xc(t) bandlimited, e.g. lowpass filtered
sampling above Nyquist rate, no aliasing

Relation between CT and DT spectra Xp(jω) = CTFT

n∈Z

xc(nT)δ(t − nT)

=
n∈Z

xc(nT)e−jωTn =

n∈Z

xd[n]e−jωTn = Xd(ejωT)

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C/D Conversion

Time domain t xc(t) t xp(t)

−2T 2T T −T

n xd[n]

−2 −1 1 2

Frequency domain ω Xc(jω)

A

ω Xp(jω)

A T 2π T

− 2π

T

Ω Xd(ejΩ)

A T

2π −2π

From xp to xd: normalization in both time and frequency

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D/C Conversion

yd[n] conversion of DT sequence to CT impulse train yp(t)

− π

T π T

T

yc(t) C/D conversion Relation between CT and DT spectra Yp(jω) = Yd(ejωT) Yc(jω) =    Yp(jω) |ω| < ωs 2 = ω T 0,

therwise

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D/C Conversion

Time domain t yc(t) t yp(t)

−2T 2T T −T

n yd[n]

−2 −1 1 2

Frequency domain ω Yc(jω)

A

ω Yp(jω)

A T 2π T

− 2π

T

Ω Yd(ejΩ)

A T

2π −2π

Reverse process of C/D conversion

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Frequency Response

xc(t)

×

p(t) xp(t) impulse train to sequence xd[n] Hd(ejΩ) yd[n] sequence to impulse train yp(t) − π

T π T

T yc(t)

Hc(jω)

1. Yd(ejΩ) = Hd(ejΩ)Xd(ejΩ)
2. Yp(jω) = Yd(ejωT) = Hd(ejωT)Xd(ejωT) = Hd(ejωT)Xp(jω)
3. Yc(jω) = Yp(jω)Hlp(jω) = Hd(ejωT)Xp(jω)Hlp(jω) = Hd(ejωT)Xc(jω)
4. Hc(jω) =

   Hd(ejωT) |ω| < ωs 2 = ω T 0,

therwise

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Frequency Response

Hc(jω) =    Hd(ejωT) |ω| < ωs 2 = ω T 0,

therwise

= ⇒ Hd(ejΩ) = Hc

jΩ

T

, |Ω| < π

Ω Hd(ejΩ)

A ωcT −ωcT 2π −2π π −π

ω Hc(jω)

ωc −ωc A

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Example: Digital Differentiator

ω |Hc(jω)| ωc −ωc ωc ω arg Hc(jω) −ωc − π

2 π 2

ωc ωs = 2ωc Ω |Hd(ejΩ)| ωc −2π 2π −π π Ω arg Hd(ejΩ) − π

2 π 2

EI331 Signals and Systems

Lecture 19 Bo Jiang

April 30, 2019

Contents

Difference Equations

Convolution Property of DTFT

(x ∗ h)[n]

← − − → X(ejω)H(ejω)

x(t)y(t)

← − − → (ˆ x ∗ ˆ y)[k]

Proof. F{x ∗ h}(ejω) =

(x ∗ h)[n]e−jωn =

x[m]h[n − m]e−jωn =

x[m]

h[n − m]e−jωn =

x[m]e−jωmH(ejω) = X(ejω)H(ejω)

Frequency Response of LTI System

LTI system T

y = T(x) = h ∗ x

if H is well-defined

◮ BIBO stable system, h ∈ ℓ1 ◮ other systems: accumulator h = u, ...

Typically, convolution property implies Y = F{y} = HX = F{h}F{x} Instead of computing x ∗ h, can do y = F−1(F{h}F{x})

Example

y[n] =

If a = b, Y(ejω) =

Or use Y(ejω) = j

Example

a sin ω0 1 − a cos ω0

Example

Method 2. Use Fourier transform of X X(ejω) =

[πδ(ω − ω0 + 2kπ) + πδ(ω + ω0 + 2kπ)] Y(ejω) =

πH(ej(ω0−2kπ))δ(ω − ω0 + 2kπ) +

πH(e−j(ω0+2kπ))δ(ω + ω0 + 2kπ) = H(ejω0)

πδ(ω − ω0 + 2kπ) + H(e−jω0)

πδ(ω + ω0 + 2kπ) y[n] = 1 2H(ejω0)ejω0n + 1 2H(e−jω0)e−jω0n

Example: Ideal Bandstop Filter

x × (−1)n Hlp(ejω) × (−1)n Hlp(ejω) + y Hhp(ejω) ω Hlp(ejω)

1 −W W π −π 2π −2π

0 < W < π/2 ω H(ejω)

1 −W W π − W π −π 2π −2π

Contents

Difference Equations

Multiplication Property of DTFT

x[n]y[n]

← − − → 1 2πX ⊛ Y = 1 2π

X(ejθ)Y(ej(ω−θ))dθ

x ⊛ y

← − − → Tˆ xˆ y Proof. F{xy}(ejω) =

x[n]y[n]e−jωn =

1 2π

X(ejθ)ejθndθ

= 1 2π

X(ejθ)

y[n]e−j(ω−θ)n

= 1 2π

X(ejθ)Y(ej(ω−θ))dθ

Example

x[n] = x1[n]x2[n], where x1[n] = sin(3πn/4) πn , x2[n] = sin(πn/2) πn X =

X1 ∗ X2 periodic aperiodic ω X1(ejω)

1 − π

π −π 2π −2π

ω X2(ejω)

1 − 3π

π −π 2π −2π

ω ˜ X1(ejω)

1 − π

Example

x[n] = x1[n]x2[n], where x1[n] = sin(3πn/4) πn , x2[n] = sin(πn/2) πn X =

X1 ∗ X2 =

τ2kπ(˜ X1 ∗ ˜ X2) periodic aperiodic aperiodic ω ˜ X1(ejω)

1 − π

ω ˜ X2(ejω)

1 − 3π

ω Y =

X1 ∗ ˜ X2)

1/2 − π

−π π −2π 2π

Example

x[n] = x1[n]x2[n], where x1[n] = sin(3πn/4) πn , x2[n] = sin(πn/2) πn X =

X1 ∗ X2 =