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Digital Signal ProcessingYCJ

1

A Typical DSP System

Sampling&

A/D

AnalogPre-filterHPRE(f)

x(t) x[n]xa(t)Analog

inDSPFilter

HDSP(z)

Analogout

AnalogPost-filterHPOST(f)

D/AConverter

HDAC(f)

ya(t) y(t) y[n]


2

Analog Signal Processing

xa(t) : an analog signal

Xa(Ω) = ∫ xa(t)e-jΩt dt is its Fourier transform

The inverse Fourier transform is

xa(t) = (1/2π)∫ Xa(Ω) ejΩt dΩ


3

Analog Signal ProcessingAnalogFilterHa(Ω)

ya(t)xa(t)

Ha(Ω) is the Fourier transform of the impulse response, ha(t), of the filter

ya(t) = ∫ ha(t-τ) xa(τ) dτ

OR, equivalently, in frequency domain

Ya(Ω) = Ha(Ω) Xa(Ω)


4

A Typical DSP SystemA Typical DSP System


Sampling&

A/DDSPFilter

HDSP(z)D/A

ConverterHDAC(f)

AnalogPost-filterHPOST(f)

xa(t) x(t) x[n]

ya(t) y(t) y[n]


5

A digital Spectral Analyzer


Sampling&

A/D

FFTProcessor

Displaythe

Spectrum

xa(t) x(t)


6

Linear Time-Invariant Systems•The delta function: δ(t)

•Linearity

•Time-invariant

•The Impulse Response

•The Convolution Integral

•The Convolution Theorem


7

Sinusoidal Response of LTI System

ejΩt Ha(Ω) ejΩtHa(Ω)

Ha(Ω) = |Ha(Ω)| e jARGHa(Ω)

Ha(Ω1) A1ejΩ1t

+ Ha(Ω2) A2ejΩ2tA1ejΩ1t

+A2ejΩ2tHa(Ω)


8

Linear FilteringXa(Ω)

Ω=2πf

Ya(Ω)

Ω=2πf


9

Sampling a Signal

x[n] = x(nT)x(t)

T = 1/fs

T: sampling period, (second)

fs: sampling frequency (sample/second)

x(t) x[n]

t n0 1 2 3 4


10

Discrete-Time Fourier Tranform (DTFT)

X(ω) = Σn x[n] e-jnω

•X(ω) is a continuous function of ω

•X(ω) is periodic on ω with the period 2π

-4π -2π 0 2π 4π

X(ω)

ω


11

Fourier Transform and The DTFTIf Xa(Ω) = ∫ xa(t)e-jΩt dt and x[n] = xa(nT)Then X(ΩΤ) = (1/T) Σk Xa(Ω - k2π/T)

-4πfs -2πfs 0 2πfs 4πfs

1/T

0

Xa(Ω)

Ω

X(ΩΤ)

Ω


12

Aliasing

Ω-4πfs -2πfs 0 2πfs 4πfs

X(Ω)

0-4πfs -2πfs 2πfs 4πfs Ω


13

AliasingCos(Ω t) = ½(ejΩt + e-jΩt )

-4πfs -2πfs 0 2πfs 4πfs Ω

-4πfs -2πfs 0 2πfs 4πfs Ω


14

Aliasing

0 10 20 30 40 50 60 70 80 90 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1


15

Anti-Aliasing Filter

A(f)

ffpass fs/2 fstop

Attenuation A(f) = -20 log |H(f)/H(0)|

dB per Octave or dB per Decade

A single-pole filter has 6 dB/Oct. or 20 dB/Dec.


16

DAC & Analog Reconstruction

D/AConverter

HDAC(f)y[n] ya(t)

ya(t) = Σn y[n] hDAC(t-nT)

Ya(Ω) = Y(ω) |ω=ΩΤ HDAC(Ω)


17

Staircase Reconstructor

-100 -80 -60 -40 -20 0 20 40 60 80 100-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

h(t)1

T/2-T/2

f = 1/T = fs

1, -T/2 ≤ t ≤ T/20, otherwise

h(t) = u(t) - u(t-T) =

H(f) = (T/j) [Sin(πfT)/(πfT)]t

f


18

Staircase Reconstructor

-100 -80 -60 -40 -20 0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

fs-fs


19

Anti-Image Post Filter

-100 -80 -60 -40 -20 0 20 40 60 80 100-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

fs/2f

-fs/2

•Removing high frequency images•Equalizing D/A converter’s in-band distortion

•(may be done digitally)• -20 log[H(fs/2)/H(0)] ~ 3.9 dB


20

Ideal Reconstructor

TH(f)

f-fs/2 fs/2

-100 -80 -60 -40 -20 0 20 40 60 80 100-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

h(t) = Sin(πt/T)/(πt/T)

tT 2T


21

A/D Converter & Quantization

R: fu

ll sc

ale

R/23Q2QQ0

-Q-2Q-3Q-R/2

R/Q = 2BQ: step size


22

Quantization Noise PowerP(e)

1/Q

e-Q/2 Q/2

E(e2) = (1/Q) ∫ e2 de = Q2/12

or

erms = Q/121/2


23

Signal to Noise Ratio: SNR

Noise Power = Q2/12Signal Power = ?Assuming Sine Wave -> Signal Power = A2/2Assuming Full Scale -> A = R/2 -> Signal Power = R2/8Assuming B bits -> Q = R/2B

Signal to Noise Power Ration = (R2/8) / (Q2/12)

SNR (dB) = 10 log(22B) + 10log(12/8)= 6.02*B + 1.76 dB


24

Discrete Time Systems

x[0], x[1], x[2], … y[0], y[1], y[2], …H

Example: y[n] = 2x[n] + 3x[n-1] + 4x[n-2]

Sample by Sample ProcessingBlock Processing


25

Sample by Sample Processingy[n] = 2x[n] + 3x[n-1] + 4x[n-2]

y[n] = 2x[n] + 3w1[n] + 4w2[n]

w2[n+1] = w1[n] w1[n+1] = x[n]

For each new input xDO y = 2x + 3w1 + 4w2

w2 = w1w1 = x

CONTINUE


26

Block Processingy[n] = 2x[n] + 3x[n-1] + 4x[n-2]

y[0] 2 0 0 0 0y[1] 3 2 0 0 0 x[0]y[2] 4 3 2 0 0 x[1]y[3] = 0 4 3 2 0 x[2]y[4] 0 0 4 3 2 x[3]y[5] 0 0 0 4 3 x[4]y[6] 0 0 0 0 4


27

Overlap and Add

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,

14, 15, 16, 17, 18, 19,


28

FIR and IIR Systems

FIR: Finite Impulse Response -- non-recursive

y[n] = 2x[n] + 3x[n-1] + 4x[n-2]

IIR: Infinite Impulse Response -- recursive

y[n] = 0.5y[n-1] + 2x[n] + 3x[n-1]


29

Sample by Sample Processingy[n] = 0.5y[n-1] + 2x[n] + 3x[n-1]

y[n] = 0.5w[n] + 3x[n] + 4v[n]

w[n+1] = y[n] v[n+1] = x[n]

For each new input xDO y = 0.5w + 2x + 3v

w = yv = x

CONTINUE


30

Linearitya1

y[n] = a1y1[n]+ a2y2[n]

+

a2

x1[n]

x[n]x2[n]

y[n]H

a1

+

a2

x1[n]

x2[n]

Hy1[n]

a1y1[n]+ a2y2[n]

Hy2[n]


31

Time-Invariance

x[n] y[n]

yD[n] = y[n-D]

H D y[n-D]

x[n]D

xD[n]H yD[n]

x[n-D]


32

Impulse Response of an LTI System

δ[n] = 1, for n=0, and 0, otherwise

h[n]δ[n]

H nn

For any input x[n], the output y[n] is given by

y[n] = Σm x[m]h[n-m]= Σm h[m]x[n-m] (direct form)


33

Convolutionh[n]δ[n]

H nnh[n-1]

δ[n-1]

H nnh[n-2]

δ[n-2]

H nn

x[n] = x[0]δ[n] + x[1]δ[n-1] + x[2]δ[n-2] + …

y[n] = x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] +


34

FIR and IIR FiltersAn FIR filter has impulse response, h[n], that extends only over a finite time interval, say 0 ≤ n ≤ M, and is identically zero beyond that:

h[0], h[1], h[2], …h[M], 0,0,0,0,0,0…..

M is the filter order, and the length of the filter impulse response is M+1.

An IIR filter, on the other hand, has impulse response, h[n], of infinite duration, 0 ≤ n ≤ ∞ , and is identically zero beyond that:

h[0], h[1], h[2], …h[M], h[M+1], …………..

The convolution sum is not computationally feasible, hence it is normally represented by a constant-coefficient linear difference equation.


35

I/O Difference Equation

Impulse Response: h[n]

FIR filter: convolution sum

IIR Filter: difference equation

y[n] = ay[n-1] + x[n]

h[n] = anu[n]


36

Causality and StabilityA causal sequence is a right-sided sequence, and an anti-causalsequence is a left-sided sequence. A two-sided sequence is mixed.

An LTI system is causal if its impulse response, h[n], is causal.

An LTI system is stable if for any bounded input sequence,the output sequence is also bounded.

An LTI system is stable if and only if its impulse responseis absolutely summable, i.e.,

Σn |h[n]| < ∞


37

Linear Convolution: Flip & Slide

y[n] = Σm x[m]h[n-m]

If h is of order M, i.e., h = h[0], h[1], …, h[M] is of length M+1and x = x[0], x[1], …,x[L-1] is of length L, then

y[n] = Σm=max(0,n-M)min(n,L-1) x[m]h[n-m]

and y = y[0], y[1], …,y[L+M] is of length (L+M)


38

Linear Convolution: Flip & Slide

h3 h2 h1 h0 → h3 h2 h1 h0 → h3 h2 h1 h0

0 0 0 x0 x1 x2 x3 … xn-3 xn-2 xn-1 xn xL-1 0 0 0

y0 yn yL-1+M

h = M+1x = L

y = h*x = L M


39

Transient & Steady State

y[n] = Σm=0n h[m]x[n-m] if 0 ≤ n < M

y[n] = Σm=0M h[m]x[n-m] if M ≤ n ≤ L-1

y[n] = Σm=n-L+1M h[m]x[n-m] if L ≤ n < L+M

output y[n]

n0 M L-1 L-1+M


40

Convolution of Infinite Sequencey[n] = Σm=max(0,n—L+1)

min(n,M) h[m]x[n-m]

Infinite filter, finite input: M = ∞, L < ∞y[n] = Σm=max(0,n—L+1)

n h[m]x[n-m]

Finite filter, infinite input: M < ∞, L = ∞y[n] = Σm=0

min(n,M) h[m]x[n-m]

Infinite filter, infinite input: M = ∞, L = ∞y[n] = Σm=0

n h[m]x[n-m]


41

Finite filter, Infinite inputBlock Processing

y0 = h*x0, y1 = h*x1, y2 = h*x2,………

y0 = L M

x = block x0 block x1 block x2

y1 = L M

y2 = L M


42

Finite filter, Infinite inputSample Processing

y[n] = Σm=max(0,n—L+1)n h[m]x[n-m]

= h[0]x[n] + h[1]x[n-1] + …. h[M]x[n-M]

FOR each input x DOw0 = xy = h0w0 + h1w1 + h2w2 +….. hMwM

wM = wM-1wM-1 = wM-2||w1 = w0


43

Circular Buffers

yx h0

h1

w0

w1

h2

h3

w2

w3

XMAC: Multiply & Accumulate

Σ


44

Z-Transform

x[n], n = -∞, ∞

X(z) = Σn=-∞∞ x[n]z-n

= … +x[-2]z2 +x[-1]z1 +x[0] +x[1]z-1 +x[2]z-2 +…

Z-transform of the impulse response h[n]H(z) = Σn=-∞

∞ h[n]z-n

is called the Transfer function of the LTI system.


45

Basic Properties of Z-transformLinearity:

Za1x1[n] + a2x2[n] a1X1(z) + a2X2(z)

Delay:Z

x[n-D] z-DX(z)

Convolution:Z

y[n] = h[n]*x[n] Y(z) = H(z) X(z)


46

Z-transform Examples

δ[n]Z

1

δ[n-M]Z

z-M

u[n]Z

1/(1- z-1)for |z|> 1

anu[n]Z

1/(1- a z-1)for |z|>|a|


47

ROC: Region of Convergence

ROC: Region of z ∈ C (complex plane) where X(z) ≠ ∞

Examples: (0.5)nu[n] → X(z) = 1/(1-0.5z-1) if |z| > 0.5-(0.5)nu[-n-1] → X(z) = 1/(1-0.5z-1) if |z| < 0.5

In generalanu[n] → X(z) = 1/(1-a z-1) if |z| > |a|-anu[-n-1] → X(z) = 1/(1-a z-1) if |z| < |a|


48

Region of Convergence: Example

X(z) = 1/(1-0.8z-1) + 1/(1-1.25z-1) = (2-2.05z-1)/(1-2.05z-1+ z-2)

Z-plane

0.8 1.25

III

III


49

Causal Sequence & ROC: PolesX(z) = A1/(1-p1z-1) + A2/(1-p2z-1) + ….. Ak/(1-pkz-1) + …

Z-plane

Causal ROC

ROC: z > max|pk|

causal sequence = right-sided sequence = ROC is outside of a circle


50

Anti-Causal Sequence & ROCX(z) = A1/(1-p1z-1) + A2/(1-p2z-1) + ….. Ak/(1-pkz-1) + …

Z-plane

Anti-Causal ROC

ROC: z < min|pk|

Anti-causal sequence = left-sided sequence = ROC is inside of a circle


51

Two-sided Sequence & ROCX(z) = A1/(1-p1z-1) + A2/(1-p2z-1) + ….. Ak/(1-pkz-1) + …

Z-plane

Mixed ROC

mixed sequence two-sided sequence = ROC is inside of a ring


52

Stability of a Sequence and ROCStable Sequence = ROC contains the UNIT CIRCLE

If H is stable , i.e., Σn=-∞∞ |h[n]| < ∞

Then on the unit circle, i.e., |z| =1|H(z)| = |Σn=-∞

∞ h[n]z-n| ≤ Σn=-∞∞ |h[n]z-n| = Σn=-∞

∞ |h[n]|<∞Therefore, its ROC contain the unit circle.

If H(z)’s ROC contains the unit circleThen Σn=-∞

∞ h[n]z-n ||z|=1= Σn=-∞∞ |h[n]| < ∞

Therefore, H is stable.


53

Stable and Causal SequenceStable and Causal: All poles are inside the unit circle

Z-plane

Causal ROCunit circle


54

Stable and Anti-Causal Sequence

Stable and Anti-Causal: All poles are outside the unit circle

Z-plane

Anti-Causal ROC

unit circle


55

Stable and Two-sided SequenceStable and Two-sided:Some poles are outside the unit circle

Some poles are inside the unit circle

Z-plane

Mixed ROC

unit circle


56

Stable System

An LTI system is stable if and only if

H(z)’s ROC contains the unit circle.


57

Inverse Z-transformBy inspection:• First order sequence/system

X(z) = 1/(1-a z-1) if |z| > |a| → x[n] = anu[n] X(z) = 1/(1-a z-1) if |z| < |a| → x[n] = -anu[-n-1]

• Second order sequence/systemX(z) =A1/(1-p1 z-1) + A1

*/(1-p1* z-1) if |z| > |p1|

where A1 = B1ejα1 and p1 = R1ejω1

→ x[n] = 2B1R1ncos(nω1+α1)u[n]

X(z) = [2B1cos(α1)-2B1R1cos(α1-ω1) z-1]/[1-2R1cos(ω1) z-1+R12z-2 ]


58

Inverse Z-transformPartial Fraction Expansion:

X(z) = N(z)/D(z)• N(z) is of degree N & D(z) is of degree M, M<N, Real Roots

= N(z)/(1-p1 z-1)(1-p2 z-1)….. /(1-pM z-1)= A1/(1-p1 z-1)+ A2/(1-p2 z-1)+…. AM/(1-pM z-1)Ak = (1-pk z-1)X(z)|z=pk

• N(z) is of degree N & D(z) is of degree M, M<N, Complex Roots

X(z) = Σk [Ak/(1-pk z-1) + Ak*/(1-pk

* z-1)] • N(z) is of degree N & D(z) is of degree M, M<N, Multiple Roots• N(z) is of degree N & D(z) is of degree M, M≥N, Complex Roots


59

Frequency Response & Z-Transform

The DTFT (Discrete-Time Fourier Transform)

X(ω) = Σn=-∞∞ x[n]e-jωn X(z) = Σn=-∞

∞ x[n]z-n

X(ω) = X(z)|z=ejω

The Inverse DTFT x[n] = (1/2π)∫−π

π X(ω)ejωndω

The Frequency Response of the LTI SystemH(ω) = Σn=-∞

∞ h[n]e-jωn

⇒ Y(ω)= H(ω)X(ω)


60

Parseval Theorem

Σn=-∞∞ |x[n]|2 = (1/2π)∫−π

π |X(ω)|2dω

Digital measurement of energy of an analog signal

Σn=-∞∞ |x[n]|2 = (1/2π)∫−π

π |X(ω)|2dω= (1/2π)(1/T)∫−∞

∞ |X(Ω)|2dΩ=(1/T)∫−∞

∞ |x(t)|2dt

Total energy = (T)Σn=-∞∞ |x[n]|2


61

Frequency Response & Pole/Zero Locations

X(z) = (1-z1z-1)/(1-p1z-1)X(ω) = (ejω -z1)/(ejω -p1)|X(ω)| = |(ejω -z1)|/|(ejω -p1)|

p1 =|p1|ejφ

z1 =|z1 |ejθ

|X(ω)|

ejω

z1

θ φω

p1


62

Transfer Function of an LTI SystemBlock Processing

ImpulseResponse h[n] I/O Convolution

Equation

TransferFunction H(z) Pole/zero

PatternI/O Difference Equation

FrequencyResponse

Specification

FilterDesign Block Diagram

Realization

Sample Processing


63

An Example

Transfer Function: H(z) = (5+2z-1)/(1-0.8z-1)

Impulse Response: H(z) = -2.5 + 7.5/(1-0.8z-1)h[n] = -2.5δ[n] + 7.5(0.8)nu[n]

Difference Equation: H(z) = (5+2z-1)/(1-0.8z-1) = Y(z)/X(z)Y(z) –0.8z-1Y(z) = 5X(z) +2z-1X(z)y[n] = 0.8y[n-1] + 5x[n] + 2x[n-1]

Block Diagram: Y(z)/X(z) = (5+2z-1)/(1-0.8z-1)


64

An Example - Continued

Block Diagram: Y(z)/X(z) = (5+2z-1)/(1-0.8z-1)

5x[n]

0.8

z-1 z-1

+ y[n]

y[n-1]x[n-1]2


65

An Example - ContinuedFrequency Response: H(z) = (5+2z-1)/(1-0.8z-1)

=5(1+0.4z-1)/(1-0.8z-1)H(ω) =5 (1+0.4e-jω)/(1-0.8e-jω)

Using the identity: |1-ae-jω | = [1-2a cos(ω)+a2]1/2

|H(ω)| = 5 [1-0.8 cos(ω)+0.16]1/2/[1-1.6 cos(ω)+0.64]1/2

|H(ω)|

-0.4 0.8

355/3

ωπ


66

Direct & Canonical FormsH(z) = N(z)/D(z)

= (b0+ b1z-1+ b2z-2 +…+ bLz-L)/(1+ a1z-1+ a2z-2 +…+ aMz-M)

Y(z)/X(z) = (b0+ b1z-1+ b2z-2 +…+ bLz-L)/(1+ a1z-1+ a2z-2 +…+ aMz-M)

Y(z)= X(z)(b0+ b1z-1+ b2z-2 +…+ bLz-L)/(1+ a1z-1+ a2z-2 +…+ aMz-M)

= X(z)(b0+ b1z-1+ b2z-2 +…+ bLz-L)[1/(1+ a1z-1+ a2z-2 +…+ aMz-M)] → Direct Form

= X(z)/(1+ a1z-1+ a2z-2 +…+ aMz-M)[(b0+ b1z-1+ b2z-2 +…+ bLz-L)] → Canonical Form


67

Direct FormY(z)= X(z)(b0+ b1z-1+ b2z-2 +…+ bLz-L) [1/(1+ a1z-1+ a2z-2 +…+ aMz-M)]

b0

b1

b2

bL

y[n]

y[n-1]x[n-1]

x[n]

-a1

-a2

-aM

y[n-2]x[n-2]

y[n-M]x[n-L]

z-1 z-1

+

z-1 z-1

z-1 z-1


68

Canonical FormY(z)= X(z)/(1+ a1z-1+ a2z-2 +…+ aMz-M) [(b0+ b1z-1+ b2z-2 +…+ bLz-L)]

b0

-a1 b1

b2-a2

-aM

w[n-1]

w[n]x[n] + + y[n]

z-1

z-1

bL

z-1


69

ExamplesExample 6.2.1: Impulse response: h = 1, 6, 11, 6Difference equation: y[n] = x[n] + 6x[n-1] +11x[n-2] +6x[n-3]Transfer Function: H(z) = 1+ 6z-1 +11z-2 + 6z-3

Pole/Zeros: H(z) = (1+z-1)(1+2z-1)(1+3z-1)Frequency Response, Block Diagrams

Example 6.2.3:(a) Difference equation: y[n] = 0.25 y[n-2] + x[n](b) Difference equation: y[n] = -0.25 y[n-2] + x[n]


70

Sinusoidal Steady State Response

x[n] = ejωon → DTFT: X(ω) = 2π δ(ω−ωο) + replicas→ Y(ω) = H(ω)X(ω) = H(ωο)2π δ(ω−ωο) + replicas→ y[n] = H(ωο) ejωon

i.e., ejωon → H→ H(ωο) ejωon = |H(ωο)| ej(ωon+Arg[H(ωo)])

cos(nωο ) → H→ |H(ωο)| cos(nωο + Arg[H(ωo)] )sin(nωο ) → H→ |H(ωο)| sin(nωο + Arg[H(ωo)] )

Linearity Aejω1n +Bejω2n → H→ A H(ω1) ejω1n +B H(ω2) ejω2n


71

Sinusoidal Transient Response

Input: x[n] = ejωon u[n] → : X(z) = 1/(1- ejωo z-1)

LTI system: H(z) = N(z)/D(z) (assuming real poles)= N(z)/(1-p1 z-1)(1-p2 z-1)…..(1-pM z-1)

all |pk| < 1, k = 1, 2, …., M

Output: Y(z) = X(z)H(z) = N(z)/(1- ejωo z-1)(1-p1 z-1)(1-p2 z-1)…..(1-pM z-1)

= H(ωο) /(1- ejωo z-1) + A1/(1-p1z-1) + A2/(1-p2z-1) + ….. AM/(1-pMz-1)

→ y[n] = H(ωο) ejωon + A1p1n + A2p2

n + ….. AMpMn u[n]

as n → ∞ y[n] → H(ωο) ejωon


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Sinusoidal Transient Response

Example: H(z) = (5+2z-1)/(1-0.8z-1)Y(z) = (5+2z-1)/(1- ejωo z-1)(1-0.8z-1)

Observation:1. Stability of the Filter2. Effective Time Constant: neff

ρ = maxk|pk|, ρ neff = ε neff = ln(ε )/ln(ρ)

Example 6.3.2


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Unit Step ResponseInput: x[n] = ejωon u[n], ωο = 0 → : X(z) = 1/(1- z-1)

LTI system: H(z) = N(z)/D(z) (assuming real poles)= N(z)/(1-p1 z-1)(1-p2 z-1)….. /(1-pM z-1)

all |pk| < 1, k = 1, 2, …., M

Output: Y(z) = X(z)H(z) = N(z)/(1-z-1)(1-p1 z-1)(1-p2 z-1)…..(1-pM z-1)

= H(1) /(1- z-1) + A1/(1-p1z-1) + A2/(1-p2z-1) + ….. AM/(1-pMz-1)

→ y[n] = H(1) + A1p1n + A2p2

n + ….. AMpMn u[n]

as n → ∞ y[n] → H(1)u[n]→ H(1) = Σn=0

∞ h[n] is the DC gain


74

Pole/Zero Designs

First Order FilterH(z) = G(1+bz-1)/(1-az-1)H(0), H(π), and speed of response neff

H(0) =G(1+b)/(1-a), H(π) = G(1-b)/(1+a)H(π) / H(0) = (1-b)(1-a)/(1+b)(1+a)a = ε 1/neff

ExampleH(z) = (5+2z-1)/(1-0.8z-1)

= 5(1+0.4z-1)/(1-0.8z-1)


75

Second-Order Filter & Resonator

Rejωo

ωo

Re-jωo

Z-Plane

Unit circle

H(z) = G/(1- Rejωo z-1)(1- Re-jωoz-1)= G/(1-2Rcos(ωo)z-1 + R2 z-2)

H(ω) = G/(1- Rejωo e-jω)(1- Re-jωoe-jω)

G: normalize H(ωo) = 1


76

Second-Order Filter & Resonator

|H(ω)|2 = G/(1- 2Rcos(ω−ωo) + R2)(1- 2Rcos(ω+ωo) + R2)

|H(ω)|2

∆ω ≈ 2(1-R)

∆ω (3 dB bandwidth)1

1/2ω

ωo π


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Second-Order Filter & ResonatorTransfer Function: H(z) = G/(1- 2Rcos(ωo) z-1 + R2 z-2)

Impulse Response: h[n] = [G/sin(ωo)] Rnsin(nωo)

Difference Equation: y[n] = 2Rcos(ωo)y[n-1] - R2y[n-2] Gx[n]G

-a1

-a2

y[n]x[n]

z-1

+

y[n-1]

z-1a1 = -2Rcos(ωo)

a2 = R2

y[n-2]


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Second-Order Filter & ResonatorExample: Design a 2-pole resonator filter with peak frequency fo = 500 Hz, and 3-dB bandwidth ∆f = 32 Hz. The sampling frequency is fs = 10KHz.

R = 0.99, G = 0.0062, a1= -1.8831 and a2= 0.9801

H(z) = 0.0062/(1 - 1.8831 z-1 + 0.9801 z-2)


79

Parametric Equalizer

Rejωo

ωo

Re-jωoZ-Plane

Unit circle

p1=R ejωo p1*=R e-jωo

z1=r ejωo z1*=r e-jωo

H(z) = (1- rejωo z-1)(1- re-jωoz-1) /(1- Rejωo z-1)(1- Re-jωoz-1)= (1 + b1z-1 + b2z-2 )/(1 + a1z-1 + a2z-2 )

a1 = -2Rcos(ωo), a2 = R2

b1 = -2rcos(ωo), b2 = r2

H(z) = (1 + b1z-1 + b2z-2 )/(1 + (R/r)b1z-1 + (R/r)2b2z-2 )


80

Parametric EqualizerH(z) = (1 + b1z-1 + b2z-2 )/(1 + (R/r)b1z-1 + (R/r)2b2z-2 )

b1 = -2rcos(ωo), b2 = r2

|H(ω)|2 = (1- 2rcos(ω−ωo) + r2)(1- 2rcos(ω+ωo) + r2)/(1- 2Rcos(ω−ωo) + R2)(1- 2Rcos(ω+ωo) + R2)

|H(ω)|2

ωπωo


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Notch Filter

Rejωo

ωo

Re-jωoZ-Plane

Unit circleNotch Filter:

Zeros on the unit circler = 1

H(z) = (1 + b1z-1 + z-2 )/(1 + (R)b1z-1 + R2z-2 )b1 = -2cos(ωo),


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Notch Filter

Notch polynomial N(z) = Πk(1-ejωk z-1)Real Coefficient N(z) = Πk(1-2cosωk z-1 +z-2)

Notch Filter = N(z)/N(ρ -1 z)= (1 + b1z-1 + b2z-2+… + b2z-(M-1) + b1z-M )/(1 + ρb1z-1 + ρ2b2z-2+… + ρ-(M-1)b2z-(M-1) + ρMb1z-M )

Examples 6.4.3, 6.4.4


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Comb Filter

From notch filter moves the zeros inside the poles

H(z) = (1 + rb1z-1 + r2b2z-2+… + r-(M-1) b2z-(M-1) + rMb1z-M )/(1 + ρb1z-1 + ρ2b2z-2+… + ρ-(M-1)b2z-(M-1) + ρMb1z-M )

r ≈ ρ and r < ρ


84

DTFT, DFT, and FFT• Fourier Transform Xa(Ω) = ∫ xc(t)e-jΩt dt

• Discrete Time Fourier Transform X(ω) = Σn x[n] e-jnω

x[n] = xc(nT) → X(Ω T) =(1/T)Xa(Ω)

• Discrete Fourier Transform X [k] = Σn=0(N-1)x[n] e-jnk2π/Ν

If only N points → X[k] = X(ω)|ω=k2π/N

• Fast Fourier Transform


85

Digital Spectral Analysis

Frequency ResolutionMainlobeSidelobeWindowing

Physical Resolution

Computational Resolution


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The Fast Fourier Transform

Decimation in Time (DIT) AlgorithmDIT Butterfly

Decimation in Frequency (DIF) AlgorithmDIF Butterfly

N log(N) Algorithm

Bits Reversal Indexing

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