Typical Real Time DSP System

8/11/2019 Typical Real Time DSP System

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Digital Signal Processing

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A Typical DSP System

Sampling&

A/D

AnalogPre-filter

HPRE(f)

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

Analog

inDSPFilter

HDSP(z)

Analog

out

AnalogPost-filter

HPOST(f)

D/AConverter

HDAC(f)

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


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Analog Signal Processing

xa(t) : an analog signal

Xa() =xa(t)e-jtdt is its Fourier transform

The inverse Fourier transform is

xa(t) = (1/2)Xa() ejtd


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Analog Signal ProcessingAnalog

FilterHa()

ya(t)xa(t)

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

ya(t) =ha(t-) xa() d

OR, equivalently, in frequency domain

Ya() = Ha() Xa()


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A Typical DSP SystemA Typical DSP System

AnalogPre-filter

HPRE(f)

Sampling&

A/D

DSPFilter

HDSP(z)

D/AConverter

HDAC(f)

AnalogPost-filter

HPOST(f)

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

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


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A digital Spectral Analyzer

AnalogPre-filter

HPRE(f)

Sampling&

A/D

FFT

Processor

Displaythe

Spectrum

xa(t) x(t)


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Linear Time-Invariant SystemsThe delta function: (t)

Linearity

Time-invariant

The Impulse Response

The Convolution Integral

The Convolution Theorem


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Sinusoidal Response of LTI System

ejt Ha() ejtHa()

Ha() = |Ha()| e jARG{Ha()}

Ha(1) A1ej1t+ Ha(2) A2ej2t

A1ej1t

+A2ej2t

Ha()


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Linear FilteringXa()

=2f

Ya()

=2f


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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


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Discrete-Time Fourier Tranform (DTFT)

X() = nx[n] e-jn

X() is a continuous function of

X() is periodic on with the period 2

-4 -2 0 2 4

X()


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Fourier Transform and The DTFTIf Xa() =xa(t)e-jtdt and x[n] = xa(nT)

Then X( ) = (1/T) kXa( - k2/T)

-4fs-2f

s0 2f

s4f

s

1/T

0

Xa()

X()


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Aliasing

-4fs -2fs 0 2fs 4fs

X()

0-4fs -2fs 2fs 4fs


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AliasingCos( t) = (ejt+ e-jt)

-4fs -2fs 0 2fs 4fs

-4fs -2fs

0 2fs

4fs


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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


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Anti-Aliasing Filter

A(f)

ffpass fs/2 fstop

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

dB per Octave or dB perDecade

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

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DAC & Analog Reconstruction

D/AConverter

HDAC(f)

y[n] ya(t)

ya(t) = ny[n] hDAC(t-nT)

Ya() = Y() |=HDAC()

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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/2

0, otherwiseh(t) = u(t) - u(t-T) = {

H(f) = (T/j) [Sin(fT)/(fT)]t

f

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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

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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/2

f-fs/2

Removing high frequency imagesEqualizing D/A converters in-band distortion

(may be done digitally)

-20 log[H(fs/2)/H(0)] ~ 3.9 dB

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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(t/T)/(t/T)

tT 2T

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A/D Converter & Quantization

R:fullsc

ale

R/2

3Q

2Q

Q

0-Q

-2Q

-3Q-R/2

R/Q = 2BQ: step size

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Quantization Noise PowerP(e)

1/Q

e-Q/2 Q/2

E(e2) = (1/Q)e2 de = Q2/12

or

erms = Q/121/2

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Signal to Noise Ratio: SNR

Noise Power = Q2

/12Signal Power = ?

Assuming Sine Wave -> Signal Power = A2/2

Assuming Full Scale -> A = R/2 -> Signal Power = R2/8

Assuming 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

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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 Processing

Block Processing

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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 x

DO y = 2x + 3w1 + 4w2w2 = w1w1 = x

CONTINUE

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

y[0] 2 0 0 0 0

y[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

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Overlap and Add

0, 1, 2, 3, 4, 5, 6,

7, 8, 9, 10, 11, 12, 13,

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

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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]

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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 x

DO y = 0.5w + 2x + 3vw = y

v = x

CONTINUE

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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]

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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]

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Impulse Response of an LTI System

[n] = 1, for n=0, and 0, otherwiseh[n]

[n]

H nn

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

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

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Convolutionh[n][n]H n

nh[n-1]

[n-1]

Hnn

h[n-2][n-2]

H nn

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

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

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FIR and IIR FiltersAn FIR filterhas 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 thefilter order, and the length of the filter impulse response is M+1.

AnIIR 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.

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I/O Difference EquationImpulse Response: h[n]

FIR filter: convolution sum

IIR Filter: difference equation

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

h[n] = anu[n]

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Causality and StabilityA causal sequence is a right-sided sequence, and an anti-causal

sequence 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.,

n |h[n]| <

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Linear Convolution: Flip & Slide

y[n] = mx[m]h[n-m]If h is of order M, i.e., h = {h[0], h[1], , h[M]} is of length M+1

and x = {x[0], x[1], ,x[L-1]} is of length L, then

y[n] = m=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)

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Linear Convolution: Flip & Slide

h3

h2

h1

h0

h3

h2

h1

h0

h3

h2

h1

h0

0 0 0 x0 x1x2x3 xn-3xn-2xn-1xn xL-1 0 0 0

y0 yn yL-1+M

h = M+1

x = L

y = h*x = L M

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Transient & Steady State

y[n] = m=0n

h[m]x[n-m] if 0 n < M

y[n] = m=0M

h[m]x[n-m] if M n L-1

y[n] = m=n-L+1M

h[m]x[n-m] if L n < L+M

output y[n]

n

0 M L-1 L-1+M

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Convolution of Infinite Sequencey[n] = m=max(0,nL+1)

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

Infinite filter, finite input: M = , L <

y[n] = m=max(0,nL+1)n

h[m]x[n-m]

Finite filter, infinite input: M < , L =

y[n] = m=0min(n,M)

h[m]x[n-m]

Infinite filter, infinite input: M = , L =

y[n] = m=0n

h[m]x[n-m]

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Fi i fil I fi i i


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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

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Finite filter, Infinite inputSample Processing

y[n] = m=max(0,nL+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 DO

w0 = x

y = h0w0 + h1w1 + h2w2 +.. hMwM

wM= wM-1wM-1 = wM-2|

|

w1 = w0

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Circular Buffersyx h0

h1

w0

w1

h2

h3

w2

w3

XMAC:Multiply & Accumulate

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Z-Transform{x[n], n = -, }

X(z) = n=-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) =n=-

h[n]z

-n

is called the Transfer function of the LTI system.

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Basic Properties of Z-transformLinearity:

Z

a1x1[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)

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Z-transform Examples

[n]

Z

1

[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|

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ROC: Region of ConvergenceROC: 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 general

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

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

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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

I

IIIII

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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

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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

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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

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

If H is stable , i.e., n=-

|h[n]| <

Then on the unit circle, i.e., |z| =1

|H(z)| = |n=-h[n]z-n| n=-|h[n]z-n| =n=-|h[n]|


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Stable and Causal Sequence

Stable and Causal: All poles are inside the unit circle

Z-plane

Causal ROC unit circle

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Stable and Anti-Causal Sequence

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

Z-plane

Anti-Causal ROC

unit circle

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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

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Stable System

An LTI system is stable if and only if

H(z)s ROC contains the unit circle.

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Inverse Z-transform

By 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/system

X(z) =A1/(1-p1z-1) + A1

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

where A1

= B1

ej1 and p1

= R1

ej1

x[n] = 2B1R1ncos(n1+1)u[n]

X(z) = [2B1cos(

1)-2B

1R

1cos(

1-

1) z-1]

/[1-2R

1cos(

1) z-1+R

1

2z-2]

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Inverse Z-transformPartial Fraction Expansion:X(z) = N(z)/D(z)

N(z) is of degree N & D(z) is of degree M, M


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Frequency Response & Z Transform

The DTFT (Discrete-Time Fourier Transform)

X() = n=-x[n]e-jn X(z) =n=-x[n]z-n

X() = X(z)|z=ej

The Inverse DTFTx[n] = (1/2)

X()ejnd

The Frequency Response of the LTI SystemH() =n=-

h[n]e-jn

Y()= H()X()

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Parseval Theoremn=-

|x[n]|2 = (1/2)

|X()|2d

Digital measurement of energy of an analog signal

n=-|x[n]|2 = (1/2)

|X()|2d= (1/2)(1/T)

|X()|2d

=(1/T)

|x(t)|2dt

Total energy = (T)n=-|x[n]|2

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Frequency Response & Pole/Zero Locations


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Frequency Response & Pole/Zero Locations

X(z) = (1-z1z-1)/(1-p1z

-1)

X() = (ej-z1)/(ej-p1)|X()| = |(ej-z1)|/|(ej-p1)|

p1 =|p1|ej

z1 =|z1 |ej

|X()|

ej

z1

p1

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Transfer Function of an LTI System


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Transfer Function of an LTI System

Block ProcessingImpulse

Response h[n] I/O Convolution

Equation

Transfer

Function H(z) Pole/zeroPatternI/O Difference

Equation

Frequency

Response

Specification

Filter

Design Block DiagramRealization

Sample Processing

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An Example


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p

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] = -2.5[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)

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An Example - Continued


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a p e Co ued

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

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An Example - Continued


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p

Frequency Response: H(z) = (5+2z-1)/(1-0.8z-1)

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

H() =5 (1+0.4e-j)/(1-0.8e-j)

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

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

|H()|

-0.4 0.8

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Direct & Canonical Forms


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Direct & Canonical Forms

H(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

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Direct Form


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Y(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

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Canonical Form


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Y(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

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ExamplesExample 6.2.1: Impulse response: h = {1, 6, 11, 6}

Difference 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]

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Sinusoidal Steady State Response


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y p

x[n] = ejon DTFT: X() = 2 () + replicas

Y() = H()X() = H()2 () + replicasy[n] = H() ejon

i.e., ejon H H() ejon = |H()| ej(on+Arg[H(o)])

cos(n)H|H()| cos(n+ Arg[H(o)] )sin(n) H|H()| sin(n+ Arg[H(o)] )

Linearity

Aej1n +Bej2n H A H(1) ej1n +B H(2) ej2n

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Sinusoidal Transient Response


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Input: x[n] = ejon u[n] : X(z) = 1/(1- ejoz-1)

LTI system: H(z) = N(z)/D(z) (assuming real poles)= N(z)/(1-p1z

-1)(1-p2z-1)..(1-pMz

-1)

all |pk| < 1, k = 1, 2, ., M

Output: Y(z) = X(z)H(z) = N(z)/(1- ejoz-1)(1-p1z-1)(1-p2z

-1)..(1-pMz-1)

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

y[n] = {H() ejon + A1p1n + A2p2n + .. AMpMn}u[n]

as n y[n] H() ejon

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

Y(z) = (5+2z-1

)/(1- ej

oz-1

)(1-0.8z-1

)

Observation:

1. Stability of the Filter

2. Effective Time Constant: neff= maxk|pk|, neff = neff= ln()/ln()

Example 6.3.2

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

LTI system: H(z) = N(z)/D(z) (assuming real poles)= N(z)/(1-p1z-1)(1-p2z

-1).. /(1-pMz-1)

all |pk| < 1, k = 1, 2, ., M

Output: Y(z) = X(z)H(z) = N(z)/(1-z-1)(1-p1z-1)(1-p2z

-1)..(1-pMz-1)

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

y[n] = {H(1) + A1p1n + A2p2n + .. AMpMn}u[n]as n y[n] H(1)u[n]

H(1) = n=0

h[n] is the DC gain

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g

First Order Filter

H(z) = G(1+bz-1)/(1-az-1)

H(0), H(), and speed of response neff

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

H() / H(0) = (1-b)(1-a)/(1+b)(1+a)a = 1/neff

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

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

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Second-Order Filter & Resonator


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Rejo

o

Re-j

o

Z-Plane

Unit circle

H(z) = G/(1- Rejoz-1)(1- Re-joz-1)

= G/(1-2Rcos(o)z-1 + R2z-2)

H() = G/(1- Rejo e-j)(1- Re-joe-j)

G: normalize H(o) = 1

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|H()|2 = G/(1- 2Rcos(o) + R2)(1- 2Rcos(+o) + R2)

|H()|2

2(1-R)

(3 dB bandwidth)1

1/2

o

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Transfer Function: H(z) = G/(1- 2Rcos(o) z-1 + R2z-2)

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

o)

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

G

-a1

-a2

y[n]x[n]

z-1+

y[n-1]z-1

a1 = -2Rcos(o)

a2 = R2

y[n-2]

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Second-Order Filter & ResonatorExample:

Design a 2-pole resonator filter with peak frequencyfo = 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)

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Parametric Equalizer


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Rejo

o

Re-joZ-Plane

Unit circle

p1=R ejo p1

*=R e-jo

z1=r ejo z1

*=r e-jo

H(z) = (1- rejoz-1)(1- re-joz-1) /(1- Rejoz-1)(1- Re-joz-1)

= (1 + b1z-1 + b2z

-2)/(1 + a1z-1 + a2z

-2)

a1 = -2Rcos(o), a2 = R2b1 = -2rcos(o), b2 = r2

H(z) = (1 + b1z-1 + b2z

-2)/(1 + (R/r)b1z-1 + (R/r)2b2z

-2)

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H(z) = (1 + b1z-1 + b2z

-2)/(1 + (R/r)b1z-1 + (R/r)2b2z

-2)

b1 = -2rcos(o), b2 = r2

|H()|2 = (1- 2rcos(o) + r2)(1- 2rcos(+o) + r2)/(1- 2Rcos(o) + R2)(1- 2Rcos(+o) + R2)

|H()|2

o

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Notch FilterRejo

o

Re-joZ-Plane

Unit circle

Notch Filter:

Zeros on the unit circle

r = 1

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

-1 + R2z-2)

b1 = -2cos(o),

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Notch FilterNotch polynomial N(z) = k(1-ejkz-1)

Real Coefficient N(z) =k(1-2coskz-1 +z-2)

Notch Filter = N(z)/N( -1z)

= (1 + b1z-1 + b2z-2+ + b2z-(M-1) + b1z-M)/(1 +b1z-1 +2b2z-2+ +-(M-1)b2z-(M-1) +Mb1z-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 +b1z-1 +2b2z-2+ +-(M-1)b2z-(M-1) +Mb1z-M)

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DTFT, DFT, and FFTFourier Transform Xa() =xc(t)e-jtdt

Discrete Time Fourier Transform X() = nx[n] e-jn

x[n] = xc

(nT) X( T) =(1/T)Xa

()

Discrete Fourier Transform X [k] = n=0(N-1)

x[n] e-jnk2/

If only N points X[k] = X()|=k2/N

Fast Fourier Transform

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Digital Spectral Analysis


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Frequency Resolution

Mainlobe

Sidelobe

Windowing

Physical Resolution

Computational Resolution

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The Fast Fourier Transform


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Decimation in Time (DIT) Algorithm

DIT Butterfly

Decimation in Frequency (DIF) Algorithm

DIF Butterfly

N log(N) Algorithm

Bits Reversal Indexing

Typical Real Time DSP System

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