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Lecture 2: Wednesday Jan 11, 2017

Lecture:

• Determinstic DSP (Chapter 2)

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Fourier and Inverse Fourier Transforms

X( ej) =∞Xk= –∞xk e

–jk

xk = X( ej)ejk d

0 k

For a discrete-time signal xk:

1 2

x0

x1x2

12------

–

10

Example

X(ej) =∞Xk= 0

0.8ke–jk

=∞Xk= 0

(0.8e–j)k

= .

0 5 10

0.8

1

0.64xk = 0.8kuk

1

1 0.8e j––---------------------------

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Filtering

Time domain: yk =∞Xi= –∞xihk – i = xk hk

Freq domain: Y(ej) = X(ej)H(ej)

Terminology:

hk = impulse response H(ej) = frequency response|H(ej)| = magnitude response.

hkxk yk

FILTER

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Pop Quiz:What is the Fourier transform X(ej) of the unit step, xk = uk?

X(ej) = ?

0 5 10

1 ...

k

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Z TransformZ transform:

H(z) =∞Xk= –∞hkz

–k ,

When z = e j, H(e j ) reduces to the DTFT.

ROC = values of z for which above sum is finite.

If DTFT exists, ROC includes u. circle.

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Example

X(z) = 6 + 6z–1 + 3z–3

Pop Quiz: What is the ROC?

0

6

xk = (6, 6, 0, 3, ...):

6

0

3

1 3

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Example

X(z) =∞Xk= 0

akz–k =∞Xk= 0

(az–1)k =

0 5 10

a

1

a2

xk = akuk

a3

1

1 az 1––--------------------

when |a/z|< 1

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NotationCausal ⇒ hk = 0 for k < 0

Anticausal ⇒ hk = 0 for k > 0

Monic ⇒ {h0 = 1}AND{causal OR anticausal}

Right-sided ⇒ hk = 0 for k < some K.

Left-sided ⇒ hk = 0 for k > some K.

FIR ⇒ both right- and left-sided.

2-sided ⇒ neither right- nor left-sided.

0

0

0

1

K 0

K0

... ...

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ROCFIR ⇒ ROC = entire plane, except possibly z = 0 and/or |z| = Right-sided ⇒ ROC = | z | > r (beyond a circle)Left-sided ⇒ ROC = | z | < r (inside a circle)Two-sided ⇒ ROC = r1 < | z | < r2 (a ring)

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Stable ROCA filter is bounded-input bounded-output (BIBO) stable if

∞Xk= –∞|hk|

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Poles and ZerosA rational transfer function

H(z) = A

⇒{b1, ..., bM}arezeros ⇒{a1, ..., aN} arepoles

• # poles = # zeros [including z = 0 and |z| = ]• ROC is bounded by poles• Causal ⇒ ROC = | z | > | outermost pole|• Causal and stable ⇒ all poles must be inside the unit circle. • Anticausal and stable ⇒all poles must be outside the unit circle.

• N = 0 ⇒FIR; N > 0 ⇒ IIR

k 1=M

z bk–

k 1=N z ak–

---------------------------------

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Examples

H(z) = 1 – 0.5z–1 zero at 1/2, pole at origin.

H(z) = 1 - 0.5z zero at 2, pole at ∞

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Examples

Stable and 2-sidedStable and left-sidedStable and right-sided

1

Imz

Rez

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The Matched FilterA filter that is “matched” to hk has impulse response h–k* :

k

k

hk

h–k*

H(ej) H(z)

H*(ej). Same mag response. H*(1/z*)

Example:

The corresponding matched filter:

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What happens to poles and zeros?

b is a zero of H( z ) ⇒ H( b ) = 0 ⇒ H*(1/z*) = 0 when z = 1/b*⇒ 1/b* is a zero of H*(1/z*)

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Compare H to its MatchH(z) H*(1/z*)

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ExampleH( z ) = has a zero at 1/a*, and a pole at z = a:

What is magnitude response?

|H(e j)| = = = 1, for all

z 1– a*–1 az 1––--------------------

Stable and right-sided

a

1/a*

e j– a*–1 ae j––----------------------- e j– 1 a*e

j–1 ae j––-----------------------

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All Pass FiltersAn allpass A(z) satisfies

|A(e j )| = 1

A(e j)A*(e j) = 1

ak a*– k = k

A(z)A*(1/z*) = 1 .

any zero [or pole] must be accompanied by a matching pole [or zero] at the conjugate-reciprocal location.

Every rational allpass transfer function has form:

A(z) = e j , B(z) = 1 + b1z + + bNzN B z

zN M– B* 1/z* -------------------------------------------

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An All Pass

d

1/d*

1/c*

cRe{ z}

Im{ z}

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Minimum Phase

H(z) = A

[Strictly] minimum-phase |bk|, |ak| < 1.

[Strictly] maximum-phase |bk|, |ak| > 1.

Loosely minimum-phase |bk| 1, |ak| < 1.

Loosely maximum-phase |bk| 1, |ak| > 1.

A minimum-phase transfer function is both causal and stable.

Why?

Min ph all poles/zeros inside none at z = H(z) = khkz–k cannotcontain any positive powers of z hk is causal hk is right-sided ROC isoutside the outermost pole. But all poles inside ROC must include the unitcircle stability.

k 1=M

z bk–

k 1=N z ak–

---------------------------------

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Nonzero at time zeroA min-phase sequence must be nonzero at time zero (prevents zero at z = )

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Properties of Min Phase• Causal• Stable• Nonzero at time zero (prevents zero at | z | = )• Of all transfer functions / sequences with a given magnitude response:

• Only the minimum-phase transfer function can be inverted by a causaland stable filter

• Only the minimum-phase sequence has its energy maximallyconcentrated in the vicinity of time zero.

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ExampleFind a min-phase sequence hk with samemagn response as xk:

X( z ) = 2z–1 + 3z–2

= 2 ·

• zeros at –3/2 and • poles at 0 and 0 A( z ) = · z2

H( z ) = 2 · = 3 + 2z–1

1 20k

xk

23

z32---+

z2------------

z 1–32---+

z32---+

-----------------

1 20 k

hk

23

z 1–32---+

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ExampleFind a min-phase sequence hk with samemagn response as xk:

X( z ) = 8(z + 3/2)(z + 3/4)

• zeros at –3/2 and –3/4• poles at and

A( z ) = · z–1

H( z ) = 8 · (z–1 + 3/2)(z + 3/4)· z–1

= 12 + 17z–1 + 6z–2

–1 0–2 k

xk

98

18

z 1– 3/2+z 3/2+

-----------------------

k–10 –2

hk

8

3

10

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Two Numbers

A{x, y} =

G{x, y} = = exp(A{ logx, logy})

H{x, y} = = = =

Q1. Drive 800 miles from ATL to NYC. 80 mph there, 40 mph back.

Average speed? mph = totalmiles/totalhrs = 2(800)/(10 + 20) = 53.3 mph

Q2: QCOM gains 20%, 60%, and 10% in 3 consecutive years.

What is the average rate of return?

(X)(1.2)(1.6)(1.1) = X(1 + r)3

⇒ 1 + r = G{1.2, 1.6, 1.1} = 1.283⇒ 28.3%

x y+2

------------

xy

21x--- 1

y---+

------------- 1

A 1x--- 1y---

--------------------- 2xyx y+------------ G

2

A------

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OrderingLet S = {x, y}.

min{S} H{S} G{S} A{S} max{S}

Example: S = {5, 45}

min{S} H{S} G{S} A{S} max{S}

0

minH

G

A

max

5 9 15 25 45

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Valid PSDA transfer function S(z) is a “valid PSD” when S(e j ) 0 for all

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Arithmetic Mean

The arithmetic mean of S(ej) is

A{S} = lim iSi =

-

1N---- 1

2------ S ej d

–

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Geometric MeanThe geometric mean is

G{S} = limi = exp .

Unproven Fact:

G{|1 – cej|2} = 1 for any |c| 1.

Example:

S(ej ) = 2 – 2cos() ⇒A{S} = 2 and G{S} = 1.

Remark: G{S} is nonzero despite the zero at = 0.

Si1 N/ 1

2------ logS ej d

–

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Harmonic MeanThe harmonic mean is

H{S} = lim1/i1/Si = .

Interpret: H = 1/g0, where G( z ) = 1/S( z )

1

12------ 1

S ej ---------------- d

–

---------------------------------------

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Inequalitiesmin{S} H{S} G{S} A{S} max{S}

All inequalities collapse to equality if and only if S is a constant

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ExampleS(z) =

A{S} = 4, G{S} = 3, and H{S} = 12/5 = 2.4.

3

1 0.5z 1–– 1 0.5z– -------------------------------------------------------

A{S} = 4

G{S} = 3

H{S} = 2.4

12

0 –

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Properties1. min{S} H{S} G{S} A{S} max{S} ,

with equalities if and only if S(e j ) is constant

2. A{aS bP } = aA{S} + bA{S}

3. G = .

4. G{|1 – ce–j|2} = 1 for any |c| 1.

SPQ

------- G S G P

G Q ----------------------------

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What Exactly Limits Data Rate? 3 Key Parameters

r( t )s( t )n( t )

H( f )

bandwidth W [Hz]

signal power P [W] one-sided

The 3 combine to give the signal-to-noise ratio: SNR = P

N0W------------

noise density N0 [W/Hz]

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Solution: Use QAM

• Bandwidth determines symbol rate, 1/T = W

• SNR determines size of QAM alphabet, M = 1 +

⇒Rb = W log2(1 + )

Compare to Shannon capacity:

⇒C = W log2(1 + SNR)

DACa s( t )UP

SNR

-------------

SNR

-------------