Post on 30-Jan-2021
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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
-------------