Transcript Lecture 5: Test 1 review

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Lecture 5: Test 1 review
Instructor:
Dr. Gleb V. Tcheslavski
Contact: [email protected]
Office Hours: Room 2030
Class web site:
http://ee.lamar.edu/gleb/dsp/ind
ex.htm
ELEN 5346/4304
DSP and Filter Design
Fall 2008
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Fundamental system’s properties
(5.2.1)
1. Linearity:
x1n  y1n andx2n  y2n  ax1n  bx2n  ay1n  by2n x1n ,x2n ,a,b
2. Time-invariance:
xn  yn  xnn0  ynn0 xn n0
(5.2.2)
3. Causality:
xn  yn  yn  fct ( xk  k  n) -nooutputbeforeinput
(5.2.3)
4. Stability (BIBO):
 xn  Bx    yn  By  for every bounded input,
(5.2.4)
the output must be bounded
1,2,3 Must hold for ANY inputs and constants!
Most of properties are specified for the relaxed systems (empty states)!
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Fall 2008
3
Difference equations
LCCDELTI system
states
xn
yn
memory
No driving terms – homogeneous solution.
LTI
Whatever is stored (represents the memory of the system)  operations on the
previous inputs  initial conditions  reflected in the homogeneous solution.
Usually:
ytot ,n  yh,n  y p,n ;n  n0
(5.3.1)
Remember and understand SFGs of an LTI system.
ytot ,n  yzi,n  yzs,n
zero-input
(5.3.2)
zero-state
ytot ,n  ytr ,n  yss,n
transient
(5.3.3)
steady-state
The connection between (5.2.1), (5.2.2), and (5.2.3) is the same (in general) total response
ELEN 5346/4304
DSP and Filter Design
Fall 2008
4
Difference equations
Example: find the total solution for the system specified by the following DE:
yn  yn1  6 yn2  xn
For the following input and the initial conditions:
xn  8un
y1  1; y2  1
Let’s find the homogeneous solution first: xn = 0. The characteristic eqn.:
 n   n1  6 n2  0 n2 ( 2    6)   n2 (  3)(  2)  0
Characteristic roots are 1 = -3; 2 = 2. Therefore, the homogeneous solution is:
yh ,n  C1  3  C2  2 
n
For the particular solution, we assume:
Therefore, combining with DE for large n:
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Fall 2008
n
y p,n  k
k  k  6k  8un
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Difference equations
Which gives k = -2.
Then, the total solution is:
yn  C1  3  C2  2   2;n  0
n
n
We find the constants by satisfying the initial conditions:
2
2

y

C

3

C
2
 2  1




1
2
 2
C1  1.8;C2  4.8

1
1

 y1  C1  3  C2  2   2  1
Finally, the total solution is:
yn  1.8  3  4.8  2   2;n  0
n
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DSP and Filter Design
n
Fall 2008
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Difference equations
Alternatively, the zero-input response can be found as a form of the homogeneous
solution, where the constants are chosen to satisfy the initial conditions.
y0   y1  6 y2  1  6  7
y1   y0  6 y1  7  6  13
From the equation for the homogeneous solution, we get:
y0  C1  C2 ; y1  3C1  2C2
Therefore: C1 = -5.4; C2 = -1.6
The zero-input response is
yzi ,n  5.4  3  1.6  2  n  0
n
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Fall 2008
n
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Difference equations
The zero-state response can be found as a form of the total solution when zero
initial conditions are satisfied.
y0  x0  8; y1  x1  y0  0
We find the constants for the total solution as C1 = 3.6 and C2 = 6.4. Therefore,
the zero-state response for n  0 and with i.c.s yzs,-2 = yzs,-1 = 0 is:
yzs ,n  3.6  3  6.4  2   
n
n
Again, the total solution is:
yn  1.8  3  4.8  2   2;n  0
n
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DSP and Filter Design
n
Fall 2008
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LTI systems
If thesystem ' sinput :xn   xk nk 
(5.4.1)
k
If the system is linear:
thesystem ' soutput : yn   xk ( nk )  {and TI } xk hnk
k
since
(5.4.2)
k
n  hn ;nk  hn,k (ingeneral ).ForTI :nk  hnk
(5.4.3)
Therefore:
yn   xk hnk   hk xnk  xn  hn  hn  xn linearconvolution
k
(5.4.4)
k
hn  yzs,n whenxn   n
Alternatively:
(5.4.5)
LTI ,BIBO hk  Bh  
(5.4.6)
LTI ,Causalhn  hnun (hn  0n  0)
(5.4.7)
k
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Fall 2008
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SFG


Let a yn k   bm xn-m

(5.5.1)
m 
S n 1  AS n  bxn

In general!
(5.5.2)
  ( A)  1  BIBO
(5.5.3)
yn  cT S n  dxn
(5.5.4)
if xn  0andnislargeSn 0 yn  0
- Relaxed system
If there is no input, the output must go to zero!
What’s about unstable AND not-relaxed systems?
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DSP and Filter Design
Fall 2008
(5.5.5)
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SFG
Signal flow graph does not necessarily have to represent
Fundamental of Direct 2 forms! They can be arbitrary…
z-1
d1
A perfectly valid
form of SFG:
xn
c1
c2
d2
d3
Find the LCCDE and statespace representation
quantities!
Regrouping an SFG might be a good idea!
Unit step response: empty states, impulse at the input.
ELEN 5346/4304
DSP and Filter Design
Fall 2008
yn
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DTFT and DFT
DTFT of a finite length sequence (sampled at k 
DTFT at
k

2 k
N
2
k ) is DFT{xn}
N
   periodicxn   xn rN
IDTFT
r
k  0,...N  frequencysamples
generally, aliased result
“sampling” usually leads to “aliasing”!
IDFT{XkHk} =
xn hn
- a circular convolution
IDFT{XkHk} =
xn  hn
- a linear convolution – a practical application
N
after sufficient zero-padding
“FFT” really means “Fast DFT”.
ELEN 5346/4304
DSP and Filter Design
Fall 2008
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Suggestions
You may need proofs, derivations, properties…
Look at the examples covered in class.
Think about “good questions” for the test!
SFG  {A,b,c,d}
Need to understand and be able to test for:
Linearity, Time Invariance, BIBO, Causality.
properties?
What are and how to find:
yss,n, ytr,n, yzi,n, yzs,n, yh,n, yp,n?
Direct 1 and Direct 2 – implementation aspects
Be able to interpret results of DFT,
What are the frequency response (LTI), Impulse response, transient response?
Purpose of scaling??
ELEN 5346/4304
DSP and Filter Design
Fall 2008