1
Lecture 2: System’s description (LTI,
causal, finite dimensional)
Instructor:
Dr. Gleb V. Tcheslavski
Contact: gleb@ee.lamar.edu
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
2
The Signal Flow Graph


Let a yn k   bm xn-m

for a BIBO stable LTI finite-dimensional system
m 


Assumea0  1 yn   a yn k   bm xn m

(2.2.1)
m 
*
Δ
yn is a function of previous (and current) inputs and outputs, therefore the
system is causal.
If ak and bm are constants → Linear Constant coefficient Difference Equation
yn
+
Signal Flow
Graph (SFG):
*
+
a1
a2
z-1
z-1
yn-1
xn-1
memory
xn-2
b0
z-1
z-1
z-1
b1
b2
…
…
yn-2
z-1
xn
Remark: because we discretize the signal by amplitude, the system becomes non-linear!
ELEN 5346/4304
DSP and Filter Design
Fall 2008
+
+
Δ
3
Difference equation


yn   a ynk   bm xnm

(2.3.1)
m 
yn is a function of ak, bm, xn, memory = states.
or yn   hk xnk h0 xn   hk xnk

(2.3.2)

If we know the initial conditions, we can solve the difference equation:
x1,n  y1,n 

 predictionsof system ' sbehavior
x2,n  y2, n 

If (2.3.1) is a linear constant coefficient difference equation,
the output consists of a homogeneous solution and a
particular solution: yn = yh,n + yp,n
(2.3.3)
ELEN 5346/4304
DSP and Filter Design
Fall 2008
4
Homogeneous (complementary) solution
xn  0,n 0theoutputisduetotheinitialconditions
N
Homogeneous equation:
a y
k 0
k
n k
0
(2.4.1)
N
Let ' sassume yn    ak  nk  0
n
(2.4.2)
k 0
Then nN (a0 N  a1 N 1 
 an1  an )  0is a characteristicequation
(2.4.3)
The characteristic equation has N characteristic roots: 1, … N
• If the system is real, xn, yn, and all the coefficients (a, b) are real
• For LTI systems hn are real → roots are either real or complex conjugate pairs
N
n
• yh ,n   ci i is a form of solution, where ci depend on initial conditions.
(2.4.4)
• if 1 = 2, solution in form of
(2.4.5)
i 1
ELEN 5346/4304
DSP and Filter Design
yh,n  c11n  c2n1n
Fall 2008
5
Particular solution
“We apply an input and see what happens to the output for large n”
Assumption: yp,n has the same form as xn:
xn
yp,n
c
k
k0nM  k1nM 1 
An M
cos 0norsin 0n
k1 cos 0n  k2 sin 0n
cn
knc n
An (k0nM  k1nM 1 
An nM
5
1
1
yn 1  yn  2  xn  xn 1  y1  6; y2  6;x1  1;xn  2n ,n  0
6
6
2
Characteristic equation:
Ex.:
 kM
yn 
n
n
 kM )
(2.5.1)
5
1
1 
1
1
1

1
1
     0        01  ;2   yh,n  c1    c2   ;n  0 formof solution
6
6
2 
3
2
3

 2
 3
(2.5.2)
2
ELEN 5346/4304
DSP and Filter Design
Fall 2008
6
Alternative form
xn
Memory is empty
(relaxed system)
xn-1n
xn-1n
z-1
+
z-1
+
…
We can find a solution in form:
yn = yzi,n + yzs,n (zero input + zero state)
(2.6.1)
The output will be linear iff both: zero input and zero state are linear.
For a relaxed system:
xn = n → yn = hn
ELEN 5346/4304
DSP and Filter Design
-
zero state response!
Fall 2008
(2.6.2)
7
Example…
Let’s find yzi,n for the example in (2.5.1).
Assuming zero input, yp,n = k (a constant) and must satisfy the LCCDE
5
1
k k k 0
6
6
Difference Equation:
(2.7.1)
(2.7.2)
“large enough n” implies that all terms are “active”: in our case, n ≥ 2
n
k  0 yzi ,n
n
1
1
 c1    c2   ;n  0
2
3
In form of homogeneous solution
0
0
5
1
1
5
1
1
1
1
n  0 y0  y1  y2  x0  x1   6   6  0   0  4  c1    c2    c1  c2
6
6
2
6
6
2
2
3
5
1
1
5
1
1
1
1
1
n  1 y1  y0  y1  x1  x0   4   6  0   0  2  c1  c2
6
6
2
6
6
2
3
2
3
4 
1    7.5
c1   1
c 
 2 1    3 
c
1/
2
1/
3

 
 2 
 3
ELEN 5346/4304
DSP and Filter Design
n
(2.7.6)
yzi ,n
Fall 2008
(2.7.3)
(2.7.4)
(2.7.5)
n
1
1
 7.5    3   ;n  0
2
 3
(2.7.7)
8
Example (cont)
yn 
5
1
1
yn 1  yn  2  xn  xn 1  y1  6; y2  6;x1  1;xn  2n ,n  0
6
6
2
n
Got so far:
yh,n
is LTI
n
n
(2.8.1)
n
1
1
1
1
 c1    c2   ;n  0 (2.8.2) yzi ,n  7.5    3   ;n  0
2
 3
2
 3
Sincexn  cn  2n  y p,n  kcn  k 2n
(2.8.3)
(2.8.4)
By substituting (2.8.4) into (2.8.1) for large n, we get:
5
1
1
10
1

k 2n  k 2n 1  k 2n 2  2n  2n 12n 2  4k  k    2n  2 (4  1)
6
6
2
6
6

1

n
4

1

 k  5k  2 y p ,n  2  2
2

n
n
1
1
 
 
Totalsolution : yn  yh,n  y p,n  c1    c2    2  2n
 2
 3
5
1
y2  y1  y2   c1
6
6
DE 
y1   c1
ELEN 5346/4304
DSP and Filter Design
(2.8.5)
(2.8.6)
n
n

1
1




n
c2 
 5.5  (2.8.7) y

5.5

2

2

2
n  0

c

tot
,
n





 
2
 3
(2.8.8)
 2 
c2 
Fall 2008
9
Example (cont 2)
What’s about the zero state solution yzs,n?
xn  2n ,same y p ,n  2  2n ;n  0
(2.9.1)
n
n
1
1
initialconditions0 yzs ,n  c1    c2    2  2n ;n  0
2
 3
1
1

y

0

0

1


{

(2.9.2
)
}

c

c

2

1
1
2
 0
2
2
(2.8.1) y1  y2  0
 y  5 y  0  2  1 {(2.9.2)} c1  c2  4  3 3
 1 6 0
2
2 3
4
(2.9.2)
(2.9.3)
c1  2;c2  1
n
n
 1 1
 yzs ,n  2       2  2n ;n  0
 2  3
n
n
n
n
1
1
1
1




ytot ,n  yzs ,n  yzi ,n              2  2n 5.5    2    2  2n n  0
 2
 3
2
 3
(2.9.5)(2.8.8)
ELEN 5346/4304
DSP and Filter Design
Fall 2008
(2.9.4)
(2.9.5)
10
Example (cont 3)
n
ytot ,n
transient response
n
For large n
ELEN 5346/4304
n
steady state response


n  0


ytr ,n  0;
DSP and Filter Design
(2.10.1)
ytot ,n  ytr ,n  yss,n
Another representation:
1
1
ytr ,n 5.5    2  
2
 3
yss ,n  2  2n
n
1
1
 5.5    2    2  2n n  0
2
 3
(2.10.2)
yss,n dominates
Fall 2008
11
Example (cont 4)
To evaluate a system’s unit pulse response hn
1

when xn   n  for the relaxed system : i.c. = 0 y p ,n  k n ; yh ,n  c11  c22 ;1  ;2 
2

(2.11.1)
5
1
1
Particular: k n  k n 1  k n  2   n   n 1largenk  0  0cannot solve for k!
6
6
2
(2.11.2)
n
n
1
1
(2.11.3)
ytot ,n  c1    c2    k n ;n  0
2
 3
5
1
1
h0  h1  h2  x0  x1 hn  0n  0(causalsystem)  c1  c2  k (2.11.4)
DE:
6
6
2
5
1
1
 1
1
1
(2.11.5)
h1  h0  h1  x1  x0    c1  c2  0
6
6
2
 2
2
3
5
1
1
 8 1
1
1
h2  h1  h0  x2  x1     c1    c2    0
6
6
2
 6 6
2
 3
2
ELEN 5346/4304
DSP and Filter Design
Fall 2008
2
(2.11.6)
12
Example (cont 5)
 c1   6 
 c2    5
 k   0 
(2.12.1)
n
n
n
n

1
1
1
1
 
 
 
  
forn  0hn  6    5   6    5    un
2
 3    2 
 3  
hn  hnun  forcausalsystems
n
1
Note :6   lasts forever !thisisanInfiniteImpulseResponse(IIR)system
2
h
n
 thesystemisBIBOstable
n
ELEN 5346/4304
DSP and Filter Design
Fall 2008
(2.12.2)