of Solution Linear Constant-Coefficient Difference Equations
Solution of Linear Constant-Coefficient Difference Equations
Example: Determine the response y ( n), n 0 of the system described by the second-order difference equation
y ( n) = 0.7 y ( n - 1) - 0.1 y ( n - 2) + 2 x ( n) - x ( n - 2)
to the input
x ( n ) = 4 n u ( n)
The homogenous solution is
y ( n) - 0.7 y ( n - 1) + 0.1 y ( n - 2) = 0
n - 0.7 n -1 + 0.1 n - 2 = 0
n + 2 ( 2 - 0.7 + 0.1) = 0
1 = 0.5
and
2 = 0.2
yh (n) = c1 0.5n + c2 0.2n
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Solution of Linear Constant-Coefficient Difference Equations
Particular Solution:
y p ( n) = K 4 n u ( n )
K 4n u (n) - 0.7 K 4 n -1 u (n - 1) + 0.1K 4n - 2 u (n - 2) = (2)4n u (n) - 4n - 2 u (n - 2)
n=2
K 42 - 0.7 K 41 + 0.1K 40 = 2(4)2 - 40
16 K - 2.8K + 0.1K = 32 - 1
y p (n) = 2.33(4) n u (n)
The total solution
K=
31 = 2.33 13.3
y (n) = c1 0.5n + c2 0.2n + 2.33(4) n u (n)
Solution of Linear Constant-Coefficient Difference Equations
To find c1 and c2 For n=0: From difference equation,
y (0) = 0.7 y (0 - 1) - 0.1 y (0 - 2) + 2 x (0) - x (0 - 2) y (0) = 2
From the total solution,
,
y (0) = c1 + c2 + 2.33
For n=1:
From difference equation,
y (1) = 0.7 y (1 - 1) - 0.1 y (1 - 2) + 2 x (1) - x (1 - 2) y (1) = 1.4 + 8 = 9.4
From difference equation,
y (1) = 0.5c1 + 0.2c2 + 9.32
c1 = 0.466
Therefore,
2 = c1 + c2 + 2.33 9.4 = 0.5c1 + 0.2c2 + 9.32
c2 = -0.807
Total Solution
y ( n) = 0.466(0.5) n - 0.807(0.2) n + 2.33(4) n u ( n)
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The Impulse Response of a LTI recursive system
y zs (n) =
n k =0
h(k ) x(n - k ), n 0
x ( n) = ( n)
yzs (n) = h(n)
The impulse response can be obtained from the linear constantcoefficient difference equation. That is the solution of homogeneous equation and particular solution to the excitation function. In the case where the excitation function is an impulse function. The particular solution is zero y p (n) = 0 , since x ( n) = 0 for n>0.
The Impulse Response of a LTI recursive system
Example: Find the impulse response form the following equation,
y ( n) - 3 y (n - 1) - 4 y (n - 2) = x (n) + 2 x( n - 1)
The homogenous solution is
yh (n) = C1 (-1) n + C2 4n
The particular solution is zero when x ( n) = ( n ) To find C1and C2, we evaluate difference equation and homogenous solution for n=0 and n=1.( y ( -1) = 0, y ( -2) = 0, since the system must be relaxed)
y (0) = 1 y (1) = 3 y (0) + 2 = 5
The impulse response is
y (0) = C1 + C2
y (1) = -C1 + 4C2
C2 =
6 5
C1 = -
1 5
1 6 h(n) = - ( -1) n + 4n u ( n) 5 5
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Implementation of Discrete-Time Systems
A system can be described by a linear constant-coefficient difference equation. Let's consider the first order system
y (n) = -a1 y (n - 1) + b0 x(n) + b1 x(n - 1)
The system can be described by two systems in cascade. The first is a nonrecursive system described by the equation
v(n) = b0 x(n) + b1 x(n - 1)
The second part is recursive system
y (n) = -a1 y (n - 1) + v(n)
x(n) z-1 b0 b1 + v(n) + -a1 z-1 y(n)
This is called a direct form I structure.
Implementation of Discrete-Time Systems
Using convolution properties, we can interchange the order of the recursive and nonrecursive system
x(n) + -a1 z-1 w(n) z-1 b0 b1 + y(n)
w(n-1)
w(n-1)
w(n) = -a0 w(n - 1) + x(n)
x(n) + -a1 w(n) z-1 w(n-1) b0 b1
y (n) = b0 w(n) + b1w(n - 1)
+ y(n)
Using only one delay. It is more efficient in terms of memory requirement. It is called the direct form II structure
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Implementation of Discrete-Time Systems
In general form, The difference equation is given by
y (n) = -
N k =1 N k =0
ak y (n - k ) +
bk x (n - k )
N k =0
The nonrecursive system is The recursive system is
y ( n) = -
N k =1
v(n) =
bk x(n - k )
ak y ( n - k ) + v ( n )
Direct form II structure , the recursive system is
w(n) = -
N k =1
ak w(n - k ) + x(n) y ( n) =
N k =0
The nonrecursive system is
bk w(n - k )
Implementation of Discrete-Time Systems
x(n)
z-1
b0 b1
v(n)
+ +
z-1
y(n)
x(n)
+
w(n)
z-1
b0 b1
+
y(n)
+
+
-a1
z-1
+
-a1
+
z-1
b2
+
+
-a2
+
-a2
z-1
b2
+
z-1
bM
-aN
z-1
+
-aN-1
z-1
bM
z-1
-aN
w(n-N)
Step-1
Step-3
"
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Implementation Discrete-Time of Systems
A special case of general case
N k =0
ak = 0, k = 1,2,..., N
y ( n) =
bk x ( n - k )
which is a nonrecursive LTI system. Such a system coefficients bk
h( k ) =
bk 0 k M 0 otherwise
The second part, set M=0 to obtain the general case difference equation
y ( n) = -
N k =1
ak y (n - k ) + b0 x(n)
" "
The crosscorelation of x(n) and y(n) is s sequence rxy (l ) is defined as
rxy (l ) =
n =-
x( n) y ( n - l ), l = 0, 1, 2,...
Or
rxy (l ) =
n =-
x(n + l ) y (n), l = 0, 1, 2,...
The reverse crosscorrelation is
ryx (l ) =
n =-
n =-
y ( n) x( n - l ), l = 0, 1, 2,...
y (n + l ) x( n), l = 0, 1, 2,...
Or
ryx (l ) =
rxy (l ) = ry x ( -l )
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In special case , we have autocorrelation, which is defined as So, that
rxx (l ) =
n =-
x ( n) x ( n - l )
Or
rxx (l ) =
n =-
x(n + 1) x(n)
Example:
x ( n ) = {...0,0, 2, -1,3,7,1, 2, -3,0,0,...}
y ( n ) = {...0,0,1, -1, 2, -2, 4,1, -2,5,0,0,...}
rxy (0) =
rxy (1) =
n=-
x(n) y (n) = 0 + 0 + 2 + 6 - 14 + 4 + 2 + 6 + 0 + 0 = 7
x(n) y ( n - 1) = 0 + 0 + -1 - 3 + 14 - 2 + 8 - 3 + 0 + 0 = 13
n =-
rxy (l ) = {10, -9,19,36, -14,33,0,7,13, -18,16, -7,5, -3}
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Problem 43.a
$%
Determine the direct form II realization for the following LTI system
2 y (n) + y (n - 1) - 4 y (n - 3) = x(n) + 3x(n - 5)
y ( n) + 1 1 3 y (n - 1) - 2 y (n - 3) = x(n) + x(n - 5) 2 2 2
x(n)
+ z-1 + -1/2 z-1 z-1 2 z-1 z-1 3/2 1/2 +
y(n)
"
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#
Problem 44.a
x(n)
+
$%
+
Compute the first samples of its impulse response
y(n)
z-1
1/2
x( n) = ( n)
x( n) = {1, 0, 0,...}
y ( n) = 1 y (n - 1) + x(n) + x(n - 1) 2
y (1) =
1 3 y (0) + x(1) + x(0) = 2 2
y (0) = x(0) = 1
1 3 y (1) + x(2) + x(1) = 2 4 1 3 y (3) = y (2) + x(3) + x(2) = 2 8 y (2) =
3 3 3 3 3 3 y ( n ) = 1, , , , , , ,... 2 4 8 16 32 64
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Problem 44.b
y ( n) =
$%
Find the input output relation
1 y (n - 1) + x(n) + x(n - 1) 2
Problem 44.c The input
x( n) = {1,1,1,...}
y (1) =
1 5 y (0) + x(1) + x(0) = 2 2
1 y (n) = y (n - 1) + x(n) + x(n - 1) 2
y (2) =
1 13 y (1) + x(2) + x(1) = 2 4
y (0) = x(0) = 1
y (3) =
1 29 y (2) + x(3) + x(2) = 2 8
......
"
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#
Problem 44.d
y (n ) = u(n ) * h(n ) =
k =0
$%
u ( k )h ( n - k ) =
n k =0
Use convolution
h(n - k )
y (0) = h(0) = 1
y (1) = h(0) + h(1) = 5 2
y (2) = h(0) + h(1) + h(2) =
13 4
29 8
y (3) = h(0) + h(1) + h(2) + h(3) =
"
#
Problem 54
$%
Find the y(n) for the following equation
y (n) - 4 y (n - 1) + 4 y (n - 2) = x(n) - x(n - 1)
x( n) = ( -1) n u ( n)
The characteristic equation
2 - 4 + 4 = 0
The homogenous solution is The particular solution is
1 = 2, 2 = 2
yh (n) = c1 2n + c2 n 2n
y p (n) = k (-1) n u (n)
k (-1) n u (n) - 4k (-1) n -1 u (n - 1) + 4k (-1) n - 2 u (n - 2) = (-1) n u (n) - (-1) n -1 u (n - 1)
For n=2
k + 4k + 4k = 1 + 1
k=
2 9
" !
9
#
The total solution is
$%
2 y ( n ) = yh (n ) + y p (n ) = c1 2n + c2n 2n + ( -1) n u(n ) 9
Using the initial condition,
y (-1) = y (-2) = 0
we can obtain from difference equation at n=0
y (0) - 4 y (-1) + 4 y (-2) = x(0) - x(-1) y (0) = 1
n=1
y (1) - 4 y (0) + 4 y ( -1) = x(1) - x(0) y (1) = 2
y (1) = c1 2 + c2 2 - 2 =2 9
From the total solution
y (0) = c1 + 2 =1 9
c1 = 7 9
y (1) = c1 2 + c2 2 -
2 =2 9
c2 =
1 3
The total solution
y ( n) =
7 n 1 n 2 2 + n2 + (-1) n u (n) 9 3 9
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Problem 55
$%
Find the impulse response h(n) for the following equation
y (n) - 4 y (n - 1) + 4 y (n - 2) = x(n) - x(n - 1)
The homogenous solution is
y h ( n ) = c1 2 n + c2 n 2 n
So system response To find the constant
y (0) - 4 y ( -1) + 4 y (-2) = (0) - (-1) y (0) = 1
h(n) = c1 2n + c2 n2n u (n)
y (1) - 4 y (0) + 4 y (-1) = (1) - (0) y (1) = 3
From the system response h(n)
y (0) = c1 = 1
y (1) = c1 2 + c1 2 = 3
c2 =
1 2
1 h( n) = 2 n + n 2 n u ( n) 2
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