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Unformatted text preview: M. J. Roberts  10/7/06 Chapter 5  DiscreteTime System Properties
Solutions
Modeling DiscreteTime Systems
1. Bernoulli’s method can be used to numerically find the dominant root of a
polynomial equation (if it exists). It is an example of a discretetime system. If the
equation is of the form aN q n + aN 1q n 1 + + a1q + a0 = 0 , the method consists of
solving the difference equation
aN q n + aN 1q n 1 +
with the initial values q 1 =q + a1q n
2 = N + 1 + a0 q n
=q N +1 N =0 0 and q 0 = 1. The dominant root is the limit approached by q n + 1 / q n . Draw a DT system to
find the dominant root of a fourthdegree polynomial equation. Find the dominant
root of 2q 4 + 3q 3 8q 2 + q 3 = 0 .
a3
a4
a2
a4
a1
a4
a0
a4 %
% D
D
D
D Program to find the dominant root of a fourthdegree equation
using Bernoulli's method q = [1 0.01 0.01 0.01] ; a = [2 3 8 1 3] ;
while abs(q(1)/q(2)  q(2)/q(3)) > abs(q(1)/q(2))/10000,
qn = a(2:end)*q'/a(1) ;
q = [qn [q(1:end1)]] ;
q(1)/q(2)
end The dominant root is 2.964.
Solutions 51 M. J. Roberts  10/7/06 Properties of DT Systems
2. Show that the system of Figure E1 is nonlinear, BIBO stable, static and noninvertible. (The response of an analog multiplier is the product of its two
excitations.)
Analog
Multiplier
x[n]
y[n]
2
Figure E1 A DT system
Homogeneity:
Let x1 n = g n . Then y1 n = 2 g 2 n ( Let x 2 n = K g n . Then y2 n = 2 K g n
Multiplying the first equation by K, ) 2 = 2K 2 g 2 n . K y1 n = 2K g 2 n .
Then
y2 n . K y1 n
Not homogeneous.
Let x1 n = g n . Then y1 n = 2 g 2 n
Let x 2 n = h n . Then y2 n = 2 h 2 n
Let x 3 n = g n + h n . ( Then y3 n = 2 g n + h n ) 2 ( =2 g n 2 +h n 2 + 2g n h n ). Adding the first two equations,
y1 n + y2 n = 2 g 2 n + 2 h 2 n y3 n . Not additive.
Since the system is not homogeneous and not additive it is also not linear.
The system is also not incrementally linear.
The system is statically nonlinear because the nonlinearity is inherent in the
squared relationship between excitation and response.
Time Invariance:
Let x1 n = g n . Then y1 n = 2 g 2 n .
Solutions 52 M. J. Roberts  10/7/06 Let x 2 n = g n n0 . Then y2 n = 2 g 2 n n0 .
The first equation can be rewritten as
y1 n n0 = 2 g 2 n no = y2 n
Time Invariant
Stability:
If the excitation is bounded, the response is bounded.
Stable
Causality:
At any discrete time, n = n0 , the response depends only on the excitation at that same
time. Causal.
Memory:
At any discrete time, n = n0 , the response depends only on the excitation at that same
time. System has no memory.
Invertibility:
Inverting the functional relationship, x n = y n 1
2 2 The square root is multiple valued. Two different excitations can cause the same
response.
Not Invertible.
3. Show that a system with excitation x n and response y n described by
y n = nx n ,
is linear, time variant and static.
Homogeneity:
Let x1 n = g n . Then y1 n = n g n
Let x 2 n = K g n . Then y2 n = nK g n .
Multiplying the first equation by K,
K y1 n = nK g n .
Then
K y1 n = y2 n .
Homogeneous.
Solutions 53 M. J. Roberts  10/7/06 Let x1 n = g n . Then y1 n = n g n
Let x 2 n = h n . Then y2 n = n h n
L e t ( x3 n = g n + h n . y3 n = n g n + h n
Additive. ) = ng T h e n n + n h n = y1 n + y2 n . Since the system is homogeneous and additive it is also linear.
The system is also incrementally linear.
The system is not statically nonlinear because it is linear.
Time Invariance:
Let x1 n = g n . Then y1 n = n g n .
Let x 2 n = g n n0 . Then y2 n = n g n n0 .
The first equation can be rewritten as ( ) y1 n n0 = n n0 g n n0 y2 n Time Variant
Stability:
If the excitation is a constant, the response is unbounded as n approaches infinity.
Unstable
Causality:
At any discrete time, n = n0 , the response depends only on the excitation at that same
time. Causal.
Memory:
At any discrete time, n = n0 , the response depends only on the excitation at that same
time. System has no memory.
Invertibility:
Inverting the functional relationship,
x n =y n /n .
Invertible.
4. Show that the system of Figure E4 is linear, timeinvariant, BIBO unstable and
dynamic. Solutions 54 M. J. Roberts  10/7/06 x[n] y[n] + + D
Figure E4 A DT system
y n =x n +y n 1
y n 1 =x n 1 +y n 2
y n =x n +x n 1 +y n 2
Then, by induction,
+x n k + y n =x n +x n 1 + = x n k
k =0 Let m = n k . Then
y n = n x m =
m= n x m
m= Homogeneity:
n Let x1 n = g n . Then y1 n = g m
m= Let x 2 n n = K g n . Then y2 n = Kg m = K m= n g m = K y1 n . m= Homogeneous.
n Let x1 n = g n . Then y1 n = g m
m= Let x 2 n n = h n . Then y2 n = h m
m= Let x 3 n = g n + h n .
n Then y3 n = m= (g m +h m )= n m= g m + n h m = y1 n + y2 n . m= Additive.
Since the system is homogeneous and additive it is also linear.
The system is also incrementally linear because it is linear.
The system is not statically nonlinear because it is linear.
Time Invariance:
Solutions 55 M. J. Roberts  10/7/06 n Let x1 n = g n . Then y1 n = g m .
m= Let x 2 n = g n n0 . Then y2 n = n g m n0 .
m= The first equation can be rewritten as
y1 n n0 = n n0 g m
m= Let m = q n0 . Then
y1 n n0 = n g q
q= n0 = y2 n Time invariant
Stability:
If the excitation is a constant, the response increases without bound. () n Also the solution of the homogeneous difference equation is y h n = K 1 = K .
Therefore the eigenvalue is 1 whose magnitude is not less than 1 and the system
must be BIBO unstable.
Unstable
Causality:
At any discrete time, n = n0 , the response depends only on the excitation at that
discrete time and previous discrete times.
Causal.
Memory:
At any discrete time, n = n0 , the response depends on the excitation at that discrete
time and previous discrete times.
System has memory.
Invertibility:
Taking the first backward difference of both sides of the original system equation,
y n y n 1 = n n 1 x m
m= x n =y n x m
m= y n 1 The excitation is uniquely determined by the response.
Invertible.
5. Show that a system with excitation x n and response y n described by Solutions 56 M. J. Roberts  10/7/06 ( ), y n = rect x n
is nonlinear, time invariant and noninvertible.
Homogeneity:
Let x1 n = g n . Then y1 n = rect g n ( ) ( Let x 2 n = K g n . Then y2 n = rect K g n
Not homogeneous. ) K y1 n . ( )
= rect ( h n ) Let x1 n = g n . Then y1 n = rect g n
Let x 2 n = h n . Then y2 n
Let x 3 n = g n + h n . ( Then y3 n = rect g n + h n
Not additive. ) y1 n + y2 n . Since the system is not homogeneous and not additive it is also not linear.
The system is also not incrementally linear.
The system is statically nonlinear because the nonlinearity arises from the nonlinearity of the rectangle relation between excitation and response.
Time Invariance:
Let x1 n = g n . Then y1 n = rect g n . ( ( ) Let x 2 n = g n n0 . Then y2 n = rect g n n0
The first equation can be rewritten as ( y1 n no = rect g n no ). )= y 2 n Time invariant
Stability:
No matter what values the excitation may have the response can only have the values,
zero or one.
Stable
Causality:
At any discrete time, n = n0 , the response depends only on the excitation at that
discrete time and not on any future excitation.
Causal.
Memory:
At any discrete time, n = n0 , the response depends only on the excitation at that
discrete time and not on any past excitation.
Solutions 57 M. J. Roberts  10/7/06 System has no memory.
Invertibility:
Many different excitations can cause the same response.
Not invertible.
6. Show that the system of Figure E6 is nonlinear, timeinvariant, static and invertible.
5
x[n] 10 + y[n] Figure E6 A DT system
y n = 10 x n
Homogeneity:
Let x1 n = g n . Then y1 n = 10 g n 5, 5 Let x 2 n = K g n . Then y2 n = 10K g n
Not homogeneous.
Let x1 n = g n . Then y1 n = 10 g n K y1 n . 5 Let x 2 n = h n . Then y1 n = 10 h n 5 5 Let x 3 n = g n + h n . ( Then y3 n = 10 g n + h n
Not additive. ) 5 y1 n + y2 n . Since the system is not homogeneous and not additive it is also not linear.
The system is incrementally linear because the only deviation from linearity is caused
by the presence of the nonzero, zeroexcitation response.
The system is not statically nonlinear because it is incrementally linear.
Time Invariance:
Let x1 n = g n . Then y1 n = 10 g n 5. Let x 2 n = g n n0 . Then y2 n = 10 g n n0
The first equation can be rewritten as
y1 n n0 = 10 g n n0 5. 5 = y2 n Time invariant
Solutions 58 M. J. Roberts  10/7/06 Stability:
If the excitation is bounded, the response is bounded.
Stable
Causality:
At any discrete time, n = n0 , the response depends only on the excitation at that
discrete time and not on any future excitation.
Causal.
Memory:
At any discrete time, n = n0 , the response depends only on the excitation at that
discrete time and not on any past excitation.
System has no memory.
Invertibility:
Solving the system equation for the excitation as a function of the response,
x n = y n +5
10 Invertible.
7. Show that the system of Figure E7 is timeinvariant, BIBO stable, and causal.
x[n] 0.25 y[n] + +
+ D 1 D 2 Figure E7 A DT system
Homogeneity:
Let x1 n = g n . Then 4 y1 n
Let x 2 n = K g n . Then 4 y2 n y1 n 1 + 2 y1 n 2 = g n
y2 n 1 + 2 y2 n 2 = K g n Multiply the first equation by K. 4K y1 n
Then, equating results,
4 y2 n K y1 n 1 + 2K y1 n 2 = K g n y2 n 1 + 2 y2 n 2 = 4K y1 n If this equation is to be satisfied for all n,
y2 n = K y1 n .
Homogeneous.
Solutions 59 K y1 n 1 + 2K y1 n 2 M. J. Roberts  10/7/06 Additivity:
Let x1 n = g n . Then 4 y1 n y1 n 1 + 2 y1 n 2 = g n Let x 2 n = h n . Then 4 y2 n y2 n 1 + 2 y2 n 2 = h n Let x 3 n = g n + h n . Then 4 y3 n y3 n 1 + 2 y3 n 2 = g n + h n Add the two first two equations. ( 4 y1 n + y2 n ) (y 1 ) ( n 1 + y2 n 1 + 2 y1 n 2 + y2 n 2 )=g n +h n Then, equating results,
4 y3 n ( y3 n 1 + 2 y3 n 2 = 4 y1 n + y2 n ) (y 1 ) ( n 1 + y2 n 1 + 2 y1 n 2 + y2 n 2 ) If this equation is to be satisfied for all n,
y3 n = y1 n + y2 n .
Additive.
Since the system is both homogeneous and additive, it is linear.
Since the system is linear it is also incrementally linear.
Since the system is linear, it is not statically nonlinear.
Time Invariance:
Let x1 n = g n . Then 4 y1 n y1 n 1 + 2 y1 n 2 = g n Let x 2 n = g n n0 . Then 4 y2 n
We can rewrite the first equation as
4 y1 n n0 y2 n 1 + 2 y2 n 2 = g n n0 y1 n n0 1 + 2 y1 n n0 2 = g n n0 Then, equating results,
4 y1 n n0 y1 n n0 1 + 2 y1 n n0 2 = 4 y2 n If this equation is to be satisfied for all n, then Solutions 510 y2 n 1 + 2 y2 n 2 M. J. Roberts  10/7/06 y2 n = y1 n n0 .
Time Invariant.
Stability:
The eigenvalues of the system homogeneous solution are found from the
characteristic equation,
4 2
+2=0 .
They are
1,2 = 0.125 ± j0.696 or 1,2 = 0.7071e± j1.3931 Therefore the homogeneous solution is of the form, ( ) ( n ) n y h n = K h1 0.7071 e+ j1.3931n + K h2 0.7071 e j1.3931n and, as n approaches infinity the homogeneous solution approaches zero and the total
solution approaches the particular solution. The particular solution is bounded
because it consists of functions of the same form as the excitation and all its unique
differences and the excitation is bounded in the BIBO stability test. Therefore if x is
bounded, so is y.
Stable.
Causality:
We can rearrange the system equation into
y n = ( 1
x n + y n 1 + 2y n 2
4 ) showing that the response at time, n, depends on the excitation at time, n, and the
response at previous times. It does not depend on any future values of the excitation.
Causal.
Memory:
The response depends on past values of the response.
The system has memory.
Invertibility:
The original system equation, 4 y n
excitation in terms of the response.
Invertible. y n 1 + 2 y n 2 = x n , expresses the Modeling DiscreteTime Systems
8. At the beginning of the year 2000, the country, Freedonia, had a population, p, of 100
million people. The birth rate is 4% per annum and the death rate is 2% per annum,
compounded daily. That is, the births and deaths occur every day at a uniform
fraction of the current population and the next day the number of births and deaths
changes because the population changed the previous day. For example, every day
Solutions 511 M. J. Roberts  10/7/06 0.02
, of the total population at the end
365
of the previous day (neglect leapyear effects). Every day 275 immigrants enter
Freedonia. the number of people who die is the fraction, (a)
Write a difference equation for the population at the beginning of the nth day
after January 1, 2000 with the immigration rate as the excitation of the system.
(b)
By finding the zeroexctiation and zerostate responses of the system
determine the population of Freedonia be at the beginning of the year 2050.
The difference equation is
p n +1 = p n + 0.04
p n
365 (1 + 5.48 p n +1 10 0.02
p n + 275
365
5 )p n = 275 The eigenvalue is 1 + 5.48 10 5 = 1.0000548 and the zeroexcitation response is ( p n = 108 1.0000548 ( ) ) n n The impulse response is h n = 1.0000548 u n . The response to the
immigration rate is the convolution of the impulse response with the immigration rate,
275u n , or ( ) n p n = 1.0000548 u n 273.9 u n = 273.9 n 1 m=0 (1.0000548) m Using the summation formular for a geometric series,
p n = 275 ( ) 1 1.0000548n
1.0000548n 1
u n = 275
u n = 5018248.2 1.0000548n 1 u n
1 1.0000548
0.0000548 The total solution is ( ) ( n ) p n = 108 1.0000548 + 5018248.2 1.0000548n 1 u n ( p n = 1.05 108 1.0000548 ) n 5018248.2 , n The beginning of the year 2050 is the 18250th day. Solutions 512 0 , n 0 M. J. Roberts  10/7/06 ( p 18250 = 1.05 108 1.0000548
9. ) 18250 5018248.2 = 280,420,000 In Figure E9 is a MATLAB program simulating a DT system.
(a) Without actually running the program, find the value of x when
solving the difference equation for the system in closed form. (b) n = 10 Run the program and check the answer in part (a). by x = 1 ; y = 3 ; z = 0 ; n = 0 ;
while n <= 10,
z = y ;
y = x ;
x = 2*n + 0.9*y  0.6*z ;
n = n + 1 ;
end Figure E9
This system is a secondorder DT system. It can be modeled by the difference
equation
x n 0.9 x n 1 + 0.6 x n 2 = 2n .
The homogeneous solution is ( ) ( n x h n = K h1 0.45 + j0.63 + K h2 0.45
or ( x h n = K h1 0.7746e j0.9509 ) n ( + K h2 0.7746e ) j0.63 j0.9509 n ) n . The form of the particular solution is a linear combination of functions of the same form of
the forcing function 2n and all its unique differences. A forward difference of 2n is
2 n + 1 2n = 2 , a constant and a second forward difference is zero. Therefore the form of
the particular solution is ( ) x p n = K p1n + K p2 .
Substituting that into the difference equation and solving,
K p1n + K p2
K p1n + K p2 ( ( ) ) ( ( ) ) 0.9 K p1 n 1 + K p2 + 0.6 K p1 n 2 + K p2 = 2n ( ) 0.9K p1 n 1 ( ) 0.9K p2 + 0.6K p1 n 2 + 0.6K p2 = 2n 0.7K p1 = 2
0.3K p1 + 0.7K p2 = 0 Solutions 513 M. J. Roberts  10/7/06 K p1 = 2.857 , K p2 = 1.224
So the total solution is ( x n = 2.857n + 1.224 + K h1 0.7746e j0.9509 ) n ( + K h2 0.7746e j0.9509 ) n . From the recursion in the MATLAB program,
x 0 = 0.9 1 0.6 3 = 0.9
x 1 = 2 + 0.9 ( 0.9 ) 0.6 1 = 0.59 . Therefore
1.224 + K h1 + K h2 = 0.9 ( 2.857 + 1.224 + K h1 0.7746e j0.9509 ) n ( + K h2 0.7746e j0.9509 ) n = 0.59 or
1
0.7746e j0.9509 K h1 1
0.7746e j0.9509 K h2 = 2.124
3.491 and
K h1 = 1.062 + j2.012 = 2.2751e j 2.054 , K h2 = 1.062 j2.012 = 2.2751e j 2.054 . So the final total numerical solution is ( x n = 2.857n + 1.224 + 2.2751e j 2.054 0.7746e j0.9509 ) n + 2.2751e j 2.054 (0.7746e j0.9509 and
x 10 = 29.99 .
Properties of DiscreteTime Systems
10. n+1 A DT system is described by y n = m= x m . Classify this system as to time invariance, BIBO stability and invertibility.
Homogeneity:
Let x1 n = g n . Then y1 n = n+1 g m
m= Let x 2 n = K g n . Then y2 n = n+1 Kg m = K m= Homogeneous. Solutions 514 n+1 m= g m = K y1 n . ).
n M. J. Roberts  10/7/06 n+1 Let x1 n = g n . Then y1 n = g m
m=
n+1 Let x 2 n = h n . Then y2 n = h m
m= Let x 3 n = g n + h n .
Then y3 n = n+1 m= ( g m +h m )= n+1 g m + m= n+1 h m = y1 n + y2 n . m= Additive.
Since the system is homogeneous and additive it is also linear.
The system is also incrementally linear because it is linear.
The system is not statically nonlinear because it is linear.
Time Invariance:
n+1 Let x1 n = g n . Then y1 n = g m .
m= Let x 2 n = g n n0 . Then y2 n = n+1 g m n0 .
m= The first equation can be rewritten as
y1 n n0 = n n0 +1 g m = n+1 g q
q= m= n0 = y2 n Time invariant
Stability:
If the excitation is a constant, the response increases without bound.
Unstable
Causality:
At any discrete time, n = n0 , the response depends on the excitation at the next
discrete time in the future.
Not causal.
Memory:
At any discrete time, n = n0 , the response depends on the excitation at that discrete
time and previous discrete times.
System has memory.
Invertibility:
Inverting the functional relationship,
y n = n+1 x m .
m= Solutions 515 M. J. Roberts  10/7/06 Invertible.
Taking the first backward difference of both sides of the original system equation,
n+1 y n 1 = y n n+1 1 x m
m= x n +1 = y n x m
m= y n 1 The excitation is uniquely determined by the response.
Invertible.
11. A DT system is described by
ny n 8y n 1 = x n . Classify this system as to time invariance, BIBO stability and invertibility.
Homogeneity:
Let x1 n = g n . Then n y1 n 8 y1 n 1 = g n Let x 2 n = K g n . Then n y2 n 8 y2 n 1 = K g n Multiply the first equation by K. nK y1 n
Then, equating results,
nK y1 n 8K y1 n 1 = K g n 8K y1 n 1 = n y2 n 8 y2 n 1 If this equation is to be satisfied for all n,
y2 n = K y1 n .
Homogeneous.
Additivity:
Let x1 n = g n . Then n y1 n 8 y1 n 1 = g n Let x 2 n = h n . Then n y2 n 8 y2 n 1 = h n Let x 3 n = g n + h n . Then n y3 n 8 y3 n 1 = g n + h n Add the two first two equations. ( n y1 n + y2 n ) 8( y 1 ) n 1 + y2 n 1 = g n + h n Then, equating results, Solutions 516 M. J. Roberts  10/7/06 ( n y1 n + y2 n ) 8( y 1 ) n 1 + y 2 n 1 = n y3 n 8 y3 n 1 If this equation is to be satisfied for all n,
y3 n = y1 n + y2 n .
Additive.
Since the system is both homogeneous and additive, it is linear.
Since the system is linear it is also incrementally linear.
Since the system is linear, it is not statically nonlinear.
Time Invariance:
Let x1 n = g n . Then n y1 n 8 y1 n 1 = g n Let x 2 n = g n n0 . Then n y2 n
We can rewrite the first equation as (n ) n0 y1 n n0 8 y2 n 1 = g n n0
8 y1 n n0 1 = g n n0 Then, equating results, (n ) n0 y1 n n0 8 y1 n n0 1 = n y2 n 8 y2 n 1 This equation cannot be satisfied for all n, therefore
y1 n n0 . y2 n
Time Variant.
Stability: The homogeneous equation is
ny n = 8y n 1
or
y n = 8
y n 1 .
n Thus, as n increases without bound, y n must be decreasing because it is
its previous value and 8
times
n 8
approaches zero. Rearranging the original equation,
n
Solutions 517 M. J. Roberts  10/7/06 y n = x n
n + 8
y n 1 .
n For any bounded excitation, x n , as n gets larger, x n / n must be bounded and (8 / n ) y n 1 must be getting smaller because it is a decreasing fraction of its
previous value. Therefore for a bounded excitation, the response is bounded.
Stable.
Causality:
We can rearrange the system equation into
y1 n = g n + 8 y1 n 1
n showing that the response at time, n, depends on the excitation at time, n, and the
response at a previous time.
Causal.
Memory:
The response depends on past values of the response.
The system has memory.
Invertibility:
The original system equation, n y n
in terms of the response.
Invertible.
12. 8 y n 1 = x n , expresses the excitation A DT system is described by
y n = x n .
Classify this system as to linearity, BIBO stability, memory and invertibility.
Homogeneity:
Let x1 n = g n . Then y1 n = g n
Let x 2 n = K g n . Then y2 n =
Multiplying the first equation by K, Kg n = K y1 n = K g n K g n . y2 n . Not homogeneous.
Let x1 n = g n . Then y1 n = g n
Let x 2 n = h n . Then y2 n = h n
Solutions 518 M. J. Roberts  10/7/06 Let x 3 n = g n + h n . Then y3 n = g n + h n
Not additive. y1 n + y2 n . Since the system is not homogeneous and not additive it is also not linear.
The system is also not incrementally linear.
The system is statically nonlinear because of the squareroot relationship between
excitation and response.
Time Invariance:
Let x1 n = g n . Then y1 n = g n .
Let x 2 n = g n n0 . Then y2 n = g n n0 .
The first equation can be rewritten as
y1 n n0 = g n n0 = y2 n
Time invariant
Stability:
If the excitation is bounded, the response is bounded.
Stable
Causality:
At any discrete time, n = n0 , the response depends only on the excitation at that same
time. Causal.
Memory:
At any discrete time, n = n0 , the response depends only on the excitation at that same
time. System has no memory.
Invertibility:
Inverting the functional relationship,
x n = y2 n .
Invertible. Solutions 519 ...
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This note was uploaded on 12/10/2010 for the course DCIS 32 taught by Professor Go during the Spring '10 term at College of E&ME, NUST.
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