Unformatted text preview: Signals and Systems
Fall 2015 Issued: 9 September 2015
Due: 16 September 2015
Homework Assignment #3 1. Iterate by hand to find the first few values of y(k) for the following difference equations:
(a) y(k+1) + y(k) = x(k), y(0) = 0, x(k) = k ≥0
.
0, k < 0 1, (b) y(k+1) + y(k) = 0, y(0) = 1.
(c) y(k+2) − y(k+1) − 2y(k) = x(k+1) + x(k), y(0) = y(1) = 0, x(k) = 1, k = 0
.
0, k ≠ 0 (d) y(k+2) − y(k+1) − 2y(k) = 0, y(0) = 1, y(1) = 0.
2. For each difference equation or operational relation, draw a corresponding simulation
diagram having the minimum possible number of delay blocks:
(a) y(k+1) + y(k) = x(k)
(b) y(k+2) − y(k+1) = 1.5x(k+1) − 0.5x(k)
(c) (E2−1){y(k)} = (E+2){x(k)}
(d) {y(k)} = 6E 2 + 2E
{x(k)}
E 2 + 5E + 4 (e) {y(k)} = 6E 3 + 2
{x(k)}
E3 3. Find the operational transfer function for each of the following simulation diagrams:
(a) (b)
1
E 1
E 1
E 3 0.5 2 4
2 − (See Problem 4 on reverse.) 1
E 4. (a) Write a computer program to simulate the discretetime system shown below and plot its
output. Make your program correspond to the given simulation diagram; that is, program
coupled difference equations in variables that correspond to the outputs of the delay
blocks.
(b) Run your program to obtain the sequence y(k), assuming x(k) is the unit step sequence
and given zero initial conditions in all delay blocks. x(k) 1 q3(k)
E 0.9 1 q2(k)
E 1 q1(k)
E −0.7
−0.7 y(k) Signals and Systems
Fall 2015
Solution to Homework Assignment #3
1a) y k 1 y k x k , y 0 0, x k 1, k
0, k 0
.
0 We can rewrite the equation as follows: y k 1 y k x k By using the initial condition, we have: y 0 0 y 1 y 0 x 0 0 1 1 y 2 y 1 x 1 1 1 0 y 3 y 2 x 2 0 1 1 y 4 y 3 x 3 1 1 0 ...
After taking a few iterations, it can be easily seen that the sequence continues to oscillate
between 0 and 1.
1b) y k 1 y k 0, y 0 1. The system can be described as: y(k 1) y k Taking a few iterations: y 1 y 0 1 y 2 y 1 1 y 3 y 2 1 y 4 y 3 1 ...
Similar to (a), after a few iterations, the sequence will oscillate alternatively between 1
and 1. 1c) y k 2 y k 1 2y k x k 1 x k , y 0 The equation can be written as below: y k 2 y k 1 Taking a few iterations: 2y k x k 1 x k y 1 0, y k 1, k
0, k 0
.
0 x 0
x 1
x 2
x 3
x 4
x 5
x 6
x 7
x 8
... 1
0
0
0
0
0
0
0
0 y
y
y
y
y
y
y
y
y 0
1
2
3
4
5
6
7
8 0
0
y
y
y
y
y
y
y 1
2
3
4
5
6
7 2y
2y
2y
2y
2y
2y
2y 0
1
2
3
4
5
6 x 1
x 2
x 3
x 4
x 5
x 6
x 7 x 0 1
x 1 1
x 2 3
x 3 5
x 4 11
x 5 21
x 6 43 The sequence grows significantly without bound.
1d) y k 2 y k 1 2y k 0, y 0 1, y 1 0. We can rewrite the equation as follow: y k 2 y k 1 2y k Inserting the initial conditions and taking a few calculations: y 0 1 y 1 0 y 2 y 1 2y 0 2 y 3 y 2 2y 1 2 y 4 y 3 2y 2 6 y 5 y 4 2y 3 10 y 6 y 5 2y 4 22 y 7 y 6 2y 5 42 y 8 y 7 2y 6 86 ...
The sequence also grows unboundedly. When k>1, its values are as twice as the results
obtained in (c). 2a) y k 1 y k x k y k 1 y k x k The simulation diagram is plotted below.
y(k+1) x(k) 1
E _ 2b) y (k 2) (E 2 y k 1 y(k) 1.5x k 1 0.5x k E ){ y(k )} (1.5E 0.5){x k } ( E 2 E ){w(k )} {x k }
.
y (k ) (1.5E 0.5){w k } The simulation diagram can then be easily sketched as:
1.5
x(k) 2c) w(k+2) E2 1 y k w(k) 0.5 _ y(k) ) E 2 x k E2 1 w k
y k 1
E 1 w(k+1)
E E x k 2 w k . The corresponding simulation diagram is plotted below.
w(k+2) 1 w(k+1)
E x(k) 1 w(k)
E 2 y(k)
) 2d) 6E 2 2E
x k
E 2 5E 4 y k
E2 E2
y k 5E 4 y k 5E 4 w k
6E 2 6E 2 x k 2E w k 2E x k . The simulation diagram is then given as: 6
2
1
E w(k+2) x(k) w(k+1) 1 y(k)
w(k) E ) 5
4 2e) 6E 3 2
x k
E3 y k E3 y k E 3 {w(k )} 6E 3 x k
6E 3 y k 2 x k 2 w k . The simulation diagram of the system is:
6 x(k)
) w(k+3) 1
E w(k+2) 1
E w(k+1) 1
E w(k) 2 y(k) 3a)
x(k) 1 q2(k) 1 q1(k)
E
E 3 y(k) 4
2 By setting q1 k and q2 k after delay operators as above, we have: q1 k 1 q2 k E q1 k q2 k 1 2q1 k 4q2 k x k 4 q2 k x k E q2 k 2 q1 k
2
E E 4 q2 k
1 q2 (k ) E 4
and q1 k 2
E q2 k q1 k 2
q1 k
E 1
q2 k
E 4q 2 k x(k )
x(k ) 1
q2 k
E E E 2 2 E
x(k )
4E 2 1
x k
4E 2 Now,
y k 3q1 k
E2 Hence, y k q2 k 1 3
x k
4E 2 3q1 k Eq2 k E2 E2
x k
4E 2 E2 3
x k
E 2 4E 2 Therefore, the operational transfer function of the system is:
H (E) E2 3
.
E 2 4E 2 x k 3b)
1 q1(k)
E x(k)
0.5 y(k) 2 1 q2(k)
E By setting q1 k and q2 k as above, we have: q1 k 1 2q1 k 0.5q2 k ( E 2) q1 k 0.5q2 k 0.5
q2 k
E 2 q1 k q2 k 1 x k q1 k q2 k x k
1 E 2 x k
0.5
q2 k
E 2 E q2 k
E q2 k 0.5
E 2 1 x k 1 x k E 2 q2 k q2 k 1 1 x k x k E 2 Hence, 1 1 q2 (k ) E 2
x k
0.5
E 1
E 2
E 3
x k
2
E 3E 2.5 E 2
E E 2 1 E 2 0.5 x k On the other hand, we have:
y k q1 k
1
E 2
1 q2 k 1
E 2 0.5 q2 k E 1.5 q2 k x k E 1.5 E 3 E2 E2 E 2 3E x k 2.5 3E 2.5 q2 k x k Finally, the operational transfer function of the system can be obtained as follows: H (E) 2 E 2 7.5E 7
E 2 E 2 3E 2.5
1 2 E 3.5
E 3E 2.5
2 4a) x(k) 1 q3(k)
E
0.9 1 q2(k)
E 1 q1(k)
E y(k) 0.7 From the diagram, we obtain the following expressions: q1 k 1 q2 k q2 k 1 q3 k q3 k 1 x k y k q1 k 0.9q3 k
0.7q3 k 4b)
%Define Input
N = 100;
x = ones(1,N+1);
% Initial Conditions
time = zeros(1,N+1);
q1 = zeros(1,N+1);
q2 = zeros(1,N+1);
q3 = zeros(1,N+1);
y = zeros(1,N+1);
y(1) = q1(1)  0.7*q3(1);
%Iteration
for k = 0:N
n = k + 1;
time(n+1)= k;
q1 (n+1)= q2(n);
q2 (n+1)= q3(n);
q3 (n+1)= 0.9*q3(n) + x(n);
y
(n+1)= q1(n+1)  0.7*q3(n+1);
end
scatter(time,y,'b')
grid on
xlabel('time (k)') , xlim([0,N]), set(gca,'XTick',0:10:N)
ylabel('y') , ylim([1.5,3.5]), set(gca,'YTick',1.5:0.5:3.5)
title('H(E) = (10.7E^2)/E^2(E0.9)') ...
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Full Document
 Fall '09
 MALIKELBULUK
 SIGNALS, Recurrence relation, Fibonacci number, simulation diagram, operational transfer function

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