A3 - Signals and Systems Fall 2015 Issued 9 September 2015...

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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 discrete-time 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 1-a) 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. 1-b) 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. 1-c) 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. 1-d) 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). 2-a) 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 _ 2-b) 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) 2-c) 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) ) 2-d) 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 2-e) 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) 3-a) 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 3-b) 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 4-a) 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 4-b) %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) = (1-0.7E^2)/E^2(E-0.9)') ...
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  • Fall '09
  • MALIKELBULUK
  • SIGNALS, Recurrence relation, Fibonacci number, simulation diagram, operational transfer function

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