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by: Nhi Phan
Nhi Phan
Montgomery College
GPA 4.0

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About this Document

Study guide exam1
Dr. Omi
Study Guide
50 ?




Popular in signals

Popular in Engineering and Tech

This 7 page Study Guide was uploaded by Nhi Phan on Wednesday September 28, 2016. The Study Guide belongs to ENEE222 at Montgomery College taught by Dr. Omi in Fall 2016. Since its upload, it has received 9 views. For similar materials see signals in Engineering and Tech at Montgomery College.


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Date Created: 09/28/16
(P 1.11) ______________ (i) The phase shift (phase at t=0) equals 0.25 < pi/2. The first zero will occur when 500*pi*t + 0.25 = pi/2 or t = (pi/2 - 0.25)/(500*pi) = 0.841 ms (ii) The period of x(t) is T = (2*pi)/(500*pi) = 4.0 ms Thus time interval [0, 0.01] corresponds to 2.5 periods of the sinusoid. The zeros will occur for values of t such that 500*pi*t + 0.25 = (2*k+1)*(pi/2) where k is integer (i.e., when the LHS equals an odd multiple of pi/2. Solving for t (as in problem 1), we have t = (2*k+1)*(0.001) - 0.0005/pi The first term on the RHS above is a rational number, the second one is irrational. Therefore each value of t given by the above equation is irrational. The values of t generated by MATLAB are multiples of 0.001, hence rational. Therefore, none of the 101 values in the vector x are zero. This can be verified using the MATLAB command min(abs(x)) (P 1.16) ______________ (i) We have cos(w*(n+1)+q) = cos(w*n+q)*cos(w) - sin(w*n+q)*sin(w) and cos(w*(n-1)+q) = cos(w*n+q)*cos(-w) - sin(w*n+q)*sin(-w) = cos(w*n+q)*cos(w) + sin(w*n+q)*sin(w) Adding the two equations together, we have cos(w*(n+1)+q) + cos(w*(n-1)+q) = 2*cos(w*n+q)*cos(w) (ii) We have x[1] = cos(W*2 + q), x[2] = cos(W*3 + q), x[3] = cos(W*4 + q) The formula derived in part (i) gives (n = 2): x[1] + x[3] = 2*cos(w)*x[2] and thus cos(w) = (x[1] + x[3])/(2*x[2]) = 0.3624 and w = arccos(0.3624) = 1.200 (between 0 and pi). Therefore x[1] = A*cos(1.2 + q) = 1.7740 x[2] = A*cos(2.4 + q) = 3.1251 x[3] = A*cos(3.6 + q) = 0.4908 We have x[2]/x[1] = cos(2.4+q)/cos(1.2+q) = 3.1251/1.7740 = 1.7616 and using cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) we obtain cos(2.4)*cos(q) - sin(2.4)*sin(q) = = 1.7616*(cos(1.2)*cos(q) - sin(1.2)*sin(q)) Dividing both sides by cos(q) we obtain the following equation in tan(q): cos(2.4) - sin(2.4)*tan(q) = 1.7616*(cos(1.2) - sin(1.2)*tan(q)) or: 0.9664*tan(q) = 1.3757 leading to tan(q) = 1.4235 Since arctan(q) = 0.9584, there are two possible values for q: q = 0.9584 and q = 0.9584 + pi = 4.1270 A phase shift of pi reverses the sign of cos(.), so both values of q are acceptable solutions (same x[2]/x[1] ratio). We take the one consistent with a positive value of A: cos(1.200 + 0.9584) = -0.5544 cos(1.200 + 4.1270) = 0.5544 and thus q = 4.127, A = 1.7740/0.5544 = 3.200 (P 1.17) ______________ (i) Since w = 7*pi/9 = (7/18)*2*pi, it follows that sinusoid is periodic with period N = 18. MATLAB: n = 0:71; %(four periods) w = 7*pi/9; x = cos(w*n + pi/6); bar(n,x) %(bar graph) (ii) We have cos(pi*n) = (-1)^n, sin(pi*n) = 0 Therefore cos(a + pi*n) = cos(a)*cos(pi*n) - sin(a)*sin(pi*n) = cos(a)*cos(pi*n) Taking a = 7*pi*n/9, we have cos(16*pi*n/9 + pi/6) = x1[n]*cos(pi*n) = x2[n] To express x2[n] using a frequency w in [0,pi]: x2[n] = cos(16*pi*n/9 + pi/6) = cos(-2*pi*n/9 + pi/6) (reduce w by 2*n) = cos(2*pi*n/9 - pi/6) (cos(t) = cos(-t)) Thus w = 2*pi/9. We also have cos(2*pi*n/9 - pi/6) = cos(2*pi*n/9 + 11*pi/6) and now the phase is also in [0,2*pi). (P 1.19) ______________ (i) We know that w = W*T_s, i.e., w = 150*pi*(3/1000) = 0.45*pi (ii) w is a rational multiple of pi, therefore x[n] is periodic. The period N is determined by expressing w as (r/N)*2*pi where r and N have no common factors. In this case, w = (9/40)*2*pi so N=40. (iii) x[n] is constant for all n if w = 2*k*pi i.e., T_s = (2*k*pi)/(150*pi) = k*(13.3333...) ms or f_s = 1/T_s = 75/k samples/sec for some integer k. x[n] alternates in value between cos(q) and -cos(q) if w = pi + 2*k*pi i.e., T_s = (0.5 + k)*(13.3333...) ms or f_s = 1/T_s = 75/(k+0.5) samples/sec for some integer k.


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