EECE 200 Homework 1 Solution, Fall 2012

American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering EECE200 – Introduction to Engineering– Fall 2012‐2013 Homework 1 Solution Problem 1 [30 points] a. The amplitude of y1 is 10V [1] The amplitude of y2 is 2V [1] The amplitude of y3 is 6V [1] b. For y1 T= 0.0314-0 = 0.0314 sec [1] F= 1/T = 31.8309 Hz [1] ω = 2πF = 200 rad/sec [1] For y2 T= 0.002618-0.000524 = 0.002094 sec [1] F= 1/T = 477.464 Hz [1] ω = 2πF = 3000 rad/sec [1] For Y3 T= 0.00628-0 = 0.00628 sec [1] F= 1/T = 159.1549 Hz [1] ω = 2πF = 1000 rad/sec [1] c. The third second signal has the highest frequency (477.464 Hz)   [1] d. For Y1 At T = 0.00786 sec Y1 =10 ; 10=10sin(200t+φ1) ; sin(200t+φ1)=1 ; 200t+φ1=π/2 [1] φ1=0 rad/s [1] φ1=0 deg [1] On the graph at T=0 sec the signal has a zero amplitude. Substituting in the equation verifies the calculated results [1] For Y2 At T = 0 sec Y2 =2 ; 2=2sin(3000t+φ2) ; sin(3000t+φ2)=1 ; 3000t+φ2=π/2 [1] φ1=π/2 rad/s [1] φ1=90 deg [1] On the graph at T=0.005 sec the signal has a 0.5 amplitude. Substituting in the equation verifies the calculated results [1] For Y3 At T = 0.0015 sec Y3 =-6 ; -6=6sin(1000t+φ3) ; sin(1000t+φ3)=-1 ; 1000t+φ3=-π/2 [1] φ3=π rad/s [1] φ3=180 deg [1] On the graph at T=0 sec the signal has a zero amplitude. Substituting in the equation verifies the calculated results [1] e. 1 point for each correct column (total 3) and one point for labeling each axis (total 2) [5 points]. Problem 2 [10 points] a. 108Mhz to 88Mhz [1] b. 200KHz [1] (108Mhz‐88Mhz)/200KHz = 100 stations [1] c. Radio Monte Carlo [1]. Frequency is 1233 KHz [1] No enough bandwidth [1] d. A color TV has bigger bandwidth as it needs to transmit the different color [3] Problem 3 [25 points] a. 4x4 = 16 pixels [2] b. 2 bits are needed to represent the pixel because we have 4 levels of gray which are represented with 2 bits (2^2=4 levels) [1] for the number of bits and [1] for verification Very Dark red: 00 [2] Dark red: 01 [2] Light red: 10 [2] White: 11 [2] c. [0.5] for each correct element in the matrix 2 0 1 2 1 3 0 3 0 2 3 1 3 3 2 0 d. 16 (elements)x2(bits for each element)x25 (frames per second)x60 (sec per minute)x5 (minutes)/8 (bits per byte)= 30000 bytes. 30000/1024= 29.2968 Kbytes. 29.2968/1024= 0.02861 Mbytes [4] 0 2 4 6 8 10 12 31.83 159.154 477.464 Amplitude Frequency Frequency Domain Plot Amplitude e. No the size will not change [1] Problem 4 [35 points] a. 110111001.0110102 = 1x28+1x27+0x26+1x25+1x24+1x23+0x22+0x21+1x20+0x2-1+1x2-2+1x2- 3+0x2-4 +1x2-5+0x2-6 = 441.40625 [9] b. 548.2310 = 111001010.00111010111 [4] on the answer and [5] on the steps 458/2 229 0 229/2 114 1 114/2 57 0 57/2 28 1 28/2 14 0 14/2 7 0 7/2 3 1 3/2 1 1 1/2 0 1 0.23x2=0.46; 0.46x2=0.92; 0.92x2=1.84; 0.84x2=1.68; 0.68x2=1.36; 0.36x2=0.72; 0.72x2=1.44; 0.44x2=0.88; 0.88x2=1.76…………………….. c. 43768 = 4x83+3x82+7x81+6x80 = 230210 = 1000111111102[2] for the answer and [2] for the steps 2302/2 1151 0 1151/2 575 1 575/2 287 1 287/2 143 1 143/2 71 1 71/2 35 1 35/2 17 1 17/2 8 1 8/2 4 0 4/2 2 0 2/2 1 0 1/2 0 1 d. 1010111102 = 15EHEX [4] 11102=EHEX 01012=5HEX 00012=1HEX e. 1A8B16 = 1x163+10x162+8x161+11x160 = 6795 [9]