EECE 200 Homework 1 Solution, Fall 2011

American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering EECE200 – Introduction to Engineering– Fall 2011-­‐2012 Homework 1 Solution Problem 1 [30 points] a. The amplitude of y1 is 10V [1] The amplitude of y2 is 5V [1] The amplitude of y3 is 2V [1] b. For y1 T= 0.0209-0 = 0.0209 sec [1] F= 1/T = 47.74648 Hz [1] ω = 2πF = 300 rad/sec [1] For y2 T= 0.00733-0.00315 = 0.004188 sec [1] F= 1/T = 238.732 Hz [1] ω = 2πF = 1500 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 second signal has the highest frequency (1500 Hz) [1] d. For Y1 At T = 0.00522 sec Y1 = 10 ; 10 = 10 sin (300t+φ1) ; sin (300t+φ1)=1 ; 300t+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 = 5 ; 5 = 5 sin (1500t+φ2) ; sin (1500t+φ2)=1 ; 1500t+φ2=π/2 [1] φ1=π/2 rad/s [1] φ1=90 deg [1] On the graph at T=0.001 sec the signal has a 0 amplitude. Substituting in the equation verifies the calculated results [1] For Y3 At T = 0.00157 sec Y3 =-2 ; -2 = 2 sin (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. Radio one [1]. Frequency is 105.5 MHZ [1]. Range of FM station is 88 MHz to 108MHz [1] b. Total bandwidth of AM radio is 1700-­‐530= 1170 KHZ. Since each station is assigned a 10KHz bandwidth, total number of stations that the AM bandwidth can afford = 1170/10=117 stations.[3] c. Yes the TV station bandwidth is bigger than the FM stations because they are transmitting audio as well as video which requires more data or bandwidth [2] d. No the bandwidth needed for a black and white TV is smaller since we are sending data for one color only. [2] Problem 3 [25 points] a. 5x5 = 25 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 Black: 00 [2] Dark grey: 01 [2] Light Grey: 10 [2] White: 11 [2] c. 0 3 2 3 1 2 3 1 3 3 0 3 2 3 2 0 2 4 6 8 10 12 47.74648 159.1549 238.732 Amplitude Frequency Frequency domain plot 3 3 3 1 3 2 0 1 3 0 [0.32] for each correct element in the matrix d. 25 (elements)x2(bits for each element)x30 (frames per second)x60 (sec per minute)x2.5 (minutes)/8 (bits per byte)= 28125 bytes; 28125/1024 = 27.46582 Kbytes; 27.46582/1024 = 0.02682 Megabytes [5] Problem 4 [35 points] a. 10011101.0111012 = 1x27+0x26+0x25+1x24+1x23+1x22+0x21+1x20+0x2-1+1x2-2+1x2-3+1x2-4 +0x2-5+1x2-6 = 157.453125 [9] b. 485.2310 = 111100101.00111010111000010100011110101 [4] on the answer and [5] on the steps 485/2 242 1 242/2 121 0 121/2 60 1 60/2 30 0 30/2 15 0 15/2 7 1 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. 15748 = 1x83+5x82+7x81+4x80 = 89210 = 11011111002 [2] for the answer and [2] for the steps 892/2 446 0 446/2 223 0 223/2 111 1 111/2 55 1 55/2 27 1 27/2 13 1 13/2 6 1 6/2 3 0 3/2 1 1 1/2 0 1 d. 1000111112 = 11FHEX [4] 11112=FHEX 00012=1HEX 00012=1HEX e. 1C8B16 = 1x163+Cx162+8x161+Bx160 = 7307 [9]