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Solution1

# Solution1 - EE 200 Fall 2009(Weber Homework 1 Solutions 1...

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Unformatted text preview: EE 200 Fall 2009 (Weber) Homework 1 Solutions 1. The function definition is divided into four parts: -x - 2 2x + 1 y= 1 -2x + 1 0 b. Frequency is 1 period if -2 x -1 if -1 x 0 if 0 x 2 otherwise 2. a. The signal has 4 complete cycles in 7 msec, so the period is 7msec/4 = 1.75msec. = 1 1.75msec = 571Hz. c. Amplitude is about 3 volts 3. d. = 2f = 2 571Hz = 3590radians/sec. (f1 f2 )(x) = 3 2 sgn( sin(x)) 2 3 (f2 f1 )(x) = 2 3 sin( sgn(x)) 3 2 (f1 f2)(x) 2 (f2 f1)(x) 1 1 x -2 -1 -1 1 2 -2 -1 -1 1 2 x -2 4. a. -4 6 3 -1 = -18 b. 5 6 -4 -3 = -20 -15 -24 -18 c. -1 4 2 3 2 6 -5 -2 7 -3 = -5 8 17 23 6 1 d. 4 -1 9 5 -3 6 2 0 -4 1 = 9 -6 -55 27 32 -31 5. a. From the definition of RMS voltage given in the assignment, VRMS = = 1 T VP eak T (VP eak sin(2f0 t))2 dt 0 1 T T sin2 (2f0 t)dt 0 Using the trig identity from the assignment, the thing under the square root works out to 1 T T sin2 (2f0 t)dt 0 = = = = 1 T T 0 1 1 - cos(2 2f0 t) dt 2 2 T 1 1 - 2 2T cos(4f0 t)dt 0 t=T t=0 1 1 1 sin(4f0 t) - 2 2T 4f0 1 -0 2 1 1 = VP eak 2 2 since T = 1/f0 . Therefore VRMS = VP eak b. If the VRMS is 120 Volts, then VP eak is about 170 Volts. 2 ...
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