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Solution1 - EE 200 Fall 2008(Weber Homework 1 Solutions 1...

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EE 200 – Fall 2008 (Weber) Homework 1 Solutions 1. The function definition is divided into four parts: y = x 2 if 2 x ≤ − 1 x if 1 x 1 x + 2 if 1 x 2 0 otherwise 2. ( f 1 f 2 )( x ) = 0 . 5 signum (2 sin( πx )) ( f 2 f 1 )( x ) = 2 sin( π × 0 . 5 signum ( x )) ( f 1 f 2 )(x) 1 1 2 -1 -2 -1 x ( f 2 f 1 )(x) 1 1 2 -1 -2 -1 x -2 2 3. (a) This is a graph of a function. x X y Y a 1 b 1 c 2 (b) Not a graph of a function. For a in domain there are two possible values in the range. (c) Not a graph of a function. The value of the function is not given for c in the domain. 4. From the definition of RMS voltage given on page 7 of the text, V RMS = radicalBigg 1 T integraldisplay T 0 ( V Peak sin(2 π 60 t )) 2 dt = V Peak radicalBigg 1 T integraldisplay T 0 sin 2 (2 π 60 t ) dt 1
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The thing under the square root works out to 1 T integraldisplay T 0 sin 2 (2 π 60 t ) dt = 1 T integraldisplay T 0 parenleftbigg 1 2 1 2 cos(2 × 2 π 60 t ) parenrightbigg dt = 1 2 1 2 T integraldisplay T 0 cos(4 π 60 t ) dt = 1 2
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