hmw4_sol_sp08

hmw4_sol_sp08 - i ECE220 Homework Assignment #4 Solutions...

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i ECE220 Homework Assignment #4 Solutions Problem 4.33 Write the following signals in phasor notation: v 1 ( t ) = 4cos(2 π 3000 t + 4 π/ 5) v 2 ( t ) = 60cos(2 π 400 t - π ) v 3 ( t ) = Re [10 e j (2 π 1000 t - π/ 3) ] v 4 ( t ) = - 3 . 4cos(500 t ) v 5 ( t ) = - 3 . 5sin(400 t ) v 6 ( t ) = 100 v 7 ( t ) = - 100 Solution ˜ v 1 = 4 e j 4 π 5 , with angular frequency ω 0 = 2 π 3000 . ˜ v 2 = 60 e , with angular frequency ω 0 = 2 π 400 . v 3 ( t ) = 10cos(2 π 100 t - π 3 ); ˜ v 3 = 10 e j ( - π/ 3) , with angular frequency ω 0 = 2 π 1000 . v 4 ( t ) = - 3 . 4cos(500 t ) = 3 . 4cos(500 t + π ); ˜ v 4 = 3 . 4 e , with angular frequency ω 0 = 2 π 500 . v 5 ( t ) = - 3 . 5sin(400 t ) = 3 . 5cos(400 t + π/ 2); ˜ v 5 = 3 . 5 e jπ/ 2 , with angular frequency ω 0 = 2 π 400 . ˜ v 6 = 100 e j 0 , with angular frequency ω 0 = 0 . ˜ v 7 = 100 e , with angular frequency ω 0 = 0 . ————————————————————————————– Problem 4.36 It can be shown that the current flowing through a capacitor in an RC circuit is given by the phasor equation ˜ I c = jωRC 1 + jωRC ˜ I s
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ii where ˜ I s representsthephasorassociatedwiththeinputcurrentand ˜ I c represents the phasor associated with the capacitor current. Let i s ( t ) = 10cos(2 π 500 t ) , R = 1000 Ohm and C = 10 - 8 F. Find i c ( t ) using phasors. Plot the magnitude and phase of ˜ I c for frequencies in the range 0 to 100 KHz. Solution 1 ˜ I s = 10 e j 0 = 10 with angular frequency ω = 2 π 500 . The phasor associ- ated with the capacitor current i c ( t ) is given by: ˜ I c = j ωRC 1 + j ωRC I s = j 2 π 500 · 1000 · 10 - 8 1 + j 2 π 500 · 1000 · 10 - 8 10 = 0 . 0314 e j 1 . 5394 We can compute the capacitor current i c ( t ) in the time domain as i c ( t ) = Re (0 . 0314 e j 1 . 5394 e j 2 π 500 t ) = 0 . 0314cos(2 π 500 t + 1 . 5394) 2 Computing the magnitude spectrum of the phasor as a function of
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hmw4_sol_sp08 - i ECE220 Homework Assignment #4 Solutions...

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