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HW5_Soln

# HW5_Soln - Escuti/Masnari 2007 ECE200 Fall 2007 Homework 5...

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 5 Solutions Periodic Signals in the Time Domain Page 1 of 14 1. [16 pts total, 1/3 pts each part each quantity] Find the amplitude, peak-to-peak value, average (DC) value, frequency, period, and phase angle of the following sinusoidal waveforms: a) v ( t ) = 3 V + (12 V )cos(2 ! 100 t + ! 4 ) Amplitude = 12 V Peak-to-peak value = 24 V Average (DC) Value = 3 V Frequency = 100Hz Period = 1/100Hz =0.01s θ = π /4 radians b) v ( t ) = 3 V + (12 V )cos 2 ! 100 t + ! 4 " # \$ % & Amplitude = 12 V Peak-to-peak value = 24 V Average (DC) Value = 3 V Frequency = 0.01Hz Period = 1/0.01Hz =100s θ = π /4 radians c) v ( t ) = ! 3 V ! (12 V )cos(2 " 100 t + " 4 ) Amplitude = 12 V Peak-to-peak value = 24 V Average (DC) Value = -3 V Frequency = 100Hz Period = 1/100Hz =0.01s θ = π /4 ± π = 5 π /4 or –3 π 4 radians (to remove minus sign from cos(-) ) d) v ( t ) = 3 V + (12 V )sin(2 ! 100 t + ! 4 ) Amplitude = 12 V Peak-to-peak value = 24 V Average (DC) Value = 3 V Frequency = 100Hz Period = 1/100Hz =0.01s To find the phase, we need to convert sine to a cosine. Consider the following identity: sin( α ) = cos( α - π /2) Thus, sin 2 ! 100 t + ! 4 " # \$ % & = cos(2 ! 100 t + ! 4 - ! 2 ) = cos(2 ! 100 t - ! 4 ) θ = – π /4 radians

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Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 5 Solutions Periodic Signals in the Time Domain Page 2 of 14 e) v ( t ) = 3 V ! (12 V )sin(2 " 100 t + " 4 ) Amplitude = 12 V Peak-to-peak value = 24 V Average (DC) Value = 3 V Frequency = 100Hz Period = 1/100Hz =0.01s Using the above method, -sin 2 ! 100 t + ! 4 " # \$ % & = cos(2 ! 100 t - ! 4 ± ! ) Therefore, the phase angle is either 3 π /4 radians or - 5 π /4 radians f) v ( t ) = 3 V + (12 V )cos(2 ! 100( t + 0.001)) Amplitude = 12 V Peak-to-peak value = 24 V Average (DC) Value = 3 V Frequency = 100Hz Period = 1/100Hz =0.01s Since v ( t ) = 3 V + (12 V )cos(2 ! 100( t + 0.001)) = 3 V + (12 V )cos(2 ! 100 t + 0.2 ! ) θ = 0.2 π radians g) v ( t ) = ! 3 V ! (12 V )cos 2 2 " 100 t ( ) Since cos( A ) 2 = 1 + cos(2 A ) 2 , then v ( t ) = - 3 V - (12 V )cos 2 2 ! 100 t ( ) = - 3 V - 6 V - 6 V cos(2 ! " 200 t ) = - 9 V + 6 V cos(2 ! " 200 t ± ! ) Amplitude = 6 V Peak-to-peak value = 12 V Average (DC) Value = -9 V Frequency = 200Hz Period = 1/200Hz =0.005s θ = ± π radians
Escuti/Masnari 2007 ECE200 – Fall 2007 – Homework 5 Solutions Periodic Signals in the Time Domain Page 3 of 14 h) v ( t ) = 3 V + (12 V )sin 2 2 ! 100 t ( ) Since sin 2 ( A ) = 1 - cos(2 A ) 2 , then v ( t ) = 3 V + (12 V )sin 2 2 ! 100 t ( ) = 3 V + 6 V - 6cos(2 ! 200 t ) = 9 V + 6cos(2 ! " 200 t ± ! ) Amplitude = 6 V Peak-to-peak value = 12 V Average (DC) Value = 9 V Frequency = 200Hz Period = 1/200Hz =0.005s θ = ± π

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HW5_Soln - Escuti/Masnari 2007 ECE200 Fall 2007 Homework 5...

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