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Part A The periodic square-wave voltage seen in (Figure 1) is applied to the circuit shown in (Figure 2). Select the correct first nonzero term in the Fourier series that represents the steady-state voltage vo if Vm = 157 V and the period of the input voltage is 4n us. In the expressions below the phase angle is expressed in degrees. O vol = 26.83 cos(1500t +63.43) V O vol = 26.83 cos(500t +63.43°) V O vol = 60 cos(500t +63.43) V O vol = 60 cos(1500t +63.43°) V O vol = 26.83 sin(500t +63.43) V Figure < 1 of 2 O vol = 60 sin(500t +63.43°) V Vi Submit Request Answer Vm Part B 0 T/2 T Select the correct second nonzero term in the Fourier series that represents the steady-state voltage v, if Vm = 157 V and the period of the input voltage is 4rt us. In the expressions below the phase angle is expressed in degrees. -Vm O V.3 = 20 cos(2500t - 146.31°) V O V3 = 16.64 sin(1500t – 146.31°) V
Figure 2 2 of 2 > 1012 w + + Vi 10 mH V.
Part B Select the correct second nonzero term in the Fourier series that represents the steady-state voltage v, if Vm= 15 V and the period of the input voltage is 47 ps. In the expressions below the phase angle is expressed in degrees. Vo3 = Ο υο3 = 20 cos(2500t - 146.31°) V 16.64 sin(1500t – 146.31°) V O V3 = 20 cos(1500t – 146.31°) V 003 = 16.64 cos(1500t – 146.31") V O V.3 = 16.64 cos(2500t - 146.31°) V Vo3 = 20 sin(1500t – 146.31°) V Submit Request Answer
Part Select the correct third nonzero term in the Fourier series that represents the steady-state voltage v, if Vm = 157 V and the period of the input voltage is 47 us. In the expressions below the phase angle is expressed in degrees. V.5 = O V5 = 11.14 sin(2500t +21.80) V 12 cos (2500t + 21.80°) V O Vo5 = 12 sin(2500t +21.80°) V O vos = 11.14 cos(2500t + 21.80°) V 12 cos(1500t +21.80°) V Vos = 11.14 cos(1500t + 21.80°) V O 005 Submit Request Answer