The transfer function H(F) of a filter is shownbelow. The input signal is given by :Vin(t) 4t3 cos10 TH-305 sin (2400074)Find the expression (as a function of time) forthe output signal of the filter.# (F)२1510ZH (F)-180°510(KH₂)

Given data The expression for the input signal is Vint=4+3cos104πt-30°+5sin24000πt The transfer function of a filter is shown via the magnitude and phase graph in the question. We need to find the expression for the output signal of a filter. Calculation of the input frequency From the first part of the expression of the input signal, ω1=2πf12πf1=1000πf1=5000 Hzf1=5 kHz From the second part of the expression of the input signal, ω2=2πf22πf2=12000πf2=24000 Hzf2=12 kHzDetermination of the magnitude and phase angle of the output signal at f1=5 kHz Since the given transfer function is of low pass filter, so it will allow only a pass of frequency f1=5 kHz. At 5 kHz magnitude is 1.75 times that of Vin and phase angle is 45° So, The magnitude of the output signal at f1=5 kHz Vout 1=1.75×5=8.75 V The phase angle of the output signal at f1=5 kHz ϕ=-45°-30°=-75°Determination of the expression for the output signal So, for the 4 V Dc input voltage, frequency is 0. The magnitude of the output signal is, Vout, Dc=4×2=8 V Now on combining all of the above output calculated parameters, Voutt=8+8.75cos104πt-75° Therefore the expression for the output signal is Voutt=8+8.75cos104πt-75°. Conclusion The magnitude of the output signal at f1=5 kHz is 8.75 V. The phase angle of the output signal at f1=5 kHz is -75°. The expression for the output signal is Voutt=8+8.75cos104πt-75°.

Engineering

Electrical Engineering

The transfer function H(F) of a filter is shown below. The input signal is given by : Vin (t) = 4+3 cos (10¹ At-30°) + 5 sin (24000 nt) Find the expression (as a function of time) fure the output signal of the filter. # (f) 2 1 5 10 ZH (F) -180° 104 5 10 f² (KHz)

The transfer function H(F) of a filter is shown below. The input signal is given by : Vin (t) = 4+3 cos (10¹ At-30°) + 5 sin (24000 nt) Find the expression (as a function of time) fure the output signal of the filter. # (f) 2 1 5 10 ZH (F) -180° 104 5 10 f² (KHz)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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