1- Spectrum of the sinusoidal signals:f(t) = A cos(2лft + 0)Rules:1. The amplitude is always positive.-A cos 2πfot⇒ A cos(2лfot - 180º)2. The phase ids always measured from the real axis.A sin 2лfot A cos(2лfot - 90°)3. If the angular frequency is used (c), the amplitude is multiplied by 2π(a) f(t)=-7 sin(3πt) - 5cos (6πt + 90º)(b) f(t)=-4 sin(106πt) + 8cos (107πt + 170º)(c) f(t)==0(-0.5)" cos[n(wot + 10°)]

General Rules1. Amplitude is always positive: If you have a negative cosine, adjust by adding 180∘to…

Answered: 1- Spectrum of the sinusoidal signals:…

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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A voltage wave of 250 kHz has a Vmax-5 V and Vmin=-35 V. The duty cycle of this wave is 0.875. This voltage waveform has been applied across a 2.5 µH inductor for a very long time so that steady state has been achieved. Diagrams of the wave parameters and circuit are shown in this figure: L What is the average voltage of this wave? Average voltage = 35 V Voltage (arb, unit) 1.25 Minimum current: -3.5€ Amps 1 0.75 0.5 0.25 0 4.25 05 Have you integrated the wave over full cycle? 4.75 4 -1.25 Vmax 0.5 Vmin What is the average current flowing through the inductor? Average current = 3.56 A 15 Period/s On average, there is no power consumed by the inductor. Have you considered this? What are the maximum and minimum currents that flow through the inductor? Maximum current: 0.50 Amps D= 25 TH TH Have you used the relationship for current and voltage in an inductor? Have you considered the slope of the current wave?
:Converting the following sine function sin(3t - 76°) 184- :to a cosine function will result in the following A cos(3t + BO) ?What is the value of A
Show step by step solution and write legibly.
R = 250 and the sinusoidal alternating emf revice operates at Em = 40 v and fo = 55 Hz a) What is VRct) and amplitude Ve of UR(E) ? b) and the current
1- The wavelength of a 50Hz sinusoidal waveform is A-40ms B-100ms C-50ms D-25ms 2- The intensity of the signal waveform is A-regular waveform B-Period C-Wavelength D- Amplitude 3- The average value of an AC waveform is equal to A- 0.637* Vmax B-0.707 * Vmax C-1/√2 *Vmax - 4 Radian is Equal to A-Degree x Π 180 Π π 180 D-1.11 B-Radian x C- Degree x D- π 180
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determine which waveform is lagging in each of the following pairs. (d) -sin 5t, cos (5t+2)(e) sin 5t + cos (5t-45) here's the continuation. thank you again.
Consider the sinusoidal wave below. What is the frequency of the wave? 20 V - 15 V 10 V 5 V O V .t (μs) 10 20 30 -5 V - -10 V - -15 V – -20 V - te
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The RC circuit of the figure is fed from the constant magnitude variable frequency sinusoidal voltage sourceVin. At 100HZ, the R and L element each as voltage drop Urms · If the frequency of the source is changed to 50HZ, then voltage drop across R is. www Vin 000
A series RL circuit connected to an AC voltage sinusoidal source of 100 V peak voltage and 1 kHz frequency. If R = 1 ohm and L = 1 mH, then... a) The rms current through the circuit leads the source voltage b) The rms current through the circuit is 11.114 A 279.043° c) The rms current through the circuit is 15.718 A z-80.96° O d) The rms current through the circuit is 15.718 A 270.56°

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