Problem 3:The output signal of an AM modulator isgiven by4(+) = 5(cos (1800 πt) + cos(2200μt)]+20 cos (20007+)]☹ Determine the modulating signal m(t).and the carrier c(t)⑤ Determine the modulation indese, aDetermine the ratio of the power in thesidebands to the power in the carrier.Hint: start by using a trigonometricidentity.

Let's solve each part of this problem for the AM modulated signal given by:u(t) = 5(cos⁡(1800πt) +…

Answered: Problem 3: The output signal of an AM…

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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can you pls answer
The figure shows multiple modulation schemes for the signal that is given by the dashed line sinusoidal function, the X-axis is the time axis is seconds and the Y-axis is the amplitude. Answer the following: What is the quantization noise in a modulator that uses a 4-bits register for the analog to digital conversion of the given sinusoidal signal? Please give deatiled solution for good understanding
Q4- Drive a formula for NBEM by using Bessel function Q5- If an information signal X(t) cos(2n700t) + 10cos(2n5000t). Write the formula of the frequency modulation (FM) when Ac=20 , Fc=10® Hz, Kf= t2000 Q6-) If NBFM signal passes through a frequency multiplier. With B< 0.5 and fc=200kHz. Let fm ranges from 5OHZ to 15kHz, and let the maximum frequency deviation Aƒat the output be 75kHz. Find the required frequency multiplication n and the maximum allowed frequency deviation at the input.
Q2: A modulating signal is used to modulate a carrier signal(AM modulation). Let Pr , Pc, Psa. PusB, PLSB and u are the total power, carrier power, side band power, upper side band power, lower side band power and modulation index respectively. Prove the following: a. Pc = Pr () c. Pusp = Pasa =) P, 2+m2. b. PsB = Pr %3D 2+m2 2 (2+m2
Report: - 1. What is the advantage of angle modulation for (noise and interference), the bandwidth of transmission and the average power of the transmission? comparison with the amplitude modulation ? Eunloin briefly when the frequency modulation can be NREM and
Assume the modulated signal s(t) = 4cos(10πt)cos(1000πt). What type of modulation does this correspond to? a.DSB-SC b.SSB-SC c.FM d.AM
he FT of a signal f () as: Fle)= sin c'leı- 1005) + sinc’to + 1007). Using the modulation theorem, the signal /() is to be: lint, use the FT pair: tril-) cɔ -100 I + 1007 B. 2r A,+ 2 trit- 2 C. tri) cos(100) D. tri) sin(100) 9. The Fourier Transform of x(1) = e²" u(-t) is jas A. 2+ ja •B. 2- jo С. 2- jo D. 2-0
An FM signal has a maximum frequency deviation of 1500h and a modulating frequency of 500h, its total power is 6500w, applied to a resistive load of 1ohm, the carrier frequency is 400Mhz. Determine the voltage of the modulated signal and the modulation voltage.
IX. A carrier signal is amplitude modulated by a sinewave. The resulting AM waveform is shown below. a) b) c) d) 5.5V- 2V What is the percentage of modulation? What is the amplitude of the carrier signal? To achieve 75 percent modulation, what amplitude of the modulating signal is needed? Calculate the effective voltage of the AM signal
A sinusoidally modulated AM waveform is shown below: Determine: a) Peak amplitude of the upper and lower side frequencies. b) Peak amplitude of the unmodulated carrier. The modulation index. The carrier power and the power in the sidebands. The total transmitted power. 3 0 -3 -9 The transmission efficiency. The amplitude of the carrier which must be added or subtracted to attain a modulation index of 0.5. In AM modulation, the modulation index is kept < 100%, why?
Q2: A modulating signal is used to modulate a carrier signal(AM modulation). Let Pr , Pc, PSB, PUSB, PLSB and u are the total power, carrier power, side band power, upper side band power, lower side band power and modulation index respectively. Prove the following: c. Puss = PLSB = Pr (2+m2

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