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2. A DSB-SC AM signal is modulated by the signal m (t) = 2cos2000nt + cos6000nt. The modulated signal is u(t) = 100m(t)cos (2nf.t), where fe = 1 MHz. Determine and sketch the spectrum of the AM signal.“ Answer: 1) The modulated signal is written as u(t) = 100(2 cos(2n103t) + cos(2n × 3 × 10³t)) cos(2nf.t) = 100[cos(2n(fc + 103)t) + cos(27(f. – 103)t)] + 100[cos(2n(fe + 3 x 103)t) + cos(27(fe – 3 x 103)t)]« - - Taking the Fourier transform of the previous expression, we obtain - U(f) = 50[8(f - (fe + 103)) + 8(f + fc + 10³) + 8(f – (fc + 10³)) + 8(f + fc + 10³ )]« + 25[8(f – (fe + 3 x 103)) + 8(f + fc + 3 × 10³) + 8(f – (fc – 3 × 10³)) + 8(f + fc – 3 × 10³)« - The spectrum of the signal is depicted in the next figure |50 25 -1003 -1001 -999 -997 997 999 1001 1003 KHz

2. A DSB-SC AM signal is modulated by the signal m (t) = 2cos2000nt + cos6000nt. The modulated signal is u(t) = 100m(t)cos (2nf.t), where fe = 1 MHz. Determine and sketch the spectrum of the AM signal.“ Answer: 1) The modulated signal is written as u(t) = 100(2 cos(2n103t) + cos(2n × 3 × 10³t)) cos(2nf.t) = 100[cos(2n(fc + 103)t) + cos(27(f. – 103)t)] + 100[cos(2n(fe + 3 x 103)t) + cos(27(fe – 3 x 103)t)]« - - Taking the Fourier transform of the previous expression, we obtain - U(f) = 50[8(f - (fe + 103)) + 8(f + fc + 10³) + 8(f – (fc + 10³)) + 8(f + fc + 10³ )]« + 25[8(f – (fe + 3 x 103)) + 8(f + fc + 3 × 10³) + 8(f – (fc – 3 × 10³)) + 8(f + fc – 3 × 10³)« - The spectrum of the signal is depicted in the next figure |50 25 -1003 -1001 -999 -997 997 999 1001 1003 KHz

Introductory Circuit Analysis (13th Edition)
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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2. A DSB-SC AM signal is modulated by the signal m (t) = 2cos2000nt + cos6000nt. The modulated signal is 100m(t)cos (2nft), where fe = 1 MHz. Determine and sketch the spectrum of the AM signal. и(t) Answer: 1) The modulated signal is written as - u(t) = 100(2 cos(2n103t) + cos(2n × 3 x 10³t)) cos(2nf.t)- 100[cos(2n(f. + 10³)t) + cos(2n(fc – 103)t)] + 100[cos(27(f. + 3 x 103)t) + cos(27(fe - 3 × 10³)t)]« Taking the Fourier transform of the previous expression, we obtain - U(f) = 50[8(f - (fc + 103)) + 8(f + fc + 10³) + 8(f – (fc + 10³)) + 8(f + fc + 10³)]« + 25[8(f – (fc + 3 x 103)) + 8(f + fc + 3 × 103) + 8(f – (fe – 3 × 103)) + 8(f + fe – 3 x 103)e The spectrum of the signal is depicted in the next figure |50 25 -1003 -100i -999 -997 997 999 1001 1003 KHz
Transcribed Image Text:2. A DSB-SC AM signal is modulated by the signal m (t) = 2cos2000nt + cos6000nt. The modulated signal is 100m(t)cos (2nft), where fe = 1 MHz. Determine and sketch the spectrum of the AM signal. и(t) Answer: 1) The modulated signal is written as - u(t) = 100(2 cos(2n103t) + cos(2n × 3 x 10³t)) cos(2nf.t)- 100[cos(2n(f. + 10³)t) + cos(2n(fc – 103)t)] + 100[cos(27(f. + 3 x 103)t) + cos(27(fe - 3 × 10³)t)]« Taking the Fourier transform of the previous expression, we obtain - U(f) = 50[8(f - (fc + 103)) + 8(f + fc + 10³) + 8(f – (fc + 10³)) + 8(f + fc + 10³)]« + 25[8(f – (fc + 3 x 103)) + 8(f + fc + 3 × 103) + 8(f – (fe – 3 × 103)) + 8(f + fe – 3 x 103)e The spectrum of the signal is depicted in the next figure |50 25 -1003 -100i -999 -997 997 999 1001 1003 KHz
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