B st 50 cos2 108 t cos2 104 t 4 this signal is

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Unformatted text preview: 2(fm + ∆fmax ) = 2(10 + 20) kHz = 60 kHz. b. s(t) = 50 cos[2π 108 t] cos[2π 104 t + π /4]. This signal is in the form Am(t) cos[2π 108 t], where m(t) = cos(2π 104 t + π /4). Hence, the signal is amplitude modulated using DSB-SC. As in part a., the message signal, m(t), is a tone with frequency fm = 10 kHz. Hence, the bandwidth occupied by s(t) is BW = 2fm = 20 kHz. 1-12 Solution s(t) = 50 cos(ωc t + 5 sin(2π 1000t)). This signal is in the form A cos(ωc t + θ(t)), where θ(t) = 5 sin(2π 1000t). Hence, the signal is purely angle modulated. The message si...
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