Question

3(20%) Assume a message signal is given by m(t) = 4 cos(2π//) + cos(4π.t). Let x (t)-5m(t) cos(2t f t) + 5m(t) sin( 2 fct), where m(t) İs the Hilbert Transform of m(t). (10%) (a) Derive x(t) (10%) (b) Prove, by sketching the spectra, that x(t) is a lower-sideband SSB signal of m(t).

Answer 1

M(t) = 4 cos (2 pi F_m t) + cos (4 pi f_m t) Now Hilbert transform of cos w t rightarrow^is sin w t Therefore m^(t) = 4 sin (2 pi f_m t) + sin (4 pi f_m t) Also, we know that 2 cos A cos B = cos (A + B) + cos (A - B) 2 sin A sin B = cos (A - B) - cos (A + B) Now x_c (t) = 5 m (t) cos 2 pi f_c t + 5 m^^(t) sin 2 pi f_c + x_c (t) = 5 times 4 cos 2 pi f_c + x cos 2 pi f_m t + 5 cos 2 pi f_c t cos 4 pi f_m t + 5 sin (2 pi f_c t) times 4 sin 2 pi f_m t + 5 sin 2 pi f_c t sin 4 pi f_m t rightarrow 10 cos (2 pi (f_c + f_m) t) + 10 cos 2 pi (f_c - f_m) t + 2.5 cos 2 pi (f_c + 2 f_m) t + 2.5 cos 2 pi (f_c - 2 f_m) + = + 10 cos 2 pi (f_c - f_m) t - 10 cos 2 pi (f_c + f_m) t + 2.5 cos 2 pi (f_c - 2 f_m) t - 2.5 cos 2 pi (f_c + 2 f_m) t