An amplitude modulated (AM) signal is given by PAM (t) = [A + m(t)]cos (wet), let m(t) = A cos(wmt). For =0.75, find the power efficiency. A. 100% D. 25% B. 75% C. 22% A single tone AM (DSB+C) modulated signal is shown in the figure below. Calculate the power efficiency (in percentage). A. 12.50% B. 22.22% C. 11.11% D. 33.33% E. 16.67% E. 27.27% A. m(t) = 40 sin (500xt) +20 sin (1000mt) B. m(t) = 20000 cos (500nt) +20000 cos (1000xt) C. m(t) = cos (500nt) + cos (1000nt) D. m(t) = 8 sin (500mt) + 4 sin (1000ml) E. m(t) = 8 cos (500mt) + 4 cos (1000ml) 3 0 -3 -9 The following signal is a frequency modulated signal with Ky=20,000 radian/sec/volt. Determine the message signal m(t): PFM (t) = 5cos (2000000nt +40 sin(500nt) + 20 sin(1000nt))