Question

4. Given the following signals, a) (6 pts). Find the signal energy in the voltage x(t) = 10rect b) (6 pts). Find the average signal power in the following periodic voltage, x(t). Express your answer in dBV 4AAK x(t) 9 10 11 12 Given the following amplifier 22 kn 56 kn o ww x(t) c) (6 pts). What is the loss in decibels? d) (7 pts). If the input to the amplifier is 3cos(2T300t+25°) V, what is the output in dBmV

Answer 1

Solution:

(a) x (t) = 10 rect (t/0.002) The energy of x (t) implies E = integral^infinity_-infinity x^2 (t) dt = integral^0.001_-0.001 10^2 dt = 100 integral^0.001_-0.001 dt = 100 t integral^0.001_-0.001 = 100 (0.001 + 0.001) = 100 times 0.002 = 0.2 J x (t) = t + 1 -3 < t < -2 The period T = 59 The energy F_x = integral^2_-3 x^2 (t) dt = integral^2_-3 (t + 1)^2 dt = (t + 1)^3/3 Vertical-bar ^2_-3 = 1/3 [3^3 - (-9 + 1)^3] = 1/3 [27 + 8] = 35/3 J \r\n in dBr E_x = 10 log (35/3) = 10.66 dBr c The output y (t) = -22k/56k times x (t) = -0.39 (t + 1) -3 < t < 2 energy E_y = (0.39)^2 35/3 J = 1.7745 = 2.49 dB loss = 10.66 - 2.49 dB = 8.17 dB d if input x (t) = 3 cos (2 pi 300 t + 250) v output y (t) =