EE422 Solution to Homework #23

EE422

Solution to HW#18

1a) Note: the Dynamic range is D=9-(-5)=14 volts. N=8 bits. MSE = (D²/12)2^-2N = ((14)²/12)2^-16 = 2.4923x10^-4 Watts

x(t) = 2 + 7cos1000t

P_ave = A₀²+A₁²/2 + A₂²/2 + ... = (2)² + (7)²/2 = 28.5 Watts

SNR = P_ave/MSE = (28.5)/(2.4923x10^-4) = 1.1435x10⁵ = 50.58 dB

b) x(t) = 0.5 + (1/4)cos1000t + (1/3)sin1000t = 0.5 + (5/12)cos2(1000t - 53.13° )

Therefore, the Dynamic range is D=11/12 - 1/12 = 5/6. MSE = (D²/12)2^-2N = ((5/6)²/12)2^-16 = 8.83x10^-7 Watts

P_ave = (0.5)² + (1/2)(5/12)² = 0.337 Watts

SNR = P_ave/MSE = (0.337)/(8.83x10^-7) = 3.814x10⁵ = 55.8 dB

c) Montavani's has the higher SNR (as you would expect if you ever listened to Heavy Metal!)

d) Minimum f_s = 2*16kHz = 32 kHz

e) An ideal low pass filter does not exist!! We need to sample slightly faster to enable us to recover the original signal with a non-ideal low pass filter!!

2a) Sketch of the probability density function:

b) Mean of x:

c) Mean of x²:

d) MSE due to quantization (using midpoint):

e) MSE due to quantization (using arbitrary value k):

f) Take derivative with respect to k:

. Thus, without any other information we should chose the midpoint of the quantizing region to represent our signal in order to minimize the mean square error!