MAE 2381 CHAPTER 7 EXAM QUESTIONS 100% SOLVED

MAE 2381 CHAPTER 7 EXAM
QUESTIONS 100% SOLVED

Aliasing - ANSWERif the sample rate becomes too low, the sine wave appears to be
of a lower frequency

Sampling Theorem - ANSWERTo reconstruct the frequency content of a measured
signal accurately, the sample rate must be more than twice the highest frequency
contained in the measured signal
f_s >= 2*f_m
dt < 1/(2*f_m)

f_s - ANSWERSample Rate

f_m - ANSWERMaximum Frequency

dt - ANSWERSample Time Increment

The false frequency misinterpreted from the frequency content of the original signal
when a signal is sampled at a rate less than 2*f_m, and the higher frequency content
of the analog signal takes on the false identity of a lower frequency - ANSWERAlias
Frequency

Nyquist Frequency - ANSWER- The maximum frequency that can be detected
f_(N(yquist)) = f_max = f_s/2 = 1/(2dt)
- Represents a folding point for the aliasing phenomenon

Alias frequency, f_a, can be predicted from the ___________________ -
ANSWERFolding Diagram

f_N < f < 2*f_N or 1 < f/f_N < 2 - ANSWEROut-of-Phase

2*f_N < f < 3*f_N or 2 < f/f_N < 3 - ANSWERIn-Phase

if f_sig = 10 Hz and f_sampling = 12 Hz, what is f_N and f_a - ANSWERf_N =
f_sampling/2 = 12 Hz/2 = 6 Hz
f_sig/f_N = 10 Hz/6 Hz = 1.67
f_sig = 1.67*f_N is directly above 0.33*f_N on folding diagram so
f_a = 0.33*f_N = 0.33*6 Hz = 2 Hz