The continuous harmonic data signal is given below:
The minimum sample frequency $$f_s$$ required to properly reconstruct the continuous signal is:
A. 1 sample per 4 sec
B. 1 sample per 2 sec
C. 1 sample per 1 sec
D. 2 samples per 1 sec

1 Answer

James Dowd

Updated on December 29th, 2020

To answer this question, we must consult the Nyquist Sampling Theorem.

This theorem states that in order for a continuous time signal to be distinguishable from a sample taken at discrete time intervals, we must choose a sampling rate over twice as fast as its highest frequency component.

Let's first calculate the frequency of our continuous harmonic data signal. Let's recall frequency is the inverse of the period:

$f = \frac{1}{T}$

We observe that from $t_{n}$ to $t_{n + 2}$ , one full waveform exists, and this total time represents our period. Since we are given $Δ t = t_{n} - t_{n - 1} = 1 s e c$ , and we know the time delta between $t_{n}$ to $t_{n + 2}$ is twice this amount, our period must be equal to 2 seconds. Therefore, our frequency is:

$f = \frac{1}{T} = \frac{1}{2} = . 5 H z$

Since .5Hz is our only, and highest, frequency, we multiply this by two to get our Nyquist frequency:

$f_{n} = 2 f f_{n} = 2 (. 5) = 1 H z (H z = s a m p l e s / s e c o n d)$

Therefore, the minimum sample frequency required to reconstruct the continuous signal is 1 sample per 1 second.

The continuous harmonic data signal is given below:The minimum sample frequency $$f_s$$ required to properly reconstruct the continuous signal is:A. 1 sample per 4 secB. 1 sample per 2 secC. 1 sample per 1 secD. 2 samples per 1 sec

1 Answer

James Dowd

Related Questions

The continuous harmonic data signal is given below:
The minimum sample frequency $$f_s$$ required to properly reconstruct the continuous signal is:
A. 1 sample per 4 sec
B. 1 sample per 2 sec
C. 1 sample per 1 sec
D. 2 samples per 1 sec