﻿ Sampling sine and cosine
Spring 2019
Signals and Noise
Sampling Sine and Cosine
Learn It!
Pre-Requisite Knowledge
Goal
Find the Nyquist frequency
What is the minimum sampling rate to resolve the signal?
For a signal given by $y(t)=2+0.5 cos (40 \pi t)+0.25 sin(100 \pi t)$ where t is in seconds, the minimum sampling rate which avoids loss of information should be slightly above which value?
Part 1
Find the right frequency
$f_{Nyquist}=f_{max}*2$
The minimum sampling frequency, the Nyquist limit, is double the highest frequency in the signal itself.
$y(t)=\underbrace{2}_{f=0Hz}+\underbrace{0.5 cos (40 \pi t)}_{f=20 Hz}+\underbrace{0.25 sin(100 \pi t)}_{f=50 Hz}$
First, we need to identify the frequency of each of the components. Remember, the part inside the trig function is $2 \pi f t$.
The 2 term has frequency 0 because it is constant.
The cosine term has frequency 20, because it can be rewritten as $0.5 cos(2 \pi \cdot \underbrace{20}_{frequency!} t)$.
the sine term has frequency 50, we can rewrite it as $0.25 sin(2 \pi \cdot 50 t)$
+
?
Does this formula work for sine functions AND cosine functions
$f_{max}=50Hz$
The highest frequency among all the terms is 50Hz
$f_{Nyquist}=50Hz \cdot 2=100Hz$
The answer is twice the frequency of the highest frequency term, which is the sine term in this problem.