Spring 2018
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?
\(y(t)=2+0.5 cos (40 \pi t)+0.25 sin(100 \pi t)\)
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.