Signals and Noise

Learn It!

Goal

\(y(t)=2+0.5 cos (40 \pi t)+0.25 sin(100 \pi t)\)

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 \(0.25 sin(2 \pi \cdot 50 t)\)

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

Yes. Since a sine function is the same as a cosine function shifted a quarter of a period
to the right, the sampling rate doesn't care if it is a sine or cosine. only the frequency matters.

\(f_{max}=50Hz \)

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.