# Nyquist frequency mirroring

I need someone to check my understanding of Nyquist frequency. I understand it in the following manner.

$$F_{nyquist} = ( F_{sampling}/2 ) * n$$ where n is an integer (negative and positive). For instance:

$$F_{sampling} = 10 kHz$$ then $$F_{nyquist} = [-10, -5, 0_{(?)}, 5, 10] kHz$$ I am not sure, whether 0 is a nyquist frequency (Question 1). Then the mirroring a.k.a detected alias frequencies can be found at the following locations: ($F_s$ in the picture is $F_{sampling}$)

• green - signal being sampled (original signal)
• red - alias frequencies
• yellow - is it alias frequency ? (Question 2)

Questions:

1) is F = 0 Nyquist frequency ?

2) does the mirroring happens around F = 0 ?

3) is my understanding correct ? Are the red circles alias frequencies ? The problem is, that I cant replicate all of the alias frequencies in matlab. I can see alias at $$F_{alias} = F_{sampling}*n + F_{signal}$$. For signal at $$F_{signal} = 3 kHz$$ I can see aliases at $$F_{alias} = [13, 23, 33]$$ , but I DONT see it at $$F_{alias} = [7, 17, 27]$$ as expected due to the mirroring.

The following MATLAb code should illustrate the process I understand as "looking" for aliases.

clear all;
clc;

% sampling rate
F_sample = 10000;
T = 1 / F_sample;
% time range
t = [0:T:T*10];         % asmpling time - 10 samples
t_x = [0:T/100:T*10];   % smooth "time" to see the sinusoids of F_aliases signlas

% nyquist frequency
F_nyquist = F_sample / 2;

% arbitrary signal within sampling range (F_signal is less than F_sample / 2)
F_signal = 3000;

% when the following frequencies are sampled, they should appear as the
% F_signal
F_alias_1 = F_nyquist*2 - F_signal; % 1st mirroring - Zone 2
F_alias_2 = F_nyquist*2 + F_signal; % 2nd mirroring - Zone 3

% from F to w
w_signal = 2*pi*F_signal;
w_alias_1 = 2*pi*F_alias_1;
w_alias_2 = 2*pi*F_alias_2;

% calculate signals
x1 = sin(w_signal*t);       % samples (red circles)
x2 = sin(w_alias_1*t_x);    % blue signal
x3 = sin(w_alias_2*t_x);    % green signal

plot(t,x1,'ro');      % Sampling (red circles)
hold on;
plot(t_x,x2,'b');     % 1st frequency that SHOULD be sampled the same as original signal
hold on;
plot(t_x,x3,'g');     % 2nd frequency that SHOULD be sampled the same as original signal

xlabel('time');
ylabel('amplitude');


This is the outcome: • red circles - samples of 3 kHz signal at 10 kHz sampling rate (original signal at 3kHz is not depicted, just its samples)
• blue - alias at 7 kHz (1st above 5 kHz (Zone 2), half of sampling frequency)
• green - alias at 13 kHz (2nd above 5 kHz (Zone 3), half of sampling frequency)

Obviously, the blue doesnt "fit" the samples, while the green fit them perfectly

I understand the Nyquist frequency a bit differently - maybe this will help you.

There is only one Nyquist frequency, and it is at $$F_{n} = \frac{F_{s}}{2}$$ Then, any signals with a frequency above $+F_{n}$ or below $-F_{n}$ are "folded" - they appear in the region $[-F_{n}, F_{n}]$. Here's an example: Here's what's going on in this picture:

• Zone 1 is inside the folding frequency, so the signal there isn't modified
• Zone 2 is just above $f_n$, so it is flipped over to the other side. The aliased version of Zone 2 is identical to Zone 1.
• Zone 3 is above $f_n$, so it is flipped over. When this happens, it appears on the negative side of the frequency axis - it is a mirror image of Zone 1.
• Zone 4 follows the same process - it is flipped twice, putting it on top of Zone 1.

1. $F = 0$ is not a folding frequency. See question 2.
2. The reason why you can see signals at, say, $F = -3\text{ kHz}$ is because we need both positive and negative frequencies to make signals. For example, a cosine at $F = f_0$ is $$\cos(2\pi f_0 t) = e^{i \cdot 2\pi f_0 t} + e^{-i \cdot 2\pi f_0 t}$$ and the second term has the negative frequency you're seeing. It's impossible to make real (non-complex) signals without negative frequencies!
3. It sounds like you're describing this backwards. If you sample at $F_s = 10 \text{ kHz}$, then cosine signals at the frequencies $$F = 3, 7, 13, 17, 23, 27, \dots \text{ kHz}$$ should all appear the same. Here's a quick plot that I made to show this: This plot is $\cos(3 \cdot 2\pi t)$ and $\cos(7 \cdot 2\pi t)$. Notice that they intersect at 0, 0.1, 0.2, ..., so if you only sample at these times, you'll see the exact same signal.

• As my comms lecturer put it "negative frequencies are just a mathematical convenience". Nov 24, 2015 at 13:28
• Thank you for your answer. (I like the picture, very descriptive). Please check my first post, I have answered your question there. Nov 24, 2015 at 13:59
• I don't want to go into a ton of detail - this is going to get too big too fast - but your 7 kHz sine wave appears at 3 kHz, with a negative sign on the front. I think this is because it's a sine wave - you could try switching to a cosine to see if that acts a bit better. Nov 24, 2015 at 14:02
• no, it doesnt change the outcome (well, instead of sine, there is cosine), it still doesnt match Nov 24, 2015 at 14:04
• I added a quick plot of cosine at 3 and 7 kHz - it looks like they match at the sample points (multiples of 0.1). Not sure what's going on in your example. Nov 24, 2015 at 14:12