Electronic – MATLAB – FFT aliases occur sooner than expected

The problem has been solved.

function X = ReadTiAdcData(filename,N)
fid = fopen(filename,'r');
X = fread(fid,'int16');
fclose(fid);
X = X-2^(N-1);
X = X./2^(N-1);

Jim B's post explains the HSDC's binary file encoding and how to convert from it. N = 12. Turns out the full scale voltage didn't matter because we just needed the scaled voltage level relative to full scale, which was already encoded in the binary file, "full scale" being 4096.

I hope this is helpful for those who are struggling with converting raw ADC data using the TI HSDC.

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.

Now, I'll try to answer your questions:

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.