How do you properly cut out negative frequencies from FFT of a real signal if it reduces sample size?

If a real time signal has $$N$$ samples, then the magnitude of its FFT will also have $$N$$ samples--half of which will have positive frequencies and half of which will have negative frequencies. The distribution of these magnitudes will be symmetric about frequency $$= 0$$. If you cut out the $$\frac{N}{2}$$ negative frequency samples because they're not physically meaningful, you've cut your sample by half. What's the proper way to handle this?

If it's at all helpful, I used Python's Scipy FFT to do the calculation.

This is a computational follow-up to this conceptual question: What is the physical significance of negative frequencies?

• What’s the proper way to handle what? If you want to reconstruct your signal, you’ll need the negative frequencies. Is that what you’re asking?
– Jdip
Commented Apr 4 at 18:49
• For my application, the negative frequencies don't make sense, so my thought is to simply ignore them and only use the positive frequencies. Would that make sense to do? Commented Apr 4 at 19:04
• What is your application? If it's just analysis and you don't need to modify/synthesize back to time-domain, then yes it makes sense.
– Jdip
Commented Apr 4 at 19:27
• uhm, keep in mind that your $N$ samples going into the FFT are real values for the FFT to display the Hermitian symmetry where the "negative frequencies" (those are the bins of the FFT between $X[\tfrac{N}2+1]$ and $X[N-1]$ inclusive) are a complex conjugate reflection of the positive frequencies (between $X[1]$ and $X[\tfrac{N}2-1]$ inclusive). The number of real parameters that define an $N$-point FFT is actually $2N$ because the imaginary parts of all of those samples of $x[n]$ going in are zero. Commented Apr 4 at 19:54
• All I want to see is the distribution of frequencies in the data. I don't need to do an inverse FFT to go back to time space, although if I did, then I would, of course, use the full set of positive and negative frequencies. Commented Apr 4 at 19:55

For a real input vector, the $$0^{th}$$ bin will be real. Then for bins $$0 < n < N/2$$, the values will be complex, so each will be good for two real numbers (I.e., $$a + jb$$ contains two real numbers: $$a$$ and $$b$$). Then, if N is even, the bin at $$n = N/2$$ will be real.
I'm all, there will be $$N$$ unique values in the DTFT for the $$N$$ unique values in the input.