I'm coding the amplitude spectrum plot of a square wave signal with the following relation@
|(=3,5,7,...)|/|(=1)|∝^
where I've already calculated the constant by taking the signal FFT and fitting the peaks. However, I'm having trouble figuring out what the amplitude is to plot in this case - any insight would be appreciated.
I believe the OP may be caught on the relationship between a peak in the complex FFT output and the amplitude of the related sinusoidal component in the time domain. Each bin in the FFT of a real signal has a magnitude that is directly proportional to the sinusoidal components in the time domain for the frequency of that bin. Typically the FFT of a real waveform with $N$ samples will also have a gain of $N$, (and for a sinusoid this gain is distributed in two components with $N/2$ in each, see details below) but this scaling is dependent on actual FFT implementation.
If the OP is extracting the relative gains for the harmonics of a square wave by fitting the peaks, then the magnitude of each complex result will be proportional to the amplitude for that particular frequency component. The true harmonics of a square wave will be the odd harmonics with an amplitude that goes down at $1/n$ where $n$ is the harmonic number.
The FFT can only approximate this true result since it is circular about the sampling rate: as the sampling rate is increased so that more or more of the "true" components are in the range of $f_s/2$, the closer the approximation will be.
Further details:
First, the FFT is an algorithm that computes the DFT, which is typically given as:
$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j\omega_o nk}$$
With the time domain reconstruction, or inverse DFT as:
$$x[n] = \frac{1}{N}\sum_{n=0}^{N-1}X[k]e^{j\omega_o nk}$$
Where we have $X[k]$ as each of the bins in the DFT result as a complex magnitude, giving the magnitude and phase for that particular frequency component at index $k$. The first DFT bin where $k=0$ is the "DC bin" and represents the DC component or average value of the time domain signal. The next DFT bin where $k=1$ represents what I will refer to as the fundamental frequency. This makes it very easy to understand the DFT by first understanding the principles of the Fourier Series Expansion (FSE): in the FSE, any arbitrary analytic waveform over time duration $T$ can be expressed as an infinite sum of sinusoidal components each as a multiple of the fundamental frequency, which is $1/T$ Hz.
In the DFT, the frequency components are given as complex phasors, notably samples of $e^{j \omega t}$. We recognize from Euler's identity for a cosine (and sine has a similar relationship) that a sinusoid is made up of two such phasors:
$$2\cos(\omega t) = e^{j \omega t} + e^{-j \omega t}$$
So notice how the $N$ samples in time of $x[n]$ and the sampling rate $f_s$ sets the total time duration $T$, as $T = N/f_s$. The "fundamental frequency" $\omega_o$ is therefore $2\pi f_o = 2\pi 1/T = 2\pi f_s/N$. It is common to express the DFT in terms of the normalized sampling rate where $f_s =1$ and with that we get the expression for the DFT as:
$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$
and inverse DFT as:
$$x[n] = \frac{1}{N}\sum_{n=0}^{N-1}X[k]e^{j2\pi nk/N}$$
Also notice how the two tones (given by Euler's) for a sinusoidal components will actually appear in the DFT result: A sinusoid will have two exponential components at $\pm f$ where $f$ is the frequency of the sinusoid. For any sampled system, the frequency compoents in the first Nyquist zone which is from $\pm f_s/2$ will repeat periodically in all higher Nyquist zones. This means a single frequency component at $f$ will repeat at $Mf_s + f$ for all integers $M$. The DFT is giving us the frequencies from DC (at $f=0$) to nearly $f_s$ ($f_s$ is the periodic repetition of DC). So a sinusoid with frequency $f$, which then has components at $\pm f$ will appear in the DFT at $f$ and $f_s-f$.
From this we get the insight into the growth of $N$ (consider the DFT with a DC signal at the input where $x[n]=1$ for all samples and $k=0$, the above formula for the DFT simplifies to the sum of 1 over $N$ samples.), and why and how we would have the additional 1/2 factor given the Euler's relationship I provided.
For further understanding beyond this it is important to understand how "frequency leakage" occurs in the DFT, specifically when the sinusoidal components are not an integer sub-multiple of the sampling rate. I go into that further in this existing post.
