I am analyzing a signal in MATLAB by taking its fft. The singal has $200$ samples and I am currently taking nfft = 256. I understand that this would result in zero padding. Now, I want to try to reduce the nfft to 128, 64 etc. I wanted to ask what will happen to the fft.
Matlab says for Y = fft(X,n), If X is a vector and the length of X is greater than n, then X is truncated to length n. What would this actually result in. Can anyone explain, please?
The fast Fourier transform (FFT) is an efficient algorithm to compute the discrete Fourier transform (DFT). The DFT $X[k]$ of time-domain signal $x[k]$ of length $M$ is defined by
$$ X[k] = \sum_{\color{red}{i=0}}^{\color{red}{N-1}}x[\color{red}{i}]\cdot e^{-j2\pi k\color{red}{i}/N}, $$
with $N$ the length of the DFT. If $N>M$, i.e., the DFT length is greater than the length of the time-domain signal, it is typically assumed that $x[k]=0$ for $k\geq M$. You already mentioned that this process is usually called zero-padding. Alternatively, observe what happens in the equation if $M>N$, i.e., if the DFT length is shorter than the signal: The last values of $x$ are not taken into account for the computation, for all frequency bins $k$.
Hence, (as usual) what MATLAB says is correct. The question to your answer "What would this actually result in?", however, depends on the properties of the time-domain signal $x$. What happens if $x$ is a periodic signal has been addressed extensively in other answers here. In every case everything behaves as if the last $M-N$ values of $x$ would not exist, as described above.
Reducing the NFFT samples in your DFT will result is less resolution in the spectrum you'll be observing. The frequency resolution is defined as $$ f_k = \frac{F_s}{N}k\quad \text{with}\quad k=0, 1, \ldots, N-1\tag{1} $$ In your $N$-point Discrete Fourier Transform (DFT), decreasing $N$ increases the spacing between the $f_k$. Or put in another way, for a fixed sampling frequency $F_s$,
\begin{align} N_2 &< N_1 \implies \Delta f_{k_2} > \Delta f_{k_1}\\ \text{Where}\quad f_{k_1} &=\frac{F_s}{N_1}k_1\quad\text{and}\quad f_{k_2} =\frac{F_s}{N_2}k_2\\ \implies \Delta f_{k_1} &=\frac{F_s}{N_1}\quad\text{and}\quad \Delta f_{k_2} =\frac{F_s}{N_2} \end{align}
So say, you have some frequency components (i.e. peaks) you're observing in your spectrum with a high $N$, gradually reducing this value, you'll see that the peaks are getting less and less localized visually. Note that it is only visually, your DFT got your spectrum, you play with $N$ (i.e. NFFT) to visualise more out of your spectrum. Together with $F_s$, it determines the frequency resolution as defined in $(1)$.
EDIT:
To compute the energy, you sum the squared amplitudes spectrum values. Look for instance at the MATLAB code below for the sine wave $$ s(t) = 3\sin(2\pi 150t) $$
Fs = 500; % Sampling frequency
T = 1/Fs; % Sampling period
L = 2000; % Length of signal
t = (0:L-1)*T; % Time vector
f = Fs*(0:(L/2))/L;
S = 3*sin(2*pi*150*t);
NFFT = 2^nextpow2(L);
NFFT1 = L;
NFFT2 = NFFT/2;
NFFT3 = NFFT/4;
Y = fft(S, NFFT1)/NFFT1;
Y2 = fft(S, NFFT2)/NFFT2;
Y3 = fft(S, NFFT3)/NFFT3;
EY = sum(Y.*conj(Y));
EY2 = sum(Y2.*conj(Y2));
EY3 = sum(Y3.*conj(Y3));
Your FFT have to be normalized and the energy computed as sum of the squared amplitude spectrum. For the sample code above you get:
EY = 4.5
EY2 = 4.49644474953643
EY3 =4.49832143842713
Which is from the amplitude of the sine $3$, giving ($3^2/2$). Also showing the effect of taking different $N$ values for the FFT on resulting energy.
An FFT of a finite segment of longer data convolves that longer data with the transform of a rectangular window the length of the FFT. The transform of that rectangle is a periodic Sinc, which gets conjugate mirrored by strictly real data, The shorter the FFT the shorter the window, the “fatter” the Sinc lobes.
Added: If a short FFT window excludes a portion of a longer signal, the energy in the portion completely excluded will no longer be represented in an unnormalized FFT result. For normalized results, if the energy in the portion excluded is exactly proportional to the length excluded, then the normalized FFT result can appear to be the same. But that requires that you know or have measured the energy in the excluded portion.
For signals that are known to be exactly integer periodic in both the longer and shorter DFT lengths, the data can be wrapped around the shorter window by summation, and the DFT results normalized to the longer of the two lengths will be the same.
