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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. If the signal is not periodic everything behaves as if the last $M-N$ values of $x$ would not exist, as described above.

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.

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

$$ X[k] = \sum_{i=0}^{N-1}x[i]\cdot e^{-j2\pi ki/N}, $$

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. If the signal is not periodic everything behaves as if the last $M-N$ values of $x$ would not exist, as described above.

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. If the signal is not periodic everything behaves as if the last $M-N$ values of $x$ would not exist, as described above.

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

$$ X[k] = \sum_{i=0}^{N-1}x[i]\cdot e^{-j2\pi ki/N}, $$

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 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

$$ X[k] = \sum_{i=0}^{N-1}x[i]\cdot e^{-j2\pi ki/N}, $$

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. If the signal is not periodic everything behaves as if the last $M-N$ values of $x$ would not exist, as described above.