Poirot said:What is the Fourier transform of $f'(ax)$, where a>0 is a constant? Firstly, I reasoned that (lets say $F[f]$ is the Fourier transform of f) $F[f('x)]=\frac{1}{a}F[f](\frac{k}{a})$ by scaling theorem, then using the derivative rule we get $F[f'(ax)]=\frac{ik}{a}F[f(x)](\frac{k}{a})$. But when I did it manually (substitution and integration by parts), I got an extra factor of $\frac{1}{a}$. So which is correct and why the discrepency?
The Fourier Transform of f'(ax) is the mathematical operation that decomposes a function f'(ax) into its constituent frequencies. It is a powerful tool used in signal processing and data analysis.
Discrepancies in the results of Fourier Transform of f'(ax) can occur due to various factors such as numerical errors, sampling rate, and the chosen window function. These discrepancies can also arise if the function f'(ax) is not well-behaved or if the data is noisy.
To minimize discrepancies in the results of Fourier Transform of f'(ax), it is important to carefully choose the sampling rate and window function. It is also recommended to use multiple window functions and average the results to get a more accurate representation of the function's frequency components.
Yes, the Fourier Transform of f'(ax) can be applied to non-periodic functions. However, since the Fourier Transform assumes periodicity, the function should be extended periodically beyond its original domain to get meaningful results.
Yes, there are other transforms such as the Laplace Transform and the Wavelet Transform that can also be used for signal analysis. Each transform has its own advantages and limitations, and the choice depends on the specific application and the type of data being analyzed.