Indeed, but more importantly it should also be realised that the primitive function is incorrect for ##k = 1##.anuttarasammyak said:
Ohh alright yeah that makes sense. Ty!anuttarasammyak said:
What does this mean?Orodruin said:
That your integration is wrong when k = 1.lys04 said:
It's gonna be wrong for all k s.t e_-ikx has a real part?Orodruin said:
You are misunderstanding my comment. I said that apart from the fact pointed out by @anuttarasammyak in post #2, the integral is also wrong for ##k = 1##. This will be crucial to correct also when you have corrected the imaginary part error.lys04 said:
(my emphasis)Orodruin said:
The Fourier transform of sin(x) is given by the formula: F{sin(x)}(ω) = (1/2j) [δ(ω - 1) - δ(ω + 1)], where δ is the Dirac delta function. This indicates that sin(x) has frequency components at ω = ±1.
The non-zero terms in the Fourier transform of sin(x) correspond to the delta functions at the frequencies ω = 1 and ω = -1. These terms indicate the presence of these specific frequency components in the signal.
The Fourier transform decomposes a time-domain signal into its constituent frequencies. For sin(x), which oscillates at a frequency of 1 Hz, the transform reveals that there are significant contributions at ω = ±1, reflecting the sinusoidal nature of the signal.
The Dirac delta function in the Fourier transform signifies that the signal sin(x) contains pure frequency components at specific points (ω = ±1) with no other frequencies present. This is characteristic of idealized sinusoidal signals, which are composed of single frequencies.
Yes, the Fourier transform of sin(x) can be visualized as two spikes on a frequency spectrum at ω = 1 and ω = -1, with the height of each spike representing the amplitude of the corresponding frequency component. This visual representation highlights the discrete nature of the frequency content of the sinusoidal signal.