- #1
!kx!
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Hi,
I was wondering what would the Fourier transform of a signal like below give:
s(t) = sin(2πt*10) ; t in [0s,5s]
= sin(2πt*20) ; t in [5s,10s]
I certainly did not expect it to give me 2 sharp peaks at frequencies 10Hz and 20Hz - because I understand that the addition of sine waves of frequencies obtained from Fourier Transform will give me original signal - so I'd expect to see two sharp peaks (at 10Hz and 20Hz) for something like...
r(t) = sin(2πt*10) + sin(2πt*20) ; t in [0s,10s]
I did a fast Fourier transform [using Matlab fft] of the above two signals.. and I obtained more or less the same peak frequencies in both : 10Hz, 20Hz. In the fft(s)There are some other low frequency components, which have considerably low amplitudes than above two frequencies.
Am I missing something? Or those low amplitude frequency components are all that are required to change s(t) into r(t)?
I was wondering what would the Fourier transform of a signal like below give:
s(t) = sin(2πt*10) ; t in [0s,5s]
= sin(2πt*20) ; t in [5s,10s]
I certainly did not expect it to give me 2 sharp peaks at frequencies 10Hz and 20Hz - because I understand that the addition of sine waves of frequencies obtained from Fourier Transform will give me original signal - so I'd expect to see two sharp peaks (at 10Hz and 20Hz) for something like...
r(t) = sin(2πt*10) + sin(2πt*20) ; t in [0s,10s]
I did a fast Fourier transform [using Matlab fft] of the above two signals.. and I obtained more or less the same peak frequencies in both : 10Hz, 20Hz. In the fft(s)There are some other low frequency components, which have considerably low amplitudes than above two frequencies.
Am I missing something? Or those low amplitude frequency components are all that are required to change s(t) into r(t)?