Waves : 1 Dimensional Waves
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Examples - Problems - Solutions - Terminology
Fourier Transform[edit | edit source]
So far, you've learned how to superimpose a finite number of sinusoidal waves. However, a wave in general can't be expressed as the sum of a finite number of sines and cosines. Fortunately, we have a theorem called Fourier's theorem which basically states that under certain technical assumptions, any function, f(x) is equal to an integral over sines and cosines. In other words,
Now, if we're given the wave function when t=0, φ(x,0) and the velocity of each sine wave as a function of its wave number, v(k), then we can compute φ(x,t) for any t by taking the inverse Fourier transform of φ(x,0) conducting a phase shift, and then taking the Fourier transform.
Fortunately, the inverse Fourier transform is very similar to the Fourier transform itself.
This tells us that, since waves which are very spread out, like the sine wave, have a narrow range of wave numbers, wave functions whose wave numbers are very spread out will only be significant at a narrow range of positions.
Fourier Transform Properties[edit | edit source]
unitary, angular frequency
unitary, ordinary frequency
|2||Shift in time domain|
|3||Shift in frequency domain, dual of 2|
|4||If is large, then is concentrated around 0 and spreads out and flattens|
|5||Duality property of the Fourier transform. Results from swapping "dummy" variables of and .|
|6||Generalized derivative property of the Fourier transform|
|7||This is the dual to 6|
|8||denotes the convolution of and — this rule is the convolution theorem|
|9||This is the dual of 8|
|10||For a purely real even function||is a purely real even function||is a purely real even function|
|11||For a purely real odd function||is a purely imaginary odd function||is a purely imaginary odd function|
Fourier Transform Pairs[edit | edit source]
|Time Domain||Frequency Domain|