Why are fourier transforms so powerful?
- Thread starter K.I.L.E.R
- Start date
K.I.L.E.R said:
If it's for water rendering see (not using FFT, but NSE) : http://www.andyc.org/lecture/viewlog.php?log=Realistic Water Rendering, by Yann L
Yann Lombard talking about that, using Navier-Stokes Equations (simplified).
[Edit : Typo]
Sure, take an LRC circuit.K.I.L.E.R said:
In this circuit, we have a resistor, a capacitor, and an inductor connected in series. This circuit is connected to some power supply which supplies V(t).
Now, the voltage across the resistor is simple: deltaV = IR.
The voltage across the capacitor is also simple. Since capacitance is just the ratio of charge to voltage, deltaV = qC.
Finally, we have the inductor. This circuit is a little bit different, in that it resists changes in current. So, the voltage is: deltaV = -L(dI/dt).
Now we have to consider how q in capacitance relates to the currents in the other equations. It turns out it's very simple: current is just flow of charge, so I = dq/dt.
From this, we can develop an equation. Since all of the above are in series, the total voltage drop across all three must equal the voltage of the source: V(t) = -L(d^2q/dt^2) + R(dq/dt) + C(q).
The above is an ordinary second-order differential equation, and as such if we consider the equation: -L(d^2q/dt^2) + R(dq/dt) + C(q) = 0, we know right away that the solution is complex exponentials. So, an obvious tactic is just to take q(t) = integral(dw a(w) e^iwt). This is a Fourier transform, where we can now define everything about our function q(t) instead by talking about the function a(w), which is a function of the angular frequency w.
If we make the above change, we find that we can factor out the integral, and take the time derivatives explicitly. This can be seen by virtue that since our functions are well-behaved, one can take the time derivatives inside the integral. And since integrals are linear operators, a sum of integrals is just the integral of the sum. Finally consider that the time derivative of e^(iwt) = iw e^(iwt). Thus, we have:
integral(dw a(w) ( Lw^2 + iwR + C ) e^(iwt)) = 0
Now, we want the above equation to be true for all functions a(w). The only way for this to be true is if (Lw^2 + iwR + C = 0). Here I'd like to point out that I did something a little bit different than what is usually done: I put everything in terms of the Fourier transform of the charge as a function of time. What is usually done is to put everything in terms of the Fourier transform of the current. This leads to the same result, but it's written a little bit differently:
-iwL + R + C/(iw) = 0
Finally, if we go back to the full equation, where the right hand side was V(t) to start with, and also take the Fourier transform of the right hand side, we find pretty quickly that we can express the above equation as:
(-iwL + R + C/(iw))I(w) = V(w)
...where I(w) is the Fourier transform of the current, and V(w) is the Fourier transform of the voltage. Thus, if you give me a voltage as a function of time, I can take its Fourier transform, stick it on the right hand side of the above equation, and easily find current as a function of time.
WhiningKhan
Regular
K.I.L.E.R said:
Well...How about you sign up on a maths class on Fourier methods? Pretty much every university with maths department runs such classes. No offense, but you don't really strike to me as a fellow who would grasp the whole issue from BBS postings anyway (being the self-proclaimed "Retarded Moron" and all)...
