Changes between Version 6 and Version 7 of u/erica/CF_spectra

Timestamp:

01/05/15 16:40:10 (10 years ago)

Author:

Erica Kaminski

Comment:

—

u/erica/CF_spectra

@@ -26 +26 @@
 == Power Spectrum == 
 
-To get a power spectrum, one takes a fourier transform of the data, and then takes the complex conjugate of the resultant fourier transform. This gives the power in each of the component frequencies of the signal.  
+To get a power spectrum, one takes a fourier transform (FT) of the data, and then takes the complex conjugate of the resultant fourier transform. This gives the power in each of the component frequencies of the signal.  
 
 
 == Kinetic Energy == 
 
-There seems then to be 2 different ways of computing the power spectrum of the kinetic energy. First, we can take the fourier transform of [[latex($ \sqrt{\rho (v_x + v_y + v_z)^2}$)]] and then square this to get the power spectrum. Or, we can take [[latex($\sqrt{\rho v_x^2}$)]] and then sum up the squares of the various fourier transforms for each dimension. That is, does 
+There seems then to be 2 different ways of computing the power spectrum of the kinetic energy. First, we can take the fourier transform of [[latex($ \sqrt{\rho (v_x + v_y + v_z)^2}$)]] and then square this to get the power spectrum. Or, we can take the FT of [[latex($\sqrt{\rho} ~v_x$)]] and then sum up the squares of the various FTs for each dimension. 
 
-[[latex($(F(\sqrt{v_x^2 + v_y^2 + v_z^2} ~))^2 = (F(v_x))^2 + (F(v_y))^2 + (F(v_z))^2 ~?$)]] 
+That is, since the LHS does not strictly equal the RHS,  
 
-Further, which do we want, 
+[[latex($(FT(\sqrt{\rho(v_x^2 + v_y^2 + v_z^2)} ~))^2 = (FT(\sqrt{\rho} v_x))^2 + (FT(\sqrt{\rho}v_y))^2 + (FT(\sqrt{\rho}v_z))^2 $)]] 
 
+(but their integrals do - by Parsevel's theorem, stating the total power is equal), yet each give power spectra with the correct units for KE, which do you choose?? 
 
-[[latex($(F(\sqrt{\rho(v^2)} ~))^2 ~or~ (F(\sqrt{\rho}))^2*(F(v))^2 $)]]
+It seems to be intuitively you want to save the squaring for after taking the FT, since if you do it before, 'you lose information about direction'. For instance, consider the 1D case of the initial condition of colliding flows. If you square the initial condition, you just get a constant function, peaking about k=0. 
 
-as it seems both give the same units..
+Further, since, 
+
+[[latex($FT(\sqrt{\rho}v_x ~) ~.DNE.~ FT(\sqrt{\rho})\times FT(v_x) $)]]
+
+yet, again, each will have the correct units for KE once squared, which is proper? For the case of the initial condition -- what does the LHS and RHS look like? How about the power spectra for the LHS and RHS? What about at some late times when there is lots of structure in the density?
+