AN OPTICAL HARMONIC ANALYZER 407 



/(x) = aa -\- ^ an cos nx + Zl ^« si'i nx (1) 



= Co + E Cn COS {nx — 0„). (2) 



1 



Comparison of (1) and (2) gives the following relations between the 

 coefficients in the two forms of the expression: 



ttn ~ Cn COS <j)n, t,, — Cn sln (jl)„. (3) 



c,r = a,r + 6n^ 0,. = tan"' — . (4) 



an 



The form (2) giving amplitude and phase angle of the harmonics is 

 generally more useful, but most methods of analysis give the coefficients 

 in form (1) necessitating the computation of the amplitude and phase 

 angle from the relations (4). One of the advantages of the optical 

 analyzer is that it will give either set of coefficients directly. 



The coefficients in the Fourier Series can be determined from the 

 following expressions: ^ 



ttn — - I fix) cos nxdx, bn—- \ f{x) sin nxdx. (5) 



1 /»2 7r r*2ic 



Cn = - I fix) COS {nx — (j)n)dx, ( fix) sin {nx — 4>n)dx = 0. (6) 

 ""Jo Jo 



1 r-'^ 

 ao = Co = — I fix)dx. (7) 



^ Jo 



We will now describe two methods by which a function can be 

 represented on a photographic film. In a variable area record the 

 film, which is elsewhere opaque, contains a transparent portion whose 

 width at any point is proportional to the function. Such a record is 

 shown in the upper part of Fig. 1. In a variable density record the 



to this requirement. In fact this selection of a proper scale corresponds directly to a 

 necessary adjustment of the analyzer. 



^ The expressions given in (6) can be derived from the more familiar expressions 

 (5) as follows. From (3) 



an cos 0„ + bn sin <^„ = Cnicos'^ 4>n + sin^ <^„) = c„, 

 bn cos 0„ — a„ sin 0„ = 0. 



Substituting values of a„ and bn given by (5) in these expressions leads at once to the 

 expressions [6). 



