REGULAR COMBINATION OF ACOUSTIC ELEMENTS 281 



We choose now K' = a /-^ = e~' and K = yJ-~ = e~K 



Then 



cosh 5 = cosh r cosh /. 



To show that 5 Is the propagation constant for an infinite sequence of 

 such sections, it is necessary to show that b is the same for any two 

 sections. But equation (43) holds good for any two sections, and 

 hence b is the same, and represents a solution for an infinite sequence 

 of sections. Now 



e~^ = cosh 5 — sinh 8 = cosh T cosh / — Vsinh^ F cosh'' t + sinh- /. 

 Hence 



— = Ke~^ = e~'[cosh r cosh / — Vsinh" F cosh' / + sinh^ Q 

 pi 



and 



r C - e'e' 1 



= e'[cosh r cosh / — Vsinh^ T cosh'-^ / + sinh'^ /], 



and hence the pressure and volume velocity have the same propagation 

 constant b but an inverse multiplying factor. 



The specific impedance Zi, looking into a given section, is by equa- 

 tion (37) 



„ _ ZqB _ y r Vtanh- / + sinh' Y — tanh / 1 , . 



'^'- A -Ke-'- "^'l ih^hT J ^^^^ 



and similarly Z^, the specific terminating impedance, can be shown 

 equal to Zi. Hence the impedance per square centimeter at the junc- 

 tion points is the same for each section. 



To observe the action of a tapered filter, let us obtain the product of 

 the pressure by the volume velocity and see how these are propagated. 

 Since the specific impedance is the same from section to section, this will 

 represent also the power propagation. Now since 



cosh 5 = cosh r cosh t, 



a pass band occurs when 1 = cosh 5 = — 1 and hence the band 

 occurs only when V is imaginary, since cosh F < 1 and > — 1, or 

 when the filter repeated recurrently is in its pass band. Furthermore 

 the pass band for the tapered structure will not be as wide as that for a 

 similar recurrent structure, since for the tapered structure the band 



