TRANSIENT RESPONSE OF AN FM RECEIVER 721 



and from (17) and (18) this evidently is 



HI) = mf - fo)z{f) + m-f - fo)2(-f) (21) 



or, since Z(f) = Z(-f) 



F{f) = hZ(f) Mf - /o) + ^(-/ - /o)]. (22) 



Anyone familiar with the rules of Fourier transforms could write down this 

 frequency function in the first place and then proceed to find the time func- 

 tions by the reverse of the process just carried out. But the time functions 

 are more closely related to the physics of the problem and therefore provide 

 a more fundamental starting point for its solution. 



It will be appreciated that, although the above discussion has been phrased 

 to apply to the problem of finding the voltage drop across an impedance 

 when a frequency modulated current flows through it, the formulas also give 

 the current through an admittance when a frequency-modulated voltage is 

 applied across it. They also give the output voltage or current of a four- 

 terminal network when a frequency-modulated current; or voltage is applied 

 at the input. These various applications of the formulas obviously can be 

 made by placing definitions on Z and H appropriate to the particular 

 problem. 



The next step is to assume suitable values of the impedance Z(f) and vari- 

 ous forms of the frequency modulation function d and to employ these 

 particular values in the general formulas 8, 9, 17 and 18. 



Balanced Frequency Detector 



The impedance can be defined, in general, as a rational algebraic function, 

 viz.: 



, . (ico - ai )( i(a - ch) " ' {iw - dm) PM /^.x 



Z{l(A}) = 7-. ..- r .- r = - , . . . {^Z6) 



{tea — pl){tT — p2) " • {tOJ — pn) Q{^0)) 



Writing the polynominals P and Q in this way as the products of their factors 

 exhibits the a's as the zeros of Z and the p's as the poles. The latter deter- 

 mine the frequencies of free vibration of the network. For the network to 

 be stable, the p's must all have negative real parts. The a's and />'s are 

 either real or occur in conjugate complex pairs. By the partial fraction 

 rule, the expression can be broken up into a series of simple fractions; thus 



Z(:-co) = ^^ + ^^ + • • • . (24) 



ZCt) — Pi ICO — p2 



If the poles are all simple, the A 's are given by 



A=^^ (25) 



