ps8no4sol

ps8no4sol - Problem Set 8 #4 Solution Solution to PS8 Extra...

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Unformatted text preview: Problem Set 8 #4 Solution Solution to PS8 Extra credit #4 E. Sterl Phinney ACM95b/100b 12 Mar 2004 4. (7 × 3 points extra credit) Bessel Functions and FM radios FM ( F requency M odulated) radio works by encoding an audio signal m ( t ) (air pressure as a function of time) as frequency modulation of a radio wave about a carrier frequency f c . We consider the case when the audio signal is a pure tone of frequency f M , so S FM ( t ) = A c cos(2 πf c t + β f sin(2 πf M t )) (1) β f is known as the “ frequency modulation index ”. (a) Show that J- n ( x ) = J n (- x ) = (- 1) n J n ( x ). There are many ways to show this, e.g.(i) direct inspection of the power series for the finite at x = 0 solutions to Bessel’s differential equation (Prob 2a eq 3), (ii) noticing that Bessel’s differential equation is invariant under parity x → - x and under n → - n and then using the initial conditions (Prob 2b eq 5) to show that J is even, J 1 is odd, and then the recursion (Prob 2e, eq 8) to show that J 2 n is even, J 2 n +1 is odd, and (iii) changing signs of x and n and the limits of integration of the integral representation (Prob 2a, eq 2). Here we show this by (iv) using the generating function of Problem 2c, eq 6: exp x 2 t- 1 t = ∞ X n =-∞ J n ( x ) t n (2) In eq (2), replace x by- x and t by 1 /t (if this is confusing, let x =- ξ and t = 1 /τ in eq (2), and then rename ξ = x and τ = t ). The left hand side is unaffected by these changes, so (using a new dummy index m for the sum) exp x 2 t- 1 t = ∞ X m =-∞ J m (- x ) t- m (3) Now replace the dummy index in eq (3) by m =- n . Since the left hand side is the same as that of eq (2), we have ∞ X n =-∞ J n ( x ) t n = ∞ X n =-∞ J- n (- x ) t n (4) Since power series are unique, equating the coefficients of t n gives the first of the desired equalities J n ( x ) = J- n (- x ) . To prove the second equality, return to eq (2), and this time replace x by- x and t by- t . Again the left hand side is unaffected by this change, exp x 2 t- 1 t = ∞ X n =-∞ J n (- x )(- t ) n = ∞ X n =-∞ (- 1) n J n (- x ) t n (5) Again using the fact that power series are unique, we equate the coefficients of t n on the right sides of equations (2) and (5) , we get J n ( x ) = (- 1) n J n (- x ). Multiplying both sides by (- 1) n gives the desired equality: J n (- x ) = (- 1) n J n ( x ) . (b) Show that the Fourier transform of S FM ( t ) given by eq (1) is ˜ S FM = A c / 2 ∞ X n =-∞ J n ( β f ) ( δ ( f- [ f c + nf M ]) + δ ( f + [ f c + nf M ])) (6) Write eq (1) as S FM ( t ) = Re A c e i 2 πf c t e iβ f sin(2 πf M t ) (7) = Re A c e i 2 πf c t ∞ X n =-∞ J n ( β f ) e in 2 πf M t , (8) = A c 2 ∞ X n =-∞...
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This note was uploaded on 01/08/2011 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

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ps8no4sol - Problem Set 8 #4 Solution Solution to PS8 Extra...

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