X(t)e 2πi nt t dt + 1 T

Transcription

1 HOMEWORK 31 I) Use the Fourier-Euler formulae to show that, if X(t) is T -periodic function which admits a Fourier series decomposition X(t) = n= c n exp (πi n ) T t, then (1) if X(t) is even c n are all real, c n = c n () if X(t) is odd c n are all imaginary If X(t) is aperiodic and has a Fourier Transform pair ˆX(f), show that (1) if X(t) is real and even ˆX(f) is real and even, ˆX( f) = ˆX(f) () if X(t) is real and odd ˆX(f) is imaginary and odd, ˆX( f) = ˆX(f) Proof If X(t) is a T -periodic real function which admits a Fourier series decomposition X(t) = n= c n exp (πi n ) T t then (1) if X( t) = X(t) = 1 T = T X(t)e πi nt t dt = 1 T X( t)e πi n T t dt + 1 T X(t) cos (π n ) T t dt, X(t)e πi nt t dt + 1 T and c n = c n = c n since c n R () Similarly, if X( t) = X(t) = 1 T = i T X(t)e πi nt t dt = 1 T X(t)e πi n T t dt = 1 T X(t)e πi nt t dt + 1 T X( t)e πi n T t dt + 1 T X(t) sin (π n ) T t dt X(t)e πi n T t dt = 1 T X(t)e πi n T t dt = X(t) ( e πi n T t + e πi n T t) dt = X(t)e πi n T t dt = Similarly, if X(t) is real aperiodic and has a Fourier Transform pair ˆX(f), (1) if ˆX( f) = ˆX(f) ˆX(f) = X(t)e πift dt = () if ˆX( f) = ˆX(f) ˆX(f) = X(t)e πift dt = X(t)e πift dt + X(t)e πift dt + 1 X(t)e πift dt = X(t) ( e πi n T t e πi n T t) dt = X(t)e πift dt = i X(t) cos (πft) dt X(t) sin (πft) dt

2 II) Assuming suitable periodicity conditions, for the ramp function X(t) defined by X(t) = t for 1 t 1, (1) calculate the Fourier series coefficients, () calculate the Fourier Transform Proof (1) To construct a Fourier series representation, assume that the ramp function X(t) is T -periodic X(t + nt ) = t, with T t T and n N with T = ; the coefficients of the Fourier series representations are te iωnt dt with = π n ; setting T u = t, du = dt dv = e iωnt dt, v = i e iωnt one obtains (integration by parts) = i [ e iωn T udv = 1 T [uv]t/ 1 T ( i ] T + e i 1 T vdu = ) [ e it ] T/ = since [e iωnt ] T/ = T e iωn e iωn T T = i sin c n = ( 1) n i πn () The Fourier transform is ˆX(ω) = X(t) e iωt dt = i 1 [ ] te it T/ T i 1 T/ e iωnt dt = T i T cos = i ( 1)n = For T =, t e iωt dt The integral is similar with the previous one, with the difference that the exponent is now iωt instead of i t Integrating by parts as before we get ˆX(ω) = it [ ] ( ) e iω T + e iω T i [ ] e iωt T/ ω ω = it ω cos ωt + i ω sin ωt The transform is imaginary and odd, as expected

3 III) Show that the Fourier Transform has the following properties: (1) Shift 1: F[X(t a)] = ˆX(f) exp ( πifa); () Shift : F[X(t) exp (πif t)] = ˆX(f f ); (3) Scale: F[X(at)] = 1 a ˆX( f a ); d n (4) Derivative 1: df ˆX(f) = ( πi) n F[t n X(t)]; [ n ] d n (5) Derivative : F dt X(t) = ( πif) n ˆX(f) n Proof (1) Shift 1: Change variable from t to ξ = t a, t = ξ + a, dt = dξ, F[X(t a)] = () Shift : X(t a) e πift dt = F[e πif t X(t)] = X(ξ) e πif(ξ+a) dξ = e πifa ˆX(f) X(t) e πi(f f )t dt = ˆX(f f ) (3) Scale: Let a > ; change variable from t to ξ = at, t = ξ/a, dt = dξ/a, X(ξ) e πi f a ξ dt = 1 ( a ˆX f a F[X(at)] = X(at) e πift dt = 1 a If a <, ξ = at, t = ξ/a, dt = dξ/ a F[X(at)] = X(at) e πift dt = 1 X(ξ) e πi f a ξ dξ = 1 ( ) a a ˆX f ; a (4) Derivative 1: Proving for n = 1 is sufficient, since dn df ˆX(f) = d [ (f)] ˆX(n 1) = ( ) n df d d n 1 ˆX(f) ; formally df df n 1 d df ˆX(f) = d df F X(t)e πift dt = πi ) ; tx(t)e πift dt = πi (tx(t)) = πi F[tX(t)] (5) Derivative : Again proving for n = 1 is sufficient: [ ] d [ ] d dt X(t) = dt X(t) e πift d [ dt = ] X(t) e πift dt X(t) d dt dt e πift dt = πif ˆX(f), since lim t ± X(t) = IV) 3

4 (1) Show that the Gaussian is invariant under the Fourier Transform, (1) F[exp( πt )] = exp( πf ) where ˆX(f) = F[X] = X(t)e πift dt () Use equation (1) to show that the Fourier Transform of X(t) = A exp ( π σ t ) [ A is ˆX(f) = exp f ] (3) Use the shifting properties of the Fourier Transform to show that if X(t) = A exp ( [ ] π σ t ) A exp (πif t) then ˆX(f) = exp (f f ) Proof (1) Gaussian is invariant to FT: F [ exp ( πt )] = e πt e πft dt = Change the variable (complete the square) e πt πift dt πt πift = π ( t + ift f ) πf = π (t + if) πf to z = π (t + if), t = F [ exp ( πt )] = z if, dt = dz yields π π where the last integral becomes Note that e π(t+if) πf dt = e πf e π(t+if) dt e π(t+if) dt = 1 π e (t+if) dt = e z dz = 1 e t dt; since f is a constant, changing from one integral to the other involves just a redefinition of the origin, from + i to + fi (a translation along the imaginary axis) () Fourier transform of X(t) = A exp ( π σ t ) = A exp [ π ( t )] [ ( ) ] = A exp π t Change the variable to ξ = at with a = Then X(t) = A Y (at) with Y (t) = e πt Using the scale theorem and the result at point 1 above 4

5 (3) Fourier Transform of X(t) = A exp ( π σ t ) exp (πif t) The result derives directly from the second shift theorem (proven at III); denote so that Y (t) = A exp ( π σ t ) X(t) = A exp ( π σ t ) exp (πif t) = Y (t) exp (πif t) The second shift theorem states that ˆX(f) = F[Y (t) exp (πif t)] = Ŷ (f f ) Combine with result at point above Ŷ (f) = A [ exp f ], to obtain Ŷ (f) = An application of this result: denote by [ ] A exp (f f ) = 1 ; a gaussian group of oscillations (waves) X(t) = A exp ( t ) exp (πif t) of fre- quency f and group duration has a gaussian amplitude spectrum of spectral width σ Note that the definition of implies σ = 1 π ; the narrower the spectrum, the longer the group duration 5