Proofs of Parseval’s Theorem & the Convolution …

1 Proofs of parseval s Theorem & the Convolution Theorem (using the integral representation of the -function)1 The generalization of parseval s theoremThe result is f(t)g(t) dt=12 f( )g( ) d (1)This has many names but is often called Plancherel s key step in the proof of this is the use of the integral representation of the -function ( ) =12 e i d or ( ) =12 e i d .(2)We firstly invoke the inverse Fourier transformf(t) =12 f( )ei td (3)and then use this to re-write the LHS of (1) as f(t)g(t) dt= (12 f( )ei td )(12 g( 0) e i 0td 0)dt.(4)Re-arranging the order of integration we obtain f(t)f(t) dt=(12 )2 f( )g( 0) ( ei( 0)tdt) Use delta fn hered 0d .(5)The version of the integral representation of the -function we use in (2) above is ( 0) =12 eit( 0)dt.(6)Using this in (5), we obtain f(t)g(t) dt=12 f( )( g( 0) ( 0)d 0)d =12 f( )g( ) d.

2 (7)(7) comes about because of the general -function property F( 0) ( 0)d 0=F( ).2 parseval s Theorem (also known as the energy Theorem )Takingg=fin (1) we immediately obtain |f(t)|2dt=12 |f( )|2d .(8)The LHS side is energy in temporal space while the RHS is energy in spectral :Sheet 6 Q6 asks you to use parseval s Theorem to prove that dt(1+t2)2= integral can be evaluated by the Residue Theorem but to use parseval s Theorem you willneed to evaluatef( ) = e i tdt1+t2. To find this, construct the complex integral Ce i zdz1+z2andtake the semi-circleCin the upper (lower) half-plane when <0 (>0). The answers are e when <0 and e when >0. Then evaluate the RHS of (8) in its two Convolution Theorem and the auto-correlation functionThe statement of the Convolution Theorem is this: for two functionsf(t) andg(t) with FouriertransformsF[f(t)]=f( ) andF[g(t)]=g( ), with Convolution integral defined by1f ?

3 G= f(u)g(t u)du,(10)then the Fourier transform of this Convolution is given byF(f ? g)=f( )g( ).(11)To prove (11) we write it asF(f ? g) = e i t( f(u)g(t u)du)dt.(12)Now define =t uand divide the order of integration to findF(f ? g) = e i uf(u)du e i g( )d =f( )g( ).(13)This step is allowable because the region of integration in the uplane is infinite. As we shalllater, with Laplace transforms this is not the case and requires more normalised auto-correlation function is related to this and is given by (t) = f(u)f (t u)du |f(u)|2du.(14)1It makes no difference which way round thefand theginside the integral are placed: thus we could writef ? g= f(t u)g(u)du.(9)2