Transcription of EE120 - Fall'15 - Lecture 4 Notes - inst.eecs.berkeley.edu
1 EE120 - fall 15- Lecture4 Notes11 Licensed under a Creative Arcak9 September2015 Continuous Time Fourier TransformChapter4inOppenheim & WillskyApplicable to aperiodic signals (unlike Fourier series which is appli-cable only to periodic signals).Main idea:Treat aperiodic signalx(t)as the limit of a periodic signal x(t)as periodT (see figure below). AsTincreases, the funda-mental frequency 0=2 Tdecreases and the harmonic componentsbecome closer in frequency, forming a continuum in the limitT .Example:1 T1T1 T1T1T x(t)tx(t)AperiodicPeriodicRecall from Lecture3that the periodic signal on the right has Fourierseries coefficients:ak=2sin(k 0T1)k 0T, 0=2 T.
2 (1)This expression is not defined fork=0,but we interpreta0to be the limit ask 0, ,a0=2T1 ( T1) =k 0(2)a0T 0a1Ta2 Tenvelope forTak This envelope is the Fourier transform" ofx(t)above. In general:ak=1T T x(t)e jk 0tdt(3)Tak= x(t)e jk 0tdt=( x(t)e j tdt) ,X(j )(Fourier Transform) =k 0(4) EE120 - fall 15- Lecture 4 Notes 2X(j ) = x(t)e j tdt(Analysis Equation)x(t) =12 X(j )ej td (Synthesis Equation)(5)Derivation of the synthesis equation: x(t) = k= akejk 0t=12 k= 2 T = 0[X(j )ej t] =k 02(6)2 The summation term onthe right can be pictured as:k 0(k+1) 0 X(j )ej tTake the limit asT :x(t) =12 X(j )ej td (7) convergence :If the "Dirichlet conditions" below hold, then12 W WX(j )ej td converges tox(t)asW for allt, except at discontinuities whereit converges to the average.
3 D1)x(t)is absolutely integrable: |x(t)|dt< ;D2)x(t)has finite # of minima and maxima within any finite inter-val;D3)x(t)has finite # of discontinuities within any finite interval andthe discontinuities are :1)x(t) =e atu(t),a> (j ) = 0e ate j tdt= 0e (a+j )tdt=1a+j e (a+j )t 0 =1X(j ) =1a+j ,|a+j |= a2+ 2,]a+j =tan 1( /a)|X(j )|=1 a2+ 2,]X(j ) = tan 1( /a)1/a |X(j )| /2 /2 X(j ) EE120 - fall 15- Lecture 4 Notes 32) Dirac delta:x(t) = (t)X(j ) = (t)e j tdt=1 for all 3) Rectangular pulse:x(t) ={1|t|<T10|t| T1 T1T1tx(t)X(j ) = T1 T1e j tdtFor =0,X(j0) = T1 T1dt=2T1. For 6=0, T1 T1e j tdt= 1j e j t T1T1=ej T1 e j T1j =2sin( T1).}
4 Combining,sinc( ),{sin 6=01 = 123 1 2 3X(j ) ={2T1 =02sin( T1) 6=0=2T1sinc(T1 )4)X(j ) ={1| |<W0| | WA derivation similar to Example3gives:x(t) =12 W Wej td =W sinc(W t)Note the duality in Examples 3 and pulseFT sincsincFT rectangular pulse5)X(j ) =2 ( 0)x(t) =12 2 ( 0)ej td =ej 0tIf 0=0, then 1FT 2 ( ). Note the duality with 15- Lecture 4 Notes 4 Fourier Transform of Periodic & WillskyFrom Example5above:ej 0tFT 2 ( 0)(8)By linearity: k= akejk 0tFT k= 2 ak ( k 0)(9)Example:x(t) =cos( 0t) =12ej 0t+12e j 0tX(j ) = ( 0) + ( + 0) 0 0 X(j )Example: Impulse Trainx(t) = k= (t kT)ak=1T T/2 T/2 (t)e jk 0tdt=1 Tfor allkX(j ) =2 T k= ( k 0)t TT2Tx(t).}}}
5 X(j ) 02 03 0 Properties of the Fourier & WillskyConsiderx(t)FT X(j )andy(t)FT Y(j ).Linearity:ax(t) +by(t)FT aX(j ) +bY(j ),a,b R(10)Time-Shift:x(t t0)FT e j t0X(j )(11) EE120 - fall 15- Lecture 4 Notes 5 Proof: x(t t0 , )e j tdt= x( )e j (t0+ )d =e j t0 x( )e j d =X(j )Conjugation and Conjugate Symmetryx (t)FT X ( j )(12)Ifx(t)is real:X(j ) =X ( j )(becausex(t) =x (t)) |X(j )|=|X( j )|(even symmetry)(13)]X(j ) = ]X( j )(odd symmetry)(14)Example1above:1/a |X(j )| /2 /2 X(j )Differentiation and Integrationdx(t)dtFT j X(j )(15) t x( )d FT 1j X(j ) + X(0) ( )(16)Example:x(t) =u(t)(unit step)Notex(t) = t ( )d and (t)FT (j ) ,X(j ) =1j (j ) + (0) ( ) =1j + ( )Time and Frequency Scalingx(at)FT 1|a|X(j a),a6=0(17) EE120 - fall 15- Lecture 4 Notes 6 Proof: x(at , )e j tdt= x( )e j /ad a, ifa>0= x( )e j /ad a, ifa<0=1|a| x( )e j /ad =1|a|X(j a)Example:x(t) =sinWt tFT T1T1tx(t)We can interpret this as a scaling of:x0(t) =sint tFT 11 X0(j )x(t) =WsinWt Wt=Wx0(Wt)FT X0(j /W) =X(j )Corollary (a=-1):x( t) X( j )(18)Ifx( t) =x(t)thenX( j ) =X(j )Ifx(t)is also real:X( j ) =X (j )}X(j ) =X (j ), ,X(j )is s Relation: |x(t)|2dt=12 |X(j )|2d (19)Example.
6 X(t) =e atu(t)a>0 X(j ) =1a+j |x(t)|2dt= 0e 2atdt=12ae 2at 0 =12a |X(j )|2d = 1a2+ 2d =1atan 1( a) = a=2 12aInitial Value:x(0) =12 X(j )d (synthesis eq n witht=0)(20)DC Component:X(0) = x(t)dt(analysis equation with =0)(21)
