Sampling Example. ( ) δ ( f 1) (1/2)cos(12πt), T 0 = 1

Transcription

1 Samplig Example Le x = cos( 4π)cos( π). The fudameal frequecy of cos 4π fudameal frequecy of cos π is Hz. The ( f ) = ( / ) δ ( f 7) + δ ( f + 7) / δ ( f ) + δ ( f + ). ( f ) = ( / 4) δ ( f 8) + δ ( f 6) + δ ( f + 6) + δ ( f + 8). is 7 Hz. The Takig pairs of impulses a a ime, he iverse CTFT is The fudameal frequecy of cos 6π x = ( / ) cos( 6π) + cos( π). is 8 Hz. The fudameal frequecy of is 6 Hz. Therefore he fudameal frequecy of = ( / ) cos( 6π) + cos( π) is Hz. Sice we have ow show ha cos π x cos( 4π)cos( π) = ( / ) cos( 6π) + cos( π), x () - cos(4π), T 0 = /7 - x() Produc x () - cos(4π)cos(π), T 0 = / cos(π), T 0 = (/)cos(6π), T 0 = /8 x () - - x() Sum (/)cos(π), T 0 = /6 x () (/)cos(6π) + (/)cos(π), T 0 = / - ha meas ha he fudameal frequecy of cos( 4π)cos π fudameal period is / secod. Also, he highes frequecy i cos( 4π)cos π 8 Hz. is also Hz ad is is Now sample i a f s = 6 o ge x[ ] = cos( 4π /6)cos( π /6). This is samplig a he Nyquis rae. Therefore he samplig is o quie fas eough o avoid aliasig. The DTFT of x[ ] is X F ( F) = ( / ) δ ( F 7 /6) + δ ( F + 7 /6) / δ ( F /6) + δ ( F + /6)

2 X F ( F) = ( / 4) δ ( F 7 /6) + δ ( F + 7 /6) δ F /6 X F ( F) = / 4 + δ ( F + /6) δ ( F / ) + δ ( F 3 / 8) + δ ( F + 3 / 8) + δ ( F + / ) 0.5 X(F) The iverse DTFT of X F ( F), is x F - [ ] = X F F ca lie aywhere i F. Le he iegral be [ ] = X F F x e j π F df = / 4 0 e j π F df. The iegraio rage of oe δ ( F 8 /6) + δ ( F 6 /6) + δ ( F + 6 /6) + δ ( F + 8 /6) 0 x[ ] = ( / 4) δ ( F / ) + δ ( F 3 / 8) + δ ( F 5 / 8) x[ ] = ( / 4) e jπ 0 + e j 3π /4 j 5π /4 + e e jπ =e j 3π/4 = / 4 e j π F df e jπ + e jπ + e j 3π /4 + e x[ ] = ( / ) cos( π) + cos( 3π / 4) Therefore cos( 4π / 6)cos π /6 wih he rigoomeric ideiy cos( x)cos y = ( / ) cos( π) + cos( 3π / 4) = ( / ) cos( x y) + cos( x + y) j 3π /4. which agrees e j π F d

3 Produc Sum cos(4π/6), N 0 = 6 x [] cos(π/6), N 0 = 6 x [] (/)cos(6π/6), N 0 = x [] (/)cos(π/6), N 0 x [] cos(4π/6)cos(π/6), N 0 - x[] (/)cos(6π/6) + (/)cos(π/6), N 0 The relaio bewee he CTFT ad he DTFT is X F ( F) = f s f s F k - x[] k = ( ). If samplig is doe accordig o he samplig heorem, hese muliple aliases of ( f ) do' overlap ad we could recover ( f ) from X F ( F) by muliplyig X F ( F) by o cu off all he aliases, replacig F by f / f s, ad he dividig by f s. If we rec F do ha i his case he sages of he rasiio are Muliply by rec ( F) : ( / 4) δ F / + δ ( F 3 / 8) + δ ( F + 3 / 8) + δ ( F + / ) ( / 4) δ F / Replace F by f / f s : ( / 4) δ f / f s / + δ ( F 3 / 8) + δ ( F + 3 / 8) + δ ( F + / ) rec F + δ ( f / f s 3 / 8) + δ ( f / f s + 3 / 8) + δ ( f / f s + / ) Use he scalig propery of he impulse. Divide by f s : δ ( f 8) + δ ( f 6) + δ ( f + 6) + δ ( f + 8) f s / 4 ( / 4) δ f 8 The iverse CTFT of his las expressio is + δ ( f 6) + δ ( f + 6) + δ ( f + 8) Figures showig hese seps ( / )cos( 6π) + ( / )cos( π) = x.

4 I his case we violaed he samplig heorem by samplig a he Nyquis rae bu go he correc aswer ayway. This occurred because he highes frequecy erm i x = ( / )cos( 6π) + ( / )cos( π) is a cosie. If i had bee a sie, he recosrucio of he origial coiuous-ime sigal would have bee wrog. Le x = cos( 4π)si( π). The ( f ) = ( / ) δ ( f 7) + δ ( f + 7) j / δ ( f + ) δ ( f ) ( f ) = ( j / 4) δ ( f 6) δ ( f 8) + δ ( f + 8) δ ( f + 6). The iverse CTFT is x rigoomeric ideiy cos( x)si y Hz, x si π /6 [ ] = cos 4π /6 = ( / ) si( 6π) si( π) = ( / ) si( x + y) si( x y) ad which agrees wih he. If we sample a 6 x () - cos(4π), T 0 = /7 - x() Produc x () - cos(4π)si(π), T 0 = / si(π), T 0 = (/)si(6π), T 0 = /8 x () - - x() Sum (/)si(π), T 0 = /6 x () (/)si(6π) - (/)si(π), T 0 = / - If we sample he sigals a 6 Hz we ge x[ ] = cos( 4π /6)si( π /6) ad, usig he rigoomeric ideiy cos( x)si( y) = ( / ) si x + y ge cos( 4π /6)si( π /6) = ( / ) si( 6π /6) si π /6 =0. si( x y) we

5 cos(4π/6), N 0 = 6 x [] - - x[] Produc x [] - si(π/6), N 0 = 6 cos(4π/6)si(π/6), N 0 (/)si(6π/6), N 0 = x [] - - x[] Sum (/)si(π/6), N 0 x [] (/)si(6π/6) - (/)si(π/6), N 0 - X F ( F) = j / 4 δ ( F 3 / 8) δ ( F / ) + δ ( F + / ) δ ( F + 3 / 8) The erms δ ( F / ) + δ ( F + / ) add o zero everywhere ad X F ( F) = ( j / 4) δ ( F 3 / 8) δ ( F + 3 / 8). 0.5 X(F) F - [ ] = / which does equal because if we apply we ge The iverse DTFT is x si π /6 cos( 4π /6)si π /6 cos( x)si( y) = ( / ) si( x + y) si( x y) o cos( 4π /6)si π /6 cos( 4π /6)si( π /6) = ( / ) si( 6π /6) si( π /6). = si( π) = 0 for ay ieger value of. If we ow ry o recosruc Bu si 6π /6 he origial coiuous-ime fucio from he samples we ge hese seps Muliply by rec ( F) :

6 Replace F by f / f s : ( j / 4) δ F 3 / 8 δ ( F + 3 / 8) ( j / 4) δ F 3 / 8 ( j / 4) δ f / f s 3 / 8 Use he scalig propery of he impulse. Divide by f s : δ ( F + 3 / 8) rec F δ ( f / f s + 3 / 8) δ ( f 6) δ ( f + 6) f s j / 4 ( j / 4) δ f 6 The iverse CTFT of his las expressio is δ ( f + 6) Figures showig hese seps ( / )si( π) x. The sie a half he samplig rae is missig. This is a error due o aliasig. If we use he for aalysis, samplig a 6 Hz x = cos( 4π)cos( π) x[ ] = cos( 4π /6)cos( π /6) The fudameal period of each of he cosies ha is muliplied is 6. Bu he fudameal period of he produc is 8. If we view his as a produc of cosies ad fid he of each ad he periodically covolve he 's we mus use N = 6. Usig mn cos πq / N δ mn k mq mn / + δ mn k + mq N is he fudameal period of each cosie, 6. So i le N = 6 ad m =. I he firs cosie q = 7 ad 6 cos 4π / 6 I he secod cosie q = ad. 8 δ 6 k 7 + δ 6 k + 7

7 6 cos π / 6 We ca ow use he propery of he ad ge cos( 4π)cos π 6 cos( 4π)cos π cos( 4π)cos π. 8 δ 6 k + δ 6 k + x y / N N Y k X k 8( δ 6 k + δ 6 k + ) ( / 6)8 δ 6 k 7 + δ 6 k Now use he fac ha ( δ k + δ k + ) 4 δ 6 k 7 + δ 6 k δ 6 k 8 + δ 6 k 6 + δ k δ 6 k + 8 cos( 4π /6)cos( π /6) = ( / ) cos( π) + cos( 3π / 4) we ca use is fudameal period which is N ad ad = cos π( 4 / 8) cos π cos 3π / 4 = cos π( 3 / 8) 4 δ 8 k ( [ ] + δ 8 [ k + 4] ) ( [ ] + δ 8 [ k + 3] ) 4 δ 8 k 3 The ( / ) cos π + cos( 3π / 4) / 8 + 4( δ 8 [ k 3] + δ 8 [ k + 3] ) 4 δ 8 [ k 4] + δ 8 [ k + 4] ( / ) cos π + cos( 3π / 4) δ 8 8 k 4. [ ] + δ 8 [ k + 4] + δ 8 [ k 3] + δ 8 [ k + 3] We ca cover he las resul o he same basis as he firs, N = 6 usig he Chage of Period propery of he If x X k N he x mx k / m, k / m a ieger mn 0, oherwise Applyig i o he las resul,

8 cos( π) + cos( 3π / 4) / [ ] + δ 8 [ k / + 4] 4 δ 8 k / 4, k / a ieger 6 +δ 8 [ k / 3] + δ 8 [ k / + 3] 0, oherwise Examie δ 8 [ k / 4]. I is a periodic impulse ha occurs every ime k / 4 is a ieger muliple m of 8. If k / 4 m he k 8 = 6m. So he impulses also occur every ime k 8 is a ieger muliple of 6. Therefore δ 8 k / 4 ( / ) cos π + cos( 3π / 4) δ 6 k 8 Sice he impulses i +δ 6 k 6 sipulaio of "0, oherwise" is reduda ad ( / ) cos π + cos( 3π / 4) [ ] + δ 6 [ k + 8] [ ] = δ 6 [ k 8] ad 4 δ 6 k 8, k / a ieger 6 +δ 6 [ k 6] + δ 6 [ k + 6] 0, oherwise [ ] + δ 6 [ k + 8] [ ] + δ 6 [ k + 6] 4 δ 6 6 k 8 oly occur for eve values of k, he [ ] + δ 6 [ k + 8] + δ 6 [ k 6] + δ 6 [ k + 6] This cofirms ha he wo soluios usig N ad N = 6 are equivale. Sice we are samplig a periodic sigal over a ieger umber of periods, we ca fid he CTFT direcly from he usig I his example, 7 N / X( f ) = ( / N ) X k [ ]δ f kf s / N. k = N / X( f ) = ( /6) 4 δ 6 [ k 8] + δ 6 [ k 6] + δ 6 [ k + 6] + δ 6 [ k + 8] k = 8 7 X( f ) = ( / 4) δ 6 [ k 8] + δ 6 [ k 6] + δ 6 [ k + 6] + δ 6 [ k + 8] k = 8 δ f k δ f k I he summaio rage of -8 o +7, he periodic impulses occur oly a k values of 8, 6, - 6 ad -8. Therefore X( f ) = ( / 4) δ ( f 8) + δ ( f 6) + δ ( f + 6) + δ ( f + 8)

9 ad his is he correc exac CTFT of he origial coiuous-ime sigal, x = cos( 4π)cos( π). If we ried o do he same kid of aalysis o he sigal usig N = 6 ad mn si πq / N x = cos( 4π)si( π) δ mn k + mq jmn / he of he sampled sigal would be cos( 4π)si π 6 which ca be reduced o ( / 6)8 δ 6 k 7 + δ 6 k + 7 cos( 4π)si π 6 δ mn k mq j8( δ 6 k + δ 6 k ). j4 δ 6 k 6 δ 6 k + 6 The ryig o fid he CTFT of he origial coiuous-ime sigal we ge X( f ) = ( j / 4) δ ( f 6) δ ( f + 6) ad his is o he correc CTFT. The sie compoe a half he samplig rae is missig because of aliasig.