SIMG Solution Set #5

Transcription

1 SIMG Soluton Set #5. For each of the transfer functons, sketch H n [ξ] and evaluate and sketch the correspondng mpulse response h n []. Also classfy the flters as lowpass, hghpass, phase, etc. (a) H [ξ] =e +πξ ½ h [] = F {ep [+πξ]} = F ep +π ξ ¾ = δ + allpass (translates the functon) (b) H [ξ] =e +π = +0 h [] = F e +π = F e +π [ξ] ª = e +π δ [] = δ [] allpass (nverts the functon)

2 +π( RECT [π]) (c) H 3 [ξ] =e e +π( RECT [π]) ª h [] = F ξ < = ξ < = RECT [ξ] = = H 3 [ξ] =ep[+π ( )] = ξ > = ξ > = RECT [ξ] =0 = H 3 [ξ] =e +π( 0) = ξ = = ξ = ± = RECT [ξ] = = H 3 [ξ] =e +π ( ) =+ µ H 3 [ξ] = RECT [ξ] [ξ]+ two solated ponts wth ampltude + at ξ = ± h 3 [] = F { RECT [ξ] [ξ]} = SINC δ [] =SINC δ [] allpass (phase flter) (d) H [ξ] =RECT [ξ] H [ξ] =RECT [ξ] e +πξ H [ξ] = RECT [ξ] e +πξ = RECT [ξ] e +πξ h [] = SINC [] δ + = SINC + lowpass flter wth lnear phase

3 +π( RECT [ξ]) (e) H 5 [ξ] =RECT [ξ] H 3 [ξ] =RECT [ξ] e H 5 [ξ] = RECT [ξ] e +π( RECT [ξ]) = RECT [ξ] ( RECT [ξ] [ξ]) ³ h 5 [] = SINC [] SINC δ [] h = SINC [] SINC SINC [] δ [] = F [RECT [ξ] RECT [ξ]] SINC [] = SINC SINC [] lowpass flter 3

4 . For the functon: f [] =COMB [] RECT [] (a) Sketch F [ξ] 50% duty-cycle square wave F [ξ] = COMB [ξ] ξ SINC For each of the transfer functons lsted, sketch H [ξ], the correspondng mpulse response h [], and the output that results f f [] s appled to the nput. Also classfy the flters as lowpass, hghpass, phase, etc. (b) H [ξ] =RECT [ξ] G [ξ] =F [ξ] H [ξ] = δ [ξ] = g [] = []

5 (c) h [] =RECT [] H [ξ] =SINC [ξ] = G [ξ] =COMB [ξ] SINC ξ SINC [ξ] G [ξ] = δ [ξ] = g [] = [] (same output as #b) (d) H [ξ] = RECT [ξ +]+RECT [ξ ] µ H [ξ] = RECT ξ + µ + RECT ξ " # " # ξ + ξ = RECT + RECT " # ξ = RECT µ δ ξ + + δ ξ h [] = SINC sn π = SINC sn [π] 5

6 note that the rectangles n H [ξ] do not enclose any of the Drac delta functons n F [ξ], whch means that: (e) H [ξ] =+ξ G [ξ] =F [ξ] H [ξ] =0[ξ] = g [] =0[] ½ h [] = F ¾ π (πξ) (πξ) ª = F {} π F = δ [] π δ00 [] The transfer functon ncreases faster than the SINC functon decays, so ths transfer functon wll sgnfcantly amplfy the ampltude at large spatal frequences, and thus t sharpens the functon. Ths gves sgnfcant overshoots at the edges, as shown n the llustraton made usng the DFT. 6

7 3. Show that the autocorrelaton of the mpulse response of the constant-phase and lnearphase allpass flters are on-as Drac delta functons. transfer functon of constant-phase allpass flter s : H [ξ] = [ξ] ep [+φ] mpulse response of constant-phase allpass flter s : h [] = δ [] ep [+φ] h [] Fh [] = h [] h [ ] =(δ [] ep [+φ]) (δ [] ep [+φ]) = (δ [] ep [+φ]) (δ [] ep [ φ]) = (δ [] δ []) ep [+φ] ep [ φ] = δ [] δ [] = δ [] =h [] Fh [] transfer functon of lnear-phase allpass flter s : H [ξ] = [ξ] ep [+πξ 0 ] mpulse response of constant-phase allpass flter s : h [] = δ [ 0 ] h [] Fh [] = h [] h [ ] =(δ [(+) 0 ]) (δ [( ) 0 ]) = δ [ 0 ] δ [ 0 ] = δ [ 0 ] δ [ ( + 0 )] = δ [ 0 ] δ [ + 0 ] = δ [ ( 0 0 )] = δ [] =h [] Fh [] 7

8 . Evaluate the M-C-M or C-M-C chrp Fourer transforms (whchever s more convenent) F = ³hf [] e π ( ) e +π ( ) e π ( ) = π e ³n³f [] e +π ( ) o e π ( ) e +π ( ) for the followng functons: (a) f [] =δ [] use the M-C-M µµ = = = ep ³ π δ [] ep µ [δ [] []] ep µ δ [] ep ³ +π ep = [] (b) f [] =δ [ 0 ] use the M-C-M µ δ [ 0 ] ep µ = δ [ 0 ] ep ep ³ +π ³ +π ³ +π ³ ep π ³ ep π ³ π ³ π ³ π ep ep ³ 0 µ = ep π δ [ 0 ] ep ³ 0 Ã " µ #! 0 = ep π ep +π ³ 0 ³ = ep π e +π ( ) e +π ( 0 ) e +π 0 = ³e π( 0 ) e +π( 0 ) ³e +π( ) e π( ) ³ +π ep ³ + π ³ +π ep = ep +π 0 0 a lnear-phase eponental ³ ep π ³ π ³ π ep ³ π ³ ep π ³ ep π e +π 0 8

9 (c) f [] =[] use the M-C-M ³h[] e π( ) e +π( ) e π( ) = ³e π( ) e +π( ) e π( ) ³e π( ) e +π( ) = F n³ e π e +π ξ ³ o e + π e π ξ F ª = δ [] = ³e π( ) e +π( ) e π( ) = δ [] e π( ) = ³δ [] e π ( ) 0 = δ [] (d) f [] =ep[+πξ 0 ] use the M-C-M F = ³hf [] e π( ) e +π( ) e π( ) ³ f [] e π ( ) = ep[+πξ 0 ] ep π ³ = ep π ξ 0 ep +π (ξ 0 ) Now do the convoluton: hf [] e π ( ) e +π ( ) µ ³ = ep π ξ 0 ep +π (ξ 0 ) ³ ep +π = ep +π (ξ 0 ) µ ³ ³ ep π ξ 0 ep +π = ep +π (ξ 0 ) µ ³ ep π δ ξ ep = ep +π (ξ 0 ) µ ³ ³ ep π ep +π = ep +π (ξ 0 ) ³ δ [] δ ξ 0 = ep +π (ξ 0 ) δ ξ 0 = ep +π (ξ 0 ) ξ δ 0 = 3 ep +π (ξ 0 ) δ ξ 0 ³ +π δ ξ 9

10 Now do the postmultplcaton: ³hf [] e π ( ) e +π ( ) e π ( ) = 3 ep +π (ξ 0 ) δ ³ ξ 0 ep π = 3 ep +π (ξ 0 ) δ µ ξ 0 ep " π # ξ 0 = 3 δ ξ 0 ep +π (ξ0 ) ep π (ξ 0 ) = 3 δ ξ 0 ξ = δ 0 h = δ ξ 0 F = δ ξ0 the transform of a lnear-phase eponental s a translated Drac delta functon (e) f [] =ep[+π ] = F h =ep + π ep π use the M-C-M wth =. The premultplcaton gves: ep +π ep π () =[] Now do the convoluton: [] ep +π n h = F δ [ξ] ep + π ep πξ o n h = F δ [ξ] ep + π o h = [] ep + π Now do the postmultplcaton: h [] ep + π ep π () = F h = F [ξ] =ep + π ep π (ξ) 0

11 h 5. Use the C-M-C chrp Fourer transform to fnd an epresson for f [] ep n terms of F +π F = π e ³n³f [] e +π ( ) o e π ( ) e +π ( ) the frst convoluton has the same form as the desred functon h cross-multply : e + π F = n³f [] e +π ( ) o e π ( ) e +π ( ) now want to convolve from the rght to generate a Drac delta functon that cancels the last term e +π ( ) e π ( ) = F n³ ³ o e + π e π ξ e π e +π ξ n³f [] e +π ( ) e + π = [ξ] = e +π ( ) e π ( ) = δ [] = e+π( ) e π( ) = δ [] o e π ( ) e +π ( ) ( ) e π = ³f [] e +π ( ) e π ( ) F ( ) e π = ³f [] e +π ( ) e π ( ) now multply both sdes from the rght by e +π ( ) h f [] e +π ( ) = µ e + π F ( ) e π e +π ( ) = n³ F e π ( ) o e +π ( ) π e+ n³ f [] e +π ( ) = e+ π F e π( ) o e +π ( )