Solution Set #1

Transcription

1 Solution Set # i. Evaluate δ [SINC []] = δ sin[π] π Recall relation for Dirac delta function wit functional argument (Fourier Metods, δ [g []] = δ [ 0] (6.38) dg =0 Te SINC function as zeros at all nonzero integer values of, so our task ere is to evaluate te slopes at tose coordinates: µ d sin [π] = cos (π) sin π d π π µ d sin [π] d π = cos (πn) sin πn =n6=0 n πn = n ( )n πn 0 µ d sin [π] = d π = n d =n6=0 Since te slope decreases wit increasing n, te area of te Dirac delta functions increases wit increasing n sin [π] δ = X δ [ n] π = X n δ [ n] = n COMB [] δ [] n6=0 n n6=0 SoteresultconsistsofaCOMBfunctiontateceptforteDiracdeltaatteorigin tat as been modulated by n δ [SINC []] (add arroweads to eac spike )

2 . Prove te relation F {RECT [, y]} = SINC [ξ] SINC [η] (zzzzz) F {RECT [, y]} = F {RECT [] RECT [y]} = F {RECT []} F {RECT [y]} F {RECT []} = = = Z + Z + RECT [] ep [ πiξ] d ep [ πiξ] d ep [ πiξ] πiξ µ ep = i (πξ) =+ = πiξ + (ep [ iπξ] ep [+iπξ]) i (πξ) = ep [+iπξ] ep [ iπξ] = πξ i ep πiξ sin [πξ] πξ SINC [ξ] F {RECT []} F {RECT [y]} = SINC [ξ] SINC [η] SINC [ξ,η]

3 3. Evaluate te -D Fourier transform: F f [, y] ª ½µ ¾ = F + f [, y] y Recall te derivative teorem (Fourier Metods): ½ ¾ d n F {g []} = F d f [] =(+πiξ) n n F [ξ] =G [ξ] (9.3) F ½µ + y ½ ¾ F f [, y] ½ ¾ F f [, y] y ¾ f [, y] = (+πiξ) F [ξ] = (π) ξ F [ξ] = (+πiη) F [ξ] = (π) η F [ξ] = (π) ξ + η F [ξ] = (π) ρ F [ξ] 3

4 . Given tat: F ª ep +iπ =ep +i π i ep iπξ (a) Use te scaling teorem of te -D Fourier transform to sow tat: ³ F ½ep +iπ = α ep +i α ¾ π i ep iπ (αξ) were α is some numerical constant wit units of lengt. Direct substitution io F nf = b F [bξ] b ³ = F ½ep +iπ = α ep +i α ¾ π i ep iπ (αξ) (b) Sketcorplottecirpfunctionep +iπ i α as magnitude and pase α =+, +. as real and imaginary parts AND ep [+iπ ]

5 ep +iπ i 5

6 (c) Sketc or plot te spectrum F nep +iπ io α AND as magnitude and pase α =+, +. as real and imaginary parts F ª ep +iπ = ep +i π i ep iπξ µ = ep iπ ξ Re F ªª µ ep +iπ = cos π ξ Im F ªª µ ep +iπ = sin π ξ F ª ep +iπ = [ξ] µ Φ F [ξ] = π ξ F {ep [+iπ ]} =ep +i π ep iπξ 6

7 ³ ¾ F ½ep +iπ = ep +i π i ep iπ (ξ) µ = ep iπ ξ Re F ªª µ ep +iπ = cos π ξ Im F ªª µ ep +iπ = sin π ξ F ª ep +iπ = [ξ] µ Φ F [ξ] = π ξ F nep +iπ io = ep +i π ep iπ (ξ) 7

8 5. For te frequency-domain function H [ξ] =RECT ξ ξ ma e +iπ(αξ) were α is some numerical constant wit units of lengt. (a) Use te filter teorem to evaluate [] =F {H [ξ]} F ½ RECT ξ ξ ma [] = ξ ma SINC ³ [] = ξ ma α F ep +iπ (αξ) ª = α e+i π e iπ( α) ¾ = ξ ma SINC [ξ ma ]= ξ ma SINC ³ ξ ma ep +i π i SINC µ +i α ep π i ³ ξ ma ep ξ ma ³ ep iπ α ³ iπ α 8

9 (b) Given tat F {f []} F [ξ], substitute ³ ξ = α αξ ³ (αξ) α in te formula for te -D Fourier transform to derive a relationsip between f [α ξ] and F [ξ]. Because α as units of lengt, te representations ave te proper dimensionality. ξ = F {F [ξ]} = ³ α αξ ³ ³ (αξ) ³ = +ξ =+ +(αξ) α α α αξ Z + Z + F [ξ] e +πiξ dξ = F [ξ] e +iπ(ξ) dξ Z + f [] = F [ξ] e +( +iπ α) +(αξ) ( α αξ) dξ = e +iπ( α) Z + ³F [ξ] e +iπα ξ e iπα ( α ξ ) dξ = e +iπ( α) ³ F [ξ] e +iπ(αξ) = ³³F [ξ] e +iπ(αξ) e iπ(αξ) ξ α e iπ(αξ) e +iπ(αξ) ξ α αξ α = = α ξ f α ξ = ³³F +iπ(αξ) +iπ(αξ) [ξ] e e iπ(αξ) e f α ξ = f α ξ e iπ(αξ) = = ³f α ξ iπ(αξ) e e +iπ(αξ) = = ³³F [ξ] e +iπ(αξ) e iπ(αξ) e +iπ(αξ) ³³F +iπ(αξ) +iπ(αξ) [ξ] e e iπ(αξ) e e iπ(αξ) ³F +iπ(αξ) [ξ] e e iπ(αξ) ³F +iπ(αξ) [ξ] e ³e +iπ(αξ) iπ(αξ) e ³F +iπ(αξ) [ξ] e α δ [ξ] = F [ξ] e+iπ(αξ) α Inverse Fourier transform of bot sides: F f α ξ ª = F ½µµ α F [ξ] e+iπ(αξ) e iπ(αξ) e +iπ(αξ) ¾ α F i α = = µµµ µ α f [] π α e+i π α e+i e iπ( α) π α e i e +iπ( α) ³³³f [] e iπ( α) e +iπ( α) e iπ( α) F i = α e +i π α ³³³f [] e iπ ( α) e +iπ ( α) e iπ ( α) 9 π α e+i e iπ( α)

10 " # " # α + RECT α = RECT ξ ma ξ ma = α e +i π ³³³SINC [ ξ ma ] e iπ( α) e +iπ( α) e iπ( α) 0

11 (c) Use tis result to find a DIFFERENT epression for H [ξ] tat includes a SINC function in te frequency domain. F [ξ] e +iπ(αξ) = ³f α ξ iπ(αξ) e e +iπ(αξ) ξ RECT e +iπ(αξ) = ³ ξ ξ ma SINC ξ ma α ξ iπ(αξ) e e +iπ(αξ) ma H [ξ] = ξ ma SINC ξ ξ ma α ep iπ (αξ) ep +iπ (αξ) (d) Evaluate te impulse response resulting from (c) it will include a RECT function and compare wit te result of (a); graps will be elpful. [] = F {H [ξ]} = F ξ ma SINC ξ ³ e iπ(αξ) e +iπ(αξ) ξ ma α µ µ µ = ξ ma RECT e iπ(αξ) e +iπ(αξ) ξ ma α ξ ma α [] = µ µrect e iπ(αξ) e +iπ(αξ) α ξ ma α = ξ ma e +i π SINC ³ e iπ ( α) (from a) α µ µ RECT α ξ ma α ξ ma e iπ(αξ) e +iπ(αξ) = ξ ma α e +i π SINC ³ ξ ma e iπ( α) (e) Sow tat tese epressions yield te epected results in te limiting cases of ξ ma and ξ ma 0 for a finite α. ½ µ µ ¾ lim { []} = lim RECT e iπ(αξ) e +iπ(αξ) ξ ma 0 ξ ma 0 α ξ ma α = ½µ µ ¾ lim RECT e iπ(αξ) e +iπ(αξ) α ξ ma 0 ξ ma α = ξ ma e +i π SINC ³ e iπ ( α) (from a) α ξ ma

12 6. In tis problem, you can use some imaging or computing toolbo (e.g., IDL or MatLab or even SignalSow by Juliet Bernstein) to generate and operate on a -D array based on te -D cirp function: f [] =ep ±i µπ α φ 0 Te array sould ave N samples (N is even and could be a power of ) indeed by n suc tat = n and n = N, N +,,, 0, +,, + N. (a) Derive te relation between α and te array size N suc tat te -D sampled function is just aliased at te edges of te -D array. Recall te formula for spatial frequency in terms of te pase: ξ [] Φ [] π ξ [] = π ξ [] = α µπ + y α φ 0 = ³π = π α α For a centered array, te coordinate at te edge is = N frequency is ξ Nyquist = : N ξ N = = α = α = N ( ) = α = N. Te Nyquist

13 (b) Print images of te real and imaginary parts of te sampled comple function f [] using te value of α tat satisfies te constraint of part (a) for an array of size N. I suggest selecting N =56or 5. for α =, same as #b above: ep +iπ i n 6 3

14 (c) Evaluate te fast Fourier transform (FFT) or te discrete Fourier transform (DFT) of te sampled comple cirp (make sure tat te sample corresponding to zero frequency is centered in te array, and not at te edges as is often te case). Print out images of te function and its FFT as real part, imaginary part, magnitude, and pase. Note tat since f [, y] is an even function, so sould be its FFT. F nep +iπ io n =ep +i π 3 ep iπ i k 3

15 (d) Now construct a modulated cirp functions by multiplying f [n] by some functions s [n]. Te first modulation function sould be centered wit te ais of symmetry at te origin and binary (e.g., a rectangle); te second sould be te same function binary function after translation by an arbitrary distance. For eample, if N =5, you migt select te modulation functions to be: n i s [n] = RECT 8 n 8 s [n] = RECT 8 f [] =RECT n 8 8 ep +iπ i n 3 5

16 DFT of f [] =RECT n 8 8 ep +iπ i n sowing te modulation is approimately 3 preserved. 6

17 (e) Repeat part (d) using a modulation function of your coice tat is not binary and may even be bipolar (e.g., a Triangle or a SINC function). Submit images of te modulation function, of te real and imaginary parts of te modulated cirp functions, and of teir FFTs as real part, imaginary part, magnitude, and pase. space-domain representation of cirp function modulated by SINC function 7

18 DFT of cirp function modulated by SINC function, again sowing tat te modulation is approimately preserved. 8

19 7. (Optional) Repeat parts (b-e) of te last problem for a -D array of size N N derived from te -D function: µ µ + y f [, y] =ep ±i π φ 0 α -D cirp modulated by rectangle function and its DFT 9