Written exam Digital Signal Processing for BMT (8E070). Tuesday November 1, 2011, 09:00 12:00.

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1 Techische Uiversiteit Eidhove Fac. Biomedical Egieerig Writte exam Digital Sigal Processig for BMT (8E070). Tuesday November, 0, 09:00 :00. (oe page) ( problems) Problem. s Cosider a aalog filter with trasfer fuctio H c (s) = s +s+ ad amplitude respose H c (jω) = ω. +ω a) Is the aalog filter with trasfer fuctio H c (s) a low-pass or a high-pass filter? Explai your aswer. Cosider the amplitude respose H c (jω) for ω 0 ad for ω. lim H c(jω) = 0. ω 0 lim H c(jω) =. ω The fuctio H c (jω) is a smooth fuctio bouded betwee 0 ad. The filter with H c (s) is a high-pass filter. b) We traslate this filter ito a digital filter with the biliear trasform s = z z+, which retais the frequecy respose of the origial filter. Determie the trasfer fuctio H(z) of the digital filter ad draw the pole-zero plot. How do the poles of H(z) follow from those of H c (s)? Use the substitutio s = z z+ i H c(s). 8(z ) 8(z ) + (z ) + (z + ) 8(z ) = z z + 5 8(z ) = (z 7 + j )(z 7 j ) We see that the digital filter has two zeros = = ad two poles p = 7 j ad p = 7 + j. Im Re The aalog filter with H c (s) has poles i z ( ± j). From s = z+ we derive z = +s s ad there we substitute the (s domai) poles ( ± j). This results i the (z) domai poles 7 ± j. c) Give the formula for H(e jθ ) as a real fuctio of θ. What is the amplitude value at θ = 0 respectively at θ? At which frequecy is H(e jθ ) equal to? Sketch the amplitude respose H(e jθ ) of the digital filter.

2 We kow that H(e jθ ) = H c (jω) ω= ta θ H(e jθ ) = so 8(ta θ ) + 6(ta θ ). So we fid We also look for the poit. H(e j0 ) = 0 H(e j ) = lim θ H(e jθ ) 8(x) = lim x + 6(x) =. 8x = + 6x 8x = + 6x 6x = + x x = 8 θ 0 = arcta x = arcta /8 = of i terme va ω (is iet echt bedoeld) ω = D ta θ = 8 = H(e jθ ) θ 0 θ d) Is the digital filter with trasfer fuctio H c (s) a low-pass or a high-pass filter? Explai your aswer. Just as i questio a) we fid for the same reasos that the digital filter is a high-pass filter. Problem. Cosider a 5-poit liear phase FIR filter with impulse respose h[] = 0 for < 0 ad >. The correspodig frequecy respose H(e jθ ) is give i three poits: H(e jθ ) =, 0, 0 for θ = 0,, respectively. a) Determie the impulse respose h[] of the filter.

3 I geeral for this 5-poit liear phase filter the impulse respose looks like h[] = a, b, c, b, a ad the frequecy respose is the give as } H(e jθ ) = e {h[] jθ + h[] cos(( )θ) =0 = e jθ {a cos(θ) + b cos θ + c} H(e j0 ) = a + b + c H(e j/ ) = ( )( a + c) H(e j ) = a b + c So we solve for a, a = b, ad c i the equatios a + b + c = a + c = 0 a b + c = 0 We fid a = ad b = c =, so h[] =,,,,. b) Determie H(e jθ ) for θ = ad sketch the respose H(ejθ ). H(e j ) = e j (a cos + b cos ) + c ( = e j ) + ( ) = e j + H(e jθ ) θ c) Draw the phase respose arg{h(e jθ )}. Pay attetio to possible phase jumps of at a zero of H(e jθ ).

4 We have arg H(e jθ ) = θ because it is a 5-poit liear phase filter. Plus, there will be a + shift for every zero of H(e jθ ). So we look for the zeros. 0 = H(e jθ ) = cos θ + cos θ + With the idetity cos θ = cos θ we fid = cos θ + cos θ = cos θ(cos θ + ) So, the zeros are at = ad = [because of cos θ = 0], ad at = [because of cos θ = ]. Due to those zeros there will be a + phase shift at θ = ±.The zero at behaves as a double zero because there is o sig chage aroud θ =. φ θ d) Determie H(z). What are the four zeros of H(z)? Draw the pole-zero plot.

5 We kow that z = j (so θ = ) ad z = (so θ = ) are zeros of H(ejθ ). Sice z = j is a zero so must z = j be a zero. So we kow of the zeros ad solve for the last oe: From the impulse respose we have ad we must solve for β to get equality (z + )(z + )(z + β) z z + z + z + z + z This results i β = so (z + )(z + ) z So the filter has zeros i z = j, z = j, ad a double zero i z =, ad four poles i z = 0. Im Re e) Determie the 6-poit DFT of h[] for k = 0,,..., 5. Give the complex umerical values. Also, give a plot of oe period of the iverse DFT h[]. h[] =,,,,, 0. H[k] = 5 h[]e j 6 k =0 0 = k = 0 = j j ( + j ) = ( j ) + j ( j ) = ( j ) j + j j 5 = 0 ( + j ) = ( + j ) + j + j ( j ) = ( + j )

6 h[] = 6 5 H[k]e j 6 k k=0 k 0 5 H[k] e j e j 0 ej ej W6 k e j e j ( ) e j e j 5 0 e j e j 0 ej ej e j ej 0 e j ej 0 ej e j 0 ej e j ej ej 0 e j e j 5 ej 0 h[0] = ( + cos 6 + cos ) = h[] = ( + cos 6 + cos ) = h[] = ( cos + ) = h[] = ( + cos 6 + cos ) = h[] = ( + cos 6 + cos ) = h[5] = ( + cos + cos ) = 0 6 Problem. g[] is a N-poits discrete sigal ad G[k] is its N-poits DFT. g[] is the IDFT of G[k]. a) What is the relatio betwee g[] ad g[]? You may use a sketch i your explaatio ad/or use expressios. g[] is a periodic sigal. It cosists of the sum of all shifts over rn (for iteger r) positios of g[]. g[] g[ N] g[] r = r = 0 r = b) G [k] is a N-poits DFT that we derive from G[k] by isertig zeros i betwee the samples from G[k]. So, G [k] = G[k] ad G [k + ] = 0 for all k = 0,,..., N. Determie the IDFT g [] of the N-poits G [k] ad describe it i terms of g[].

7 We kow the followig for the IDFT s. g[] = N N k=0 G[k]e j N k g [] = N G [k]e j N k N k=0 = N N G [l]e j N l + N N = N G [l]e j N l N = N G[l]e j N l N = g[]. G [l + ]e j N (l+) g [] is periodic with period N (as it must be), but iside oe period it cotais two periods of g[] with its amplitude. Score. Problem : a) poit; b) poits; c) poits; d) poit. Total 7 poits. Problem : a) poit; b) poits; c) poit; d) poits; e) poits. Total 0 poits. Problem : ) poit; b) poits. Total poits. The fial score is determied by the sum of the poits scored, divided by ad rouded to the earest iteger (betwee ad 0).

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