UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet Eksamen i INF347/447 Digital signalbehandling Eksamensdag:. desember 5 Tid for eksamen: 9. 3. Oppgavesettet er på 6 sider. Vedlegg: Ingen Tillatte hjelpemidler: Ingen Kontroller at oppgavesettet er komplett før du begynner å besvare spørsmålene. Merknad : Alle størrelser og gurakser skal benevnes. Merknad : Les gjennom hele oppgavesettet før du begynner! Svar: Forslag til fasit, versjon-: Oppgave FIR ltre Vi studerer et realiserbart FIR lter av orden M med impulsrespons h[n]. Passbånd- og stoppbåndfrekvensene er ω p =. rad/sample og ω s =.3 rad/sample. Pass- og stopbåndets rippelnivåer er δ p =.5 og δ s =. a) Hva er bredden til lterets transisjonsbånd?.5 p. Skisser magnituden til lterets frekvensrespons og indiker δ p og δ s på den vertikale aksen..5 p. b) Utgangen y[n] fås fra inngangen x[n] ved å sette to ltre med impulsrespons h[n] i kaskade som vist i guren under. Avgjør ordenen til det eektive lteret med impulsrespons h tot [n] som gir y[n] fra x[n]. Begrunn svaret. p. c) De absolutte spesikasjonene for h[n] som tilsvarer δ p og δ s er A p =. db og A s = db. Hva er de absolutte spesikasjonene for passbånd og stoppbånd rippelnivå for det eektive lteret beskrevet i b)? p. d) Frekvensresponsene fra h[n] og h tot [n] er H(e jω ) og H tot (e jω ). Gi et utrykk for fasen til H tot (e jω ) som en funksjon av fasen til H(e jω )..5 p. (Fortsettes på side.)
Eksamen i INF347/447,. desember 5 Side Hvis H(e jω ) er lineær fase, hva kan man si om fasen til H tot (e jω )? Begrunn svaret..5 p. Svar: a) The width of the transition band is ω = ω s ω p =.3. =. rad/sample...5.875.8 H(e jω ).6.4.....3.4.5.6.7.8.9 ω/π b) First method: h[n] has length L = M + the convolution of two signals of length L and L is of length L + L. Therefore h tot has length L = M + and the eective lter is of order M Second method: The Z-transform H(z) is a polynomial of order M (M + coecients). The Z-transform of the eective lter is H tot (z) = H(z) is a polynomial of order M (M + coecients) hence the order of the eective lter is M. c) The frequency response of the eective lter is H tot (e jω ) = H(e jω ) since a convolution in time is equivalent to a product in frequency. In decibel, we then have H tot (e jω ) [db] = H(e jω ) [db]. The bandpass and stopband ripple levels of the eective lter are therefore A ptot = A p = 4.4 db A stot = A s = 4 db d) Let us denote Ψ(ω) and Ψ tot (ω) the phase of H(e jω ) and H tot (e jω ), respectively. We have H(e jω ) = H(e jω ) e jψ(ω) and H tot (e jω ) = H tot (e jω ) e jψtot(ω) = H(e jω ) e jψ(ω) which gives Ψ tot (ω) = Ψ(ω). If H(e jω ) is linear phase, this means that Ψ(ω) is proportional to ω and Ψ(ω) = Ψ tot (ω) is also proportional to ω. H tot (e jω ) is consequently also linear phase. (Fortsettes på side 3.)
Eksamen i INF347/447,. desember 5 Side 3 Oppgave Sampling og aliasing La x(t) være et signal med frekvensspektrum X(Ω) = X(Ω) = for Ω π rad/s for Ω > π rad/s a) Skisser frekvensspekteret til x[n], den ideelt samplede versjonen av x(t) med samplingfrekvens F s = 3 Hz. p. b) Hvis x[n] er en impulsrespons, hva slags lter tilsvarer det?.5 p. Er x[n] i så fall realiserbart? Begrunn svaret..5 p. c) Skisser frekvensspekteret til x[n], den ideelt samplede versjonen av x(t) med samplingfrekvens F s = 5 Hz. p. Svar: a) The maximum frequency contained in x[n] is Hz. The sampling frequency F s = 3 Hz satises the Shannon criterion (F s = 3 > Hz). There is no aliasing. See gure below. b) if x[n] was a lter it would be an ideal low pass lter. It is not realizable due to its innitely small width for the transition band and because it nulls out all frequencies between Hz and Hz (F s ). Its impulse response would also be a sinc function innitely long and not absolutely summable. c) The sampling frequency F s = 5 Hz does not satisfy the Shannon criterion (F s = 5 < Hz). There is aliasing. See gure below..8.6.4 X(e jω )..8.6 F s = 3 Hz F s = 5 Hz.4. -5-45 -4-35 -3-5 - -5 - -5 5 5 5 3 35 4 45 5 frequency [Hz] (Fortsettes på side 4.)
Eksamen i INF347/447,. desember 5 Side 4 Oppgave 3 Transformanalyse av LTI systemer Du er nylig ansatt på et forskningsprosjekt for pulsmåling av eksamenskandidater ved UiO. Dere har utviklet et kombinert sensor og dataloggingssystem som drives fra lysnettet, men dere har slurvet med hardwaredesignet og sliter veldig med mye 5 Hz målestøy fra lysnettet. Sjefen din har oppdaget at du har tatt kurs i digital signalbehandling, og vil at du skal lage et lter for å få bort denne støyen. Du synes vagt å huske at det var noe som heter notch- lter, men du var så urutinert at du solgte den gamle læreboka di. Etter desperat googling nner du følgende transferfunksjon for et notch-lter: H(z) = b [ ( cos(φ))z + z ] () a) Finn nullpunktene til H(z) som funksjon av φ. Gitt at samplingsfrekvensen i systemet er F s = khz, nn φ for å undertrykke støyen fra lysnettet. Tegn pol-nullpunktsplott for notch-lteret. p. b) Finn et utrykk for magnituderesponsen H(e jω ) til lteret og skisser denne. Du kan gjerne bruke at b = og cos(φ) =.95. Er lteret minimum fase? Begrunn svaret. p. c) Finn impulsresponsen og dierensligningen til systemet. Anta et påtrykt inngangssignal x[n]. p. d) Beregn systemresponsen y[n] for inngangssignalet x[n] = s[n] + v[n] hvor v[n] = A cos(nπ/), n >=. Kommenter det du nner. p. e) Du sliter litt med at lteret tar bort for mye av signalet i frekvensområdet rundt 5 Hz. Hva kan du gjøre for å bedre dette. Bruk polnullpunktsplottet og forklar kvalitativt hvordan din(e) endring(er) vil påvirke frekvensresponsen til systemet. p. Svar: a) The zeros are found by (second order equation): z, = cos(φ) ± 4 cos(φ) 4 z, = cos(φ) ± sin(φ) z, = cos(φ) ± j sin(φ) z, = e ±jφ () Normalized frequency is F = 5[Hz]/[Hz], which gives φ = π / = π/. We sketch the pole/zero plot and nd: (Fortsettes på side 5.)
Imaginary Part Eksamen i INF347/447,. desember 5 Side 5 Pole-zero plot of the simple FIR notch -lter at F=:=.8.6.4. -. -.4 -.6 -.8 - - -.5.5 Real Part b) This exercise can be solved in several dierent ways. We calculate it brute force as follows: We immediately recognize this as a FIR lter, so stability is not a concern. We insert z = e jω to nd the frequency response: H(z) = b [ ( cos(φ))z + z ] H(e jω ) = b [ ( cos(φ))e jω + e jω ] (3) There are several paths to nding the magnitude response. Here we use that H(e jω ) = H(e jω )H(e jω ): H(e jω ) = b [ cos(φ)e jω + e jω ][ cos(φ)e jω + e jω ] H(e jω ) = b [ cos(φ)e jω + e jω cos(φ)e jω + 4 cos(φ) cos(φ)e jω + e jω cos(φ)e jω + ] H(e jω ) = b [ + cos(ω) 8 cos(φ) cos(ω) + 4 cos(φ) ] H(e jω ) = b [4 cos(ω) 8 cos(φ) cos(ω) + 4 cos(φ) ] H(e jω ) = 4b [cos(ω) cos(φ) cos(ω) + cos(φ) ] H(e jω ) = 4b (cos(ω) cos(φ)) H(e jω ) = b (cos(ω) cos(φ)) (4) (Fortsettes på side 6.)
jh(e j! )j jh(e j! )j Eksamen i INF347/447,. desember 5 Side 6 4 Magnitude response of notch filter 3.5 3.5.5.5.5.5.5 3 3.5 Angular frequency [rad] Magnitude response of notch filter - - -3-4 -5-6.5.5.5 3 3.5 Angular frequency [rad] There was no question about the phase response, but it could have been found as the tan () of the real and imaginary parts of the frequency response. By using the trigonometric identities in the formula sheet, we (Fortsettes på side 7.)
Eksamen i INF347/447,. desember 5 Side 7 would get: ( ) H(e jω cos(φ) sin(ω) sin(ω) ) = atan ( cos(φ) cos(ω) + cos(ω) ( ) cos(φ) sin(ω) sin(ω) cos(ω) H(e jω ) = atan cos(ω) cos(φ) cos(ω) ( ) sin(ω)(cos(φ) cos(ω)) H(e jω ) = atan cos(ω)(cos(ω) cos(φ)) (5) Note that we could not have shortened this further immediately, as we have a / for ω = φ, which causes a discontinuity in ω = φ. Theoretically this lter can be considered as minimum phase. The inverse is critically stable with poles on the unit circle, so it is invertible. In practice, numerical errors would probably cause an unstable inverse. You will get a score here, as long as you explain your answer correctly. c) We use that z can be interpreted as a delay operator and nd: b, n = b cos(φ), n = h[n] = b, n =, else (6) The dierence equation gets: d) We insert for x[n] and nd: y[n] = b x[n] b cos(φ)x[n ] + b x[n ] (7) y[n] =b (s[n] + v[n]) b cos(φ)(s[n ] + v[n ]) + b (s[n ] + v[n ]) y[n] =(b s[n] b cos(φ)s[n ] + b s[n ])+ (b v[n] b cos(φ)v[n ] + b v[n ]) We cannot do much with the rst term, as s[n] is unknown. But the second term is more interesting. We denote it y v [n], and to simplify the notation, we write φ v = π/: y v [n] = b v[n] b cos(φ)v[n ] + b v[n ] y v [n] = Ab [cos(φ v n) cos(φ) cos(φ v (n )) + cos(φ v (n ))] y v [n] = Ab [cos(φ v n) cos(φ) cos(φ v n φ v ) + cos(φ v n φ v )] y v [n] = Ab [cos(φ v n) cos(φ)(cos(φ v n) cos(φ v ) + sin(φ v n) sin(φ v )) + cos(φ v n) cos(φ v ) + sin(φ v n) sin(φ v )] (8) (9) (Fortsettes på side 8.)
Eksamen i INF347/447,. desember 5 Side 8 If you answered correctly in a), you should recognize φ = φ v, opening up for further simplication: y v [n] = Ab [cos(φ v n) cos(φ v )(cos(φ v n) cos(φ v ) + sin(φ v n) sin(φ v )) + cos(φ v n) cos(φ v ) + sin(φ v n) sin(φ v )] y v [n] = Ab [cos(φ v n) cos(φ v ) cos(φ v n) sin(φ v n) sin(φ v ) + cos(φ v n) cos(φ v ) + sin(φ v n) sin(φ v )] y v [n] = Ab cos(φ v n)[ cos(φ v ) + cos(φ v )] y v [n] = () We thus get y[n] = y s [n] = b s[n] b cos(φ)s[n ]+b s[n ], which illustrates, as expected, that the signal v[n] (the 5 Hz noise) has no eect on the output signal! e) To make the stop band narrower, a pair of poles can be positioned close to the notch lter zeros, just inside the unit circle. The closer to the unit circle, the narrower the stop band would be and more of the original input signal will be left undistorted. The notch lter is now an IIR lter. Oppgave 4 Match systemene! Ligninger til 4 beskriver 4 systemer. Figurer, og 3 viser tilhørende 5 p. plott. I pol/nullpunktsplottene er det lagt til nullpunkter i origo for å få lik grad av teller og nevner i transferfunksjonen. Match systemene og gurene. Du kan anta at alle systemene er kausale. Angi i tillegg, når mulig, for hvert system: stabilitetsegenskaper (stabilt/ustabilt/kritisk stabilt). faserespons (minimum fase/blandet fase/maksimum fase). lengden på impulsresponsen (FIR/IIR). type lter (notch/ kam/ digital resonator/ allpasslter/ lavpass/ høypass/ båndpass/ båndstopp ) Merknad! Alle svar skal begrunnes. Tilfeldige kombinasjoner uten forklaring belønnes ikke. Merknad! Ved å bruke hodet og det dere har lært i kurset, er det ikke sikkert dere trenger å beregne noe som helst i denne oppgaven! Merknad 3! Det gis.33 poeng for hver riktig ligning-gur kombinasjon og. poeng for hver riktig tilleggsopplysning. Oppgaven gir maksimalt 5 poeng. (Fortsettes på side 9.)
Eksamen i INF347/447,. desember 5 Side 9 y[n] =.9 y[n ].8y[n ] + x[n] () H(z) = M (W mk M + W M mk )z k, M =, m =, W m = e jπ/m () k= H(z) = (3/4)(3/4 z ) + z (3 /4)z + (3/4) z (3) H(z) = z + (6/5)z + (6/5) z (4) Fyll ut en tabell på følgende format (Begrunnelser og beregninger tar du utenfor tabellen): Filter Karakteristikk Pol/Nullpkt (gur) Magnituderespons (gur) Impulsrespons (gur) Stabilitetsegenskaper Faserespons Impulsrespons lengde Type Ligning Ligning Ligning 3 Ligning 4 (Fortsettes på side.)
Magnitude [db] Magnitude [db] Magnitude [db] Magnitude [db] Imaginary Part Imaginary Part Imaginary Part Imaginary Part Eksamen i INF347/447,. desember 5 Side Pole/Zero (plot PZ) Pole/Zero (plot PZ).5.5 9 -.5 -.5 - - -.5.5 Real Part Pole/Zero (plot PZ3) - - -.5.5 Real Part Pole/Zero (plot PZ4).5.5 -.5 -.5 - - - -.5.5 Real Part - -.5.5 Real Part Figur : Pole/Zero plots Magnitude spectrum (plot FR) Magnitude spectrum (plot FR) - - -..4.6.8 Normalized frequency (: : rad=sample) Magnitude spectrum (plot FR3) -4..4.6.8 Normalized frequency (: : rad=sample) Magnitude spectrum (plot FR4) - -..4.6.8 Normalized frequency (: : rad=sample) -..4.6.8 Normalized frequency (: : rad=sample) Figur : Magnitude spectra (Fortsettes på side.)
Amplitude Amplitude Amplitude Amplitude Eksamen i INF347/447,. desember 5 Side Impulse response (plot IR) Impulse response (plot IR).5.5 -.5-3 n (samples) Impulse response (plot IR3) -.5 5 5 5 n (samples) Impulse response (plot IR4).5.5 - -.5-5 5 5 n (samples) - 5 5 5 n (samples) Figur 3: Impulse responses Svar: System : Plots PZ, FR4 and IR4. Stable. Minimum phase. IIR. Digital resonator (optionally also Lowpass). System : Plots PZ, FR and IR. Stable. Minimum phase. FIR. Bandpass (moving average type) System 3: Plots PZ3, FR3 and IR. Stable. Maximum phase. IIR. Allpass System 4: Plots PZ4, FR and IR3. Unstable. IIR (optionally Highpass). We can do the following reasoning to get this result. Let us consider equation rst. This is the only one having 9 zeros, and we can immediately recognize it as PZ. Using our knowledge about eect of poles/zeros on frequency response, we can also identify it as FR as zeros pull the magnitude spectrum down. Since it has no poles outside origin, it is a FIR lter, meaning the impulse response must be IR. FIR lters are always stable. The distribution of zeros suggests it is a bandpass lter centered approximately at pi/4. All zeros are on or inside the unit circle, so the lter is considered minimum phase. (Fortsettes på side.)
Eksamen i INF347/447,. desember 5 Side Second, we consider equation 3. We note that the numerator and denominator are reversed copies of each other, suggesting we are talking about an allpass lter. We recognize PZ3 as the allpass lter (or as the only plot with both zeros and poles outside the origin). The at spectrum in FR3 is also the response of an allpass lter. All poles are inside the unit circle, so it is stable. Since it has poles not at the origin, it is an IIR lter. All zeros are outside the unit circle, so it is a maximum phase lter. By exclusion, we can say the the only possible impulse responses are IR or perhaps IR4. We will come back to the correct choice. Third, let us consider equation. We can see from the impulse response that it has no zeros and two poles (second order equation in z ). At rst glance, it seems we cannot dier between PZ and PZ4. But we note that there are two zeros in the origin at PZ (to get N=M), which ts well with this system. This is a bit sketchy reasoning, and it would probably be a safer solution to solve for the poles. But after having decided on PZ, we can conclude that it is stable. The poles at approximately ±π/4, causes the spike in FR4. The impulse response is either IR or IR4. Comparing PZ and PZ3, we see that the poles at PZ is closer to the unit circle, meaning that the response will be less dampened than for PZ3. This lets us conclude that IR belongs to PZ3 or the system in equation 3. IR4 belongs to PZ, and thus, the system in equation. Another option would be to calculate the rst three values of y[n] for x[n] = δ[n]. It will then be obvious that IR4 is the correct impulse response. The system has no zeros outside the unit circle and is a minimum phase IIR lter. Based on the pole location and impulse response, we recognize the lter as a digital resonator, but it can also be considered a lter having a lowpass character. The fourth system is equation 4. This has two poles and one zero at. PZ4 has two poles and one added zero at the origin to get N=M, which ts well with this system. We see that this is an unstable system having two poles outside the unit circle. The poles look to be at an angle 3π/4, so the magnitude response should be FR. Since it is an unstable IIR lter, IR3 is the only candidate for the impulse response. It is of limited value to discuss properties of an unstable lter, but based on the magnitude response, it can be termed a highpass lter. Oppgave 5 Lineær vs. sirkulær konvolusjon Vi studerer to sekvenser med endelig lengde x [n] = {,,, 3} x [n] = {,,,,, 4} a) Beregn den lineære konvolusjonen x [n] x [n]. p. b) Beregn den 6-punkt sirkulære konvolusjonen x [n] 6 x [n]. p. (Fortsettes på side 3.)
Eksamen i INF347/447,. desember 5 Side 3 c) Hva bør den minste verdien for N være for at den N-punkt sirkulære konvolusjonen skal være lik den lineære konvolusjonen?.5 p. d) Når en ltreringsoperasjon beregnes på en datamaskin gjøres det som regel ved å bruke multiplikasjon i frekvensdomenet. Hva slags konvolusjon tilsvarer det?.5 p. Forklar hvorfor null-padding er en nyttig teknikk i dette tilfellet..5 p. Svar: a) We recall the formula for the linear convolution: h[n] = x [n] x [n] = k= x [k]x [n k] Therefore h[] h[] h[] 5 h[3] 3 4 h[4] = 7 h[5] 4 7 h[6] 4 8 h[7] 4 4 h[8] 4 b) Likewise, the formula for the N-point circular convolution is g[n] = x [n] N x [n] = N k= x [k]x [< n k > N ] For computing a 6-point circular convolution, we will need x [n] and the zero-padded version of x [n], x zp [n]: x zp [n] = {,,, 3,, } Therefore g[] 4 8 g[] 4 6 g[] 4 g[3] 4 3 = 7 4 g[4] 4 7 g[5] 4 7 (Fortsettes på side 4.)
Eksamen i INF347/447,. desember 5 Side 4 c) For the circular convolution to be equal to the linear convolution, it has to have a length of at least N + N where N and N are the length of x and x, respectively. This means that the smallest value of N so that the N-point circular convolution is equal to the linear convolution is N = 9. We can verify this by computing the 9-point circular convolution: g[] 4 g[] 4 g[] 4 5 g[3] 4 3 4 g[4] 4 = 7 g[5] 4 7 g[6] 4 8 g[7] 4 4 g[8] 4 which gives the result obtained in a). d) Using a computer for a ltering operation implies the use of the DFT and IDFT (actually FFT and IFFT). The product of the DFTs corresponds to the DFT of the circular convolution. In ltering operations and more generally in design of LTI systems, we are interested in the linear convolution between the input and the system impulse response. To make sure we get the desired result when using a computer and the DFT/IDFT (FFT/IFFT), we need to ensure the N-point circular convolution gives the same result as a linear convolution. To do that we use zero-padding and pad the input sequences to a length N with N N + N where N and N are the length of each sequence. (Fortsettes på side 5.)
Eksamen i INF347/447,. desember 5 Side 5 Formelsamling Grunnleggende sammenhenger: Lineær konvolusjon: y[n] = x[n] h[n] = Sirkulær konvolusjon: y[n] = x[n] N h[n] = sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β sin α sin β sin α = sin α cos α cos α = cos α sin α sin α + sin β = sin α + β sin α sin β = sin α + β cos α + cos β = cos α + β cos α cos β = sin α + β cos α + sin α = cos α = (ejα + e jα ) cos α β sin α β cos α β sin α β sin α = j (ejα e jα ) { N a n for a = = N n= a N a ellers ax + bx + c = x ± = b ± b 4ac a k= N k= x[k]h[n k] = x[k]h[< n k > N ] = Diskret tid-fouriertransformasjon (DTFT): Analyse: X(Ω) = = Syntese: x[n] = π Diskret fouriertransformasjon (DFT): Analyse: X[k] = (Fortsettes på side 6.) N Syntese: x[n] = N n= N k= N k= n= π π x[n k]h[k] = h[n] x[n] x[< n k > N ]h[k] = h[n] N x[n] x(n)e jωn X(Ω)e jωn dω x[n]e jπkn/n, k N k= X[k]e jπkn/n, k N
Eksamen i INF347/447,. desember 5 Side 6 z-transformasjonen: Analyse: X(z) = n= x[n]z n