Digital Signal Processing. Homework 2 Solution. Due Monday 4 October Following the method on page 38, the difference equation
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1 Digital Sigal Proessig Homework Solutio Due Moda 4 Otober 00. Problem.4 Followig the method o page, the differee equatio [] (/4[-] + (/[-] x[-] has oeffiiets a0, a -/4, a /, ad b. For these oeffiiets A(z i Equatio.9 is A 4 ( z 6z + ( z z + z z z ( 4z ( z Fidig the roots of A(z we have z/4 ad z/ so the homogeeous solutio of the form. h A / 4 + A / alwas produes a value of zero o the left side of the differee equatio. Usig the iput ad iitial oditios to defie two values of [], we a solve for the oeffiiets. [] ( ( [] 0 A + A 0 [] A ( / 4 + A ( / From this A - ad A, so the solutio is [] ( / 4 + ( / ( / ( ( / 4 [] e. Problem.6. (Note that the freque respose is omputed for a iput x, ad H e x H e e. the output is [] ( [] ( a. I the differee equatio [] (/ [-] x[] + x[-] +x[-] substitute the expressios for x[] ad [] to get H ( ( ( ( e e (/ H ( e e e + e + e. Fator out to get j e ω
2 ( ( ( ( H ( e (/ H ( e e e ( + e + e b it ad solve for H ( e as follows: e. Sie e is ever 0, divide H H ( ( e (/ H ( e e H ( e ( (/ e ( + e + e + e + e ( e (/ e b. From the expressio for the freque respose the differee equatio oeffiiets j multipl the orrespodig powers of e ω, so from the deomiator a0, a /, ad a /4, ad from the umerator b0, b -/, ad b. Thus the differee equatio is [] +(/[-] + (/4[-] x[] /x[-] +x[-]. Problem.9 [] x[][ k h a k k [] h[ m][ x m] ] h[ m][ x m] h[ m][ x m] m m 0 a. x[] u[]. For <0, [] 0. For 0 < < 5, m 0 Ad for > 5, [] 0. b. The output is the same as i part (a exept that it is delaed b 4 samples. The sequee startig at 0 would be [ ] [ ] { K K} [ ] a [ ] b [ ] oes. [] { K K} b a.. The output would be the differee whih is the respose to a sequee of Problem.
3 X ( e ae + ja si ( a os j ( a si a os + a a os will be a periodi futio of ω with a period of π. M Re Im a os { X ( e } M os φ ( { X ( e } M si φ + a ( ta a os a os a si a os ( φ + a + a a si a os M j ( ω e φ Plots are show for a-0.5 ad a-0.9. Note that with a<0 both at as high pass filters with the maximum magitude at the highest freque. As the magitude of a ireases, the differee betwee the gai at the highest freque ad the gai at zero freque ireases. (Studets ma have seleted a value of a for the plot with - < a < 0.
4 Extra: Chek some simple values: ω X(e M(ω φ(ω Re Im 0 /(-a /(-a 0 /(-a 0 π/ /(+ja /sqrt(+a ata(-a /(+a -a/(+a p /(+a /(+a 0 /(+a 0 π/ /(-ja /sqrt(+a ata(a /(+a a/(+a 5. For a movig average filter of legth. a. Fid a expressio for the freque respose. b. Fid the freque values at whih the freque respose is zero. Create a sequee of te samples of a iput sigal at that freque, ad verif b ovolutio that the output will be zero. The impulse respose of a legth movig average filter is h [] δ ( k. The k 0 freque respose a be omputed i two differet forms. The first is: ( j ω e ω ω ( j j e H e + e + e ( e + + e ( + os. This will be zero whe os(ω -0.5 whih happes whe ω +/-π/.
5 The seod is ω jω jω / jω / jω / si e e e e e ( H e. The deomiator / / / e e e e ω si is 0 at ω0. The umerator is zero at ω0, π/, ad 4π/. The pole ad zero at ω0 ael eah other. b. A siusoidal sigal with a freque of π/, ad 4π/ should produe zero output. os(π / { ,5-0.5 } si(π / { } A liear ombiatio of these two futios with a weightig or time shift will produe a zero output from ovolutio with the movig average filter beause a sum of three oseutive values is For the differee equatio [] a*x[] + b*x[-] + a*x[-], fid a expressio for the freque respose i terms of a ad b. Write the freque respose as a real valued expressio multiplied b a omplex expoetial b fatorig out e from all three terms. Evaluate the freque respose ad sketh the magitude for the followig parameter hoies. a. a, b - How does this ompare to ab? b. a, b This is ver similar to the freque respose i the previous problem. ( ( ( e j ω e H e a + be + ae ae + b + ae ( b + aos a. e e H ( e ( b + aos ( + os. This is the same as the freque respose i the previous program exept that it is shifted b p so that the maximum magitude ours at ωπ rather tha at ω0. b. e e e H ( e b + aos maximum magitude at ω0 ad a magitude of 0 at ωπ. Skethes or plots ( ω ( ( + os ( + os. This has a 7. Use the result i Equatio.40 for the impulse respose of a ideal low pass filter to ompute the values of the impulse respose from -0 to 0 for the utoff frequeies below. I eah ase sketh the impulse respose. Explai what happes to the impulse respose as the utoff freque is dereased. You ma use MATLAB to do this problem. h lp [] si π ( ω ω si π ω
6 π a. ω : zeros i h[] at, 4, 6,, 0,, π b. ω : zeros i h[] at 4,,, 4 π. ω : zeros i h[] at, 6, d. ω π : All oeffiiet are zero exept whe 0, so the filter is a sigle impulse whih does ot hage the iput at all. The utoff freque is speified as the maximum possible freque, so ot frequeies are suppressed b the filter. The periodi extesio of the freque respose is a ostat value of. 00 Sall Wood
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