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## Digital Signal Processing

by: Ms. Isobel Rau

37

0

3

# Digital Signal Processing EE 451

Ms. Isobel Rau

GPA 3.67

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
3
WORDS
KARMA
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## Popular in Electrical Engineering

This 3 page Class Notes was uploaded by Ms. Isobel Rau on Thursday October 15, 2015. The Class Notes belongs to EE 451 at New Mexico Institute of Mining and Technology taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/223653/ee-451-new-mexico-institute-of-mining-and-technology in Electrical Engineering at New Mexico Institute of Mining and Technology.

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Date Created: 10/15/15
BB 451 Fall 2000 BB 451 Solution of Difference Equations in the Time Domain The difference equation M Z aan k 2 ban k k0 k0 where a 1 has the solution Mn 01W CNARMLM D06n 4 4 DMNI3n M N 1 ypn 1 quot39 1mm yss in where the Ak s are the roots of the characteristic polynomaial of the system A a1ANquot lt aN1gtaN 0 For repeated roots use A2 nA nZAZ etc For example if A1 A2 you would use 11 A CZan instead of 1M 02W ypn is the particular steaystate solution which depends on the input Mn 9 n AMn Kn n Anna n KcWMn A cosw0n B sinw0n K1 cosw0n K2 sinw0n If the characteristic polynomial has a root at the value of an input exponential eg A1 and R R Mn you would use KnAZ for ypn eg ypn Kn There are A N Di s If A g N there are no Di s To find the impulse response there is no ypn and there are M N 1 Dis If M lt N there are no Di s for the impulse response Solve for the unknowns by finding M0 M1 M2 until you get many equations you have unknowns Solve these equations for the unknowns You can solve the difference equation for the case Mn 0 subject to the initial conditions of the system This is the natural response or zeroinput response 1 Mn of the system You can further solve for the case Mn 0 for n lt 0 for the actual input Mn This is the forced response or zerostate response 125 of the system The total response is the sum of the two Mn yMn 1Z5n A system is BIBO stable if for every bounded input the output is bounded If the input Mn is bounded then the particular solution ypn will be bounded Thus the only possible unbounded terms in Eq 1 are the A s These terms are bounded AZ gt 0 n gt so if lt 1 Hence the system is 8180 stable if lt 1 for all the k s AZ doesn t blow up if 1 However if AM 2 1 the input Mn A2 will produce the output Mn 11A KnAZ and the KnAZ term will blow up n gt so BB 451 HR Filter Design Fall 2000 BB 451 HR Filter Design One way to design HR lters is by use of the bilinear transformation To design a lowpass HR lter transform a lowpass continuoustime lter to a discretetime lter using the transformation 71 2391 12391 To design another type of lter highpass bandpass or bandstop another step is needed There are several ways to do this One of the simplest is to nd different transformations which will map a lowpass continuoustime lter in one of these other types of discretetime lters Here are a set of transformations which will do that LP HP BF BS 5 1 2 1 12 1 1 252 1z4 1 24 1z I 1 2 I 1 z 1 262 12 Ci 63l9 1 2 13 amp civc3s39 1 3 1 3 9 1 3 1 1 w w c cosw Q tan C0t 1 Si h p Si 1 11hgt Smwpgsmwph smwpgsmwph W1 0 7r cosquot1 0 or 7r A lter is usually speci ed by passband and stopband frequencies Law and wstop and by passband ripple Egg and stopband attenuation stp Bandpass and bandstop lters have low and high passband and stopband equencies To design a discretetime HR lter do the following 1 Prewarp the discretetime f I i to quot t I i using the row labeled a If lowpass or highpass transform Law and damp to RPM and 9310p b If bandpass or bandstop nd 2 Then transform wpl wph ms and cash to Q QM 93 and 93h Let 9W mm and let 9310p min93 Qsh BB 451 HR Filter Design Fall 2000 2 Design a continuous time lowpass lter using RPM and 9310p from step 1 For Butterworth design nd N and QC using the formulas 7 261 612 6 lt1 m2 1 622 5 6 N 7 log 66 10g qussQstop QC 9310p 8 1 of QC quss 6 1 where 61 1 1039R1 W20 and 62 IO39RWPgu Since N must be an integer you must round N up to an integer value This lter will have N poles at and N zeros at in nity Transform the continuoustime lter to a discretetime lter using one of the following two methods W a Replace every 5 in the continuoustime transfer function as speci ed with the row labeled 5 Do lots of algebra to get it into a usable form b Map splane poles and zeros to corresponding 2plane poles and zeros using the equations in the row labeled 2 For bandpass and bandstop lters every splane pole maps to two zplane poles and every spland zero maps to two zplane zeros Don7t forget splane zeros at in nity For lowpass lters an splane zero at in nity maps into a zplane zero at 1 For highpass lters an splane zero at in nity maps into a zplane zero at 1 For bandpass lters an splane zero at in nity maps into an zplane zero at 1 and an 2plane zero at 1 For bandstop lters and splane zero at in nity maps into zplane zeros at 6351 where we cos391c Then nd G by The discretetime transfer function is N 2 2 i1 2 Pi Using MATLAB steps 1 and 2 can be done with the function buttord The entire design process can be done with the function butter

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