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Session4

# Session4 - EE362-DSP I University of Saskatchewan Session 4...

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EE362-DSP I University of Saskatchewan a39 a38 a36 a37 Session 4 Frequency Response, Eigenfunctions, Suddenly Applied Complex Exponentials Brian Daku Page 51

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EE362-DSP I University of Saskatchewan a39 a38 a36 a37 Black Box Filter A black box LTI filter has been generated in MATLAB. The question is: What type of filter is in this black box? The filter is implemented as a MATLAB function, which is in the file bbfilter.p (this file is on the website). Given the input x the output of the filter can be generated using y=bbfilter(x); , as shown in the following figure. x y bbfilter(.) What approach should be used to determine the type of filter in the black box? Brian Daku Page 52
EE362-DSP I University of Saskatchewan a39 a38 a36 a37 Magnitude/Phase Response of a LTI System If the input to a LTI system is a sinusoid, the output is also a sinusoid, having the same frequency, but the magnitude and phase could be altered. x [ n ] = cos( ωn ) y [ n ] = a cos( ωn + φ ) (3) Expanding the input and output gives x [ n ] = cos( ωn ) = 1 2 e jωn + 1 2 e jωn y [ n ] = a cos( ωn + φ ) = a 2 e j ( ωn + φ ) + a 2 e j ( ωn + φ ) Since this is a LTI system, if the input is x 1 [ n ] = e jωn then the output will be y 1 [ n ] = ae j ( ωn + φ ) = ae jωn e The frequency term can be removed by multplying the output term by the complex conjugate, ( e jωn ) = e jωn , z 1 [ n ] = y 1 [ n ] e jωn = ae jωn e e jωn = ae Thus | z 1 | = a and z 1 = φ at frequency ω . e jωn is referred to as an eigenfunction for the LTI system. Using this eigenfunction, the magnitude and phase for each frequency can determined (ie a frequency response). Example 4.1: Determine the magnitude and phase of the filter in bbfilter.p at the frequency f = 0 . 4 cycles/sample. Brian Daku Page 53

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EE362-DSP I University of Saskatchewan a39 a38 a36 a37 What are Eigenfunctions for LTI Systems? In general, any function f [ n ] is an eigenfunction if it satisfies the relationship, y [ n ] = T { f [ n ] } = λf [ n ] , (4) where λ is the eigenvalue for the system, which does not depend on the time variable n . If the system is LTI with impulse response h [ n ] then equation (4) simplifies to, y [ n ] = f [ n ] h [ n ] = λf [ n ] . Example 4.2: Is e jωn (for −∞ < n < ) an eigenfunction of a LTI system with impulse response h [ n ] ? y [ n ] = h [ n ] e jωn = summationdisplay k = −∞ h [ k ] e ( n k ) (5) = parenleftBigg summationdisplay k = −∞ h [ k ] e jωk parenrightBigg e jωn = λ ( ω ) e jωn (6) Yes, e jωn is an eigenfunction, since y [ n ] is the product of the eigenvalue λ ( ω ) , which does not depend on n , and the original eigenfunction e jωn .
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Session4 - EE362-DSP I University of Saskatchewan Session 4...

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