Lesson_25 - Lesson25 Q2BriefReview...

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Lesson 25 Lesson 25 Q2 Brief Review Lesson 25  -  Continuous-time signals and systems Challenge 25
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  Lesson 25 Quiz #2 Q1.  FIRs, Linear Convolution and z-Transforms You a given an FIR with impulse response  h =[0,2,0,2,0] and signal  x [ k ]=[1,-1,0,0,0]. a.  Using the techniques developed in Chapter 5, compute  y [ k ] =  h * x [ k ].  k   0 1 2 3 4 5 6 7 8 x [k-0] * h    0 2 0 2 0 x [k-1] * h 0 -2 0 -2 0 x [k-2]* h     0 0 0 0 0 x [k-3]* h       0 0 0 0 0 x [k-4]* h       0 0 0 0 Σ   0 2 -2 2  -2 0 0 0 0
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  Lesson 25 Quiz #2 b. Using the techniques developed in Chapter 6, determine the FIR’s steady- state frequency response. Specify the magnitude and phase responses.  Compute the steady state frequency response: H ( e j ω )= 2  e -j ω  + 2  e -j3 ω Simplifying {force into Euler form} H ( e j ω ) = 2  e -j2 ω  ( e j ω  +  e -j ω  )    = 4  e -j2 ω  ( e j ω  +  e -j ω  )/2  e -j2 ω  (4cos( ω ) )
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  Lesson 25 Quiz #2 You now have H ( e j ω ) =  e -j2 ω  (4cos( ω ) ) Magnitude response: | H ( e j ω )|=4sin( )   Phase response H ( e j ω )= -2 . As a check, the N=5 impulse is symmetric which guarantees lilnear phase  behavior where the slope of the phase is –(N-1)/2 = -2.
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  Lesson 25 Quiz #2 c. At what (normalized) frequencies  ω  are FIR's nulls or zeros located (if they  exist)? The nulls are located at  points where 4cos( ω )=0, or at  ω ± π /2, which also  corresponds to  ±  f s /4.   Alternatively, the zero can be obtain form H(z)= 2z -1  + 2z -3  = 2z -1  (1+z -2 ) which  has roots (zeros) at z =   ,  ± j1, where  ± j1 correspond to the normalized unit  circle frequencies  ω = ± π /2 (again).  ± j π /2  =  ± j1 
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Lesson_25 - Lesson25 Q2BriefReview...

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