FIR_Filters_2010

FIR_Filters_2010 - S Y S C - 4 4 0 5 : D i g i t a l S i g...

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SYSC-4405:Digital Signal Processing FIR_Filters_2010.fm MOHAMMED EL-TANANY, PROFESSOR SYSTEMS & COMPUTER ENGINEERING, CARLETON UNIVERSITY, OTTAWA, ONTARIO 1 / 32 Finite Impulse Response (FIR) Filters An FIR filter is described by the difference equation The unit sample response of such a system is given by In other words, y(n) can be written as follows: y n () b k xn k k 0 = M = hn b k k 0 = M xn () δ n = b k δ nk k 0 = M == y n hk k 0 = M =
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SYSC-4405:Digital Signal Processing FIR_Filters_2010.fm MOHAMMED EL-TANANY, PROFESSOR SYSTEMS & COMPUTER ENGINEERING, CARLETON UNIVERSITY, OTTAWA, ONTARIO 2 / 32 The Problem is: Given a complex frequency response function . Determine the filter coefficients {b 0 , b 1 ,.... b L }. How to Obtain the filter coefficents Recall that is periodic with period 2 π . An Example of a 5-Tap FIR Filter z 1 b 2 z 1 z 1 z 1 b 1 b 0 b 4 b 3 xn () yn y n b 0 b 1 1 b 2 2 b 3 3 b 4 4 ++++ = He j ω j ω
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SYSC-4405:Digital Signal Processing FIR_Filters_2010.fm MOHAMMED EL-TANANY, PROFESSOR SYSTEMS & COMPUTER ENGINEERING, CARLETON UNIVERSITY, OTTAWA, ONTARIO 3 / 32 Also recall that the frequency response is defined as: We wish to solve the relation above to express h(n) in terms of the frequency response. Multiply both sides of the relation above by and integrate over a period: where Therefore He j ω () hn z n n = ze j ω = e j ω n n = == e jm ω j ω e ω ω d π π e j ω n n = e ω ω d π π e jm n ω ω d π π n = e ω ω d π π 2 π m equal n 0 otherwise = 1 2 π ------ j ω e jn ω ω d π π =
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SYSC-4405:Digital Signal Processing FIR_Filters_2010.fm MOHAMMED EL-TANANY, PROFESSOR SYSTEMS & COMPUTER ENGINEERING, CARLETON UNIVERSITY, OTTAWA, ONTARIO 4 / 32 Knowing that , the result above may be rewritten as follows: Design of an Ideal Lowpass Filte r For simplicity we assume an ideal lowpass response with zero phase. The cut-off frequency is ω c . The impulse response of such a filter is described by: ω 2 π ff s () = hn 1 f s --- He j 2 π s e j 2 π fn f s f d f s 2 f s 2 = ω c −ω c −2π Magnitude response of an ideal lowpass filter (zero delay) 1 1 2 π ------ j ω e jn ω ω d π π 1 2 π 1 e ω ω d ω c ω c ω c π ω c n sin ω c n ---------------------- × == = =
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SYSC-4405:Digital Signal Processing FIR_Filters_2010.fm MOHAMMED EL-TANANY, PROFESSOR SYSTEMS & COMPUTER ENGINEERING, CARLETON UNIVERSITY, OTTAWA, ONTARIO 5 / 32 Remarks: h(n) has an infinite number of terms -15 -10 -5 0 5 10 15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 h(n) for an Ideal LPF Response of an Ideal LPF with a cut-off frequency π /4
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SYSC-4405:Digital Signal Processing FIR_Filters_2010.fm MOHAMMED EL-TANANY, PROFESSOR SYSTEMS & COMPUTER ENGINEERING, CARLETON UNIVERSITY, OTTAWA, ONTARIO 6 / 32 h(n) is non-causal Therefore, h(n) is non-realizable.
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FIR_Filters_2010 - S Y S C - 4 4 0 5 : D i g i t a l S i g...

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