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Unformatted text preview: DIGITAL SIGNAL PROCESSINGI HOME WORK #2 Submitted by Sriram Yarlagadda Sid # 2196944 LOW PASS FIR DIGITAL FILTER A low pass FIR digital filter is m point average filter 1 1 1 1 ) ( ) ( ) ( ) ( 1 ) ( ] [ 1 ] [ N k k N K k N k z N z H z X z Y Therefore z z X N z Y k n x N n y Here The transfer function=H(z)= ) ( ) ( z X z Y For impulse response X(z) =1 Substituting z= jw e 1 1 ) ( N k jwk jw e N e H H ( jw e ) is the frequency response PROBLEM 1 1a) GIVEN: Frequency response of an ideal Low Pass filters as follows: H ( jw e ) = otherwise dB dB 1 1 3 3 The frequency response for ideal LPF is represented graphical as follows: REQUIRED: To compute the Impulse Response of a LPF Solution The impulse response of an ideal LPF is given by dw e e H n h jwn jw ) ( 2 1 ] [ dB w dB w jwn jwn dB w dB w jwn dw e dw e dw e n h 3 3 3 3 ) ( ) ( ) 1 ( { 2 1 ] [ n dB n n h j e e n n h nj e n h dB jw dB jw dB w dB w jwn ) 3 sin( ] [ } 2 { 1 ] [  } { 2 1 ] [ 3 3 3 3 The impulse response of an ideal LPF is n 3dB) sin(n h[n] Here 3dB =2 c f denotes the normalized cut off frequency If 3dB is very small then we apply LHospital rule. Therefore the impulse response at very small 3dB is given by dB n h 3 ] [ 1b) REQUIRED: To plot the time domain impulse response Solution The time domain impulse response is plotted with the help of MATHCAD by making the following declarations Cut off frequency The graph is to be plotted for 201 points The impulse response for LPF is given by Plot w 0.08 n 100 100 h n ( ) sin n w ( ) n n if w n if 1c)...
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 Spring '10
 Dr.RobertoJAcousta
 Digital Signal Processing, Signal Processing

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