Chapt 4 Design of Digital Filters_FIR

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    2/2/2012Watumull Institute of Technology, Worli

    Prepared by Chandrashekhar Padole, DSP-BE Computer, Mumbai Uni 1

    Chapt 04

    Design of Digital Filters ( FIR)B.E. Comps, Mumbai Uni

    PrePrepared by Chandrashekhar Padole

    Lecturer

    Watumull Institute of Tech , Worli

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    Chapt 04 Design of Digital Filters

    Design of FIR filters

    Design of IIR filters from analog filters

    Frequency transformation

    Design of digital filters based on least-squares method

    Digital filters from analog filters

    Properties of FIR filters

    Design of FIR filters using windows

    Comparison of IIR and FIR filters

    Linear phase filters

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    What is filter ?

    Basic building block of DSP:Input is given to filter and output of the system would be

    signal obtained from input and filter's impulse response.

    FilterInput Output

    [ 1 1]Input Output

    [ 1 -1]Input Output

    Integrator ( Lowpass filter)

    Differentiator(High pass filter)

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    Contd..

    Different impulse response different characteristics

    Characteristic of the filter Frequency domain characteristics

    FrequencyCharacteristic

    PhaseCharacteristic

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    Frequency Characteristic

    H(w)

    W

    Frequency characteristic plot between frequencies and magnitude

    For given system if particular frequency is passed thorough the system,its magnitude at output can be obtained from corresponding magnitudefrom frequency characteristicNote: most of the time input wont be single pure sinusoidal/frequency but

    would be a set of frequencies

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    Phase Characteristic

    (w)

    W

    Phase characteristic plot between frequencies and phase shift

    For given system if particular frequency is passed thorough the system,its delay/phase shift at output can be obtained from corresponding phsefrom phase characteristicNote: most of the time input wont be single pure sinusoidal/frequency butwould be a set of frequencies

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    Ideal Characteristics

    Frequency Phase

    Sharp cutoffLinear

    phase

    Sharp cutoff is requirement foralmost all the applications

    Non-linear phase characteristicdistorts the signal.Linear phase characteristic has

    constant group delay.

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    Effect of Linear Phase Characteristic

    Let us pass signal , made up of summing frequency 1 Hz, 2Hz & 3Hz, through filters with different phase characteristics

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    Contd..

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    Contd..

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    Contd..

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    Contd..

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    Contd..

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    Challenge ..

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    Verify .

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    M file used in MATLAB for this simulation

    clear all;close all;figure,

    subplot(4,1,1);plot(sin(2*pi*1*[0:0.01:3]));title(' Inoput Signal Composition');subplot(4,1,2);

    plot(sin(2*pi*2*[0:0.01:3]));subplot(4,1,3);plot(sin(2*pi*3*[0:0.01:3]));

    subplot(4,1,4);plot(sum([sin(2*pi*1*[0:0.01:3]);sin(2*pi*2*[0:0.01:3]);sin(2*pi*3*[0:0.01:3])]));

    figure,subplot(4,1,1);plot(sin(2*pi*1*[0:0.01:3]+pi/8));title(' Filtered Signal with phase characteristic pi/8');subplot(4,1,2);plot(sin(2*pi*2*[0:0.01:3]+pi/8));subplot(4,1,3);plot(sin(2*pi*3*[0:0.01:3]+pi/8));

    subplot(4,1,4);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/8);sin(2*pi*3*[0:0.01:3]+pi/8)]));

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    Contd..

    figure,subplot(4,1,1);plot(sin(2*pi*1*[0:0.01:3]+pi/4));title(' Filtered Signal with phase characteristic pi/4');subplot(4,1,2);plot(sin(2*pi*2*[0:0.01:3]+pi/4));subplot(4,1,3);

    plot(sin(2*pi*3*[0:0.01:3]+pi/4));

    subplot(4,1,4);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/4)]));

    figure,subplot(4,1,1);

    plot(sin(2*pi*1*[0:0.01:3]+pi/2));title(' Filtered Signal with phase characteristic pi/2');subplot(4,1,2);plot(sin(2*pi*2*[0:0.01:3]+pi/2));subplot(4,1,3);plot(sin(2*pi*3*[0:0.01:3]+pi/2));

    subplot(4,1,4);

    plot(sum([sin(2*pi*1*[0:0.01:3]+pi/2);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/2)]));

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    Contd..

    figure,subplot(4,1,1);plot(sin(2*pi*1*[0:0.01:3]+pi/8));title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');subplot(4,1,2);plot(sin(2*pi*2*[0:0.01:3]+pi/4));subplot(4,1,3);

    plot(sin(2*pi*3*[0:0.01:3]+pi/2));

    subplot(4,1,4);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/2)]));

    figure,subplot(4,1,1);

    plot(sin(2*pi*1*[0:0.01:3]+pi/4));title(' Filtered Signal with phase characteristic 2w(linear phase with group delay 2)');subplot(4,1,2);plot(sin(2*pi*2*[0:0.01:3]+pi/2));subplot(4,1,3);plot(sin(2*pi*3*[0:0.01:3]+pi/1));

    subplot(4,1,4);

    plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/1)]));

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    Chapt 04

    figure,

    subplot(6,1,1);plot(sum([sin(2*pi*1*[0:0.01:3]);sin(2*pi*2*[0:0.01:3]);sin(2*pi*3*[0:0.01:3])]));title('I am Input Signal , which of followings looks like me ?');

    subplot(6,1,2);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/8);sin(2*pi*3*[0:0.01:3]+pi/8)]));%title(' Filtered Signal with phase characteristic pi/8');

    subplot(6,1,3);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/4)]));%title(' Filtered Signal with phase characteristic pi/4');

    subplot(6,1,4);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/2);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/2)]));%title(' Filtered Signal with phase characteristic pi/2');

    subplot(6,1,5);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/2)]));%title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');

    subplot(6,1,6);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/1)]));%title(' Filtered Signal with phase characteristic 2w(linear phase with group delay 2)');

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    Chapt 04

    figure,

    subplot(6,1,1);plot(sum([sin(2*pi*1*[0:0.01:3]);sin(2*pi*2*[0:0.01:3]);sin(2*pi*3*[0:0.01:3])]));title(' Input Signal ');

    subplot(6,1,2);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/8);sin(2*pi*3*[0:0.01:3]+pi/8)]));title(' Filtered Signal with phase characteristic pi/8');

    subplot(6,1,3);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/4)]));title(' Filtered Signal with phase characteristic pi/4');

    subplot(6,1,4);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/2);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/2)]));title(' Filtered Signal with phase characteristic pi/2');

    subplot(6,1,5);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/2)]));

    title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');

    subplot(6,1,6);plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/1)]));title(' Filtered Signal with phase characteristic 2w(linear phase with group delay 2)');

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    Linear Phase Conclusion

    If filter doesn't have linear phase characteristic( constant groupdelay) then signal shape gets distorts.e.g. First 3 filtered outputs

    If filter has linear phase characteristic( constant group delay) thensignal shape would be preserved.e.g. Last 2 filtered outputs

    Linear phase can be obtained easily by FIR filter by having

    symmetric/anti-symmetric property in its impulse response.

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    FIR Filter

    FIR filter has finite number of samples in its impulse response

    Advantages:

    The question of stability and realizability never arise for FIR filters ( it isalways stable and realizable) It gives linear phase relationship with frequency , which can beachieved by having symmetric or anti-symmetric impulse response of thefilter

    Disadvantages :

    Long sequences for h(n) are generally required to adequatelyapproximate sharp cut-off filters

    Hence , higher computation complexityThe delay of linear phase FIR filters need not always be an integer

    number of samples . This non-integral delays can lead to problems insome signal processing applications

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    Characteristic of FIR filter

    Let { h(n)} be a causal finite durations sequence of length N( 0 to N-1).Its z-transform

    =

    =

    1

    0

    )()(N

    n

    nznhzH --------(1)

    Its Fourier transform,

    =

    =

    1

    0

    )()(N

    n

    jwnjwenheH

    Which is periodic in frequency with period of 2

    )()()2( mjj

    eHeH +

    = ......3,2,1,0 =mfor

    --------(2)

    Consider h(n) be real and its magnitude and phase ,

    )()()(

    jjwjweeHeH =

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    Contd..

    From eq (2) , since wnjwne jwn sincos =

    symmetricAnti-

    symmetric

    Magnitude of Fourier transform is symmetric and the phaseis an anti-symmetric function

    )()(jwjw

    eHeH

    =

    0)()( =

    Consider , we have to have linear phase responsei.e. =)(

    where is constant ( phase delay in samples)

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    Contd..

    We would find , condition/restriction on impulse response such that itwill give linear phase response

    Mathematically,

    =

    =

    1

    0

    )()(N

    n

    jwnjwenheH

    jjweeH

    = )( Requirement

    =

    =

    1

    0)()(

    N

    n

    jwnjw

    enheH

    Given

    =

    =

    =

    1

    0

    1

    0 )sin()()cos()(

    N

    n

    N

    n wnnhjwnnh

    =

    =

    =

    1

    0

    1

    01

    )cos()(

    )sin()(

    tanN

    n

    N

    n

    wnnh

    wnnh

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    Since n=0, sin(wn)=0Hence , limit starts

    from 1

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    Contd..

    =

    =

    =

    1

    0

    1

    0

    )cos()(

    )sin()(

    )tan(N

    n

    N

    n

    wnnh

    wnnh

    +

    =

    =

    =

    1

    1

    1

    1

    )cos()()0(

    )sin()(

    )tan(N

    n

    N

    n

    wnnhh

    wnnh

    There are two possible solutions

    1) =0 , h(0) can be arbitrary& h(n)=0 for n0

    In this case filter will have order 0 and it would bejust amplifier and not a filter. This is not toouseful result

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    Contd..

    2) 0

    ==

    =

    =

    1

    0

    1

    0

    )cos()(

    )sin()(

    )cos(

    )sin()tan(

    N

    n

    N

    n

    nnh

    nnh

    =

    =

    =

    1

    0

    1

    0

    )cos()sin()()sin()cos()(N

    n

    N

    n

    nnhnnh

    0)}cos()sin()sin()){cos((

    1

    0=

    = nnnh

    N

    n

    0)sin()(1

    0

    =

    =

    nnhN

    n

    0)(sin)(1

    0

    =

    =

    nnhN

    n

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    Contd..

    From this equation , unique solution is obtained for the set ofconditions

    1) = (N-1)/2i.e. for any value of sequence , there is only one value of phase delay

    , which is the condition to obtain linear phase.2) h(n)=h(N-1-n) for 0 n N-1i.e. impulse response sequence must have a special kind of symmetry

    for the value of

    For linear phase filter , impulse response should be either symmetric oranti-symmetric.As impulse response can be of odd or even length, there are four

    possible types of impulse response which will have linear phasecharacteristic.

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    Linear Phase Filters Impulse Responses

    1) Symmetric and Even N

    2) Anti-symmetric and Even N

    3) Symmetric and Odd N

    4) Anti-symmetric and Odd N

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    Symmetric and Even & Odd N

    0 1 2 3 4 5 6 7 8 9 10

    N=11 ( Odd)= 5

    Center ofsymmetry

    0 1 2 3 4 5 6 7 8 9

    N=10 ( Even)= 4.5

    Center ofsymmetry

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    Anti-Symmetric and Even & Odd N

    0 1 2 3 4 5 6 7 8 9 10

    N=11 ( Odd)= 5

    Center of

    symmetry

    0 1 2 3 4 5 6 7 8 9

    N=10 ( Even)= 4.5

    Center ofsymmetry

    P f f Li h h i i f ll f

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    Proofs of Linear phase characteristics for all four types

    Refer Rabinar and Gold

    P iti f i Li h filt

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    Position of zeros in Linear phase filters

    =

    =

    1

    0

    )()(N

    n

    nznhzHWe have

    )1()2()3(

    4321

    )0()1()2(

    )4()3()2()1()0(

    ++++=

    NNN zhzhzh

    zhzhzhzhh

    Symmetry/ anti-symmetry in impulse response

    2/)1()(

    =

    NzzH [ ]

    [ ]

    [ ]})2(

    )1(

    )0({

    2/)5(2/)5(

    2/)3(2/)3(

    2/)1(2/)1(

    +

    +

    +

    NN

    NN

    NN

    zzh

    zzh

    zzh

    -----------(1)

    C td

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    Contd..

    If z is replaced by z-1 in eq 1 we get

    2/)1(1)(

    =

    NzzH [ ]

    [ ][ ]

    }

    )2(

    )1(

    )0({

    2/)5(2/)5(

    2/)3(2/)3(

    2/)1(2/)1(

    +

    +

    +

    NN

    NN

    NN

    zzh

    zzh

    zzh

    -----------(2)

    Comparing eq 1 & 2 we get

    )()()1(1 zHzzH N =

    H(z) and H(z

    -1

    ) are identical within adelay of (N-1) samples and multiplier 1(r,-)

    (r,)

    (1/r,)

    (1/r,-)

    Id l Filt

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    Ideal Filter

    Consider ideal low-pass filter characteristic

    H(w)

    Wc w

    =0

    1)(H

    )(

    ~

    kH2/3