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Windowing
Disadvantage of F.S is abrupt truncation of FS expansion of the freq response. This truncation result in a poor convergence of the series.
The abrupt truncation of infinite series is equivalent to multiplying it with the rectangular
sequence.
W_{R}(n) = 1 n ≤ M
= 0 else where
W_{R}(e^{jw}) => FT of Rectangular Window
The desined window chts are
1. Small width of main lobe of the fre response of the window containing as much as of the total energy as possible.
2. Side lobes of the frequency response that decrease in energy as w tends to π
3. even function about n=0
4. zero in the range
Let us consider the effect of tapering the rectangular window sequence linearly from the middle to the ends.
Triangular Window:
= 0 else where
In this side lobe level is smaller that that of rectangular window, being reduced from 13 to 25dB to the maximum. However, the main lobe width is now 8π/N There is trade off between main lobe width and side levels.
General raised cosine window is
= 0 else where
If α=0.5 Hanning Window
If α =0.54 Hamming Window
Kaiser Window
= 0 else where
β is constant that specifies a freq response trade off between the peak height of the side lobe ripples and the width or energy of main lobe and Io(x) is the zeroth order modified Bessel function of the first kind. Io(x) can be computed from its power series expansion given by
Window  Peak amplitude of side lobe dB  Transition width of main lobe  Minimum stop band deviation dB 
Rectangular  13  21  
Triangular  25  25  
Hanning  31  44  
Hamming  41  53  
BlackMan  57  74  
Kaiser  variable  variable   
If we let K_{1},W_{1} and K_{1},W_{1} represent cutoff (pass band) * stop band requirements for the digital filter, we can use the following steps in design procedure.
1. Select the window type from table to be the one highest up one list such that the stop band gain exceeds K_{2}.
2. Select no. of points in the windows function to satisfy the transition width for the type of window used. If Wt is the transition width, we must have Wt = W_{2}W_{1≥ }
where K depends on type of window used.
K=1 for rectangular , k=2 triangular…..
Therefore
If analog freq are given, it must be converted in to Digital using w= Ω T
Ex:
Apply the Hamming Window to improve the low pass filter magnitude response ontained
in ex1:
= 0 else where
N = 2M+1 = 21
W_{H}(0) = 1 W_{H}(6) = 0.39785
W_{H}(1) = 0.97749 W_{H}(7) = 0.26962
W_{H}(2) = 0.91215 W_{H}(8) = 0.16785
W_{H}(3) = 0.81038 W_{H}(9) = 0.10251
W_{H}(4) = 0.68215 W_{H}(10) = 0.08
W_{H}(5) = 0.54
Next these window sequence values are multipled with coefficients h(n), obtained in ex1, to ontain modified F.S Co eff h’(n)
h’(0) =0.25
h’(1) =0.22
h’(2) =0.14517
h’(3) =0.0608
h’(4) =0
h’(5) =0.02431
h’(6) =0.02111
h’(7) =0.0086725
h’(8) =0
h’(9) =0.00256
h’(10) =0.00255
bi' = h’(iM) 0≤ i ≤ 20 h’(n) = h’(n)
Ex:
Find a suitable window and calculate the required order the filter to design a LP digital filter to be used A/DH(Z)D/A structure that will have a 3dB cutoff of at 30π rad/sec and an attenuation of 50dB at 45 π rad/sec. the system will use a sampling rate of 100 sample /sec
Sol:
The desired equivalent digital specifications are obtained as
Digital …..
k_{2}≤ 50dB
1. to obtain a stop band attenuation of 50dB or more a Hamming window is shosen since it has the smallest transition band.
2. the approximate no. of points needed to satisfy the transition band requirement (or the order of the filter ) can be found for w1 =0.3 π rad &w2 = 0.45 π rad, using Hamming window (k=2), to be
N = 27 is selected
Therefore: solving above eq forδ , we get
Calculate As using the shosen values
Aso=20logδ
Step 3:
Calculate the parameter β as follows for
β= 0 for As_{o} ≤ 21 dB
= 0.5842(As_{o} 21)^{0.4} + 0.07886(As_{o} 21) for 21< As_{o} ≤ 50 dB
= 0.1102(As_{o} 8.7) for As_{o} >50 dB
Step 4:
Calculate D as follows
D = 0.9222 for As_{o}≤ 21 dB
for As_{o }>21 dB
Step 5:
Select the lowest odd value of N satisfying the inequality
Wsam : Angular Sampling frequency
Ω sam : Analog Freq
Ω t = Ω s Ω p for LPF
= Ω s Ω p for HPF
3dB cutoff freq Ω c can ve considered as follows
for LPF & HPF
Calculate the Kaiser parameter and the no. of points in Kaiser window to satisfy the following lowpass specifications.
Pass band ripple in the freq range 0 to 1.5 rad/sec ≤ 0.1 dB
Minimum stop band attenuation in 2.5 to 5.0 rad /s ≥ 40 dB
Sampling frequency : 10 rad/s
Sol:
The impulse response samples can be calculated using h(n) =
And the no. of points required in this sequence can be found as follows
Step1:
Therefore we choose, δ = 5.7564*10^{ 3}
Step 2:
As_{o} = 20 log( 5.7564*10^{3} ) = 44.797 dB
Step 3 & 4:
β = 0.5842 ( 44.797 21)^{0.4} + 0.07886 ( 44.797 21) = 3.9524
D = 2.566
Step 5:
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