Filter Design using Windowing Techniques | Signals and Systems - Electronics and Communication Engineering (ECE) PDF Download

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

•  Main lobe width =  & it can be reduced by increasing N, but area of side lobe will be constant.
• For larger value of N, transition region can be reduced, but we will find overshoots & undershoots on pass band and non zero response in stop band because of larger side lobes. So there overshoots and leakage will not change significantly when rectangular window is used. This result is known as Gibbs Phenomenon.

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 K₁,W₁ and K₁,W₁ 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 K₂.

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 = W₂-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’(i-M)  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/D-H(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 …..

 k₂≤ -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

• the attractive property of the Kaiser window is that the side lobe level and main lobe width can be varied continuously by simple varying the parameter β . Also as in other window, the main lobe width can be adjusted by varying N.
• we can find out the order of Kaiser window, N and the Kaiser parameters β to design FIR filter with a pass band ripple equal to or less that Ap, a minimum stop band attenuation equal to or greater than As, and a transition width Wt, using the following steps:

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: