The Kaiser Window (Not Keyser Soze) - PowerPoint PPT Presentation

1 / 38
About This Presentation
Title:

The Kaiser Window (Not Keyser Soze)

Description:

We will take the ideal bandpass filter impulse response, and truncate, shift and taper it. ... The truncated, shifted, tapered impulse response is given by: ... – PowerPoint PPT presentation

Number of Views:358
Avg rating:3.0/5.0
Slides: 39
Provided by: PeteGo2
Category:

less

Transcript and Presenter's Notes

Title: The Kaiser Window (Not Keyser Soze)


1
The Kaiser Window(Not Keyser Soze)
The Kaiser window is another tapered window. It
is given by
2
The Kaiser Window
I0() is the zero-order modified Bessel function
of the first kind (whew!), and is given by this
series
b sets the amount of tapering. Increasing b
increases the amount of tapering. If b 0, the
window is rectangular.
3
The Kaiser Window
Cartinhour gives a Fortran subroutine for
computing the Bessel function, but Bessel
functions are built into Matlab. So is the
Kaiser window. To generate a Kaiser window with
N30 and b6 w kaiser(30, 6) The result is
plotted on the next slide, along with the result
for b 3 This is followed by the Fourier
transform of both windows. Notice that the more
tapered window has a wider mainlobe, but the
sidelobes are smaller.
4
(No Transcript)
5
(No Transcript)
6
The Window Method
If were given a filter specification, it will
usually include the cutoff frequency. This may
be wc (radians per sample) or fs (Hertz). It
will also give the width of the transition band,
and the minimum stopband attenuation. We must
choose the window function, and the filter length
in order to meet the specification.
7
The Window Method
Now lets take the impulse response of our
ideal lowpass filter (N 29, wc 0.46p) and
apply the Kaiser window to it. First, heres the
ideal impulse response, followed by its
Fourier transform
8
(No Transcript)
9
(No Transcript)
10
The Window Method
Next, the impulse response after windowing (b
3), and the resulting frequency response
11
(No Transcript)
12
(No Transcript)
13
The Window Method
Finally, we increase b to 6
14
(No Transcript)
15
(No Transcript)
16
The Window Method
If we plot the frequency response of the ideal
filter and the two Kaiser-windowed filters
together for comparison, we see that reducing the
ripple also reduces the transition steepness.
For a fixed filter length, that is
17
(No Transcript)
18
The Window Method
Finally, we plot the three frequency responses in
dB
19
(No Transcript)
20
Empirical Formulas
Were given a filter specification in the form of
cutoff frequency (fc or wc), stopband attenuation
(d), and transition band width (Df). Presumably,
we also know fs. From these, we have to obtain N
and b. First, we translate d into the stopband
attenuation in dB
21
Empirical Formulas
Next, select b according to the following
formulas
Finally, select N
22
Empirical Formulas
For example, lets design a filter to satify the
following requirements fs 44,100 Hz (CD
sample rate) fc 20 KHz Df 2200 Hz d .001
(A 60)
In terms of normalized frequency, fc 0.4535 and
Df 0.4988, so wc 0.9070p and Dw
0.9976p The empirical formulas give b 5.605 and
N 23.8119 (round off to 24) Heres the
impulse response
23
(No Transcript)
24
Empirical Formulas
And the Amplitude response
25
(No Transcript)
26
Empirical Formulas
And the Amplitude response in dB
27
(No Transcript)
28
Bandpass, Highpass, Etc.
This is great for designing lowpass filters. What
about bandpass, highpass, and bandstop
filters? We can still use the window method.
Heres the amplitude response of an ideal
lowpass filter
-wc
wc
-p
p
29
Bandpass, Highpass, Etc.
Heres the amplitude response of an ideal
bandpass filter We can get this by taking the
frequency response of an ideal lowpass filter
which cuts off at wu, and subtracting from it the
response of a second filter which cuts off at wl.
wl
wu
-wu
-wl
-p
p
30
Bandpass, Highpass, Etc.
While were talking about ideal bandpass filter,
the impulse response is obtained by subtracting
the impulse response of the second filter from
the impulse response of the first
This is the difference of two noncausal
functions, and is therefore noncausal. We had
this problem with lowpass filters, too. We can
overcome it in a similar way. We will take the
ideal bandpass filter impulse response, and
truncate, shift and taper it.
31
Bandpass, Highpass, Etc.
The truncated, shifted, tapered impulse response
is given by
Where w(n) can be any window function, but well
use the Kaiser window.
32
Bandpass, Highpass, Etc.
Lets say were given the following filter
specification fs 44,100 Hz fl 8000 Hz (wl
0.3628p) fu 10,000 Hz (wu 0.4535p) Df
2205 Hz (Dw 0.1p) d 0.001 (A 60) Using
the empirical formulas, we get b 5.6533 and N
73.4 (73). Heres the truncated, shifted,
tapered impulse response
33
(No Transcript)
34
Bandpass, Highpass, Etc.
Heres the amplitude response
35
(No Transcript)
36
Bandpass, Highpass, Etc.
Heres the amplitude response in dB
37
(No Transcript)
38
Problems
Write a Comment
User Comments (0)
About PowerShow.com