206554: Digital Signal Processing

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20-6554 Digital Signal Processing

• Chapter 6
• Design of Recursive Digital Filters, Part b

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Filters Derived from Analogue Designs
Conversion of pre-existing successful analogue
filter designs. In analog signal/system theory,
the Laplace transform plays a similar role to the
z-transform.
H(s) is the analog transfer function and s is the
Laplace variable. The poles and zeros can be
plotted, as before, in a complex plane, called in
this case the s-plane.
Although the forms of H(s) and H(z) are similar,
the conversion to frequency response is not
carried out in the same way s is not the same
as z. Whereas for z-transforms we use
zexp(j?), for Laplace transforms we use sj?.
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The imaginary axis in the s-plane therefore
corresponds to the unit circle in the z-plane
the interpretation of poles and zeros will be
different in each case. Also, the frequency
response of an analogue filter is not periodic.
we must convert a transfer function H(s) into
H(z), so that the frequency response over the
range 0lt?lt? approximates the analogue filter over
the range 0lt?lt?.
The Bi-Linear Transformation Butterworth and
Chebyshev filters
These both approximate the ideal rectangular
response characteristic, but in different ways
see Figure 6.6 on p176.
Butterworth Gain falls off gradually towards
passband edge, passing through 3dB level at the
nominal cutoff frequency ?1. Increasing the
order improves the passband/stopband transition
(it becomes sharper)
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n is the filter order, and ?1 is the nominal
cutoff frequency.
See the Excel spreadsheet, analogfilt.xls
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Chebyshev Gives an equi-ripple performance in
the passband. The response oscillates between 1
and ?(1?2), where ? is the ripple parameter. For
a given filter order, the performance is better
than the Butterworth filter.
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F(?) is purely imaginary and periodic. Its
magnitude varies between 0 and ? as ? varies
between 0 and ?.
Next, we take the transfer function H(s) of a
desirable analogue filter, obtaining the
frequency response by replacing s with j?
As shown above, if we now substitute ? on the RHS
with tan(?/2), we get a function H(?) in which
the complete frequency response of the analogue
filter is compressed into the range 0lt ?lt ?.
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Making the substitution, we get
See p179 for details of the pole-zero locations
for these filters
Program 21 in dsp.exe allows you to choose the
filter characteristics required, and uses the
formulae on page 179 to calculate and plot the
corresponding pole-zero locations.
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As you can see, there is a button to transfer
this information directly to Program 20, which
will show the frequency response..
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Program 20
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Example 6.3
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A Butterworth low-pass filter is required with a
cut-off frequency of ?10.2?. Its response
should be at least 30dB down at ?0.4?.

• Estimate the minimum order of the filter required
• Use Program 21 to find the z-plane poles and
  zeros, and sketch them
• Derive the filters difference equation
• Use Program 20 to produce a dB plot of its
  frequency response

(a) For a cutoff frequency of 0.2?, the response
magnitude at ?0.4? is
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So
i.e.
or
Since the order has to be an integer, we must
choose n5.
(b) Program 21. Select Butterworth and lowpass
option Filter order 5 Cutoff frequency 36
(0.2?)
Real zero, order 5, at z-1 5 poles, as shown in
the figure
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(c). Derive the filters difference equation
Can obviously multiply this out and find the
difference equation, but it gets complicated with
a larger number of poles and zeros. Use the
idea of cascaded first and second order
sub-filters instead.
First-order filter
Second-order filter
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Inserting the pole values, we get
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(d) Transfer the pole/zero information to
Program 20
Note that, as required the 3dB point occurs at
?0.2? and the response is more than 30dB down
at ?0.4?
Max gain K780 (approx)
For a high-pass filter, we simply change the sign
of the real parts of all poles and zeros Gives
a mirror image in the imaginary axis in the
z-plane
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Example 6.4
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• Use programs 20 and 21 to plot the frequency
  response of a 3rd-order Chebyshev lowpass filter
  with a 3dB passband ripple and a cutoff frequency
  of 0.2? (36). Does the filter meet the cutoff
  specification of the previous example?
• Plot the pole-zero configuration, and frequency
  response magnitude function, of a 6th-order
  Butterworth high-pass filter with a cut-off
  frequency of 0.7?.

(a) 3dB passband ripple corresponds to a
fractional ripple of 0.2929 ---
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(b) Choosing the Butterworth and high-pass
options, Program 21 gives..
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Band-Pass Transformation
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From the poles and zeros of a lowpass prototype,
a suitable transformation will provide a bandpass
filter. This will have twice as many poles and
zeros, so its order will always be even.
For a bandpass filter with a lower cutoff of ?2
and an upper cutoff of ?3, the program first
finds the poles and zeros of a low-pass filter
with a cutoff of ?1 ?3 - ?2 . A pole (or zero)
located at z? then gives two poles (or zeros) in
the band-pass design, at locations
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Example, p185. Chebyshev, bandpass (50-94º),
10th order, 2dB passband ripple