IIR Filters

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IIR Filters
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The general form for z-transform transfer
functions for a digital filter is
The inverse of this z-transform transfer function
hn is the impulse response for this digital
filter.
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As an example if
we have
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The inverse z-transform of this transfer function
(the impulse response) is
hn
n
1
2
3
4
5
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This impulse response function goes on forever.
That is, the impulse response is infinite in
duration. Such filters are referred to as
infinite impulse response or IIR filters.
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Another example would be
The corresponding impulse response is
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hn
n
1
2
3
4
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This impulse response is finite in duration. Such
filters are referred to as finite impulse
response or FIR filters.
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Given a transfer function, how do we know whether
this transfer function corresponds to an impulse
response that is infinite or finite in duration?
In our previous two examples we had
(IIR)
(FIR)
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In general, any filter whose transfer function
has a denominator (that does not factor-out) will
have an impulse response that is infinite in
duration corresponding to an IIR filter.
Any filter whose transfer function does not have
a denominator will have an impulse response that
is finite in duration corresponding to an FIR
filter.
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The distinction between IIR filter transfer
functions and FIR filter transfer functions
becomes more clear if we look at the
corresponding difference equations. In our
examples the corresponding difference equations
are
(IIR)
(FIR)
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The IIR filter difference equation is recursive
in nature the current output depends upon the
previous output
Since the current output depends upon the
previous output and the previous output depends
upon its previous output, the output depends upon
the infinite past.
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The FIR filter difference equation is depends
only upon the input
If the input is finite in duration (such as an
impulse) then the output is finite in duration.
The output depends upon the finite past.
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The difference between IIR and FIR filters can
also be seen by looking at the transfer functions
and noting that the IIR transfer function can be
expanded using a geometric series that is
infinite
The FIR filter is already in a finite series form.
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Most analog filters have an impulse response
which is infinite in duration. IIR filters are
generally designed by emulating an analog
prototype filter. There are two methods for
doing this analog filter emulation
(1) the matched z-transform or impulse invariant
transform
(2) the bilinear transformation.
In both cases, we are given an analog transfer
function H(s), and we transform this function
into a digital transfer function H(z).
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The Matched z-Transform
In the matched z-transform digital filter design
method we try to match the impulse response of
the analog filter with that of the digital filter
being designed.
To match the impulse responses, we take the
inverse Laplace transform of the analog filter
H(s)?h(t), then sample the impulse response
h(t)?hn, then take the z-transform of the
sampled impulse response to get the z-transform
transfer function hn?H(z).
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Analog Prototype
Digital Filter
L-1
Z
sample
Once we have our z-transform transfer function
H(z), we apply the definition of the transfer
function to write our digital filter equations
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Example Use the matched filter design method to
design the digital equivalent of an integrator.
Solution The analog transfer function is
The inverse Laplace transform is
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We then sample the impulse response to get hn
Finally, we take the z-transform of the impulse
response to get the digital filter transfer
function.
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Finally, we apply the definition of the
z-transform transfer function to get the
relationship between the input of the digital
filter xn and the output of the digital filter
yn.
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Applying this formula to an arbitrary input xn,
we have
n xn yn
0 x0 x0
1 x1 x0x1
2 x2 x0x1x2
3 x3 x0x1x2x3
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We see that the output is the summation of the
input. Thus the digital filter accurately
represents the analog filter.
The digital filter does not always accurately
represent the analog filter as will be seen in
the next example.
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Example Use the matched filter design method to
design the following transfer function
where Wc Ws/4, and Ws is the sampling frequency.
Solution First, we find the impulse response
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Then we sample the impulse response
Then we take the z-transform of the
(discrete-time) impulse response
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We then find the frequency response of the filter
w ejw H(ejw) H(ejw)
0 1 1.26
p/2 j 0.98
p -1 0.83
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As we can see, it is not much of a low-pass
filter the frequency rolloff is not very great.
The reason for this small rolloff is aliasing
error the frequency response of H(z) is composed
of copies of the frequency response of H(s) at 0,
Ws, 2Ws, etc. The copy at 2Ws overlaps the copy
at 0.
Aliasing error is an inherent problem in matched
filter design. Unless the cutoff frequency is
very low compared to the sampling frequency, we
will get substantial error due to aliasing. This
aliasing problem is solved using the bilinear
transformation.
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The Bilinear Transformation
The bilinear transformation is a fairly direct
method of converting H(s) to H(z). Rather than
map the analog frequencies W0 to Ws/2 to the
digital frequencies w0 to wp (as we had done
with the matched z-transform), we will map the
analog frequencies W0 to ? to the digital
frequencies w0 to wp
Matched z-transform W(0,Ws/2) ? w(0,p)
Bilinear Transformation W(0,?) ? w(0,p)
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What kind of function maps W(0,?) ? w(0,p)?
How about this
As we can see from the graph on the following
slide, this function does perform the necessary
mapping.
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We know the relationship between W and w what is
the relationship between s and z?
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Since sjW and zejw, we have
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Because of the warped nature of our
transformation, it is necessary to pre-warp our
analog prototype critical frequencies so as to
coincide with the critical frequencies of the
corresponding digital filter.
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Example Use the bilinear transformation method
to find the digital equivalent to the following
transfer function
where Wc Ws/4, and Ws is the sampling frequency.
Solution First, we must pre-warp the analog
frequency
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We then substitute our pre-warped frequency,
and apply the bilinear (s?z) transformation
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The frequency response of the filter is
w ejw H(ejw) H(ejw)
0 1 1.000
p/2 j 0.707
p -1 0.000
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As can be seen, the frequency response is much
improved. At w0, the response is the same as
the analog filter at W0, and at wp, the
frequency response is the same as the analog
filter at W?.
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Example Repeat the previous example
where Wc Ws/6.
Solution Our prewarping is slightly different
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Substituting our pre-warped frequency, we have
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Applying the bilinear transformation, we have
The frequency plot is given on the following
slide.
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Example find the digital equivalent of a
second-order Butterworth filter using the
bilinear transformation. Let Wc Ws/4.
Solution The second-order Butterworth filter has
the following form
(Enter b a butter(2,1,'s') in MATLAB.)
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The prewarping is the same as in the first
example
So,
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Applying the bilinear transformation we have
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The frequency response is on the following slide.
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