3f3 6 design of iir filters
TRANSCRIPT
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Design of IIR Filters
Elena Punskayawww-sigproc.eng.cam.ac.uk/~op205
Some material adapted from courses by Prof. Simon Godsill, Dr. Arnaud Doucet,
Dr. Malcolm Macleod and Prof. Peter Rayner
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IIR vs FIR Filters
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IIR as a class of LTI Filters
Difference equation:
To give an Infinite Impulse Response (IIR), a filter must be recursive, that is, incorporate feedback N ≠ 0, M ≠ 0
the recursive (previous output) terms feed back energy into the filter input and keep it going.
(Although recursive filters are not necessarily IIR)
Transfer function:
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IIR Filters Design from an Analogue Prototype
• Given filter specifications, direct determination of filter coefficients is too complex
• Well-developed design methods exist for analogue low-pass filters
• Almost all methods rely on converting an analogue filter to a digital one
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Analogue filter Rational Transfer Function
( )
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Analogue to Digital Conversion
Im (z)
Re (z)
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Impulse Invariant method
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Impulse Invariant method: Steps
1. Compute the Inverse Laplace transform to get impulse response of the analogue filter
2. Sample the impulse response (quickly enough to avoid aliasing problem)
3. Compute z-transform of resulting sequence
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Example 1 – Impulse Invariant Method
Consider first order analogue filter
= 1 -
Corresponding impulse response is
δ - v
The presence of delta term prevents sampling of impulse response which thus cannot be defined
Fundamental problem: high-pass and band-stop filters have functions with numerator and denominator polynomials of the same degree and thus cannot be designed using this method
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Consider an analogue filter
Example 2 – Impulse Invariant Method
Step 1. Impulse response of the analogue filter
Step 2. Sample the impulse response
Step 3. Compute z-transform
The poles are mapped as
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Impulse Invariant Method
Indeed, in the general case the poles are mapped as
since any rational transfer function with the numerator degree strictly less than the denominator degree can be decomposed to partial fractions
and similarly it can be shown
kk
k
kk
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Impulse Invariant Method: Stability
Since poles are mapped as:
stable analogue filter is transformed into stable digital filter
kk
k < 1
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Summary of the Impulse Invariant Method
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• Advantage: – preserves the order and stability of the
analogue filter
• Disadvantages:– Not applicable to all filter types (high-pass,
band-stop)– There is distortion of the shape of
frequency response due to aliasing
Summary of the Impulse Invariant Method
not in common use
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Matched z-transform method
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Example of Impulse Invariant vs Matched z transform methods
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Backward Difference Method
delay
The analogue-domain variable s represents differentiation.
We can try to replace s by approximating differentiation operator in the digital domain:
Thus,y(t) = y(n) ≈
Which suggests the s-to-z transformation:
Y(z) ≈ X(z)
backward difference operator
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Backward Difference Operator
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- 0.5 = 0.5 - 0.5 = 0.5
The Left half s-plane onto the interior of the circle with radius 0.5 and centre at 0.5 in the z-plane
Stable analogue filters become stable digital filters.However, poles are conned to a relatively small set of frequencies,no highpass filter possible!
Backward Difference method - Stability
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Summary of the Backward Difference method
• Since the imaginary axis in the s domain are not mapped to the unit circle we can expect that the frequency response will be considerably distorted
• An analogue high-pass filter cannot be mapped to a digital high-pass because the poles of the digital filter cannot lie in the correct region
method is crude and rarely used
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Bilinear transform
• Bilinear transform is a correction of the backwards difference method
• The bilinear transform (also known as Tustin's transformation) is defined as the substitution:
• It is the most popular method
• The bilinear transform produces a digital filter whose frequency response has the same characteristics as the frequency response of the analogue filter (but its impulse response may then be quite different).
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The bilinear transform
• Note 1. Although the ratio could have been written (z-1)/(z+1), that causes unnecessary algebra later, when converting the resulting transfer function into a digital filter;
• Note 2. In some sources you will see the factor (2/T) multiplying the RHS of the bilinear transform; this is an optional scaling, but it cancels and does not affect the final result.
The bilinear transform
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Where is the Bilinear Transform coming from?
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To derive the properties of the bilinear transform, solve for z, and put
( )( ) 22
222
11 hence ;
11
11
ωω
ωω
+−++
=−−++
=−+
=aaz
jaja
ssz
Properties of the Bilinear Transform
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Properties of the Bilinear Transform
Look at two important cases:
1. The imaginary axis, i.e. =0. This corresponds to the boundary of stability for the analogue filter’s poles.
With =0, we have
the imaginary (frequency) axis in the s-plane maps to the unit circle in the z-plane
2. With <0, i.e. the left half-plane in the s-plane we have
left half s-plane maps onto the interior of the unit circle
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Thus the bilinear transform maps the Left half s-plane onto the interior of the unit circle in the z-plane:
This property allows us to obtain a suitable frequency response for the digital filter, and also to ensure the stability of the digital filter.
s-planez-plane
1
1
Properties of the Bilinear Transform
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Properties of the Bilinear Transform
z - 1s = z +1
=- 1+ 1
( )=
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Hence the Bilinear Transform preserves the following important features of the frequency response:
1. the Ω↔ ω mapping is monotonic, and
2. Ω= 0 is mapped to ω = 0, and Ω = ∞ is mapped to ω = π (half the sampling frequency). Thus, for example, a low-pass response that decays to zero at Ω = ∞produces a low-pass digital filter response that decays to zero at ω = π.
3. Mapping between the frequency variables is
Properties of the Bilinear Transform
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Properties of the Bilinear Transform
If the frequency response of the analogue filter at frequency Ω is H(jΩ), then the frequency response of the digital filter at the corresponding frequency ω = 2 arctan(ω) is also H(j Ω). Hence -3dB frequencies become -3dB frequencies, minimax responses remain minimax, etc.
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Proof of Stability of the FIlter
Suppose the analogue prototype H(s) has a stable pole at , i.e.
Then the digital filter is obtained by substituting
Since H(s) has a pole at , has a pole at because
However, we know that lies within the unit circle. Hence the filter is guaranteed stable provided H(s) is stable.
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Frequency Response of the FilterThe frequency response of the analogue filter is
The frequency response of the digital filter is
Hence we can see that the frequency response is warped by a function
Analogue Frequency Digital Frequency
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The steps of the bilinear transform method are as follows:
1. “Warp” the digital critical (e.g. band-edge or "corner") frequencies ωi , in other words compute the corresponding analogue critical frequencies Ω i = tan(ωi/2).
2. Design an analogue filter which satisfies the resulting filter response specification.
3. Apply the bilinear transform
to the s-domain transfer function of the analogue filter to generate the required z-domain transfer function.
Design using the bilinear transform
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Design a first order low-pass digital filter with -3dB frequency of 1kHz and a sampling frequency of 8kHz using a the first order analogue low-pass filter
which has a gain of 1 (0dB) at zero frequency, and a gain of -3dB ( = √0.5 ) at Ω c rad/sec (the "cutoff frequency ").
Example: Application of Bilinear Transform
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• First calculate the normalised digital cutoff frequency:
• Calculate the equivalent pre-warped analogue filter cutofffrequency (rad/sec)
• Thus, the analogue filter has the system function
Example: Application of Bilinear Transform
3dB cutoff frequency
sampling frequency
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Example: Application of Bilinear Transform
Apply Bilinear transform
Normalise to unity forrecursive implementation
As a direct form implementation:Keep 0.2929 factorised to saveone multiply
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Note that the digital filter response at zero frequency equals 1, as for the analogue filter, and the digital filter response at ω = π equals 0, as for the analogue filter at
Ω = ∞. The –3dB frequency is ωc = π/4, as intended.
ωc = π/4
Example: Magnitude Frequency Response
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X Re(z)
Imag(z)
1-1O
Note that:• The filter is stable, as expected• The design process has added an extra zero compared to the prototype
- this is typical of filters designed by the bilinear transform.
Example: Pole-zero diagram for digital design
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There is a Matlab routine BILINEAR which computes the bilinear transformation.
The example above could be computed, for example, by typing[NUMd,DENd] = BILINEAR([0.4142],[1 0.4142],0.5)
which returnsNUMd =
0.2929 0.2929
DENd =
1.0000 -0.4142
Example: Pole-zero diagram for digital design
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Designing high-pass, band-pass and band-stop filters
• The previous examples we have discussed have concentrated on IIR filters with low-passcharacteristics.
• There are various techniques available to transform a low-pass filter into a high-pass/band-pass/band-stop filters.
• The most popular one uses a frequency transformation in the analogue domain.
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Designing filters uding frequency transformation
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Frequency transformations
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Example. Frequency transformation
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Example. Frequency transformation
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Thank you!