# Design IIR Filters Using Cascaded Biquads

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In an earlier post on IIR Butterworth lowpass filters [3], I presented the pole-zero form of the lowpass response H(z) as follows:

$$H(z)=K\frac{(z+1)^N}{(z-p_1)(z-p_2)...(z-p_N)}\qquad(1)$$

The N zeros at z = 1 (ω= π or f = f_{s}/2) occur when we transform the lowpass analog zeros from the s-domain to z-domain using the bilinear transform. Our goal is to convert H(z) into a cascade of second-order sections. If we stipulate that N is even, then we can write H(z) as:

$H(z)=K_1\frac{(z+1)^2}{(z-p_1)(z-p_2)}\cdot K_2\frac{(z+1)^2}{(z-p_3)(z-p_4)}\cdot...\cdot K_N\frac{(z+1)^2}{(z-p_{N-1})(z-p_N)} \qquad(2)$

Each term in equation 2 is biquadratic – it has quadratic numerator and denominator. It is not necessary to use a separate gain K for each term; we could also use just a single gain for the whole cascade.

The filter is even order, so all poles occur in complex-conjugate pairs. We’ll assign a complex-conjugate pole pair to the denominator of each term of equation 2. We can then write each term as:

$$H_k(z)=K_k\frac{(z+1)^2}{(z-p_k)(z-p^*_k)},\quad k=1:N/2$$

where $p^*_k$ is the complex conjugate of $p_k$. Expanding the numerator and denominator, we get:

$$H_k(z)=K_k\frac{z^2+2z+1}{z^2+a_1z+a_2}$$

where a_{1}= -2*real(p_{k}) and a_{2}= |p_{k}|^{2}. Dividing numerator and denominator by z^{2}, we get:

$$H_k(z)=K_k\frac{1+2z^{-1}+z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}\qquad(3)$$

We want the gain of each biquad section to equal 1 at ω=0. Letting $z= e^{jω}$, we have z= 1. Then:

$$H_k(z)=1=K_k\frac{\sum{b}}{\sum{a}}$$

so

$$K_k=\frac{\sum{a}}{4}\qquad(4)$$

where a = [1 a_{1} a_{2}] are the coefficients of the biquad section. Summarizing the coefficient values, we have:

b = [1 2 1]

a= [1 -2*real(p_{k}) |p_{k}|^{2}]

$K= \sum{a}/4$

A biquad lowpass block diagram using the Direct form II
structure [4,5] is shown in Figure 1. We
will cascade N/2 biquads to implement an N
^{th} order filter (N
even). Note that the feed-forward
coefficients b have the same value for all N/2 biquads in a filter. This is evident from Equation 3.

Figure 1 Biquad (second-order) lowpass all-pole filter

Direct form II

## Example

In this example, we’ll use biquad_synth to design a 6^{th} order Butterworth lowpass filter with -3 dB frequency of 15 Hz and f_{s}= 100 Hz. Note biquad_synth uses the bilinear transform with prewarping [3] to transform H(s) to H(z). The filter will consist of three biquads, as shown in Figure 2. biquad_synth computes the denominator (feedback) coefficients a of each biquad. The gains K are computed separately. Note biquad_synth contains code developed in an earlier post on IIR Butterworth filter synthesis [3]. Here is the function call and the function output:

N= 6; % filter order fc= 15; % Hz -3 dB frequency fs= 100; % Hz sample frequency a= biquad_synth(N,fc,fs) a = 1.0000 -0.6599 0.1227 1.0000 -0.7478 0.2722 1.0000 -0.9720 0.6537

Each row of the matrix a contains the denominator coefficients of a biquad. As we already determined, the numerator coefficients b are the same for all three biquads:

b= [1 2 1];The gains for each biquad are, from equation 4:

K1= sum(a(1,:)/4; K2= sum(a(2,:)/4; K3= sum(a(3,:)/4;

Now
we can compute the frequency response of each biquad. The overall response is their product.

[h1,f] = freqz(K1*b,a(1,:),512,fs); [h2,f] = freqz(K2*b,a(2,:),512,fs); [h3,f] = freqz(K3*b,a(3,:),512,fs); h= abs(h1.*h2.*h3); H= 20*log10(abs(h));

The magnitude response of each biquad and the overall
response are plotted in Figure 3. The
sequence of the biquads doesn’t matter in theory; however, placing the biquad
with the peaking response (h3) last minimizes the chance of clipping.

Figure 2. 6^{th} order lowpass filter using
three biquads

Figure 3. 6^{th} order lowpass Butterworth cascaded-biquad response. f_{c}= 15 Hz, f_{s}= 100 Hz.

Top: response of each biquad section (blue= h1, green= h2, red= h3).

Bottom: overall response

## Coefficient Quantization

As I stated at the beginning, the cascaded-biquad design is less sensitive to coefficient quantization than a single high-order IIR, particularly for lower cut-off frequencies. To illustrate this, we’ll first look at how quantizing coefficients effects z-plane pole locations of a 6^{th} order IIR filter. The following code finds the unquantized poles of the 6^{th} order Butterworth filter with -3 dB frequency f_{c} = 5 Hz. (Note butter [6] is a function in the Matlab signal processing toolbox that synthesizes IIR Butterworth filters).

fc= 5; fs= 100; [b,a]= butter(6,2*fc/fs); % Matlab function for Butterworth LP IIR p= roots(a); % poles in z-plane

The poles are plotted as the red x’s on the left side of Figure 4. We have also plotted the poles for fc= 12 Hz (blue-ish x’s). Each set contains 6 poles. If we plot the poles of filters having fc from 1 Hz to 25 Hz in 1 Hz increments, we get the plot on the right, where only the right side of the unit circle is shown. The lower values of f_{c} are on the right, near z = 1.

Figure 4. Unquantized poles of 6^{th}-order Butterworth IIR filter.

Left: f_{c} = 5 Hz (red) and 12 Hz (blue). Right: f_{c} = 1 Hz to 25 Hz.

Now let’s quantize the denominator coefficients and see how this effects the pole locations of Figure 4. Let nbits = the number of bits per unit of coefficient amplitude:

nbits= 16;

Here is the code to find the quantized poles for a single value of f_{c}:

fs= 100; [b,a]= butter(6,2*fc/fs); a_quant= round(a*2^nbits)/2^nbits; p_quant = roots(a_quant);

Letting fc vary in 0.5 Hz increments from 0.5 to 25 Hz, we get the poles shown on the left of Figure 5. As you can see, as f_{c} decreases, quantization causes the poles to depart from the desired locations. The right side of Figure 5 shows the effect of 10-bit quantization.

Figure 5. Effect of Quantization on poles of 6^{th}-order Butterworth IIR filter.

Left: nbits = 16 Right: nbits = 10

We can do the same calculation for the biquads that make up the 6^{th} order cascaded implementation. For example, here is the code to find the quantized poles of the second biquad for a single value of f_{c} (recall that the matrix a has three rows containing the coefficients of 3 biquads).

nbits= 10; a= biquad_synth(6,fc,fs); a2= a(2,:); % 2nd biquad a_quant= round(a2*2^nbits)/2^nbits; p_quant = roots(a_quant);

This time, letting fc vary in 0.25 Hz increments
from 0.25 to 25 Hz, we get the poles shown in Figure 6, which includes only
quadrant 1 of the unit circle. The
biquad performs much better than the 6
^{th} order filter, only
departing dramatically from the unquantized curve for fc = 0.25 Hz. So we expect better performance from
cascading three biquads vs. using a single 6
^{th}-order filter.

Figure 6. Effect
of Quantization on poles of one biquad, nbits = 10.

Now we’re finally ready to compare frequency response of a biquad-cascade filter vs. a conventional IIR filter when the denominator coefficients are quantized. The cutoff frequency and quantization level are chosen to stress the conventional filter. We’ll leave the numerator coefficients of the conventional filter as floating-point. Interestingly, when implementing the biquad filter we get exact numerator coefficient values “for free”: since b = [1 2 1], we can implement b_{0} and b_{2} as no-ops and b_{1} as a bit shift.

For the biquad filter, we use biquad_synth to find the coefficients for f_{c} = 6.7 Hz:

fc=6.7; % Hz -3 dB frequency fs= 100; % Hz sample frequency a = biquad_synth(6,fc,fs) % a has 3 rows, one for each biquad a = 1.0000 -1.3088 0.4340 1.0000 -1.4162 0.5516 1.0000 -1.6508 0.8087

For the conventional filter, we again use the Matlab function butter:

[b,a]= butter(6,2*fc/fs); a = 1.0000 -4.3757 8.1461 -8.2269 4.7417 -1.4761 0.1936

In each case, we quantize coefficients to 10 bits per unit of coefficient amplitude:

nbits= 10; a_quant= round(a*2^nbits)/2^nbits; % quantize denom coeffs

First, we’ll look at the quantized pole locations. For the conventional filter, the quantized poles are:

p_quant= roots(a_quant);

For the biquad implementation, the quantized poles are:

p1= roots(a_quant(1,:))’; p2= roots(a_quant(2,:)’; p3= roots(a_quant(3,:)’; p_quant= [p1 p2 p3];

Figure 7 shows the z-plane poles for the floating-point and quantized coefficients. Quantization has little effect on the biquad version, but has a large effect on the conventional filter.

Now let’s compare the magnitude responses for quantized coefficients. We compute the response of the biquad version in the same way used to obtain Figure 3. Figure 8 shows the magnitude responses. As you would expect from the pole plots, the conventional implementation has poor performance, while the biquad implementation shows no noticeable effect due to quantization.

So how low can we go with this N= 6 Butterworth cascaded-biquad filter? As we reduce f_{c}, the quantization of 1024 steps per unit of coefficient amplitude eventually takes a toll. Figure 9 shows the z-plane poles and magnitude response for fc= 1.6 Hz. As you can see, the magnitude response is sagging. If we stay above f_{c} of 2.5 Hz = f_{s}/40, the response error is less than 0.1 dB.

Besides adding coefficient bits, there are other ways to improve performance of narrow-band IIR filters. See for example the post in reference [7].

Figure 7. Z-plane pole locations with quantized denom. coeffs. N= 6, f_{c}= 6.7 Hz, f_{s}= 100 Hz.

Blue = floating point, Red = 10 bit quantization.

Left: biquad implementation Right: conventional implementation.

Figure 8. Magnitude response with quantized denom coeffs. N= 6, f_{c}= 6.7 Hz, f_{s}= 100 Hz, nbits = 10.

Blue = biquad implementation, green= conventional implementation.

Figure 9. Z-plane poles and magnitude response of biquad-cascade filter with quantized denominator coefficients. Blue x’s = floating-point and red x’s = quantized. N= 6, f_{c}= 1.6 Hz, f_{s}= 100 Hz, nbits= 10.

## References

1. Oppenheim, Alan V. and Shafer, Ronald W., __Discrete-Time Signal Processing__, Prentice Hall, 1989, sections 6.3.2 and 6.8.1.

2. Lyons, Richard G., __Understanding Digital Signal Processing__, 2^{nd} Ed., Pearson, 2004, section 6.8.2

3. Robertson, Neil , “Design IIR Butterworth Filters Using 12 Lines of Code”, Dec 2017 https://www.dsprelated.com/showarticle/1119.php

4. Sanjit K. Mitra, __Digital Signal Processing__, 2^{nd} Ed., McGraw-Hill, 2001, section 6.4.1

5. “Digital Biquad Filter”, https://en.wikipedia.org/wiki/Digital_biquad_filter

6. Mathworks website, ”butter”, https://www.mathworks.com/help/signal/ref/butter.html

7. Lyons, Rick, “Improved Narrowband Lowpass IIR Filters”, https://www.dsprelated.com/showarticle/120.php

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Neil Robertson February 11, 2018 revised 2/20/18

## Appendix Matlab Function biquad_synth

This program is provided as-is without any guarantees or warranty. The author is not responsible for any damage or losses of any kind caused by the use or misuse of the program.

% biquad_synth.m 2/10/18 Neil Robertson % Synthesize even-order IIR Butterworth lowpass filter as cascaded biquads. % This function computes the denominator coefficients a of the biquads. % N= filter order (must be even) % fc= -3 dB frequency in Hz % fs= sample frequency in Hz % a = matrix of denominator coefficients of biquads. Size = (N/2,3) % each row of a contains the denominator coeffs of a biquad. % There are N/2 rows. % Note numerator coeffs of each biquad= K*[1 2 1], where K = (1 + a1 + a2)/4. % function a = biquad_synth(N,fc,fs); if fc>=fs/2; error('fc must be less than fs/2') end if mod(N,2)~=0 error('N must be even') end %I. Find analog filter poles above the real axis (half of total poles) k= 1:N/2; theta= (2*k -1)*pi/(2*N); pa= -sin(theta) + j*cos(theta); % poles of filter with cutoff = 1 rad/s pa= fliplr(pa); %reverse sequence of poles – put high Q last % II. scale poles in frequency Fc= fs/pi * tan(pi*fc/fs); % continuous pre-warped frequency pa= pa*2*pi*Fc; % scale poles by 2*pi*Fc % III. Find coeffs of biquads % poles in the z plane p= (1 + pa/(2*fs))./(1 - pa/(2*fs)); % poles by bilinear transform % denominator coeffs for k= 1:N/2; a1= -2*real(p(k)); a2= abs(p(k))^2; a(k,:)= [1 a1 a2]; %coeffs of biquad k end

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Design IIR Highpass Filters

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Coefficients of Cascaded Discrete-Time Systems

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