# IIR Bandpass Filters Using Cascaded Biquads

In an earlier post [1], we implemented lowpass IIR filters using a cascade of second-order IIR filters, or biquads. This post provides a Matlab function to do the same for Butterworth bandpass IIR filters. Compared to conventional implementations, bandpass filters based on biquads are less sensitive to coefficient quantization [2]. This becomes important when designing narrowband filters.

A biquad section block diagram using the Direct Form II structure [3,4] is shown in Figure 1. It is defined by two feedback coefficients *a*, three feed-forward coefficients *b*, and gain *K*. There can be one or more cascaded biquad sections in a bandpass filter. The function biquad_bp (listed in the Appendix) computes biquad coefficients and gains given the order of the lowpass prototype, center frequency, bandwidth, and sampling frequency. For a lowpass prototype of order N, the bandpass filter has order 2N and requires N biquads.

This article is available in PDF format for easy printing.

Figure 1. Biquad Section Block Diagram.

## Example 1.

Here is an example function call and results for a bandpass filter based on a 3^{rd} order lowpass prototype:

N= 3; % order of prototype LPF (number of biquads in BPF) fcenter= 20; % Hz center frequency bw= 4; % Hz -3 dB bandwidth fs= 100; % Hz sample frequency [a,K]= biquad_bp(N,fcenter,bw,fs) a = 1.0000 -0.3835 0.8786 1.0000 -0.5531 0.7757 1.0000 -0.7761 0.8865 . K = 0.1254 0.1122 0.1114

The function generates N = 3 sets of biquad feedback coefficients in the matrix a, plus a gain K for each biquad section. The gains K are computed to make the response magnitude of each section equal 1.0 at approximately f_{center}. We can use a and K to find the frequency response of the biquads. As we’ll show later, the feed-forward coefficients of the biquads are all the same: b = [1 0 -1]. We assign the parameters of each biquad as follows:

b= [1 0 -1]; % feed-forward coeffs of each biquad a_bq1= a(1,:); % biquad 1 feedback coeffs = [1 -.3835 .8786] a_bq2= a(2,:); % biquad 2 feedback coeffs = [1 -.5531 .7757] a_bq3= a(3,:); % biquad 3 feedback coeffs = [1 -.7761 .8865] K1= K(1); K2= K(2); K3= K(3); % biquad gains

Now we can compute the frequency response of each biquad.

[h1,f]= freqz(K1*b,a_bq1,2048,fs); % frequency response [h2,f]= freqz(K2*b,a_bq2,2048,fs); [h3,f]= freqz(K3*b,a_bq3,2048,fs); H1= 20*log10(abs(h1)); % dB-magnitude response H2= 20*log10(abs(h2)); H3= 20*log10(abs(h3));

These responses are plotted in Figure 2. Computing the response of the cascaded biquads:

h= h1.*h2.*h3; H= 20*log10(abs(h));

This overall response is shown in Figure 3. Given feed-forward coefficients b= [1 0 -1], the block diagram of each biquad is as shown in Figure 4. The coefficient a_{0} = 1.0 is not used. Note that the feedback gains have negative signs, so given a_{1} = -.3835 and a_{2}= .8786, the feedback gains are .3835 and -.8786.

Looking at Figure 2 again, note that the first and last biquads have response magnitude that is greater than 1.0 (0 dB) at some frequencies, even though the response is 0 dB near f_{center} = 20 Hz. When implementing a fixed-point filter, we need to take this into account to avoid overflow. For this example, overflow is most likely if the input has a large component at the peak of the response of the first biquad, which occurs at about 22 Hz. To prevent overflow, we could allow extra headroom in the first biquad; swap the 2^{nd} biquad with the first biquad; or reduce the gain K_{1} while maintaining the product of the biquad gains constant.

Figure 2. Response of each biquad section for N= 3, center frequency = 20 Hz,

bandwidth = 4 Hz, and f_{s}= 100 Hz.

Figure 3. Overall response of bandpass filter.

Figure 4. Biquad section for a bandpass filter.

## How biquad_bp Works

In an earlier post on IIR bandpass filters [5], I presented the pole-zero form of the bandpass response as:

$$
H(z)=K\frac{(z+1)^N(z-1)^N}{(z-p_1)(z-p_2)...(z-p_{2N})}\qquad(1)$$

where N is the order of the prototype lowpass filter. Calculation of the p_{k }was described in the post. For a bandpass filter, the 2N poles occur as complex-conjugate pairs. For example, the poles of the filter of Example 1 are shown in Figure 5. There are also N zeros at z= -1 and N zeros at z= +1. Our goal is to convert H(z) into a cascade of second-order sections (biquads). We can do this by writing H(z) as the product of N sections with complex-conjugate poles:

Where p^{*}_{k} is the complex conjugate of p_{k}. Note we have assigned a zero at z= -1 and a zero at z = +1 to each biquad. We could assign the zeros differently, but this method is straightforward and gives reasonable results. Expanding the numerator and denominator of the k^{th} biquad section, we get:

$$ H_k(z)=K_k\frac{z^2-1}{z^2-(p_k+p_k^*)z+p_kp_k^*}$$

$$ =K_k\frac{z^2-1}{z^2+a_1z+a_2}$$

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

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

Since we assigned the same zeros to each biquad, the numerator (feed-forward) coefficients b = [1 0 -1] are the same for all N biquads. Summarizing the biquad coefficient values, we have:

b = [1 0 -1]

a = [1 -2*real(p_{k}) |p_{k}|^{2}] (4)

To find the gains K_{k}, we let each biquad have gain of 1 at the filter geometric mean frequency f_{0}. We do this by evaluating H_{k}(z) at f_{0} and setting K_{k} = 1/|H(f_{0})|. To find f_{0}, we define f_{1}= f_{center} – bw/2 and f_{2}= f_{center}+bw/2. Then $f_0=\sqrt{f_1*f_2} $. Note that for a narrowband filter, f_{0} is close to f_{center}.

The Matlab code to find K_{k} is:

f0= sqrt(f1*f2); h= freqz(b,a,[f0 f0],fs); % frequency response at f = f0 K= 1/abs(h(1));

The biquad section with coefficients b, a, and gain K is shown in Figure 4. Here is a summary of the filter synthesis steps used by biquad_bp (The first four steps are detailed in [5]):

- Find the poles of a lowpass analog prototype filter with Ω
_{c}= 1 rad/s. - Given upper and lower -3 dB frequencies of the digital bandpass filter, find the corresponding frequencies of the analog bandpass filter (pre-warping).
- Transform the analog lowpass poles to analog bandpass poles.
- Transform the poles from the s-plane to the z-plane, using the bilinear transform.
- Compute the feedback (denominator) coefficients of each biquad using Equation 4 above.
- Compute the gain K of each biquad.

Figure 5. Z-plane poles and zeros of the bandpass filter of Example 1.

## Example 2 – Narrow bandpass filter with quantized coefficients

Let’s try implementing a narrow band-pass filter with quantized denominator coefficients. We’ll use a bandwidth of 0.5 Hz with sampling frequency of 100 Hz. With lowpass prototype of 3^{rd} order and center frequency of 22 Hz, the function call and results are:

N= 3; % order of prototype LPF fcenter= 22; % Hz center frequency bw= .5; % Hz -3 dB bandwidth fs= 100; % Hz sample frequency [a,K]= biquad_bp(N,fcenter,bw,fs) a = 1.0000 -0.3453 0.9844 1.0000 -0.3690 0.9691 1.0000 -0.3983 0.9845 . K = 0.0157 0.0155 0.0155

Now assign the numerator coefficients and quantize the
denominator coefficients to 12 bits:

b= [1 0 -1]; % numerator coeffs nbits= 12; % bits per unit amplitude a_quant= round(a*2^nbits)/2^nbits; % quantize denominator coeffs

Computing the frequency response as was done in Example 1, we get the response shown in Figure 6. Figure 7 shows the response at 1 dB per division for 12-bit and 8-bit denominator coefficients. As you can see, we get good results using 12-bit coefficients, and some response degradation using 8-bit coefficients. Figure 8 shows three of the six poles of the filter, which are very close to the unit circle compared to the less narrow filter of Example 1 (Figure 5). A conventional bandpass implementation with quantized coefficients could easily become unstable due to the poles falling outside the unit circle.

Figure 6. Response of bandpass with N = 3, bw = 0.5 Hz, and denominator coefficients = 12 bits.

Figure 7. Bandpass filter response vs. denominator coefficient quantization.

Figure 8. Three of six z-plane poles of the filter of Example 2.

## Appendix Matlab Function biquad_bp

Some of the formulas in this program were developed in an earlier post on bandpass IIR filters [5]. 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.

%function [b,a]= biquad_bp(N,fcenter,bw,fs) 4/7/19 Neil Robertson % Synthesize IIR Butterworth Bandpass Filters using biquads % % N= order of prototype LPF = number of biquads in BPF % fcenter= center frequency, Hz % bw= -3 dB bandwidth, Hz % fs= sample frequency, Hz % a= matrix of denominator coefficients of biquads. % each row contains the denom coeffs of a biquad. % There are N rows. % K= vector of gain blocks for each biquad.Length = N. % % Note:numerator coeffs of each biquad = K(k)*[1 0 -1] % function [a,K]= biquad_bp(N,fcenter,bw,fs) f1= fcenter- bw/2; % Hz lower -3 dB frequency f2= fcenter+ bw/2; % Hz upper -3 dB frequency if f2>=fs/2; error('fcenter+ bw/2 must be less than fs/2') end if f1<=0 error('fcenter- bw/2 must be greater than 0') end % find poles of butterworth lpf with Wc = 1 rad/s k= 1:N; theta= (2*k -1)*pi/(2*N); p_lp= -sin(theta) + j*cos(theta); % pre-warp f0, f1, and f2 (uppercase == continuous frequency variables) F1= fs/pi * tan(pi*f1/fs); F2= fs/pi * tan(pi*f2/fs); BW_hz= F2-F1; % Hz -3 dB bandwidth F0= sqrt(F1*F2); % Hz geometric mean frequency % transform poles for bpf centered at W0 % pa contains N poles of the total 2N -- the other N poles not computed % are conjugates of these. for i= 1:N; alpha= BW_hz/F0 * 1/2*p_lp(i); beta= sqrt(1- (BW_hz/F0*p_lp(i)/2).^2); pa(i)= 2*pi*F0*(alpha +j*beta); end % find poles of digital filter p= (1 + pa/(2*fs))./(1 - pa/(2*fs)); % bilinear transform % denominator coeffs for k= 1:N; a1= -2*real(p(k)); a2= abs(p(k))^2; a(k,:)= [1 a1 a2]; % denominator coeffs of biquad k end b= [1 0 -1]; % biquad numerator coeffs % compute biquad gain K for response amplitude = 1.0 at f0 f0= sqrt(f1*f2) % geometric mean frequency for k= 1:N; aa= a(k,:); h= freqz(b,aa,[f0 f0],fs); % frequency response at f = f0 K(k)= 1/abs(h(1)); end

## References

- Robertson, Neil, “Design IIR Filters Using Cascaded Biquads”, Feb, 2018, https://www.dsprelated.com/showarticle/1137.php
- Oppenheim, Alan V. and Shafer, Ronald W.,
__Discrete-Time Signal Processing__, Prentice Hall, 1989, section 6.9. - Sanjit K. Mitra,
__Digital Signal Processing__, 2^{nd}Ed., McGraw-Hill, 2001, section 6.4.1 - “Digital Biquad Filter”, https://en.wikipedia.org/wiki/Digital_biquad_filter
- Robertson, Neil, “Design IIR Bandpass Filters”, Jan, 2018,

https://www.dsprelated.com/showarticle/1128.php

April, 2019 Neil Robertson

**Previous post by Neil Robertson:**

Demonstrating the Periodic Spectrum of a Sampled Signal Using the DFT

- Comments
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i remember when i was an undergrad in the 1970s taking a class in television engineering and learning how to make a sharp cutoff and flat passband filter by cascading 2nd-order BPF sections. so my question to Neil is, how might this example (Figs. 2 and 3) compare to a 6th-order Butterworth BPF (which would be mapped from a 3rd-order Butterworth LPF)? otherwise, i dunno how one chooses the Q or bw of each 2nd-order section to give you the flat response that you spec'd .

the other question i was going to ask is about the coefficient quantization. when does that come into play? in an FPGA? especially 8-bit quantization. i dunno of any modern machine that would impose such coarse quantization as that.

Hi RBJ,

I wrote a previous post on IIR BPF's synthesized directly from the LP prototype. Figure 3 shows the response of different order Butterworth bandpass filters, including a 6th order BPF based on a 3rd-order LPF. The response is the same as in Figure 3 of this post (plot scale is different).

You ask how one chooses the Q and BW of each biquad section: the answer is you don't have to. If you read the derivation in the post, there is no mention of Q or bandwidth of biquads. I calculate the biquad coefficients directly from the poles and zeros.

Regarding the quantization, the example given is just to illustrate the concept. I don't have a digital background, so I don't know what quantization is typical in different applications. Of course, bits get expensive as sample rate increases, so perhaps at 100's of MHz, you could benefit from 8 or 12 bit quantization.

regards,

Neil

I appreciate this article, because it's so fundamental and easy to understand, that it can be used for communication purposes to make people understand the most crititcal issues.

Or as an introductory receipt for a newcomer. Usually, most of this is in the brains, and thus difficult to show :-)

Concerning poles/zeros: one critical aspect is stability. This can be shown with the poles/zeros position: if the filter gets narrower and/or the relation of "sample frequency" / "center frequency of the BPF" increases, poles/zeros are getting very close to the unit circle.

If (especially because of coefficient word length) they happen to cross this line, instability occurs.

Instability could also happen if the word length during calculation is not big enough. But in this case, it's very difficult to detect, because Matlab (working with high resolution floating point math) might show a wonderful filter. But if implemented in µC/FPGA/HW, working with lower resolution fixed point or integer math, it might fail.

It could be a good idea to add a paragraph about stability and word length issues.

Nevertheless, I like the article. Thanks for putting it online.

Hi bholzmayer,

Thanks for these useful insights. I did mention stability in the post:

"Figure 8 shows three of the six poles of the filter, which are very close to the unit circle compared to the less narrow filter of Example 1 (Figure 5). A conventional bandpass implementation with quantized coefficients could easily become unstable due to the poles falling outside the unit circle."

regards,

Neil

One thing that I miss, is Biquad ordering and pole/zero pairing - this has an impact on performance.

The basic idea is that poles provide gain, and poles that are closest to the unit circle provide the highest gain. Zeros on the otherhand, provide attenuation, and can be used to 'tame' this gain. Therefore, group the poles closest to the unit circle with the zeros closest to them, and so on until all poles and zeros have been allocated. Finally, arrange the biquad cascade with the pole/zero pair closest to the unit circle (i.e. with the largest gain error) at the end of the cascade.

Here's an example for an Elliptic filter:

Regards,

Sanjeev.

Hi asn,

Thanks for this tip, which is particularly useful for filters with finite zeros. Note that for the butterworth bpf analyzed here, the zeros are all at z = -1 and z = +1. In the post, I assigned a zero at each of these locations to each biquad. I think this approach will usually work well for Butterworth bpf's.

regards,

Neil

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