I hope to design 1st order highpass filter from transfer function. In the example of 1st order lowpass filter, I first get the coefficients of numerator and denominator in the variable 'b' and 'a'. In case of first order filer,

b = 1/tau;
a = [1, 1/tau];
h = freqs(b, a, w) % LPF
[digital_b, digial_a] = bilinear(b, a, fs) % Analog to Digital transformation. 

The order is like above. First, we obtain the coefficients of the transfer function and then make LPF by the 'freq' function in MATLAB. After that, I can convert the analog filter to a digital filter by using the 'bilinear' function.

However, in case of HPF, I cannot transform it using the 'bilinear' function. When I run the 'bilinear' function in MATLAB for 1st order HPF, it gives the following error

Numerator cannot be higher order than denominator.

I used the code below to make digital HPF.

b = [1, 1/tau];
a = 1;
h = freqs(b, a, w) % LPF
[digital_b, digial_a] = bilinear(b, a, fs) % Analog to Digital transformation. 

What is the problem? and how should I design HPF on MATLAB?

  • $\begingroup$ by this method, you're not designing HPF filter, by this you are designing inverse filter of your LPF which has gain one for low frequencies and amplifying high frequencies. $\endgroup$
    – Mohammad M
    Commented Jun 11, 2017 at 8:28

2 Answers 2


Low Pass

Note: Scroll down for High Pass.

I have ran your code by fixing parameters such as $\tau, f_s, w$ and i got no error. However, the following is low pass and using 'fvtool' we see that

tau = 1e-3;
fs = 1e3;
w = 0:1e-2:pi;
b = 1/tau;
a = [1, 1/tau];
h = freqs(b, a, w) % LPF
[digital_b, digital_a] = bilinear(b, a, fs) 

This gave me

digital_b =
    0.0050    0.0050
digital_a =
    1.0000   -0.9900

The figure is as enter image description here

High Pass

On the other hand, changing your $a,b$ vectors to a transfer function that corresponds to a High Pass as follows for instance

a = [1 1/tau];
b = [1/tau 0];

gives the desired high pass

enter image description here


Transfer function of a 1st order HP filter is:

$$H(s)=\frac s{s+wc}=\frac{{\displaystyle\frac s{wc}}}{1+{\displaystyle\frac s{wc}}}\;,where\;wc=2\mathrm\pi\ast\mathrm{fc}$$


a = [1 w0];
b = [w0 1e-48]; % ! two coefficients
  • $\begingroup$ why not set b = [w0, 0] ? $\endgroup$
    – Ben
    Commented Jan 15, 2018 at 2:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.