IIR Digital Filter

IIR Digital Filter to Design Low Pass Filter

Introduction: IIR or Infinite Impulse Response Filters are a type of recursive digital filter whose output are based on previous output and current inputs (or sometimes previous inputs). The name 'Infinite Impulse Response' is owing to presence of infinite train of impulses on an impulse response in the frequency domain. These filters are widely used to design of low pass, high pass or band pass digital filters. The relevant mathematical expressions can be derived from the transfer functions of an analog filter, analogous to the digital filter to be designed. Here we will make a low pass digital filter. We will proceed from analog domain and then convert it to s-domain and z-domain and finally arrive at the difference equation which can be computed by a microcontroller.

Prerequisites:

Basic electronics and mathematics
Laplace and Z transformation
IIR Digital Filters
Recursive Programming

Step 1: Parts Required

1. ESP32
2. USB Cable
3. Function Generator / Arduino and Resistors
4. DSO
5. Jumper wires

Step 2: Theory

ESP32 microcontroller is a very powerful microcontroller with large flash memory. It has 12bit ADC and 8bit DAC which makes it a perfect choice to be used to implement a digital filter. To make a digital filter, we need to arrive at a Difference equation which can be implemented by software. The necessary steps are described below:

First formulate the analogous analog filter circuit using R and C. For my case I have a low pass filter of 1st order so a single RC is used. Determine the input x(t) - output y(t) relationship in time domain
For my case it is $$x(t)-y(t) = RC.{dy(t)\over dt}$$
Find out the transfer function in s-domain using Laplace Transformation: $$H(s) = {Y(s) \over X(s)}$$
For my case, $$H(s) = {1\over RC.(s+ {1\over RC})}$$
Use Bilinear Transformation (Tustin Method) to transform from s-domain to z-domain. It is a conformal mapping which transforms jw axis of s-plane to unit circle in z-domain, only once, thus avoiding aliasing of frequency components. We need to substitute the following in H(s) to make it H(z):$$s = { {2.(1 - z^{-1})\over T.(1 + z^{-1})}}$$
Where T is sample time
There are other methods for this like Forward Difference Method,Backward Difference Method, Impulse Invariant transformation, but Bilinear transformation proved to be the best among these.

So after substituting s in H(s), we are in z domain now ,where $$H(z) = {Y(z) \over X(z)}$$ On simplification, we get: $$ 2RC.Y(z)-2RC.z^{-1}Y(z)+T.Y(Z)+T.z^{-1}Y(z) = T.X(z) + T.z^{-1}X(z)$$
Perform inverse z-transform, using the below formula $${ z^{-1}.H(z)} \xrightarrow{\text{inv. z-trans}} h(n-1)$$ to make the output as a difference equation which can be implemented in software. It will be recursive difference equation, therefore an IIR filter.
In our case, we get $$ 2RC.y(n)-2RC.y(n-1)+T.y(n)+T.y(n-1) = T.x(n) + T.x(n-1)$$ After furthur simplifcation, we arrive at the final output expression y(n): $$ \therefore y(n) = { T \over T+2RC}.x(n) + { T \over T+2RC}.x(n-1) - { T-2RC \over T+2RC}.y(n-1)$$

It is clear that the output y(n) is in discrete domain and it is dependent on previous output y(n-1), present input x(n) and previous input x(n-1). So this recursive difference equation is to be implemented in the software.
Note that the constant terms R,C and T are to be determined from the beginning.
Here, I'm using R as 1kohms and C as 1uF, so putting the formula $$f_c = {1 \over 2\pi RC}$$ we get fc = 160 Hz. So cut off frequency of the low pass filter is 160Hz.
Let take sample time T as 100uS (always follow Nyquist sampling theorem to determine T). Means that input x(t) will be sampled at 10KHz to get x(n).
Therefore putting R,C and T values, we calculate the constants for y(n-1),x(n) and x(n-1) in the difference equation shown above.
$${ T \over T+2RC} = 0.047$$
$${ T-2RC \over T+2RC} = 0.9$$
So as the constants are determined, we are ready to implement the following difference equation in software. You may also like to define these constants in the software itself and let the microcontroller calculate it for you.

ESP32 has 12bit ADC to take inputs and 8bit DAC. We will use ADC to take sampled input, filter it using the difference equation and finally use the DAC to generate the low pass output and visulalize it on a DSO.
Now to generate the input, I have used Arduino to generate square waves of varying frequency (best will be to use a Function Generator) and used voltage divider to convert 5v signals into 3.3v signals, since ESP32 is a 3.3v device, and that signal is input to the ESP32 at I/O pin 34 (ADC). Output is obtained at I/O pin 25 (DAC).

Step 3: Code

I used Arduino IDE for writing the code in ESP32. Necessary comments are given beside each statement in the code.


/********************************************
Sketch by Watchdog
IIR Example
Created On: August 22, 2021

Cutoff Frequency of Low Pass Filter 160Hz
Sampling Time T = 100uS
R = 1kohm , C = 1uF
********************************************/
#define DACOUT 25
#define ADCIN 34
void setup() {
  Serial.begin(115200); //debug purposes
  // put your setup code here, to run once:
}

double output=0,outputPrev=0,voltagePrev=0;
void loop() {
  //put your main code here, to run repeatedly:
  
  //Take ADC input
  double adcval = analogRead(ADCIN);
  
  //convert it to Voltage readings
  double voltage = adcval / 4096;
  voltage *= 3.3;   

  //Determine output voltage
  output = 0.047*voltage + 0.047*voltagePrev + 0.9*outputPrev;
  
  //Serial.print(voltage); //debug statements
  //Serial.print(",");
  //Serial.println(output);
  
  //Scale it as per DAC Resolution and output to DAC
  double out = output*256;
  out /= 3.3;
  int DACoutput = (int)out;
  dacWrite(DACOUT,DACoutput);
  
  
  //Assign variables for next iteration
  outputPrev=output;
  voltagePrev=voltage;

  //sample time T
  delayMicroseconds(100); 
}

Step 4: Results

I attached DSO's one probe at input signal to ESP32 (ADC) [yellow trace] and other probe at DAC Out of ESP32 [violet trace]. Applied various frequencies (from arduino, you may use a function generator) at the ADC input and then observed the corresponding output.Results are as follows.

5Hz

50Hz

500Hz

You may note that this low pass filter acts as an integrator which integrates a square wave to a triangular wave.
Therefore our filter works as expected and performs the low pass function. In this way we can also design and implement high pass or band pass digital filters and filter out various sensor data.