Introduction To Filters

Filtering Pulse-Like Noise

Why Worry About Pulse Noise?

There are often situations in which you have noise that looks like short random pulses.

An example of pulse-like noise might come about when you are trying to measure pressure in a tank while you are filling the tank with liquid. If the pump motor is a DC motor, it will have a commutator which produces a small noise pulse every time the brush moves over another section of the commutator.

When you have a signal that is corrupted with pulse-like noise, you can often mitigate the effects of the pulse noise by using a low pass filter. A low-pass filter is a filter that passes low frequency signals and selectively attenuates higher frequency signals. (Or, in other words, it makes the higher frequency components of the signal smaller relative to the lower frequency components.)

The reason you need a low-pass filter is that short pulses are exceptionally rich in high frequency components, and a low pass filter can be used to filter out those high frequency components. However, to understand that approach you need to know about Fourier Series.

The simplest low-pass filter is an RC circuit.

In another lesson, we found the following information for this circuit.

The circuit satisfies a differential equation:

RC(dV_out(t)/dt) + V_out(t) = V_in(t)

If the input voltage is a pulse, then we can use the differential equation and what we know of the solution to the differential equation to help us figure out what happens in this circuit when a short pulse comes along.

In another lesson, we found the solution for the differential equation when there is a step input to a first order system of this nature. If we want to figure out how the circuit responds to a short pulse we can get there by applying what we know about the solution to what happens when the system has a step input. Let's imagine the following situation.

Let us assume that we have a sensor with a 1.0 volt output.
Let us assume that we have a short pulse that comes on top of the sensor output signal.
Let us assume that the sum of those two signals is applied to the input of the filter circuit above.

We can start the analysis by examining what happens when the pulse occurs. To figure that out we being by looking at a step input added to the 1.0 volt sensor signal, and examining the response of the filter circuit. Here is the input we imagine.

In this picture, the input is the dotted blue line, and a step of 0.5 volts occurs on top of a steady input of 0.1 volts. We can move from this situation to a pulse by bringing input back to 1.0 a short time after the step is applied. In the meantime, we see that as long as the step exists, it drives the output towards 1.5.

Now, we can examine what happens when the step drops back to zero forming a pulse. We expect the response to look exactly the same until the pulse drops back. Here is what we get.

That's not totally satisfying. We can expand the vertical scale to see some details of what happens.

You can see that, when the pulse comes along, the pulse is so short that the response only begins to rise. If the time constant of the filter is long enough, the pulse doesn't produce much change in the output. However, that small amount of change does take a while to dissipate. Even so, the peak of the noise is attenuated. Let's see if we can compute how the peak of the noise is attenuated.

During the pulse the output signal rises. (See the plot above.) The rise in the output is almost a straight line. It is really the start of a rising exponential, but it looks like it could be well approximated with a straight line. (We wanted to show you it was the start of a rising exponential, and that's why we used a step first and only later looked at the exponential!)

We can separate the response to the step/pulse from the response to the constant value of the inpout.

If we have a step input to a system, and

The magnitude of the step is DV,
Then the derivative of the output (initially) is DV/t.

If the width of the pulse is DT,

The total change in voltage is DTDV/t.

We can conclude:

DV is the size of the input pulse.
DVDT/t is the size of the output pulse.
Using height of the pulse input and height of the output change due to the pulse input, the height is reduced by a factor of DT/t.
The reduction factor is the ratio of the pulse width to the time constant. To get good reduction, the time constant should be much larger than the pulse width.

Now, we can step back and reflect on what happens in this filter because there is something a little more general going on here, and we can take advantage of what is happening to get an idea of what will happen when the pulse is not a "square" pulse. Here is the filter circuit again. We will refer to it below.

Consider the following.

We want the effect of the pulse on the output to be small. Let's just assume that will happen, and examine the consequences. Oh, and we will examine just what happens because of the pulse and neglect any other signals in the circuit. (Since this is a linear system, we can examine each signal separately and add together the effects of all the signals, and we are only interested in the effect of the noise anyway.
Once we assume that the pulse is short, and that the output is small, we can assume that the capacitor never charges every much, and the current flowing through the resistor is just V_noise/R.
If that is the current, then the rate of change of voltage across the capacitor is:

dV_out/dt = V_noise/RC.
or, dV_out = (V_noise/RC)dt

So, we can see, in general, that the change in the output voltage is given by:

DV_out = Area under the noise pulse/RC

In general, the "blip" in the output due to a noise pulse is just the area under the noise pulse divided by the filter time constant as long as the pulse is short compared to the time constant.

Example

Say that you have noise pulses that are exponential. Here is a typical noise pulse for this situation.

This particular noise pulse has the form of Ae^-t/^t1. Doing the integral indicated above, the area of this pulse is:

or Pulse Area = At₁

Of course this pulse exists indefinitely even though it gets smaller and smaller. However, for all practical puroses, the pulse dies out in around 5 time constants (of the pulse, i.e. t₁). Then, as long as 5t₁ is much longer than the time constant of the filter, this approximation will be OK.

Summary

If there is pulse noise in a signal, a filter will reduce the effect of a pulse by a factor of V_pDT/t. Stating that more specifically, we have:

If the input pulse is of height V_p, then the output pulse will have an amplitude of A/t where:

A is the area of the input pulse.
tis the time constant of the filter.
The filter has a gain of 1.0
And, for this to be true, the input pulse has to be short (in time duration) compared to the time constant of the filter.

Certain simple pulses have areas that can be computed easily.

The area of a rectangular pulse is V_pDT.

V_p is the height of the pulse.
DT is the width of the pulse

The area of a a pulse that is a decaying exponential is At₁.

A is the height of the pulse
t₁ is the time constant of the decaying exponential.