Why is aliasing important to A/D converters?

To minimize aliasing an anti-aliasing filter is needed before an A/D converter. How do you design this pre-filter? What is this aliasing? Shannon, and Nyquist later, said that the maximum signal we can reconstruct is half of the sampling rate. If the signal we sample has components higher than 1/2 sampling rate then we get a phenomena called 'aliasing. If I were to send an aliased signal back out of a computer to a D/A converter it would look like the signal was 'shifted' in frequency.

Suppose a signal of F_s was put into an A/D converter and the sampling rate was F_s. It might appear on a D/A converter as a DC signal .

As an everyday example of aliasing, have you ever seen a moving fan appear still under a strobe light? When the strobe is blinking at the same rpm's as the fans rotational speed, the fan will appear motionless. The aliasing phenomena can occur if the sampled signal is at or above 1/2 the sampling rate OR IF IT HAS (Fourier) COMPONENTS AT OR ABOVE THIS RATE. In this case just those components above the Nyquist rate would be aliased, but the total signal, if reconstructed, will be distorted.

How do we design to this? Typically we use an anti-aliasing filter on the front end of the A/D. This filter is designed to attenuate signal above the Nyquist rate to below some noise level. It might be as simple as an RC or more complicated as a multi-pole active filter in a Chebychev or Butterworth configuration. The key is to understand how much attenuation we need at the stop band. Suppose our A/D is 12 bits. The ideal SNR is 72dB (= 20* log(4096)). The filter should be designed to provide 72 db of attenuation at the stop band. Now we note that a single pole filter provides 6db per octave of attenuation.

A single pole low pass filter designed to attenuate a signal to this noise specification would be designed to have a cutoff 12 octaves below half the A/D sampling rate!

How to find a short on a PCB

When I am troubleshooting new PCB's I have a unique technique for finding shorts on PCB's. I actually have to hand the credit for this technique to a business partner, Dr. Ed Kafrissen. He came up with this idea.  So let me set up a scenario for you, and then show you Ed's solution. Suppose you have a client and you need to deliver 100 PCB's to him/her quickly. You have someone stuff the boards for you, now you have to verify they work.

One of the first tests we would do is to apply power to the new PCB and bring the voltage up slowly. Perhaps without any chips installed. If the power supplies load down, then you got a problem to solve. You got a short on the PCB.

Ed came up with simple and unique solution. He would current limit a power supply to some small current, and at the rated voltage of the pcb. Ed would then hook up the PCB, which immediately loaded down the current limited supply. He would turn up the current on the supply as he felt on the PCB for a hot spot! Invariably some IC would be stuffed in backwards that our visual inspection did not find, or we would get a PCB which was not completely etched.  Alternatively, you can use a handheld IR imaging thermometer to pinpoint the hot-spot.


Anyway, that was Ed's solution to quickly find shorts on a PCB...and it worked well as long as you were careful enough to not burn yourself! Yeow!

The Exponential Digital Lowpass Filter

In the last lab note we looked at the boxcar filter structure. That was an example of an FIR filter. The exponential filter is an IIR filter.  If I need a little filtering I usually use this filter structure. In a simple implementation of this filter we calculate the filter output by taking the current sample, adding it to the previous output then dividing by 2. Mathematically y=(x+yz^-1) ; where y is the current output, x is the current input, and yz^-1is the previous output.  The z^-1 term shifts back in time 1 sample period.  Since the output is a function of the previous output this is an Infinite Impulse Response filter, or IIR filter.

In a typical embedded system we work with a fixed point math.

For the above function the characteristic equation is H(z) = Y/X = 1/(2-z^-1)= z/(2z^-1); There is a zero at 0 and a single pole at z=1/2. This is within the unit circle, thus the filter is stable. To plot the frequency response we set z=e^jωT. Then we normalize by setting T=1. We can use Wolfram Alpha to plot the equation H(ω) = 1/(2-e^-jω) . Click this link to see plot.  I copied the plot here for convenience.

Exponential Lowpass Filter response.

Here the horizontal axis is in Hz This filter attenuates to 0.7 at a frequency of 0.14Hz (remember this equation is normalized to a sampling frequency of 1 Hz).

If this is too much filtering, or too little filtering you can try adjusting the weighting of the two terms in the exponential filter. For instance, what if you choose the weighting y = ¼ (3x + yz^-2)?  Here I tried to use a power of 2 divisor to make the math dead nuts simple to implement on a fixed point processor. On an FPGA you would not even need to divide or shift, just pick off the right bits. The FPGA would need just two 'add' operations. Can you calculate the frequency domain function and plot it?

If you have any other common filter questions, leave a comment for future discussion.