Designing an Integrator circuit

[Carl-NC]

Well, I suggested you post this question and then I forgot to answer!

The purpose of the integrator is to constructively add successive positive samples (resulting from a target) while averaging out everything else. That is, random noise (which is both positive and negative) will average to zero. So the integrator needs to have a charge rate that is fast enough (compared to the sample rate) to build up a signal, and a discharge rate that will get rid of the signal fairly quickly one the target is gone, so you don't get an audio response long past the target.

So let's make up some numbers. Sample rate is 1kHz, so sample period is 1ms. Sample width is 10us or 1%. Let's say a weak target signal from the preamp is 10mV. A single sweep of the coil takes 1 second.

We get 1000 pulses per swing, so as we go over a target we might get 10-20 positive samples. Let's say exactly 10. Without an integrator, this would result in 10 samples that are 10mV high with a 1% duty cycle.
This is a time-averaged signal of 100uV and probably will never trigger the audio.

Let's say we use a single-ended integrator shown below; R1=1k, R2=100k, C1=0.1uF. When the switch is closed vin (=10mV) is converted to a 10uA current (i1) by R1 and charges the cap. dv/dt = i/C1 which gives 10uA/0.1uF = 100V/s, so in 10us the cap charges up by a whopping 1mV. Wow, a whole millivolt? Yup.

When the switch opens the integrator is left with R2 discharging C1, at a time constant τ = R2*C1 = 10ms. We have 990us of discharge time (sample switch open) so the signal drops to e^(-990us/10ms) = 90% of whatever the sampled value is. So when the switch is closed, the cap charges by 1mV, and when the switch is open the cap discharges by 10%.

We assumed a 10mV signal for 10 samples, so at the end of each of the 10 samples we have:

1: 1.00mV
2: 1.90mV
3: 2.71mV
4: 3.44mV
5: 4.10mV
6: 4.68mV
7: 5.22mV
8: 5.69mV
9: 6.12mV
10: 6.51mV

6.5mV doesn't sound like much, but it's 65 times better than the average 100uV without the integrator. And the integrator is usually followed by a gain stage so if it has a gain of 100 (e.g., HH) then the final voltage applied to the audio is 650mV. That's much better.

Finally, when there is no more target signal, the integrator cap will discharge in about 3τ (95%) so a time constant of 10ms means the integrator output dies off in about 30ms.

- Carl

[1843]

Thank you for the reply,
Cheers

[Reg]

Hi 1843,

This is a good question to ask. I am not sure I can give you an ideal answer except to say, you need to try different values to see what happens.

Now, as Carl mentioned in an earlier post, the values of the different resistors, the capacitors, the sample time, and the pulse rate all determine the charge and discharge of the capacitors. This charge and discharge on the capacitor is what is averaged out and determines the signal heard.

Now, a person needs to look at the signal out of the integrator carefully. They will see a signal that is somewhat like a sawtooth that occurs at the rate of the pulse rate. Increase the pulse rate and one will see the sawtooth change.

Ok, what happens if you change values of the caps and the resistors? Well, one can go from a much smoother output to one that that changes quite dramatically. This will result in different effects, that can range from a sluggish detector to one that sounds like it is warbling. Responses to targets will also change, meaning the average gain will change as will the ability to easily recognize small or very weak targets.

In many cases, small changes are not easily recognized and it is difficult to see just what is happening. So, carefully testing needs to be done.

As I said, if the values are such that the time constants are long, then the signal may seem smoother, but small and weak targets may suffer. If the time constants are too short, then the signal may be raspy, or somewhat more harsh or noisy, which makes it harder to hear subtle changes also.

So, one simply needs to find the ideal combination that works for their needs.

In other words, the answer isn't as simple as one might think and the final solution may hinge on just what you want the detector to do.

Reg

[1843]

How I must select feedforward and feedback resistors and caps for best integrator performance?
(They depend on the PI frequency and sampling pulse width)

Please tell me its principle.

Best regards