The Mystery of Low-Pass Filters in Lock-in Amplifiers

In summary, the low-pass filter in a lock-in amplifier acts as an "integrator" by removing high-frequency components and integrating the signal over time. This is achieved through the use of a simple RC circuit with a transfer function of 1/(1+JwRC). A differentiator can also be used to eliminate DC noise, with a transfer function of JwRC/(JwRC+1). By selecting the appropriate break frequency and resistor and capacitor values, the desired filtering effect can be achieved. Understanding the concept of transfer functions is key in understanding filters.
  • #1
Niles
1,866
0
Hi

I am reading about lock-in amplifiers, and I read that the low-pass filter in a lock-in amplifier acts as an "integrator", since much of the noise that ends up in the DC output of the multiplier will integrate to zero. How is it that the low-pass filter can do that?

I would appreicate any help, since I can't find any other source where this is stated.


Niles.
 
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  • #2
Yes, even though a filter can do things an integrator can't do and an integrator can do things a filter can't do, there is some overlap. You might think of a simple low pass filter as a lossy integrator.
 
  • #3
But how does it integrate the signal? I thought it only removed high-frequency components.
 
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  • #4
Niles said:
But how does it integrate the signal? I thought it only removed high-frequency components.

Would you consider a low pass filter as a moving averager of the input signal?

Say you apply a sine wave input of any frequency with a DC offset, and then at your output, if the frequency is greater than the cutoff frequency, you should end up with something close to the DC offset, which is the average of the sine wave.

Similarly, consider an integration

[itex]\frac{1}{T} \int_{0}^{T} f(t)dt[/itex]

This is the average of the signal over the period of integration.

I think a good way to accept that its integrating is understanding that the capacitor is acting as a storage of past inputs. As the signal changes over time, it take or adds charge from the capacitor, much as an integrator adds and subtracts infitesmal changes in values of the function over time (or whatever you're integrating with respect to).
 
  • #5
DragonPetter said:
Say you apply a sine wave input of any frequency with a DC offset, and then at your output, if the frequency is greater than the cutoff frequency, you should end up with something close to the DC offset, which is the average of the sine wave.

That I accept, but say that the DC-term is given by [itex]A\cos(\phi)[/itex], where A is the DC-constant and [itex]\phi[/itex] is some time-varying phase due to, e.g., some noise in the system. Then your explanation doesn't work.
 
  • #6
The response function of a low-pass filter has poles, and it multiplies the signal spectrum (multiplication in the frequency domain is of course the same as convolution in the time domain of the signal with the filter impulse response). A look at your Laplace transforms will verify that poles like 1/s correspond to integration.

More intuitively, an RC network is the simplest lowpass (it's a one pole filter). The output is the voltage across the capacitor which is proportional to its charge, which equals the integral of the input signal current. The RC integrates.
 
  • #7
Niles said:
That I accept, but say that the DC-term is given by [itex]A\cos(\phi)[/itex], where A is the DC-constant and [itex]\phi[/itex] is some time-varying phase due to, e.g., some noise in the system. Then your explanation doesn't work.

[itex]A\cos(\phi)[/itex] is not a DC term like you say. A is not a DC constant either, its the amplitude of your AC term.

[itex]A\cos(\phi) + B[/itex], where B represents the DC term, and A represents the amplitude of your time varying system noise, would be the right way to describe that.
 
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  • #8
Thanks to all for replying.Niles.
 
  • #9
need to edit: had an error
 
  • #10
If you are trying to remove low frequencies...you will need a differenator.

If you are trying to remove high frequencies...you will need a integrator.

Like has been said...a simple RC circuit will work. A resistor in series with a capacitor.

If you want a low pass filter...take the output across the capacitor.
If you want a high pass filter...take the output across the resistor.

In regards to how this happens...you simply take a voltage divider of each one and find it's TRANSFER FUNCTION.

I assume you know the reactance of a capacitor is 1/JwC.

Using simple voltage dividers for each...

Low pass filter transfer function: 1/(1+JwRC)

W= the frequency...or more specifically...2*pi*f.

So as you can see...if you plug in "0" for w...you get "1" for your transfer function. In other words...it will let DC values through 100%

If you plug in infinity for w looking at the other end of the spectrum...you will get zero output or gain.. So it attentuates all high frequencies and "passes" all low frequencies.

The differentiator is just the opposite. Take the voltage across the resistor...voltage divider...

Transfer function is JwRC/(JwRC+1)

Again...when w = 0...or DC...this case we get a output of zero...
When w = infinitey...we get an output or gain of one. It passes all high frequencies...hence high pass filter.

So if you are trying to eliminate DC noise...a differentiator is what you want.

Also...you can select your break frequency. The frequency of interest you want the filter to operate.

Break frequency = 1/(2*pi*RC)

Selector a break frequency you want...select a resistor size you want...and solve for C...or the capacitor size.

This leads us into bode plots, zeros and poles, s domain, differential equations...etc.

Pretty cool stuff actually?

Any questions?
 
  • #11
Thanks for that detailed explanation. I have to admit that I didn't understand 100% of it, but that is mainly due to my lack of knowledge of EE. I will have to brush up some of the terms you mention, but I will certainly do it in the near future.

Thanks for taking the time to describe it in such detail.Niles.
 
  • #12
Niles said:
Thanks for that detailed explanation. I have to admit that I didn't understand 100% of it, but that is mainly due to my lack of knowledge of EE. I will have to brush up some of the terms you mention, but I will certainly do it in the near future.

Thanks for taking the time to describe it in such detail.


Niles.

No problem.

The trick to filters is what my old hardcore electronics teacher used to ask.

WHAT'S THE TRANSFER FUNCTION?

You plug in the particular frequency you want...and find your gain. Simple as that.
Although I understand you will have to struggle a bit do to your experience. But plugging in values for when "w" equals zero (DC)...and when "w" equals infinity is a great way to get a feeling for a filter.

And incidentally...when your "w" = 1/RC...this when you "break frequency" will happen. Plug in the 1/RC for "w" in any of the above formulas I wrote and you will get 1/1+J for the low pass filter for example. If you do the math...you will see that the gain is .7 at phase angle 45 degrees. This is why the break frequency drops approximately 3 db at the break.

To calculate gain...20 log (.7)= -3 db! The 45 degrees will coincide exactly with your bodi plot.

Also, in a low pass filter...after the break frequency, the gain will drop 20 db for every decade you go up in frequency. In other words...if your break is at 100...the next 20 db drop is 1,000...then 10,000, then 100,000. It's all in the math...try it out.
 
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  • #13
Ok, I corrected what I was going to say.

I hope this doesn't confuse you, but if you choose the proper integration constants, which are what the time constant in an RC filter corresponds to, you can see how your above example (after corrected) integrated will have the same effect as a low pass filter.

First, your cosine AC term phase variable[itex]A\cos(\phi)[/itex] needs to be modified to show the frequency (your example gives a frequency of 1) and time dependence. So let's say [itex]A\cos(w*t)[/itex] instead, where w is the frequency of the noise signal and t is the time. RC is the time constant of the filter, and so it represents how long you are integrating for . . for example a small RC will fill up fast and so it is a smaller time that you integrate for.

Now we can show how integration can filter out higher frequency signals:

[itex]\frac{1}{RC} \int_{0}^{RC}(A\cos(w*t) + B) dt[/itex]

if we integrate this we get

[itex]\frac{1}{RC} A (\frac{sin(w*RC)}{w}) + \frac{1}{RC}*RC*B[/itex]

=

[itex]A*(\frac{sin(w*RC)}{w*RC}) + B[/itex]

for [itex] w*RC >> 1 [/itex], [itex]A*(\frac{sin(w*RC)}{w*RC}) + B \approx B[/itex]


So, looking at that equation, we see no matter what RC is, the DC component B will be preserved completely. But, the noise term is a function of RC and its frequency w. The sine will always be between 0 and 1, and if we let RC be really big, which corresponds to a really low cut off frequency, or if we if we say the frequency of the noise is very high relative to RC, the denominator gets very big and the noise sine term approximates to 0, leaving only the DC component.

So if you followed that, you can see how an averaging integration operation can act like a low pass filter (and vice versa, a low pass filter acts like an averaging integration operation).
 
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  • #14
DragonPetter said:
Ok, I corrected what I was going to say.

I hope this doesn't confuse you, but if you choose the proper integration constants, which are what the time constant in an RC filter corresponds to, you can see how your above example (after corrected) integrated will have the same effect as a low pass filter.

First, your cosine AC term phase variable[itex]A\cos(\phi)[/itex] needs to be modified to show the frequency (your example gives a frequency of 1) and time dependence. So let's say [itex]A\cos(w*t)[/itex] instead, where w is the frequency of the noise signal and t is the time. RC is the time constant of the filter, and so it represents how long you are integrating for . . for example a small RC will fill up fast and so it is a smaller time that you integrate for.

Now we can show how integration can filter out higher frequency signals:

[itex]\frac{1}{RC} \int_{0}^{RC}(A\cos(w*t) + B) dt[/itex]

if we integrate this we get

[itex]\frac{1}{RC} A (\frac{sin(w*RC)}{w}) + \frac{1}{RC}*RC*B)[/itex]

=

[itex]A*(\frac{sin(w*RC)}{w*RC}) + B[/itex]

So, looking at that equation, we see no matter what RC is, the DC component B will be preserved completely. But, the noise term is a function of RC and its frequency w. The sine will always be between 0 and 1, and if we let RC be really big, which corresponds to a really low cut off frequency, or if we if we say the frequency of the noise is very high relative to RC, the denominator gets very big and the noise sine term approximates to 0.

So if you followed that, you can see how an averaging integration operation can act like a low pass filter.

Wow...that's totally cool...all your formulas and so forth. I'm pretty sure we are doing and talking about the exact same thing...but clearly we had different teachers. Two different ways to get the same result. Amazing.
 
  • #15
psparky said:
Wow...that's totally cool...all your formulas and so forth. I'm pretty sure we are doing and talking about the exact same thing...but clearly we had different teachers. Two different ways to get the same result. Amazing.

I hope its correct, I'm having trouble trying to figure out an equivalent example for high pass and derivative that includes the RC constant.
 

Related to The Mystery of Low-Pass Filters in Lock-in Amplifiers

1. What is a low-pass filter?

A low-pass filter is a type of electronic circuit that allows low-frequency signals to pass through while blocking or attenuating high-frequency signals. It is commonly used in lock-in amplifiers to remove noise and unwanted signals from the measured signal.

2. How does a low-pass filter work in a lock-in amplifier?

In a lock-in amplifier, a low-pass filter is used to remove high-frequency noise from the measured signal. This is achieved by passing the signal through a series of electronic components, such as resistors and capacitors, that selectively attenuate high-frequency components of the signal.

3. What is the purpose of using a low-pass filter in a lock-in amplifier?

The main purpose of using a low-pass filter in a lock-in amplifier is to improve the signal-to-noise ratio of the measured signal. By removing unwanted high-frequency noise, the signal can be amplified and measured more accurately.

4. How do I choose the right cut-off frequency for a low-pass filter in a lock-in amplifier?

The cut-off frequency of a low-pass filter is the frequency at which the signal starts to be attenuated. To choose the right cut-off frequency for a lock-in amplifier, you should consider the frequency range of the signal you want to measure and the frequency range of the noise you want to remove. Generally, the cut-off frequency should be set slightly above the highest frequency of the signal to be measured.

5. Can I adjust the cut-off frequency of a low-pass filter in a lock-in amplifier?

Yes, most lock-in amplifiers allow you to adjust the cut-off frequency of the low-pass filter. This can be done manually by changing the values of the electronic components in the filter circuit or by using software to control the cut-off frequency. Adjusting the cut-off frequency can help optimize the filter for different signals and noise levels.

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