# How Do You Calculate the Signal Gain of a Low Pass Filter?

• Bluestribute
In summary: AC reactances can be confusing without knowing how to relate voltage and current.In summary, the schematic shows a low-pass filter identical to the one you built in studio. This filter uses a 60 k resistor and a 250 pF capacitor, and the input signal is at 30 kHz. The signal gain is ?.
Bluestribute

## Homework Statement

The schematic shows a low-pass filter identical to the one you built in studio. This filter uses a 60 k
resistor and a 250 pF capacitor, and the input signal is at 30 kHz. (1 pF=10−12 F).
http://loncapa.mines.edu/res/csm/csmphyslib/P200_Materials/TestBank/AC_Circuits/LowPassFilter.jpg

What is the signal gain? Remember, gain means the ratio of the output voltage (the voltage across some indicated component) to the input voltage (the voltage of the power supply).

?

## The Attempt at a Solution

Nothing we've ever learned let's us compare voltage without knowing current. If it helps, the reactance is 2.12×104 Ohm. I can't use V=IZ, C=Q/V, everything equation we've been taught can not be used because we're missing information.

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There is enough information to solve the problem. You basically have a voltage divider using impedances rather than resistances. Are you familiar with the the manipulation of complex numbers?

I am absolutely not familiar with that. And a quick Google search just made me even more confused on applying that to this problem . . . Teach me please teacher.

Well, if you haven't been introduced to complex numbers then you're stuck with using reactance and the related math. You should have seen examples (in your lectures, notes, or text) where voltages are calculated for circuits containing resistance and reactance.

Oh boy, and that's where you're wrong! I actually don't know why I'm so excited at my professors not actually teaching me the necessary tools for the exam. Anywho, uh, no, we've never been taught how to use reactance to relate voltage without knowing current, and we've never been taught how to calculate current when given just the numbers required for reactance and a capacitance value.

Is there a way to relate my given quantities to voltage?

Hmm. Tricky. Have you been introduced to the concept of phase angle for current versus voltage for components with reactance (capacitors, inductors)? For example, with a capacitor the current waveform leads the voltage waveform by 90°.

Briefly. We haven't had to use it in calculations. It was more of a, "and this symbol is that phase angle" type of dealio. I have no doubt though they expect us to know everything about it though.

Okay, well also briefly then, the phase shift of 90° means that you can treat the voltages across resistances and reactances like components of a vector. The resistance represents the "x" component and the reactance the "y" component. You add them like you add vector components (square root of the sum of the squares).

When adding reactances, capacitive reactance is taken to be negative while inductive reactance is positive (this is related to how the voltage either lags or leads the current in the component). So if you have both capacitive and inductive reactances in series their sum is X = XL - XC. And with a resistor in series with those it would be ##\sqrt{R^2 + (XL - XC)^2}##.

In your problem you have only a capacitive reactance and a resistance forming a voltage divider. You should be able to write the expression for the voltage divider using the above.

So I tried 60,000^2 + 21200^2 and square rooted it to get 63635, using that above equation. Now this isn't the gain though (or at least according to the website)? It's also going to be positive 'cause I'm squaring it. I'll try to attach the picture since I don't see it up above. If not, it has a Vapp on the left side, R on top, C in between the left and right sides, and Vout on the right side (and R before C).http://loncapa.mines.edu/res/csm/csmphyslib/P200_Materials/TestBank/AC_Circuits/LowPassFilter.jpg
I mean, I think I'm missing something, since it wants the ratio from the outputted voltage to the inputted voltage.

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Your images don't display because the site referenced in the url requires authorization for access. Why don't you upload the pics (use the "UPLOAD A FILE" button) to the PF site.

Assuming that the image will display a voltage divider scenario, suppose that the two components were resistors (say R1 and R2). What would be the output voltage expression?

R1 + R2? So basically (reactance + resistance)^2 + (reactance)^2 then square rooted? And that's the output voltage? My ratio ended up being 1.32 . . . aka wrong =( (and 0.782, the voltages reversed, was also wrong).

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• LowPassFilter.jpg
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No no no. I'm asking you if you know what the voltage divider equation looks like for just resistors.

Oh. No. This is the first time I've ever heard anything about voltage dividers (and googling it didn't bring up many simple, easy to follow websites).

Bluestribute said:
Oh. No. This is the first time I've ever heard anything about voltage dividers (and googling it didn't bring up many simple, easy to follow websites).
I find it hard to fathom that you're being introduced to AC reactances without first being given the foundation of resistor circuits. You can't write the expression for Vout for the following circuit?

I've never seen that before in my life.

Maybe I'm jumping ahead (we still have like, 1-2 lectures before the exam depending on if ones a review), but yeah, we've never done anything like that.

Bluestribute said:
I've never seen that before in my life.

Maybe I'm jumping ahead (we still have like, 1-2 lectures before the exam depending on if ones a review), but yeah, we've never done anything like that.
Sounds like it'll be an intense lecture or two oo)

What parts of circuit theory have you covered? I'm assuming Ohm's law at least. How about Kirchhoff's laws?
I think you might want to ask your instructor about what's to be covered in the exam and whether voltage dividers and current dividers are expected to be already understood (they can both be understood from scratch by applying the basic Laws, but I would have thought they'd be old hat by the time AC theory is introduced).

Ohm's and Kirchoff's we've learned. Obviously Gauss and the electric field stuff as well. Maybe we did learn voltage dividers just we didn't formally call them that? How would you do this problem? We've already moved on to magnets so no more new circuit stuff . . . But they might just use different names.

I have no Vin, only a Z value and Xc value. I don't know how I'm supposed to do this without a voltage because every equation has a Vin component which I do NOT have. Is there anyway to calculate this without a Vin without having to dissect and rearrange twenty equations? This is only one part of multiple things they neglected to teach us and I need to learn all by Thursday while still going to work. I'd like to know the reasons behind how and why (seems like I could use this one day when I graduate), but I just need an equation at the moment that I could use with just a brief explanation so I'm not plugging in numbers blindly . . . If I had time and not so much pressure I'd do that dirty work myself . . .

In your first post you described a circuit which had a resistor and a capacitor and apparently an input signal. None of your images are viewable by people without a logon on whatever site they are hosted on. Upload your images to PF via the "UPLOAD A FILE" button.

If no voltage was specified for the source then just assign a variable to it. Call it Vin or whatever your choice is. Your results will then be symbolic and will contain that variable.

Going back to the purely resistive voltage divider that I depicted in post #14, how would you find the output voltage, Vout? You can use Kichhoff and Ohm's law.

Wow. This explains why this problem was angering me so much.

I was using all the information I've been given to get the correct answer. They just wouldn't accept 0.24, no, they needed 0.245 in order for it to be correct (I've done it multiple times with different number sets).

So this should have ended a while ago. You know, if only they would be consistent and either round ALL numbers or NO numbers (sometimes if you type like 2, and the answer is 2.463 you'll get it because eh, close enough. Not so in this case). Though the equation Vout/Vin = X/Z is VERY new to me.

So thank you to everybody.

Also, that open circuit looks like a "thevenin" circuit? I just came across this term trying to solve this problem (If a current I=5 A exists between points a and b, R1=7 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, R2=9 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, V1=16 V, and V2=15 V,the voltage difference Vb−Va is:).

Can you use Thevenin circuits (I hope I'm using this term right) to further explain the voltage dividers and help with this problem?

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• circuitSegment.gif
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Bluestribute said:
Also, that open circuit looks like a "thevenin" circuit? I just came across this term trying to solve this problem (If a current I=5 A exists between points a and b, R1=7 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, R2=9 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, V1=16 V, and V2=15 V,the voltage difference Vb−Va is:).

Can you use Thevenin circuits (I hope I'm using this term right) to further explain the voltage dividers and help with this problem?
You have another thread open with a very similar question in it (same circuit, slightly different component values). Please don't try to create a parallel thread for the same problem. That would go against forum rules. And you don't want that.

A Thevenin equivalent can be created for the voltage divider. The divider is a source with a network of components that feeds a load (not specified here, but is implied by the open terminals). So the source and network could be rendered into a Thevenin model.

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I didn't see that someone replied to that so tried to connect it to this, but yes, I do. Ah, gotcha

## 1. What is a low pass filter?

A low pass filter is an electronic circuit that allows low frequency signals to pass through while attenuating high frequency signals. It is commonly used to remove unwanted noise from a signal or to limit the bandwidth of a signal.

## 2. How does a low pass filter affect signal gain?

A low pass filter can decrease the signal gain by attenuating high frequency components of the signal. This can result in a smaller output signal compared to the input signal.

## 3. What is the cutoff frequency of a low pass filter?

The cutoff frequency of a low pass filter is the frequency at which the filter starts to attenuate the signal. It is typically defined as the frequency at which the signal is attenuated by 3 dB (half of its original amplitude).

## 4. How is the signal gain affected by the cutoff frequency?

The signal gain is inversely proportional to the cutoff frequency. This means that as the cutoff frequency decreases, the signal gain increases and vice versa.

## 5. How do I choose the right low pass filter for my application?

The choice of a low pass filter depends on the specific requirements of your application. Factors such as the desired cutoff frequency, signal gain, and input/output impedance should be considered when selecting a filter. It is also important to choose a filter with a suitable frequency response and attenuation characteristics for your signal.

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