Construct a low-pass filter and measure the gain

In summary, the low-pass filter has a gain of Vin/Vout at 20 Hz to 10 kHz. The half power point is located at about 900 Hz.
  • #1

Homework Statement


Construct a low-pass filter and measure the gain (Vin/Vout) for a range of frequencies 20 Hz to 10 kHz using an oscilloscope. Plot your data along with the theoretical curve. Use a semi-log graph. What is the half power point for this filter? Does it match the theoretical value based on the components used? (R = 2200 ohm, C = 0.12uF)

The Attempt at a Solution


I plotted the graph and it looks like the one I attached. But I am not too sure what the "half power point" is, or where to find it on my graph. Also how can I calculate a theoretical value for the half power point using my R and C?

Thanks!

PS. Sorry for not having labels on the graph, it's just a rough copy... Ratio of Vin/Vout is on the y-axis and frequency is on the x-axis.
 

Attachments

  • graph.jpg
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  • #2
Make the x-axis (frequency) a log scale (the problem did ask for a semi-log graph). For more insight, plot the power ratio in dB on the y-axis. That's 20*log(Vo/Vin).

The half power point is where the power available to the load on the output side of the filter is 1/2 the power available at the input. Suppose your load was purely resistive (i.e. there was a very large valued resistor connected across the output of the filter). What would be the power developed by that resistor in terms of the output voltage Vo?

To produce a theoretical curve or find any theoretical values for your filter, you're going to have to derive the expression for |Vo/Vin|. Think impedance.

FYI, the half-power frequency is also called the filter's cutoff frequency, or its -3dB frequency, or its bandwidth.
 
  • #3
Well I know that |Vout/Vin| = Xc/sqrt(R^2+Xc^2) where Xc= 1/(2pi*f*C)
would calculating that with all my frequencies give me the theoretical curve?

Also is the half power point always -3dB? Which would be on my graph ~900Hz?
 

Attachments

  • graph2.jpg
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  • #4
theuniverse said:
Well I know that |Vout/Vin| = Xc/sqrt(R^2+Xc^2) where Xc= 1/(2pi*f*C)
would calculating that with all my frequencies give me the theoretical curve?

It should.

Also is the half power point always -3dB? Which would be on my graph ~900Hz?

Power varies as the square of the voltage. So when you plot 20*log(x) you're also plotting 10*log(x2). In this case x is your voltage ratio. If the ratio x2 is 1/2, then 10*log(1/2) = ?

Also, you might want to invoke a few more grid lines for your frequency axis, the log scale can make it tricky to estimate a frequency with any accuracy. I think you'll find that your -3dB frequency is a tad lower than your 900Hz estimate.
 
  • #5
10*log(1/2) = -3.01 but I found the explanation a little bit confusing at this point "when you plot 20*log(x) you're also plotting 10*log(x2)".

I also plotted my theoretical curve using the formula from my previous post. At -3dB the frequency is close to the experimental values (I'll change the grid and find a more accurate result later on), does it look right?

Also, is the half power point actually measured in Hz (ie. frequency)? so that if I am being asked to find the hpp for a similar curve all I have to do is look for the -3dB (log graph) and its associated frequency?
 
  • #6
theuniverse said:
10*log(1/2) = -3.01 but I found the explanation a little bit confusing at this point "when you plot 20*log(x) you're also plotting 10*log(x2)".

I also plotted my theoretical curve using the formula from my previous post. At -3dB the frequency is close to the experimental values (I'll change the grid and find a more accurate result later on), does it look right?

Also, is the half power point actually measured in Hz (ie. frequency)? so that if I am being asked to find the hpp for a similar curve all I have to do is look for the -3dB (log graph) and its associated frequency?

log(ba) = b*log(a), so 20*log(x) = 10*log(x2).

10*log(x) is the basis of the dB, decibel. The "deci" means 10. 10 bels cover an order of magnitude (power of 10).

The half power point coincides with a certain frequency for a given filter. Yes, you can find it by looking at the power ratio graph (20*log() graph). Just be careful of the frequency axis; sometimes the angular frequency (ω) used.
 
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  • #7
I understand it now! Thank you so much gneill!
 

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. This means that it filters out or blocks high-frequency signals and only allows low-frequency signals to pass through.

2. How does a low-pass filter work?

A low-pass filter works by using a combination of resistors, capacitors, and inductors to create a frequency-dependent impedance. This impedance allows low-frequency signals to pass through with little resistance while attenuating high-frequency signals.

3. Why would you need to construct a low-pass filter?

A low-pass filter is often used in electronic circuits to remove unwanted high-frequency noise from a signal. This can improve the quality and clarity of the signal, making it easier to analyze or use in other electronic devices.

4. How do you measure the gain of a low-pass filter?

The gain of a low-pass filter can be measured by inputting a signal of known frequency and amplitude, and then measuring the output voltage. The gain is determined by dividing the output voltage by the input voltage. This process can be repeated for different frequencies to determine the frequency response of the filter.

5. Are there any limitations or drawbacks to using a low-pass filter?

One limitation of a low-pass filter is that it can introduce phase shifts in the signal, which can affect the accuracy of the output. Additionally, the filter may not completely eliminate all high-frequency noise, depending on the design and components used.

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