Bode plot for active low pass filters

In summary: So the question is trying to ask you to look at a graph of a simple low pass circuit and see what is important. If you take a look at a graph of a simple low pass circuit, you'll see that the most important thing is the frequency at which the gain drops by 3dB - this is the "critical frequency". Beyond this frequency, the gain remains constant (or rises slightly) and all the other frequencies are just noise.
  • #1
peasngravy
72
6
Homework Statement
Derive a transfer function for each of the filter circuits and sketch a bode plot
Relevant Equations
Av = Vout/Vin
Hi

Can anyone help with this? I understand what the bode plots should look like but I am not sure how to produce a plot in excel for them without any values

The transfer functions are:

Vout/Vin = 1/(SQRT(1+(ω/ωc)^2)
and
Vout/Vin = -G/(1+j(ω/ωc))

As both are first order filters they will have a roll-off of 20db per octave
1607457378973.png1607460211644.png

Thanks!

edited to add transfer function screenshot (still no good with Latex :) )
 
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  • #2
peasngravy said:
As both are first order filters they will have a roll-off of 20db per octave decade
:smile:

What kind of plots have you done in Excel before? Also, have you downloaded a free copy of LTSpice yet? That can be a good way to check your work, plus it's a good skill to learn to be comfortable simulating circuits in Spice.
 
  • #3
peasngravy said:
I am not sure how to produce a plot in excel for them without any values
You need to select a number of appropriate frequencies and calculate the gain (in dB) and the phase (typically degrees) for each frequency. Plot the results using linear (gain, phase, vertical axis) and log (frequency, horizontal axis) to show the "straight lines". Be sure to select frequencies above and below the 3dB frequencies - say plus and minus two decades as below..

As the frequency is plotted on a log axis it is useful to choose, say, four "logarithmically-equally-spaced" frequencies per decade as 1.78, 3.16, 5.62 and 10 Hz.

See Bode Plot in wiki.

If you must use Excel try a Google search with bode plot excel

bode.png
 
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  • #4
It is common, when values aren't given, to plot these after normalizing the frequency. You can define the normalized frequency ωn = ω/ωc, then you generate values of ωn around 1 to plot the filter response. Then label your frequency axis "ω/ωc".
 
  • #5
Do I just make up values for the resistors and capacitors though? I don't understand how to calculate the gain at a certain frequency without replacing the values in the transfer function with actual numbers?
 
  • #6
You have the transfer functions which you need to plot.

You don't need R1 and R2 because you are told G = R1/R2. G is therefore a scaling factor on the Gain axis - label it 1G, 2G, 3G etc.

You don't need C because you have ω/ωc. = see DaveE's post.

It says "sketch a bode plot" so do it by hand labelling the axes to show the 3dB point. You don't need Excel.

The question is testing your understanding of the shape of a Bode plot and the importance of the 3dB frequency.

Every circuit with a 3dB breakpoint and 20 dB/decade fall off has the identical shape Bode plot.

It is only the markings on the axis which change with different circuits. One circuit, say, has a 3dB point at 10kHz and a gain of 35dB; another circuit, say, has a 3dB point of 27MHz and a gain of 19dB. The Bode plots are identical - only the markings on the axes are different.

So the question is solved by

a) sketch a generic Bode plot - without axis markings they are all identical
b) now mark the axes with the graduations for the specific circuit

In my example plots, replace 100 by ωc, 1000 by 10ωc and 10000 by 100ωc.
 
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  • #7
Ok thank you - I had actually already sketched it by hand and asked my lecturer if this was enough, but he didn't reply and I assumed that meant he was unhappy with the question :D

I will go with the hand sketch as I know how they should look anyway. Thanks everyone for your time.
 
  • #8
peasngravy said:
asked my lecturer if this was enough, but he didn't reply and I assumed that meant he was unhappy with the question
Not a great assumption. He may have just been occupied with other things. Give the guy a break, he has other things to do too.
 
  • #9
Yeah maybe but this was 2 weeks ago :biggrin:

No big deal anyway and it's not like I harassed him or anything, just asked him and moved on when he didn't reply.
 
  • Like
Likes DaveE
  • #10
peasngravy said:
I will go with the hand sketch as I know how they should look anyway.
Please let us know how you get on so that if my advice of sketching was bad :frown: I don't give it again.

In principle, I think what the question is aiming at is trying to get you to see what is important and what isn't. A simple low pass filter is characterised by:

- a flat gain response up to a given critical frequency
- a straight line (on a log/linear graph) falling at 20 dB/decade beyond the critical frequency (so be sure to plot frequency on a log scale!)
- a 3dB drop in gain at the critical frequency
- a 45 degree phase shift at the critical frequency.

Hence the circuit is completely characterised by only two things: the critical frequency and the gain.
 
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1. What is a Bode plot for active low pass filters?

A Bode plot is a graph that represents the frequency response of a system. In the case of active low pass filters, it shows how the output voltage of the filter changes in relation to the input frequency.

2. How is a Bode plot for active low pass filters different from a passive low pass filter?

An active low pass filter uses an operational amplifier to amplify and filter the input signal, while a passive low pass filter uses only passive components like resistors and capacitors. This results in a steeper cutoff slope and a higher gain in the passband for active low pass filters.

3. What information can be obtained from a Bode plot for active low pass filters?

A Bode plot can provide information about the gain, phase shift, and bandwidth of the active low pass filter. It can also show the location of the cutoff frequency and the filter's frequency response characteristics.

4. How is a Bode plot for active low pass filters useful in circuit design?

By analyzing the Bode plot, engineers can determine the filter's performance and make adjustments to optimize its frequency response. It can also help identify any potential stability issues in the circuit.

5. How can the Bode plot for active low pass filters be used in troubleshooting a circuit?

If there are unexpected changes in the circuit's frequency response, the Bode plot can be used to identify the cause. For example, if there is a sudden change in the gain at a certain frequency, it could indicate a faulty component or incorrect circuit design.

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