2 j-terms in a transfer function?

In summary, you need to rationalize the denominator, calculate the magnitude of the transfer function, and plot the results.
  • #1
peasngravy
72
6
Homework Statement
Design and PSPICE model a bandpass filter of ‘Sallen and Key’ topology, having the following characteristics:
fo = 3.3khz
Q=2.2
Verify your design by means of a Bode plot.
Relevant Equations
V2/V1 = -1/ (2R1/R2) + jwCR1 + 1/(jwCR2)
I have the above transfer function for this filter design
1607593717528.png


The transfer function becomes
1607593746306.png

which then becomes
1607596767416.png

Now when I try to use this transfer function to plot a bode plot, I always end up with negative terms which i can't get the square root of

For example, at 1hz, where
w = 6.283
R1 = 1705
R2 = 33000
C = 6.43nF

I end up with

1607597103817.png

Is anyone able to explain where I am going wrong with this?

Thanks
 
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  • #2
At first, we must clarify the question about the structure (topology) of the filter.
In the text, a "Sallen-Key"-topology is mentioned, but the shown circuit represents another structure (Multi-feedback.) So - what is really your task?
 
  • #3
LvW said:
At first, we must clarify the question about the structure (topology) of the filter.
In the text, a "Sallen-Key"-topology is mentioned, but the shown circuit represents another structure (Multi-feedback.) So - what is really your task?

my task is to design a sallen key band pass filter with an fo value of around 3.3k and a Q factor of 2.2. I believe my main objective is not the design of the circuit itself but to come up with the capacitor and resistor values for those requirements

this topology appears in my course notes under the heading “sallen key bandpass filter”.
 
  • #4
In this case, your course notes are not correct. The Sallen-Key topology contains a fixed-gain opamp (very often with unity gain, voltage follower) and a second-order RC-network within the feedback path.
Such filters are non-inverting.
Are you required to derive the transfer function by yourself or can you use it as given in relevant books?

Hint: In case of a bandpass, a fixed gain of "2" has some advantages if compared with unity gain (much smaller component spread).

Comment/Correction: When you add in your diagram a resistor R3 in series to C2 , the opamp with the resistors R2 and R3 can be regarded as an INVERTING fixed gain amplifier. In this case, the whole circuit resembles a (less known) "inverting Sallen-Key architecture". So - you now have the choice between 3 alternatives:
* Multi-feedback (your diagram)
* Inverting Sallen-Key (your diagram plus R3)
* Non-inverting (classical) Sallen-Key (new circuit with fixed pos. gain).
 
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  • #5
A Bode plot plots gain in dB versus frequency, either angular (radians/sec) or Hz, your choice. So you don't just plot ##V_2/V_1##. Instead you want to plot:
$$U_{dB}(f) = 20 log\left| \frac{V_2}{V_1} \right| $$

Looking at your transfer function and comparing it to the given circuit it seems to me like you've swapped the two resistor values in the denominator terms.
 
  • #7
I propose to consult some other knowledge sources - and then you will see that the shown circuit is NOT a "Sallen-Key" type. I suppose, in the book, they have placed a wrong circuit diagram (by accident).

Have a look on Fig. 13.7 and Fig.13-12. Here you see how a Sallen-Key topolgy looks like (unity-gain amplifiers)
 
  • #8
I found this paper on designing them https://www.irjet.net/archives/V2/i5/IRJET-V2I5143.pdf

This may be very helpful to me but I need to fully understand the relationship between s=jw. Imaginary numbers are very new to me and have caused me quite a bit of trouble so far.
 
  • #9
That could very well be my issue - i am glad I know what I did wrong but unfortunately I only know I did wrong on the wrong circuit anyway :smile: :headbang:
 
  • #10
This paper looks good. And as far as I can see - they have selected a fixed gain of "2" (as I have proposed). So you can use this paper - and the shown design process - as a good reference for solving your task.
 
  • #11
Great - thank you for pointing out that this was incorrect or i'd have used the wrong circuit.

When I have the s term, how do I derive a value for this for a bode plot if it is based on an imaginary term?
 
  • #12
@LvW - I didn't realize you had added further comments to your post above, apologies for not acknowledging them
 
  • #13
peasngravy said:
Now when I try to use this transfer function to plot a bode plot, I always end up with negative terms which i can't get the square root of

For example, at 1hz, where
w = 6.283
R1 = 1705
R2 = 33000
C = 6.43nF

I end up with

View attachment 274059
Is anyone able to explain where I am going wrong with this?

Thanks

You don't plot the transfer function itself, which is a complex value at each frequency--you plot the magnitude of the transfer function.

If you're going to keep doing this sort of stuff you really should acquire a facility with doing complex arithmetic.

Your transfer function is a fraction with a denominator that has an imaginary part. To evaluate the magnitude of the transfer function you need to get the imaginary parts out of the denominator. This is done by rationalizing the transfer function.

See: http://www.solving-math-problems.com/algebra-rationalize-denominator-with-complex-numbers.html

Then you need to calculate the magnitude of the transfer function at each frequency and plot that magnitude. If you want to plot db you take the base 10 logarithm and multiply by 20. Strictly speaking this would not be a Bode plot, but rather a frequency response plot. See the following links for more about a Bode plot.

See here (also part way through deals with "But wait! We still have j in the equation"): https://blanco.io/education/principles-of-ee-2/how-to-bode-plot/

Additional discussion:

http://faculty.mercer.edu/jenkins_he/documents/bodeplots1.pdf

https://lpsa.swarthmore.edu/Bode/BodeRules.html
 
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  • #14
Last edited:
  • #15
@The Electrician - thank you for that information.

I got there now, i am quite happy with my result
1607685967730.png


I also realized I was trying to resolve the magnitude of the complex numbers instead of just using the excel functions available (i don't have any software like MATLAB or mathcad whch would've made this far easier)
 

1. What is a transfer function?

A transfer function is a mathematical representation of the relationship between the input and output of a system. It is used to describe the behavior of a system in terms of its frequency and time domain characteristics.

2. What is a j-term in a transfer function?

A j-term in a transfer function refers to the imaginary part of a complex number. It is typically represented by the letter "j" and is used to describe the frequency response of a system in the frequency domain.

3. Why are there 2 j-terms in a transfer function?

There are 2 j-terms in a transfer function because most systems have both a real and imaginary component in their frequency response. The first j-term represents the imaginary part of the numerator, while the second j-term represents the imaginary part of the denominator.

4. How do j-terms affect the behavior of a transfer function?

J-terms affect the behavior of a transfer function by modifying the amplitude and phase of the output signal. The magnitude of the j-term determines the amount of amplification or attenuation of the signal, while the angle of the j-term determines the phase shift of the signal.

5. How do I interpret the j-terms in a transfer function?

To interpret the j-terms in a transfer function, you need to convert them into polar form. The magnitude of the j-term represents the distance from the origin on the complex plane, while the angle represents the direction. This can help you understand how the system will respond to different frequencies.

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