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i was doing some homework and one thing I had to do was to derivate the transfer function of a given bandPass. I have done it myself but i dont know if it is correct nor if i followed the right path / right assumptions. I'd be happy if someone of you could look at it and give me some feedback. Important: In my notes Ua = V(out) and Ue = V(in) the rest should be (I Hope) self explanatory.This is how I've done it: enter image description here And this is the bandpass: enter image description here

Thank you in advance!

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  • \$\begingroup\$ I suggest using laplace variable s=jw as it's more common and easier to do basic checks. For example, if all coefficients of the denominator are positive we know that the poles are on the left hand side (basic check). Then you should run all other basic sanity checks, for example: if R1=0, you should have only a high-pass. If C1 = 0, you should have a high-pass + voltage divider, etc etc. Do that first, and see if anything looks fishy \$\endgroup\$ – Andrés Jul 12 '18 at 12:37
  • \$\begingroup\$ And make sure if you write "C" it means either C1 or C2. Other than that it's a sloppy mess and you should consider writing it in Latex rather than posting an easily ambiguously-read photo of a page from your book. \$\endgroup\$ – Andy aka Jul 12 '18 at 12:47
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I have done it myself but i dont know if it is correct nor if i followed the right path / right assumptions.

Yes, it appears to be correct and you have followed the right path. Apart from the number "1" missing on the first line - you put \$j\omega C\$ rather than \$j\omega C_1\$. You also omitted brackets around the top line numerator but corrected that in the denominator.

I derived it my way (to avoid me being influenced) and got the same answer.

I'd be happy if someone of you could look at it and give me some feedback.

Feel Happy!

Here's my scrappy mess: -

enter image description here

This is why you should take the time and use Latex!! The very bottom line is me just checking the j terms match yours.

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  • \$\begingroup\$ Oh I'm sorry, I didnt notice that I forgot to name the C's in the beginning. Ok I will! Thanks alot for your time, your answer made my day! \$\endgroup\$ – chaBing Jul 12 '18 at 15:24
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These answers are obviously correct but they represent the brute-force approach. Nothing wrong here but considering all these lines, it is likely that mistakes or typos get in the way of the correct answer. Furthermore, the final expression does not give insight in case a design goal exists: what is the mid-band gain?

An easier approach consists of using the fast analytical techniques or FACTs. Just draw little sketches and read the resistance driving capacitors \$C_1\$ and \$C_2\$ to determine the time constants of the circuit. The below drawing shows the steps, you can't beat this in terms of simplicity : )

enter image description here

Once you have the time constants, assemble them following the below Mathcad file and you obtain the transfer function. Unveil the resonant frequency and the quality factor, rearrange and you have a transfer function in which the mid-band gain clearly appears.

enter image description here

Finally, the dynamic response is given below, confirming the 12-dB attenuation in the flat magnitude portion:

enter image description here

Determining a transfer function is one thing and many paths lead to the correct expression. However, what matters at the end is to shape the expression into a meaningful form that you can exploit to fulfill your design goals. The FACTs are truly unbeatable in terms of execution speed and the delivery of low-entropy expressions. Vive les FACTs ! : )

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