# Preemphasis circuit, calculating H(s) and Bode plot

I have the following circuit: (which should represent a "new" design for preemphasis block in a FM transmitter circuit)

I need to calculate the transfer function H(s) and draw a Bode plot. My question is, how to do it fast and efficient with respect to time since I won't have a computer to use in my exam.

Here's an example from NI Multisim bode plot:

I need help to calculate the H(s) and bode plot manually. Thanks.

• Can you calculate that at DC the gain is -42.35 dB (a good place to start)? Commented Jun 29, 2021 at 12:53
• Maybe, The question is how to do the equations to calculate the transfer function. I am expected to have: Two Node equations + solving TWO equations with numbers and s as a parameter
– user288805
Commented Jun 29, 2021 at 13:03
• Maybe isn't sure enough. If you don't see the importance of this first calculation then what help can I give that is meaningful? Commented Jun 29, 2021 at 13:34
• What do you mean with "how to do it fast and efficient with respect to time"? I thought you need H(s)? Otherwise you seem to have a shelf highpass and a (loaded) lowpass. Commented Jun 29, 2021 at 14:34

If you want to be fast and be efficient to determine transfer functions, there is no other way than resorting to the fast analytical circuits techniques or FACTs that I describe in my book on the subject. However, and it could be the issue for an exam, I am not sure if the person who reviews your contribution will recognize it as a valid method as he/she perhaps expects the classical KVL/KCL approach.

The methods relies on determining the time constants when the stimulus - $$\V_{in}\$$ - is reduced to 0 V or replaced by a short circuit. You first start with $$\s=0\$$ and opens the caps (and short the inductors if any), just like what SPICE does during its bias point analysis. You determine the dc gain $$\H_0\$$ in this mode. Then, you temporarily remove each capacitor and "look" through its connecting terminals to determine by inspection (no equations) the resistance $$\R\$$ you see. Then, multiply the resistance by the involved capacitor to form the time constant $$\\tau=RC\$$. So determining the transfer function is mostly about splitting the circuit into small drawings that you individually solve and eventually fix if you spot a mistake in the end.

Once you have the natural time constants, you can form the denominator. For the zero, you have to identify a specific impedance in the circuit which could block the stimulus propagation if tuned at the zero frequency. As highlighted in the below picture, it is when the parallel combination of $$\C_1\$$ and $$\R_1\$$ would become infinite. And this is true for $$\s=s_z=-\frac{1}{R_1C_1}\$$.

This is it, we have everything to be happy and a Mathcad sheet can check these results. I usually build a high-entropy expression obtained with Thévenin in this case and make sure both answers are rigorously similar in magnitude and phase. This is the case here. As I said, should you spot a small deviation, then review the individual sketches and fix the guilty one without restarting from scratch as any other analysis would require:

Because the roots are real in this circuit (the quality factor is low), you can try to model the transfer function with two cascaded poles and a zero. As shown in the below plots, all three expressions deliver the exact same response:

If I count the time needed to determine the expression using FACTs, it does not exceed a few minutes. Should I now expand the Thévenin expression, first it's likely that I make mistakes while expanding the equations but, second, I will need more time to format the expression in a nice low-entropy format.

• This answer is more than fine.
– user288805
Commented Jun 30, 2021 at 16:27
• Thank you, I hope it will motivate you to discover and master the FACTs! Have a look at my APEC 2016 seminar for a smooth introduction on the subject. Commented Jun 30, 2021 at 17:45