Network Function of a Band-Pass Filter

Question

So I've been trying to derive a the network function for a Band-Pass filter. I use voltage division to get a relation for the voltage output across the inductor.

^{simulate this circuit – Schematic created using CircuitLab}

$$ H(\omega)=\frac{V_{out}}{V_{in}}= \frac{Z_{out}}{Z_{tot}}\\ H(\omega)= \frac{\frac{Z_L Z_C}{Z_C + Z_L}}{R + \frac{Z_L Z_C}{Z_C + Z_L}}\\ $$ where $$ \frac{Z_L Z_C}{Z_C + Z_L} = \frac{-j\omega L}{\omega^2 LC -1}\\ $$ All said and done I'm supposed to get something of the form $$ H(\omega)=\frac{k}{1+jQ(\frac{\omega}{\omega_o}-\frac{\omega_o}{\omega})} $$ Where I know $$ \omega_o = \frac{1}{\sqrt{LC}} $$ and where I thought my values for k and Q should be $$ k=\frac{1}{R}\\ Q=\frac{1}{R}\sqrt{\frac{C}{L}}\\ $$ Once the dust settles on my current derivation, I get $$ k=1\\ Q=R\sqrt{\frac{C}{L}}\\ $$

Now, I'm having trouble simulating my result, so my usual verification method isn't working. I would love to know which of the above (if either) are correct and where I may have gone wrong. Thank you!

Answer 1

First, k=1 from circuit inspection: The parallel combination of L and C is an open circuit (infinite impedance) at wo=1/sqrt(LC). At this frequency, no current will flow through R and hence Vo=Vi.

Second, the Q factor is RCwo or, equivalently, the second expression you gave.

If you work further on the original expression you will certainly get it.

Answer 2

With a simple passive network like this one, no need to write a single line of algebra. Just go and "inspect" the circuit by splitting it into several small sketches. This is a second-order system so the denominator follows the form \$D(s)=1+sb_1+s^2b_2\$. First, you start with \$s=0\$ by opening the caps and shorting the inductors. You see that at this moment, the dc gain \$H_0=0\$. Then you reduce the excitation voltage \$V_{in}\$ to 0 V (replace it by a short circuit) and you determine the time constants involving the capacitor and the inductor as shown in the below picture. Once you have the time constants, you assemble them according to:

\$D(s)=1+s(\tau_1+\tau_2)+s^2(\tau_2\tau_{21})\$

For the zeros, you make use of the generalized transfer function for the numerator implying to alternatively set the energy-storing elements in their high-frequency state (short circuit for the cap, open circuit for the inductor) while "looking" at the resistance offered by the other element in this configuration. Once this is done - see how simple the drawings are - you have your complete transfer function determined in less than 1 minute with some habit!

The Mathcad file showing the low-entropy well-factored form is given below:

The FACTs are truly an excellent way of deriving transfer functions in a swift and efficient manner. Very often, in particular with passive circuits, the polynomial expressions can be formed by inspection without writing a single line of algebra: just draw small sketches and determine the \$a_i\$ and \$b_i\$ terms for \$N\$ or \$D\$ individually. I truly encourage students and engineers to dig up this technique and master the skill because it is of invaluable help if you need to solve complicated circuits quickly and want to obtain a nice canonical form, also called a low-entropy form by Dr. Middlebrook.

If you want to know more about FACTs, have a look at the seminar taught at APEC 2016

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

but also the numerous transfer functions derived in the book

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf

Answer 3

user3738697 - I recommend to use another "normal" form for a second-order bandpass function with a denominator D(jw) consisting of a second-order polynominal:

D(jw)=[wo² + jw*wo/Q + (jw)²]

For your circuit we can find

H(jw)=(jw/RC)/[1/LC + jw/RC + (jw)²] .

After comparison of both denominators you can find the expressions for wo and Q.