Consider the two filter in picture


simulate this circuit – Schematic created using CircuitLab

Are the cutoff frequencies (i.e. the two frequencies at which \$V_{out}=\frac{1}{\sqrt{2}} V_{in}\$) the same for the two filters? (And therefore also the bandwidth is the same for the two filters?)

I think they should be the same since in both cases the condition \$V_{out}=\frac{1}{\sqrt{2}} V_{in}\$ reduces to

$$R^2=(\omega L-\frac{1}{\omega C})^2$$

Nevetheless it is strange that there is a frequency for which both \$V_{out}\$s taken at resistor and at LC series have the same value \$\frac{1}{\sqrt{2}} V_{in}\$.

Doesn't this violates Kirckoff voltage law? (At any time it should be \$V_{in}=V_{R}+V_{LC\ series}\$ but if \$V_{R}=V_{LC\ series}=\frac{1}{\sqrt{2}} V_{in}\$ then this is not true).

enter image description here

  • 1
    \$\begingroup\$ When - at a certain frequency - the voltage across R is 70.7% of the total voltage (bandpass cut-off), you can calculate by yourself which voltage will be across the L-C combination at the same time. It is up to you to decide if it makes sense to declare this frequency as the band-stop cut-off. \$\endgroup\$
    – LvW
    Mar 20, 2017 at 15:46
  • \$\begingroup\$ @LvW Thanks for the comment, the voltage across LC would be \$\sim 30 \%\$ but isn't the definition of cutoff frequenciies for band stop filter "the frequencies when \$V_{out}=\frac{1}{\sqrt{2}} V_{in}\$"? (as in the picuture I added in the question) \$\endgroup\$
    – Sørën
    Mar 20, 2017 at 15:55

1 Answer 1


We can apply the fast analytical techniques here (FACTs) and they show that if you reduce the excitation voltage \$V_{in}\$ to 0 V in your example (replace the source by a short circuit), the structure remains unchanged regardless where you observe \$V_{out}\$. Therefore, the denominator \$D(s)\$ is the same between the two schematics you have drawn. The principle is always the same, observe the circuit for \$s=0\$ first and then determine the time constants combining the various energy-storing elements to form the denominator \$D(s)\$. Look at the below schematic, it is quite easy to follow:

enter image description here

The denominator is obtained by assembling the time constants the following way:


You can then rearrange it under the classical canonical form


The zero are obtained by observing what impedance combination could become a transformed short circuit nulling \$V_{out}\$ when \$s=s_z\$? Obviously this is when the series impedance made of \$C_1\$ and \$L_2\$ becomes a transformed short circuit. If you solve \$Z_1(s)=0\$, then \$1+s^2L_1C_1=0\$ and you have immediately \$\omega_{0N}=\frac{1}{\sqrt{L_1C_2}}\$. The final transfer function is given in the below Mathcad sheet for the bandstop filter:

enter image description here

For the bandpass filter, we already have \$D(s)\$ so no need to re-derive it. It is easier to slide resistor \$R_1\$ and redraw the schematic. This time, capacitor \$C_2\$ places a pole at the origin and blocks dc. To obtain the zero, we can apply the generalized transfer function expression


In our case, \$H_0\$ is 0 since \$C_2\$ blocks the dc. Therefore, we can calculate the remaining gains quite easily as shown below:

enter image description here

Once this is done, assemble the calculated terms as shown in the Mathcad screenshot below

enter image description here

I even rearranged the transfer function in a low-entropy form showing a 0-dB gain at the peak.

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. If you see a mistake, just correct the guilty term without re-starting from scratch. Of course, when you tackle circuits featuring active sources like current- or voltage-controlled sources, you often need to resort to KVL and KCL but the obtained result is always expressed in a meaningful polynomial form and it is easy to correct in case a typo is detected. If you want to know more about FACTs, have a look at the seminar taught at APEC 2016


but also the numerous transfer functions derived in the book


When you go FACTs, you don't want to be back : )


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.