Transfer function for a Passive Second order Low-pass Filter and to determine the components values

Question

I am working on an assignment about Passive, second order low-pass filters. Cool little things if you get to know them.

We are given the placements of the passive components (Inductor(L1), Resistor(R1) and Capacitor(C1)) without their values.

The only given specification is that the cut-off frequency (fc) is 78 kHz.

The circuit is shown below.

If the Photo is unclear here's the Schematic:

^{simulate this circuit – Schematic created using CircuitLab}

My questions are:

1.How to make the predicted low-pass filter Transfer function for this specific circuit?

(In other words...the expression for the Output Voltage Phasor to Input Voltage Phasor? ) (Using sinusoidal steady-state analysis and answer in modulus argument form)

AND

2.Secondly how do I calculate or choose the component values for the filter? ...

The components must work out so that:

2.1) the Phase Angle between the output voltage and input voltage is -90 degrees at the cut-off frequency (78 kHz).

(A suggestion given to me was to choose a Inductor value and a Capacitor value that is at my disposal and to calculate the required Resistor value to build the circuit after calculations ).

AND

2.2) that the Ratio of the amplitudes of the Output Voltage to the amplitude of the Input Voltage is 0.7079 at the cut-off frequency(fc).

PS: I have a 1mH Inductor at my disposal,- thats why it is the only value I entered on the circuit for now :) - And the Capacitors at my disposal is 1uF, 10uF, 22uF, 100uF or 470uF ).

I can't wait to here from you!

Answer 1

You have the choice between several approaches to determine the transfer function of this 2nd-order circuit (this is a second order because you have two energy-storing elements with independent state variables). The one that immediately comes to mind is the simple impedance divider: \$Z_1(s)=R_1+sL_1\$ and \$Z_2(s)=\frac{1}{sC_2}\$. Then, you can write \$H_{ref}(s)=\frac{Z_2(s)}{Z_2(s)+Z_1(s)}\$. You can further develop this expression and rearrange it in a form such that you unveil a quality factor \$Q\$ and a resonant frequency \$\omega_0\$. These two parameters will represent a design goal letting you determine the components values. If this is a simple exercise here, you realize that adding more elements like a loading resistance across \$C_1\$ for instance or other parasitics start making things more difficult to manage.

A simpler method consists of using the fast analytical circuits techniques or FACTs. Rather than writing algebraic lines, why not looking at simple sketches as shown below? The principle lies in determining the time constants of a circuit when the excitation \$V_{in}\$ is reduced to 0 V (a voltage source here). A 0-V voltage source is a short circuit so you will replace the source symbol by a short circuit and will "look" at the resistance offered by the energy storing elements connecting terminals when \$L\$ and \$C\$ alternatively set in two different states: dc state and high-frequency state. In the first state, a cap is an open circuit while an inductor is a short. In the second state, a cap is a short circuit and an inductor is an open circuit.

The exercise is quite simple. First, determine the gain of the circuit at dc: short the inductor and open the cap: \$H_0 = 1\$ because the capacitor is unloaded. Then, reduce the excitation to zero and determine the time constants involving \$L_1\$ and \$C_2\$ in this mode. The first drawing shows an infinite resistance "seen" from \$L_1\$'s terminals implying a 0-s time constant while the second sketch shows a time constant \$\tau_2=R_1C_2\$. The second time-constant product is obtained by shorting \$C_2\$ while "looking" at the resistance from \$L_1\$'s terminals: \$\tau_{21} = \frac{L_1}{R_1}\$. This is it, we can write the denominator following \$D(s) = 1+sb_1+s^2b_2 = 1 + s(\tau_1+\tau_2) + s^2(\tau_2\tau_{21})\$. Then, you can rearrange this polynomial form to make it fit the classical canonical form found in the literature. This is what the below Mathcad sheet shows:

The cool thing is that I could determine the transfer function just by going through simple sketches, without writing a line of algebra. If I made a mistake, it is easy to return to the guilty drawing and fix it. There is a tutorial here which I encourage those involved in studying transfer functions to review. Once you've acquired the skill, you won't go back to classical analysis!

Answer 2

Here are some hints/help: -

How to make the predicted low-pass filter Transfer function for this specific circuit?

You need to define the Q of the circuit i.e. the amplitude of the output at the natural resonant frequency (\$\omega_n\$). For a butterworth response Q = 0.7071. Q always defines the amplitude of the transfer function at \$\omega_n\$ : -

Picture from Here.

the Phase Angle between the output voltage and input voltage is -90 degrees at the cut-off frequency (78 kHz)

And, if you read the section a bit further down from the picture link entitled "Quality factor, amplitude and phase at \$\omega_n\$" you will see a proof that the filter phase is always 90 degrees at the cut-off frequency.

The cut-off frequency is defined as f = \$\dfrac{1}{2\pi\sqrt{LC}}\$

But you need to choose a resistor value that satisfies Q and this also depends on the ratio of L and C. Try using this site for help in choosing values. Okawa electric design is the site name and it's very useful.