Confused how to find the transfer function of a low-pass filter

I know that $H(\omega) = \frac{V_o}{V_i}$ but I don't quite understand why this equals $\frac{R}{R+j\omega L}$.

I understand that $V_o = IR$ but don't quite understand how $V_i = I(R+j\omega L)$.

• Can you do it if L=R2 ? It is the impdance divider relationship. Nov 22, 2017 at 21:12
• Do you know how to calculate the magnitude of R+jwL? Nov 22, 2017 at 21:21
• In the 1800's people just used the differential equations and solved them. This "new" idea of using complex numbers dates to Steinmetz's presentation in 1897. You can look up the 3rd edition, which was retitled as "theory and calculation of alternating current phenomena" and published in 1900, I think. It's worth reading the 500+ pages. Also, for fun, see Euler's formula -- 3d visualization.
– jonk
Nov 22, 2017 at 22:50
• (Well, make that 1893 in Chicago. I just grabbed my copy of the 1900 publication and he says so in the Preface. Sorry about getting the year wrong!)
– jonk
Nov 22, 2017 at 23:24
• Funkster do you understand anything yet? Nov 23, 2017 at 2:07

simulate this circuit – Schematic created using CircuitLab

In the circuit shown, there are two elements, A and B, that have impedance. The element might be a resistor or a capacitor or even an inductor.

Lets call the impedance of element A, $Z_A$, and element B, $Z_B$.

Now if both elements are resistors instead of talking about impedance we usually say resistance. We know we can use the handy voltage divider formula to find the output voltage V2.

$$V2 = \frac{Z_B}{Z_A+Z_B} \cdot V1$$

We can also use KVL to relate the voltage source to the current in the circuit.

$$V1 = i \cdot (Z_A + Z_B)$$

This is still true even if they are not both resistors. The current flowing through each element will create a voltage drop across it due to the impedance it sees. So, we must use the appropriate impedance model for the element to get the correct answer.

For an inductor - $Z = j \omega L$. We can substitue that value in to the above equations to get the expected result. $$V1 = i \cdot (R + j \omega L) \\ \\ V2 = \frac{R}{R+ j \omega L} \cdot V1$$

• It would be better to use $Z$ as the variable for impedance, since $X$ is most often used for reactance. Nov 22, 2017 at 22:06
• Really good answer (+1). Though I think Z is more conventional for impedance than X.
– user103380
Nov 22, 2017 at 22:07
• Ugh... @ThePhoton Just beat me by 10 seconds LOL...
– user103380
Nov 22, 2017 at 22:07
• Good point - Changing X to Z .. Nov 22, 2017 at 22:08
• X is more for reactance, which is the imaginary part of impedance :)
– user103380
Nov 22, 2017 at 22:09

If there is no load on the filter, then Vi is equal to the voltage across R and L. The voltage across R which is defined as Vo, as you already determined, is equal to the current through R (i1) multiplied by R which results in Vo = Ri1. Similarly, the voltage across L is equal to jwL times the same current since R and L are in series. Thus it is jwLi1. Then Vi is the sum of these 2 voltages so: V1 - jwLi1 + Ri1 = (jwL + R)i1. The transfer function is Vo/Vi which is given by: Vo/Vi = Ri1/(jwL + R)i1 which reduces to R/(jwL + R). As pointed out in a comment, you can also derive this result by considering the circuit as a voltage divider.

The mathematical process for finding the values:

Ohms Law across the inductor and resistor gives:

$V_{ac}=L*(di/dt)$

$V_o=i*R$

use KCL:

$V_o=V_i-V_ac$

so substitution gives:

$V_i=i*R+L*(di/dt)$

taking the laplace transform of both sides gives:

$V_i(s)=i*R+i*L*s$

simplify:

$V_i(s)=i(R+L*s)$

where s=jw for freqency w