What is the correct way to find the transfer function of this circuit using the Laplace transform?


Can I add the impedance values of \$R\$ and \$C\$ (\$R\$ and \$1/(Cs)\$)?


2 Answers 2


Two ways I can think to do this off the bat are this but they may not be the simplest:

  1. Treat the first stage as a voltage divider so $$ V_{mid} = V_{in}* \frac{Ls||(R+1/Cs)}{R+Ls||(R+1/Cs)} $$ and then treat the second stage as another voltage divider so $$ V_{out} = V_{mid}*\frac{1/Cs}{1/Cs+R} $$ and then put the two equations together to get your Vout/Vin which is your transfer function.
  2. You could use the pi delta transform to convert both R's and L to 3 alternate elements and then one of them would be in parallel with C which can be simplified. The first vertical transposed element wouldn't matter anymore because it's an applied voltage so what you have left is a single voltage divider, which directly relates Vin and Vout.

Personally I'd go with option 1 because I've done almost all the work for you there.

  • \$\begingroup\$ Check your first stage voltage divider equation... \$\endgroup\$ Commented May 2, 2014 at 20:24
  • \$\begingroup\$ Ah good catch. Fixed it. \$\endgroup\$
    – horta
    Commented May 2, 2014 at 20:38

Using the impedance of each circuit element, write the transfer function by inspection using voltage division in two stages:

$$\frac{V_{out}}{V_{in}} = \frac{Z_L||(Z_R + Z_C)}{Z_R + Z_L||(Z_R + Z_C)}\cdot\frac{Z_C}{Z_R + Z_C}$$

The first term is the transfer function from the input node to the intermediate (middle) node and the second term is the transfer function from the intermediate node to the output node.

Now, you can simplify this expression and then substitute the actual impedances, e.g., \$Z_L = sL\$ or reverse that order.

The trick will be to get the transfer function into a standard form.


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