This is a high pass T filter:

enter image description here

According to me, if input voltage is \$V_{\text{in}}\$, then voltage across the inductor \$L\$ should be $$V_x=\frac{j\omega L}{j\omega L + \frac{1}{2j\omega C}}V_{\text{in}}$$

Now, I think \$V_{\text{out}} = V_x\$ since we can only measure the EMF across the output terminals unless there's some load resistance places across those two output terminals.

In that case if we want the cut-off frequency, we'll need to set \$|V_{\text{out}}/V_{\text{in}}|=\frac{1}{\sqrt{2}}\$. I get \$\omega_{\text{cut-off}} = \sqrt{\frac{1}{2(\sqrt{2}+1)LC}}\$ from here. However, the actual cut-off (angular) frequency \$\omega_{\text{cut-off}}\$ (according to what we were told in class) should have been \$\frac{1}{2\sqrt{LC}}\$. I'm not sure why I'm not getting the same result.

Also, another question is: What is the use of the right hand side \$2C\$ capacitor of the T-section filter in determining the cut-off frequency? As long as no load is there, we should solely consider the voltage across the inductor \$L\$ to be the output voltage, isn't it? Or no?


2 Answers 2


What you see in the picture is the textbook representation of a 3rd order section, as Tony mentions in his answer, which does not include the terminating I/O impedances, which should exist in a real-life scenario. This means the transfer function becomes:


The output voltage is not across the inductor, otherwise that would no longer be a 3rd order cell, so the output voltage is taken across the terminating impedance, after the second capacitor.The cutoff frequency, or pulsation, is simply the denominator's \$s^3\$ term, adjusted for its power:


And here's a quick verification with LTspice, where I used unity-based values, and I/O resitances instead of impedances, for simplicity:


And the capacitors have the value 2C so that the transfer function becomes flat across the passband, otherwise, with only C, it becomes (linear amplitude, for better viewing):



With no load , the right hand 2C might as well be a short circuit and the 3 reactive parts only behave as a 2nd order high pass filter with high Q at the breakpoint.

With a resistive load the CR LPF adds a 3rd order HPF result.

( In theory with 0 ohm source and infinite load , the breakpoint has infinite Q , but in practice limited to Q = few hundred from ratings of components e.g. 100 is not hard, 1000 is not likely) With a load R this reduces by Z(f)/R at resonance.


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