# LC lowpass filter

1. Oct 25, 2015

### etf

Hi!
Lowpass filter circuit is given:

Input signal to this circuit is:

L=36.4mH, C=85.4nF.
Calculate and draw output signal.

My idea is to:
1) Represent input signal in term of Fourier series (complex form),
2) Calculate transfer function for given circuit (for n-th harmonic),
Expression for output signal would be product of Fn of input signal (complex fourier coefficient) and transfer function of circuit.

1)
Here is input signal in term of Fourier series (complex form):

2)
Here is transfer function for n-th harmonic:
(L'=L/2)

Output signal is now:

First question: what is purpose of second inductor? We don't take into account when calculating transfer function.
Second question: If I plot output voltage I calculated, it doesn't match with voltage I got in simulation software? Why?
Does it have something with location of pole of transfer function? Pole of our transfer function is on imaginary axes.

2. Oct 25, 2015

### Staff: Mentor

The inductor on the right achieves nothing without there being a load of finite impedance. With an open-circuit load, as you show, the second inductance does nothing.

Are you sure there isn't supposed to be a load of some specified resistance?

3. Oct 25, 2015

### etf

Task is to calculate cutoff frequency, equivalent resistance and output signal. I used formula fc=1/(pi*sqrt(L*C)) to calculate cutoff freq., R=sqrt(L/C) to calculate equivalent resistance. But I'm not sure about output signal. My calculations are related to steady state response, but this circuit never goes to steady state (I would say)

4. Oct 25, 2015

### Staff: Mentor

Your input signal here, is it just these two pulses then nothing, or is this waveform going to be repetitive?

For the record, you haven't yet learnt to use Laplace Transforms, I gather?

I have to leave this for others, I'm afraid I'm just too rusty with Transform application.

5. Oct 26, 2015

### Staff: Mentor

I'd be tempted to use the non-complex version of the Fourier series for the squarewave, expanding it as a sum of sines.

For a squarewave of unit amplitude (+/- 1) and a period of T:
$$f(t) = \frac{4}{\pi} \sum_{n = 1,3,5,...} \frac{1}{n} sin \left( \frac{n 2 \pi}{T} t \right)$$
Each term is a sinewave with a frequency that's an odd multiple of the fundamental. Standard circuit analysis methods can then be applied for the transfer function, treating the magnitudes of the sine terms as phasors.

6. Oct 26, 2015

### rude man

Is the input a pulse, as shown, or does it repeat ad infinitum?
For a pulse you don't use a Fourier series. You can use the Fourier integral or the Laplace transform.

7. Oct 27, 2015

### etf

Input signal is periodic...

8. Oct 27, 2015

### rude man

So then I'd go with post 5 but you should understand why
(1) there are no cosine terms; and
(2) why only odd harmonics.
Because a general periodic function has sin and cos terms and all harmonics, odd and even.
Still, drawing it isn't all that easy. Best would be software, even something simple as excel.

9. Oct 27, 2015

### etf

But it isn't about which "type" of Fourier series I use (complex like in my first post or noncomplex like in post #5). Whatever type I use, I will get response of circuit in steady state, which actually never appears in my case (ideal LC circuit), right?

10. Oct 27, 2015

### rude man

In Suze Orman's parlance, "here's the problem": your circuit has no dissipative elements like resistors. So no matter when the waveform was first applied to the network, there will always be two waveforms with separate Fourier series: the steady-state one you'd get if there was even a small amount of R in the circuit, dictated by the input waveform, and the other which is dictated by the L-C time constant only.

But wait, there's more! Since there's no R the network "remembers" what the exact state of the input, i.e. the phasing, was when it was first applied. So the steady-state solution depends on when exactly the waveform was applied. This makes your problem not only very difficult but actually undefined.

This problem IMO should never have been assigned. No network has zero dissipation.