Designing an LPF with a 17kHz Cut-off Frequency: Question and Guidance

[fran1942]

Hello, I have been asked to:
- design an LPF with a cut off frequency of 17kHz and a roll-off of 40dB Decade. I have attached a picture of the circuit diagram I came up with.
Can someone please confirm if I have this correct ?

The second part of the question was:
- prove both time domain and frequency domain plots for the output waveform for the above filter, if a 5kHz square wave was applied to it.
For this question I am rather lost. Can someone please give me some guidance on how I would display these 2 graphs. I can't find anything in my notes directly explaining how to do this.

Thanks kindly for any help.

 

[NascentOxygen]

Hi fran1942. Were you told to use only R-C passive components in your design? You can achieve a sharper response (i.e., closer to what most would consider the ideal) if you use an active element, e.g., an op-amp with feedback.

Can you simulate this with SPICE to determine its Bode plot?

 

[moogull]

Like Nascent said, SPICE will allow you to simulate a circuit for various input signals that you specify: its free! http://www.5spice.com/

 

[gneill]

Hello, I have been asked to:
- design an LPF with a cut off frequency of 17kHz and a roll-off of 40dB Decade. I have attached a picture of the circuit diagram I came up with.
Can someone please confirm if I have this correct ?

The second part of the question was:
- prove both time domain and frequency domain plots for the output waveform for the above filter, if a 5kHz square wave was applied to it.
For this question I am rather lost. Can someone please give me some guidance on how I would display these 2 graphs. I can't find anything in my notes directly explaining how to do this.

Thanks kindly for any help.

First, perhaps a small problem with your circuit. It looks as though you've simply cascaded two identical low pass filters with a cutoff frequency of about 17.5 kHz. The problem with doing this in this manner is that the two filters will interact and change both of their cutoff frequencies. One way to avoid this is to make the input impedance of the second filter section much higher than that of the output impedance of the first. Make the second resistor about 10x that of the first (adjust the second C accordingly).

For the plots, if you are familiar with Laplace transforms then you can find the response curve using the transform of a square wave to "drive" the transfer function. Working in the time domain you could decompose the square wave into its first few harmonics (until the harmonic frequency is well above the cutoff), find the amplitude of each harmonic at the output of the filter, then sum the resulting waveforms.