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I have a low pass filter with 3db cut off frequency of 200Hz and a high pass passive filter with 3db cut off frequency of 2kHz.

I need to design a cascaded band pass filter which will isolate signals for each range. The low-frequency components (20 to 150 Hz) will control Blue LED lights, the mid-frequency components (200 Hz to 2 kHz) will control Green LED lights, and the high-frequency components (2.5 kHz to 20 kHz) will control Red LED lights.

My transfer function for cascaded band pass filter is as below $$ H(j\omega)= \frac{sR_2C_2} {1+sR_2C_2+sR_1C_1+s^2R_1R_2C_1C_2} $$ where \$s=j\omega\$ . In order to isolate the signals, my design will have to be low pass-band pass-high pass filter. Is it correct? I would like to know how to determine lower and upper cut off frequency. Is it 200Hz and 2kHz?

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  • \$\begingroup\$ For every pole, be it high pass or low pass, it is simply 1/(2*pi)*R*C. I could design this easily for you, but that might be cheating... You could scale using these values for a near perfect 1.0 KHZ: 1/(2*pi)*15.9154529658302045760 Kohm*10.00 nF = 1000.00000000000045 KHZ \$\endgroup\$ – Sparky256 May 17 '18 at 2:32
  • \$\begingroup\$ i can only use 10k ohm resistance in my design. so i found out that C for my high pass filter is 7.96nF and low pass filter is 79.6nF. Is it correct? How do I get the lower and upper 3db frequency from the transfer function? \$\endgroup\$ – sterstar May 17 '18 at 3:02
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I need to design a bandpass filter by cascading passive high pass filter and passive low pass filter

And...

I have a low pass filter with 3db cut off frequency of 200Hz and a high pass passive filter with 3db cut off frequency of 2kHz.

If you cascaded those filters you won't get what you want - the low pass would start to remove frequencies above 200 Hz and the high pass would start to remove frequencies below 2 kHz. In between those two frequencies you would get very little signal i.e. this is a band-reject filter.

If you had a low-pass filter that had a cut-off of 2 kHz cascaded with a high-pass filter that has a cuto-off of 200 Hz, that would work: -

enter image description here

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You can scale these absurdly accurate values to get what you want. Multiply R or C by 5 and 1.000 KHZ becomes 200 HZ. Multiply R or C by 10 and you have 100 HZ. Divide R or C by 2 and you have 2.000 KHZ.

Instead of 10.0 nF and 16 K being the constant, you can transpose them. 10 K and 16 nF (15 nF - off the shelf values start with '15' in multiples of ten) will also get you 1 .0 KHZ. 10K and 75 nF will get you 200 HZ. Capacitor values are in steps so you will have to approximate the roll-off points.

If you are fussy about accuracy you will have to parallel capacitors to get more exact values. Example: Buy the 15 nF plus a 1.0 nF to get the 16 nF. Buy 10 K 1% or even .1% resistors, which are cheap in bags of 200 pcs.

68 nf/82 nF and other '68/82' values are as close to anything in the 70 nF range you can get. So buy the lower value and parallel what you need to get 72, etc. For 79.6 nF just buy the 82 nF and accept the small error. This does not have to meet Military or NASA specifications.

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  • \$\begingroup\$ sorry, i just want to know whether is my design with low pass-bandpass-high pass filter is correct to isolate the signals and is the lower and upper frequency 200Hz and 2kHz? As I am required to find the capacitance regarding to these frequencies \$\endgroup\$ – sterstar May 17 '18 at 5:01
  • \$\begingroup\$ Yes your design values are correct, based on a 10.0 K resistor. You may have to do the math as a test and come up with precise values, but I was just pointing out the common capacitor values you can buy. 1.5 nF, 15 nF 150 nF, 1.5 uF, 820 pF, 8.2 nF, 82 nF, 820 nF, etc. \$\endgroup\$ – Sparky256 May 17 '18 at 6:02
  • \$\begingroup\$ Thank you! I had chose 0.082uF for my low pass filter and 8200 pF for my high pass filter. I am aware that if I use those capacitance, my lower 3db frequency and upper 3db frequency will change as well. So is my new lower 3db frequency is 194.1 Hz and 1.94kHz? \$\endgroup\$ – sterstar May 17 '18 at 6:48
  • \$\begingroup\$ Yes. 194.091392073592601 HZ and 1.94091392073592601 KHZ. I wrote a program to do all this for me, and it gives me 16 to 18 decimal places. \$\endgroup\$ – Sparky256 May 17 '18 at 7:06
  • \$\begingroup\$ after i get all the values, I am not sure how to get the transfer function that can used to draw Bode plot. I was able to get the centre frequency is 613.77 Hz. But at the denominator, i dont now how to get the quadratic pole equation \$\endgroup\$ – sterstar May 17 '18 at 7:22

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