I am trying to design a filter which has the following Bode plot, enter image description here

So according to this site I designed the filter as shown below,

enter image description here

As the two cross over frequencies are f1 = 50 Hz and f2 = 200 Hz, I did the following calculation for R1 and R2

Assuming C= 0.1uF and using f1 = 1/(2*Pi*R1*C), I calculated R1= 31.83 Kohm

In order to calculate R2 I first calculated the Gain= f1/f2 = 0.25

So, now again using Gain= R2/(R1+R2), I calculated the value of R2 = 10.61 Kohm

So, in order to check i did a quick simulation in LTspice, here is what I did.

enter image description here

But I am not sure if it is right,in question it says that the sloop between f1 and f2 is -6dB but in the simulation the magnitude plot is started from -12dB, I would expect it to start from -6dB.

Also about the input and output impedance, how do I make sure the o/p impedance is greater than 10 Kohm and input impedance is less then 100 ohm? Does this has something to do with the value of R1 and R2?

  • \$\begingroup\$ Just like to add that it says the slope should be -6dB/octave. It doesn't have to start at -6dB, it just means that when the frequency decreases by a factor of eight, the magnitude of the gain should decrease by 6dB. \$\endgroup\$ – ACD Aug 15 '14 at 13:04
  • 2
    \$\begingroup\$ @ACD: An octave means a factor of two, not eight. The name derives from musical notation, in which the 2:1 interval is divided into eight notes to make a musical "scale". \$\endgroup\$ – Dave Tweed Aug 15 '14 at 13:10
  • \$\begingroup\$ TIL. Also, I hate it. \$\endgroup\$ – ACD Aug 15 '14 at 13:43
  • \$\begingroup\$ The calculations seem a bit hand-wavy to me. The circuit you show comes across to me as a high-pass filter between C and R2, with R1 providing the necessary pathway for a finite ~DC gain (lower frequencies, anyway). The first formula you used was that relating to the -3dB point of the HPF (200Hz) - which seems to be what C and R2 form. However, you use this to calculate the parallel resistor, R1 which doesn't really make sense. Then you seem to do this nonsense with frequency ratios to find a gain that is only valid for DC (at least significantly < -3dB frequency). Yet, the plot seems correct! \$\endgroup\$ – sherrellbc Aug 15 '14 at 14:49
  • \$\begingroup\$ Nevermind, I missed the fact that f1 = 50Hz, so that calculation was for the lower frequency "cut-in". I am not quite sure why that calculation is valid though .. What is the relationship betweem the DC gain +3dB point (that is, -12dB + 3 = -9dB) and the resistor in parallel with C? \$\endgroup\$ – sherrellbc Aug 15 '14 at 15:29

The slope is -6db/octave, where an octave is a doubling/halving from wherever you start (see the wikipedia page for more info). In your case, 50Hz is 2 octaves away from 200Hz, so 2*(-6db) == -12db.

As far as the output impedance, you can buffer the output with an op-amp to get it very low (much less than the 100 ohms specified).

  • \$\begingroup\$ Thank you for your answer sbell, So i just need to add opamp at the end of my circuit right? but I wanted to know if the calculation I have done is correct to obtain the bode plot as asked in the question. \$\endgroup\$ – Electronic Curious Aug 15 '14 at 13:28

Using Thevenin Equivalent circuits you can derive the output impedance of your filter to be (hasty, but confident in this result):

$$R_{out} = \frac{R_2R_1}{R_1+R_2(R_1sC + 1)}$$

What I would do would be to figure out what the gain is at -12dB (2 octaves below 0dB at a slope of -6dB/octave).

$$10^{\frac{-12}{20}} = 0.25$$

So you then know that the DC gain of this filter will attenuate by a factor of 4. You could simply conclude then than we must have R1 = 3R2.

But proceeding as you have done, after finding R2 from the -3dB cut-off frequency:

$$f_{-3dB} = \frac{1}{2 \pi R_2 C}$$ The other resitor can then be found using the DC gain calculated above:

$$Gain_{DC} = 0.25 = \frac{R_2}{R_2+R_1}$$

Your solution is valid, but you went about it in an interesting way. I've always realized that a HPF was the inverse of a LPF, but I did not realize that the "cut-in" frequency (50Hz in this case) could be used to calculate a resistor in parallel with R1 as you have shown.

I still am not sure why that works, but it appears to cosidering the plot you show.

What is the relationship between the DC gain +3dB point (that is, -12dB + 3 = -9dB), i.e. the"cut-in" frequency so to speak, and the resistor in parallel with C?

Apparently it's:

$$f = \frac{1}{2 \pi R_1 C}$$

But how?

  • \$\begingroup\$ Thanks for your answer Sherrellbc, You are right about how I did the calculation, I refereed to the link in my question and to this one [linkwitzlab.com/filters.htm] which says it is a Shelving highpass filter. \$\endgroup\$ – Electronic Curious Aug 15 '14 at 16:26
  • \$\begingroup\$ @user2280164, Thanks for the link. Be sure to upvote helpful answers. \$\endgroup\$ – sherrellbc Aug 15 '14 at 18:38

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