enter image description hereI have designed my analog low pass filter using LM358 op amp. I used r1=r2=330 Ohm and c1=c2=100nF. The cut off frequency =4.8 KHz. But i dont know where its going wrong , my attenuation which should start after 4.8KHz is starting from 1kHz itself. DC bias of +3.3v is given and input voltage of 500mv peak to peak from the function generator with 250mv offset is provided as input. I am trying it from last week with the attenuation is not achieved after 4.8KHz. Please help.

  • \$\begingroup\$ Attenuation can start before the cut-off. Measure the -3dB frequency and compare with the designed cut-off frequency. \$\endgroup\$ – nidhin Mar 2 '15 at 7:23
  • \$\begingroup\$ If you are just tinkering, you can try changing (in your case decreasing ) the R or C. \$\endgroup\$ – Plutonium smuggler Mar 2 '15 at 7:23
  • \$\begingroup\$ hello nidhin, can u just say how to measure in my case -3db frequency \$\endgroup\$ – shilpa Mar 2 '15 at 7:27
  • \$\begingroup\$ hello plutonium, i have tried doing that also but it doesn't work. \$\endgroup\$ – shilpa Mar 2 '15 at 7:27
  • \$\begingroup\$ Doesnt work ?? You mean cutoff freq doesnt change on changing the component values ? \$\endgroup\$ – Plutonium smuggler Mar 2 '15 at 7:33

If you were expecting the output to look like a typical straight line Bode plot this is not going to happen. A Bode plot is just an approximation of the actual frequency response curve.

When testing a filter circuit the corner frequency, (-3db point, or half power point) can be determined by finding the point where the output voltage is 70.7% of the original input voltage.

If you want a filter with a sharp cutoff at 4.8khz you would need to use another type of filter, or add additional cascaded sections to the original circuit. To achieve a flatter response near 4.8khz you might even add a section that gives a small amount of gain near 4.8khz.


The cut-off frequency is the frequency where the attenuation is 3dB.

This means that at 4.8kHz the amplitude of the output signal is x 0.7. Of course it starts to get attenuated at smaller frequencies.

  • \$\begingroup\$ thank u, i have also came to such a conclusion. But i dont know how to resolve it. Can you help me. \$\endgroup\$ – shilpa Mar 2 '15 at 7:30
  • 2
    \$\begingroup\$ 3dB is when the output voltage has reduced to 70.71% not 50%. \$\endgroup\$ – Andy aka Mar 2 '15 at 8:44
  • \$\begingroup\$ I think you got mixed up with Power. At 3dB, power is halved. But voltage is reduced by \$ \frac{1}{ \sqrt{2}} \$ or 0.707 \$\endgroup\$ – efox29 Mar 2 '15 at 10:47

From your description (two equal C and two equal R) I derive that you are using a Sallen-Key equal comonenet topolgy. Please note that this stucture is VERY sensitive to gain tolerances. What is the desired value of the gain value (below 3)? What is the desired response (Butterworth, Chebyshev,...)?

It would be best to show the circuit - including bias circuitry.

EDIT: OK - finally, it is NOT a multi-feedback topology, but a gain-of-two Sallen-Key structure. You have chosen equal components - leading to a pole-Q of Q=1. This gives a Chebyshev-like response with a slight gain peaking (1 dB) around the cut-off frequency. Do you observe this behaviour?

  • \$\begingroup\$ hello LvW, no im using a multiple feedback loop with Rin =Rf=10kohm . I have not included any desired reponse in my design. \$\endgroup\$ – shilpa Mar 2 '15 at 9:15
  • \$\begingroup\$ A multi-feedback LOWPASS with two R and two C only? Or is it a bandpass? Can you show the circuit? \$\endgroup\$ – LvW Mar 2 '15 at 10:07
  • \$\begingroup\$ Please read my answer (EDIT). \$\endgroup\$ – LvW Mar 2 '15 at 10:29
  • \$\begingroup\$ @shilpa - you say in your question that you are using 2x 330 ohm resistors and not 10kohm as per your comment above. A circuit would greatly help. \$\endgroup\$ – Andy aka Mar 2 '15 at 10:51
  • \$\begingroup\$ hello LvW, i have attached my filter designed(EDIT). I have mentioned all the value in it. If i want to design the same filter design with variable resistor , then how to do?? What shall be the fL and fH??? \$\endgroup\$ – shilpa Mar 2 '15 at 11:06

It sounds like you are implementing a Sallen key low pass filter and, with the values chosen for the capacitors it will have a damping ratio of 1. Here is the same Sallen Key filter frequency response with different ratios of capacitor values. Note that nearly all have an effect at 1kHz despite the resonant frequency being approximately 4.8kHz: -

enter image description here

If you are precise in choosing the capacitor ratio you can achieve a response that is largely unaltered at 1kHz. This is called a "Butterworth" response. Your design is likely to be sub-optimal if you have used equal-capacitor values and a unity gain op-amp.

EDIT - whether using a sallen-key or any other type of 2nd order filter topology, the theory remain the same and the graphs above remain identical for different component value ratios. Different component ratios produce changes in the Q factor of the circuit and, Q factor affects how the frequency response looks i.e. peaky, flat or gradual.

EDIT 2 - the circuit the OP uses is a sallen-key filter with gain and if R4 was varied from 10k to about 16kohm, the frequency response would be about flat at 1kHz: -

enter image description here

This resistor also affects the overall gain of the amplifier and if this is needed to be kept constant at 6dB (10k and 10k) then I suggest changing the ratio of the resistors R1 and R2 whilst keeping R1 x R2 (the product) the same.

Also note that using a signal generator with 50 ohms impedance is going to add errors to the filter because, in simple terms, the 50 ohm impedance adds to R2 to take it from 330 ohms to 380 ohms.

  • \$\begingroup\$ Hello Andky, im nt using a Sallen key LPF. I have attached my filter design. Please let me know what can i do?? whether to use a variable resistor or use a filter response in my breadboard design?? \$\endgroup\$ – shilpa Mar 2 '15 at 11:11
  • \$\begingroup\$ Yes you are using a sallen key filter. See this: google.co.uk/…. \$\endgroup\$ – Andy aka Mar 2 '15 at 11:18
  • \$\begingroup\$ Also, what is the signal source you are using - if this has significant output impedance your filter will give a significantly poorer spectral shape. Even a 50 ohm signal generator will make things worse. \$\endgroup\$ – Andy aka Mar 2 '15 at 11:19
  • \$\begingroup\$ Yes, Im using a 50ohm signal generator and giving 500mv peak to peak of offset of 250mv. Using a dc bias of +3.3v to LM358 opamp at its V+ terminal. \$\endgroup\$ – shilpa Mar 2 '15 at 11:26
  • \$\begingroup\$ You are also running close to the output limits that the op-amp is capable of producing for a 3V3 power supply. You should also check the function generator is producing 250mV DC. Check with a meter - sometimes a function generator will state what its output is assuming a DC load of 50 ohms. It might be that it is producing 500mV DC and this could be another problem. You ought to consider a decent rail-to-rail op-amp. \$\endgroup\$ – Andy aka Mar 2 '15 at 11:50

As already mentioned that at the -3dB or the (cut-off frequency) is where the voltage is at 70.7% of the input. You can simply measure your voltage output when it's at 4.8kHz and if it does satisfy that it is 70.7% of your input then things are working as they should.


With the circuit values given, the circuit will behave like a perfect double-integrator, with a gain of +6dB at 4.823kHz, a gain of 0dB at 6.821kHz and a gain of -3db at 9.646kHz, i.e. the amplitude frequency response will decrease linearly (against log frequency) at a roll-off of 40dB/decade. (I think I've got these values correct!). Theoretically, DC gain will be infinite.

The transfer function is: Vo(s)/Vi(s) = 2/(sCR)^2

The response is extremely sensitive to component values, as the R's and C's (330ohm and 100nF) are equal.

  • \$\begingroup\$ Sorry to disagree but the DC gain is not infinite and the circuit will not behave like a perfect double integrator. \$\endgroup\$ – Andy aka Mar 2 '15 at 12:33
  • \$\begingroup\$ Yogi, the circuit shows a second-order Sallen-Key lowpass which, of course, behaves NOT like a "perfect double-integrator". What is the reason (background) of your assumption? And the DC gain is "2" (in words: two). This can be derived by a simple visual inspection of the circuit. \$\endgroup\$ – LvW Mar 2 '15 at 12:36
  • \$\begingroup\$ Yogi , plz can you explain why its at +6db , 0db and -3db?? \$\endgroup\$ – shilpa Mar 3 '15 at 4:22
  • \$\begingroup\$ According to my circuit value given, its giving 0db at 2khz and -3db at 6khz. I applied the simple log principle for calculating db. \$\endgroup\$ – shilpa Mar 3 '15 at 6:20

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