I 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.
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.
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?
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: -
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: -
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.
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.