Trouble designing a low pass filter

Question

I was trying to design a simple 3rd order low pass filter by cascading 3 first order low pass filters along with an amplifier at the end. The transfer function is straightforward:

$$ H(j\omega )=4\bigg(\frac{1}{j\big(\frac{\omega}{\omega_c}\big)+1}\bigg)^3 $$

where $$\omega_c=\frac{1}{C_f10k\Omega}$$ is the cut-of frequency. Thus when I want to make the filter have a cut off frequency at 1KHz I simply replace values and figure out Cf:

$$ C_f=\frac{1}{2\pi\times f_c\times10k\Omega}=\frac{1}{2\pi\times10^3\times10\times10^3}\approx16nF $$

However the circuit below when simulated gives me a cut-off frequency of 500Hz. What am I getting wrong?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

You cascaded 3 filters each with a cutoff frequency of 1 kHz. Since the cutoff frequency is defined as the frequency where the filter is 3 dB down, your cascaded filters will be 9 dB down at 1 kHz. The actual cutoff frequency of your cascaded filters is that for which the gain of each individual filter is 1 dB down which will be lower than 1 kHz. Your simulation shows it to be 500 Hz. If you really want a cascaded filter cutoff frequency of 1 kHz. you will have to increase the cutoff frequency of each of the 3 separate filters.

Answer 2

If you think for a moment, single filter will have -3db cut-off point at 1kHz, then next filter added to it will also have -3dB point at 1kHz the total attenuation is -6dB at 1kHz, and same goes when the third filter is applied.

Answer 3

What am I getting wrong?

Either your definition of cut-off frequency, or the relationship between the cut-off frequencies of individual elements of a critically-damped filter and the overall cutoff frequency.

Each element has a 3dB cutoff frequency of 1kHz. Because you're cascading identical transfer functions, the effect of these filters add in log-amplitude (i.e., their responses expressed as dB add). So you're seeing a gain of -9dB at 1kHz, which is exactly what you should expect given how you designed the thing.

If you want to define the cutoff frequency as the 3dB-down point, then adjust the capacitor values. Otherwise, define "cut off" as 9dB down, declare victory, and hide where they can't find you.

Answer 4

What the other answers say it's true, I'll just add that what you have there is a repeated convolution between an input signal and the same kernel, three times, and repeated convolution converges towards a Gaussian response, exp(-x²).