# Trouble designing a low pass filter

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

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.

• That makes a lot of sense... Any tips on a straightforward way to get the overall cut off frequency? (The -3dB frequency) – Bidon Dec 6 '19 at 0:04
• You can simply plot out of the response of a single stage filter and note the relationship between the 1 and 3 dB down frequencies. For example, assume the 1 dB down frequency is 60% of the 3 dB down frequency. Then, if you want a 3 stage filter to have a 1 kHz cutoff, you should design each stage to have a cutoff of 1/(0.6) or 1.67 kHz. – Barry Dec 6 '19 at 1:01

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.

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.

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).