# Calculate the formula of RC low pass filter

In RC filter, the low pass filter formula is

$$H(s) = \frac{V_\text{o}(s)}{V_\text{in}(s)} = \frac{1}{RCs + 1}$$ $$H(j\omega) = \frac{1}{j\omega RC + 1}$$

and the absolute value is

$$|H(j\omega)| = \frac{\frac{1}{RC}}{\sqrt{\omega^2 + \left(\frac{1}{RC}\right)^2}}$$ $$\theta(j\omega) = -\tan^{-1}(\omega RC)$$

To make as the absolute value, do I need to square each part (real part and imaginary part) individually and then take the square root of it together?

• That's a general math question, how to take absolute value of a complex number. It does not otherwise relate to low pass filters or electronics. It might be faster to look up how complex numbers work on Wikipedia. Commented Mar 8, 2023 at 14:02

## One approach

$$\\left.\middle| H \middle|\right.=\left.\left.\sqrt{ H\cdot H^* }\right.\right.\$$, where $$\H^*\$$ is the complex conjugate of $$\H\$$.

If $$\H=\frac{N}{D}\$$, where $$\N=a_1+j\,a_2\$$ and $$\D=b_1+j\,b_2\$$ (and therefore $$\N^*=a_1-j\,a_2\$$ and $$\D^*=b_1-j\,b_2\$$), then $$\\left.\middle| H \middle|\right.=\left.\middle| \frac{N}{D} \middle|\right.=\frac{\left.\middle| N \middle|\right.}{\left.\middle| D \middle|\right.}=\frac{\left.\left.\sqrt{ N\,\cdot\, N^* }\right.\right.}{\left.\left.\sqrt{ D\,\cdot\, D^* }\right.\right.}=\frac{\sqrt{a_1^2+a_2^2}}{\sqrt{b_1^2+b_2^2}}\$$.

If $$\H=\left. \frac{1}{j\omega R C+1} \right.\$$ then $$\\left.\middle| H \middle|\right.=\left.\middle| \frac{1}{j\omega R C+1} \middle|\right.=\frac1{\sqrt{\left(j\omega R C+1\right)\cdot\left(-j\omega R C+1\right)}}=\frac1{\sqrt{R^2C^2\omega^2+1}}\cdot\frac{\frac1{RC}}{\frac1{RC}}= \frac{\frac1{RC}}{\sqrt{\omega^2+\frac1{R^2C^2}}}\$$

## Another approach using Euler's

For both phase and magnitude for complex values like the above, it's perhaps a little easier (if you know about Euler's, anyway) is to set $$\r_a=\left.\middle| N \middle|\right.=\sqrt{a_1^2+a_2^2}\$$, $$\r_b=\left.\middle| D \middle|\right.=\sqrt{b_1^2+b_2^2}\$$, $$\\phi_a=\arg\left(N\right)\$$, and $$\\phi_b=\arg\left(D\right)\$$ and then to recast $$\N\$$ and $$\D\$$ as $$\N=r_a\cdot\exp\left(j\,\phi_a\right)\$$ and $$\D=r_b\cdot\exp\left(j\,\phi_a\right)\$$. Now, everything just kind of falls out so that $$\H=\frac{r_a\,\cdot\,\exp\left(j\,\phi_a\right)}{r_b\,\cdot\,\exp\left(j\,\phi_b\right)}=\frac{r_a}{r_b}\exp\left(j\left[\phi_a-\phi_b\right]\right)\$$.

The magnitude is just $$\\frac{r_a}{r_b}\$$ and you can just read that straight off from the result. The phase is similarly easy to read as $$\\phi_a-\phi_b\$$, also just read off directly.

That's one of the nice things about Euler's in this context. It's easier to see why the magnitude is the way it is and how to get the right value for the phase, too.