# How we determine type of filter with pole(s), zero(s)? [duplicate]

Let's say we have this Laplace transform:

$$H_{1}(s)=\frac{1}{(s+1)(s+3)}\;, \; \Re{e} (s)>-1$$

So, we know that there is a poles at $$s=-1$$ and $$s=-3$$.

With these informations, we found that to be a low-pass filter in class. How? How do we sketch the frequency response? What are the conditions to determine filter types when we only know pole(s) and zero(s)?

The frequency response of such a transfer function is given by letting $$s=j\omega$$:

$$H(\omega) = H(s)\Big|_{s=j\omega} = \frac{1}{(j\omega+1)(j\omega+3)} = \frac{1}{(3-\omega^2)+4j\omega}$$

Rough estimation of the filter type:

$$H(0) = \frac{1}{3}, \ \ \ H(\infty) \to 0$$

and we know it's a lowpass filter. Also further maths can be done

$$|H(\omega)|^2 = \left|\frac{1}{(3-\omega^2)+4j\omega}\right|^2 = \frac{1}{(3-\omega^2)^2+(4\omega)^2}=\frac{1}{\omega^4+10\omega^2+9}$$

The squared magnitude response is illustrated as follows

Supplemental answers. ZRHan has worked out the math already. Here is just a different way to plot it so you can really see the lowpass character.

The graphical interpretation can be derived by looking at very high and very low frequencies.

For $$\omega \ll 1$$ we assume $$( 1 + j\omega) \approx 1$$ and the transfer function becomes

$$H(\omega) \approx \frac{1}{(1)\cdot (3)} = \frac{1}{3}$$

so it's constant horizontal line.

For $$\omega \gg 3$$ we assume $$( 3 + j\omega) \approx j\omega$$ and the transfer function becomes

$$H(\omega) \approx \frac{1}{(j\omega)\cdot (j\omega)} = -\frac{1}{\omega^2}$$

which (on a log/log scale) is a falling line with a slope of -40dB per decade.

The 3dB point is in the vicinity of the poles, around $$\omega_c \approx 0.91$$.

If the poles were further apart you would actually see another line with a slope of -20dB per decade between the poles. But in this case the poles are too close together.