Say I have the following transfer function:

$$\mathcal{H}(s) = \frac{Bs^2}{a^2 + \frac{a}{Q}s +s^2}$$

I want to prove it is a Low-Pass filter. In order to do it I'd like to change it to phasor form so I can calculate the gain \$|G|\$ when \$\omega \rightarrow 0\$ and \$\omega \rightarrow \infty\$ (as I do with impedance). But how could I transform it to the phasor form?

On the other hand, would it be correct if I just did \$\mathcal{H}(s \rightarrow \infty) = B\$ and \$\mathcal{H}(s \rightarrow 0) = 0\$ to assume it is a Low-Pass filter? That way I would also say that for

$$ \mathcal{H}_1(s) = \frac{B}{a^2 + \frac{a}{Q}s +s^2} $$

we have \$\mathcal{H}_1(s \rightarrow \infty) = 0\$ and \$\mathcal{H}_1(s \rightarrow 0) = \frac{B}{a^2}\$ so it is a High-Pass filter?

Is my approach correct?

  • \$\begingroup\$ Just for information: This i.stack.imgur.com/0XtT7.png can be a low pass filter transfer function. This i.stack.imgur.com/jKh3k.jpg can present a high pass filter , all assuming that the coefficients are not zero. Substitute to s number jω if you want the phasor form. The j is the imaginary unit. Gain is the absolute value of the calculated complex H(jω) \$\endgroup\$
    – user136077
    Oct 2, 2022 at 22:16

2 Answers 2


For all 2nd order of forms:


Set \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$ and \$Q=\frac{\sqrt{b_2 \,b_0}}{b_1}\$ and find the following equivalent:

$$\begin{align*} \frac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0} = \overbrace{\left[\frac{a_2}{b_2}\right]}^{\text{gain}}&\left[\frac{\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+\frac1Q\left(\frac{s}{\omega_{_0}}\right)+1}\right]\tag{high pass term} \\\\ + \overbrace{\left[\frac{a_1}{b_1}\right]}^{\text{gain}}&\left[\frac{\frac1Q\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+\frac1Q\left(\frac{s}{\omega_{_0}}\right)+1}\right]\tag{band pass term} \\\\ + \overbrace{\left[\frac{a_0}{b_0}\right]}^{\text{gain}}&\left[\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+\frac1Q\left(\frac{s}{\omega_{_0}}\right)+1}\right]\tag{low pass term} \end{align*}$$

It's very easy to understand why. Set \$\sigma=0\$ and examine each term for \$\omega=0\$, \$\omega=\omega_{_0}\$ and \$\omega=\infty\$, ignoring the gain factors and examining the magnitudes:

$$\begin{align*} &&\omega&=0&\omega&=\omega_{_0}&\omega&=\infty\\\\ &\left|\quad\left[\frac{-\left(\frac{\omega}{\omega_{_0}}\right)^2}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1Q\left(\frac{\omega}{\omega_{_0}}\right)}\right]\quad\right| &&0&& Q&&1\tag{high} \\\\ &\left|\quad\left[\frac{j\frac1Q\left(\frac{\omega}{\omega_{_0}}\right)}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1Q\left(\frac{\omega}{\omega_{_0}}\right)}\right]\quad\right|&&0&& 1&&0\tag{band} \\\\ &\left|\quad\left[\frac{1}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2+j\frac1Q\left(\frac{\omega}{\omega_{_0}}\right)}\right]\quad\right|&&1&&Q&&0\tag{low} \end{align*}$$

I'll leave it to you to examine the arguments/arctangents (phase.)

  • \$\begingroup\$ What is \$\sigma \$? \$\endgroup\$
    – ludicrous
    Oct 3, 2022 at 11:02
  • 1
    \$\begingroup\$ @ludicrous I discussed it here with you. But in general, you need to understand that multiplication with complex numbers involves two things (from our meager minds' point of view): rotation and scaling. With real numbers, it's just scaling. But with complex numbers it is two things at once. \$\omega\$ represents rotation. \$\sigma\$ represents the scaling part. If \$\sigma\$ is negative and multiplied by time, then it leads to an inward-going spiral. If zero, just a circle. If positive, then an outward-going spiral. \$\endgroup\$
    – jonk
    Oct 3, 2022 at 11:37
  • 1
    \$\begingroup\$ @ludicrous Broadly speaking, we don't like positive \$\sigma\$ values when multiplied by time as it implies bad things. Negative is okay because it means something is damping out and won't go crazy. But if we just want to focus on the rotation part of things, sinusoidal stuff, then we set \$\sigma=0\$, which eliminates all the inward and outward spiraling bits and leaves us with just the circular motion part -- frequency. Look up 3blue1brown on youtube and find his videos on complex numbers. They are very good. \$\endgroup\$
    – jonk
    Oct 3, 2022 at 11:39
  • 1
    \$\begingroup\$ @ludicrous \$s=\sigma+j\omega\$ is complex. It has a real part, \$\sigma\$, and an imaginary part, \$\omega\$. If you raise \$e\$ to the \$s\$ power, so \$e^s\$, then you can use Euler's to see what that looks like. It's fixed in time, though. You can also now introduce time, by multiplying \$s\$ by \$t\$ and get \$e^{s\,t}\$ and now some real action takes place over time. If you don't want the spiraling part (scaling with time, inward and/or outward) then set \$\sigma=0\$ to get rid of that part and just focus on rotation with \$\omega\$. Bigger \$\omega\$ just means faster rotation rates. \$\endgroup\$
    – jonk
    Oct 3, 2022 at 11:47

The correct way to do the transformation is to set \$s=j\omega \$.

$$s=\sigma +j\omega$$ So $$H_{1}(s)=H_{1}(\sigma, j\omega)$$

Signals that are represented where \$\sigma<0\$ are called exponentially damped sinusoids, stable.

Signals where \$\sigma>0\$ are called exponentially growing sinusoids, unstable.

Signals where \$\sigma=0\$ are called sinusoids.

Signals where both \$\sigma=0\$ and \$\omega=0\$ are step, ramp et cetera.

So given a transfer function, then to find the sinusoidal frequency response we need the values of \$s\$ that represent sinusoids, so as mentioned, set \$s=j\omega\$.

So $$H_{1}(s)\rightarrow H_{1}(j\omega)=\frac{-B\omega^{2}}{\omega ^2-a^2+j\frac{a}{Q}\omega }$$

I leave the rest to you. If you plot \$|H_{1}(j\omega)|\$, then you will find that your transfer function is a high pass filter.

\$\mathcal{H}_1(s \rightarrow \infty) = 0\$ and \$\mathcal{H}_1(s \rightarrow 0) = \frac{B}{a^2}\$ so it is a High-Pass filter?

Don't do this. \$s\$ is two dimensional variable. Letting \$s\rightarrow \infty\$ will place the transfer function in instability because \$\sigma\rightarrow \infty\$ also.

  • 1
    \$\begingroup\$ I think there is a typing error (minus signs in the denominator for H1). Why should it not be allowed to let s approach infinity? Instability? For very large s values (infinity) we get H(s)=B (highpass response). Remember: For the "initial value theorem" we always assume s approaching infinity. \$\endgroup\$
    – LvW
    Oct 3, 2022 at 9:00

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