Phasor of transfer function

Question

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?

Answer 1

For all 2nd order of forms:

$$\frac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0}$$

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.)

Answer 2

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.