Second order system - find -3dB frequencies and magnitude response analytically

Question

Let's take some simple second-order system like
$H(s) = \frac{j\omega T}{(1+ j \omega T)^2} $.
I know that the magnitude response is simply the absolute of the function and the -3dB frequencies can be found by solving for $|H(s)| = \frac{1}{\sqrt{2}}$. But I can't figure out how to analytically solve for $\omega$. No matter how I try to solve, I find myself in a dead end with $\omega ^3$ or $\omega ^4 $.

So how would be the correct approach look like here?

Answer 1

Notice:

For the $-3$ dB points, you know that $\left|\text{H}(\omega)\right|=\frac{1}{\sqrt{2}}$ and we can say that:

$$\color{red}{20\log\left(\left|\text{H}(\omega)\right|\right)=20\log\left(\frac{1}{\sqrt{2}}\right)=-10\log(2)\approx-3.0103}$$

Where $\log$ is the base $10$ logarithm.

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First, when $z\in\mathbb{C}$:

$$|z|=\left|\Re[z]+\Im[z]i\right|=\sqrt{\Re^2[z]+\Im^2[z]}$$

So, in your problem (assuming $j^2=i^2=-1$ and $\text{T}\space\wedge\space\omega\in\mathbb{R}$):

• $$\left|\text{H}(\omega)\right|=\left|\frac{j\omega\text{T}}{\left(1+j\omega\text{T}\right)^2} \right|=\frac{\left|j\omega\text{T}\right|}{\left|1+j\omega\text{T}\right|^2}=\frac{\omega\text{T}}{1+\left(\omega\text{T}\right)^2}$$

Now:

$$\left|\text{H}(\omega)\right|=\frac{1}{\sqrt{2}}\Longleftrightarrow\frac{\omega\text{T}}{1+\left(\omega\text{T}\right)^2}=\frac{1}{\sqrt{2}}\Longleftrightarrow$$ $$\sqrt{2}\omega\text{T}=1+\left(\omega\text{T}\right)^2\Longleftrightarrow-\text{T}^2\omega^2+\sqrt{2}\omega\text{T}-1=0$$

To solve this, use the quadratic formula:

$$-\text{T}^2\omega^2+\sqrt{2}\omega\text{T}-1=0\Longleftrightarrow\omega=\frac{1\pm i}{\text{T}\sqrt{2}}$$

So, we can conclude: there are no $-3$ dB points in the real plane!

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Now, to solve $\omega$ out of $\left|\text{H}(\omega)\right|$ :

$$\left|\text{H}(\omega)\right|=\frac{\omega\text{T}}{1+\left(\omega\text{T}\right)^2}\Longleftrightarrow$$ $$\omega\text{T}=\left|\text{H}(\omega)\right|\left(1+\left(\omega\text{T}\right)^2\right)\Longleftrightarrow$$ $$\omega\text{T}-\left|\text{H}(\omega)\right|\omega^2\text{T}^2-\left|\text{H}(\omega)\right|=0$$

Again, use the quadratic formula:

$$\omega=\frac{\text{T}\pm\sqrt{\text{T}^2\left(1-4\left|\text{H}(\omega)\right|^2\right)}}{2\left|\text{H}(\omega)\right|\text{T}^2}$$

Assuming that $\text{T}\in\mathbb{R}^+$:

$$\omega=\frac{1\pm\sqrt{1-4\left|\text{H}(\omega)\right|^2}}{2\text{T}\left|\text{H}(\omega)\right|}$$

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The quadratic formula

$$ax^2+bx+c=0\Longleftrightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$