How to graph, by hand, a log-log plot of the amplitude of $\frac{10j\omega+1}{(j\omega+10)(j\omega+1)}$?

Question

I'm doing a practice final exam for my signals and systems engineering class. One question asks for an "amplitude Bode plot" of the function

$$H(\omega)=\frac{10j\omega+1}{(j\omega+10)(j\omega+1)}$$

where $j$ represents the complex unit $\sqrt{-1}$.

That just means I need to sketch the graph of $20\log_{10}|H(\omega)|$ versus $\log_{10}\omega$. But I have no idea how to do this by hand. I do know that

$$20\log_{10}|H(\omega)|=20\log_{10}\sqrt{100\omega^2+1}-20\log_{10}\sqrt{\omega^2+100}-20\log_{10}\sqrt{\omega^2+1}$$

but I don't know how I'm supposed to sketch the graph of this as a function of $\log_{10}\omega$. (The book is awful and doesn't explain how it could be done by hand, without a calculator or graphing software.)

Answer 1

Here is a simple method to draw asymptotic bode plot (briefly mentioned by Andrei).

1. Write the transfer function as $$H(j\omega) = \frac{1}{10}\frac{j\frac{\omega}{1/10}+1}{(j\frac{\omega}{10}+1)(j\omega+1)}.$$
2. Write $20\log |H(j\omega)|$ and decompose it as follows using log and complex modulus properties $$20\log |H(j\omega)| = 20\log \frac{1}{10} + 20\log |j\frac{\omega}{1/10}+1| - 20\log|j\frac{\omega}{10}+1| - 20\log|j\omega+1|.$$
3. Analyse each term separately and, if needed, derive their asymptotic behavior. The first term is simply a constant. The second term will be 0 for $\omega \ll 1/10$ and a +20dB/dec line for $\omega \gg 1/10$. The third term will be 0 for $\omega \ll 10$ and a -20dB/dec line for $\omega \gg 10$. The fourth term will be 0 for $\omega \ll 1$ and a -20dB/dec line for $\omega \gg 1$. Putting it all together, this gives the following asymptotic bode plot (amplitude only).

This can easily be verified to be correct using Matlab for example.

Answer 2

Bode plots can be produced by approximations at low and large frequencies. In fact, the wikipedia page about Bode plots has a section about making this plots by hand https://en.wikipedia.org/wiki/Bode_plot#Rules_for_handmade_Bode_plot

Use the fact that $20\log_{10}{\sqrt{f(x)}}=10\log_{10}{f(x)}$. At large $\omega$, you can neglect the $+1$ and $+10$ parts in $H(\omega)$, so you get $H(\omega)\approx\frac{10}{j\omega}$. At small $\omega$, just neglect the denominator dependence on $\omega$, so $H(\omega)\approx0.1+j\omega$. If you want to be more correct, you can do a Taylor expansion or something like: $$H(\omega)=\frac{10j\omega+1}{(j\omega+10)(j\omega+1)}\approx\frac{(10j\omega+1)(1-j\omega)}{10}\approx0.1-0.9j\omega$$

Answer 3

Well, we have:

$$\underline{\mathscr{H}}\left(\omega\right):=\frac{1+10\cdot\omega\cdot\text{j}}{\left(10+\omega\cdot\text{j}\right)\cdot\left(1+\omega\cdot\text{j}\right)}\tag1$$

Where $\text{j}^2=-1$ and $\omega\in\mathbb{R}$

So, for the absolute value we get:

$$\left|\underline{\mathscr{H}}\left(\omega\right)\right|=\left|\frac{1+10\cdot\omega\cdot\text{j}}{\left(10+\omega\cdot\text{j}\right)\cdot\left(1+\omega\cdot\text{j}\right)}\right|=\frac{\left|1+10\cdot\omega\cdot\text{j}\right|}{\left|10+\omega\cdot\text{j}\right|\cdot\left|1+\omega\cdot\text{j}\right|}=$$ $$\frac{\sqrt{1^2+\left(10\cdot\omega\right)^2}}{\sqrt{10^2+\omega^2}\cdot\sqrt{1^2+\omega^2}}=\frac{\sqrt{1+100\cdot\omega^2}}{\sqrt{100+\omega^2}\cdot\sqrt{1+\omega^2}}\tag2$$

Take the $20\log_{10}$ of both sides:

$$20\log_{10}\left(\left|\underline{\mathscr{H}}\left(\omega\right)\right|\right)=20\log_{10}\left(\frac{\sqrt{1+100\cdot\omega^2}}{\sqrt{100+\omega^2}\cdot\sqrt{1+\omega^2}}\right)=$$ $$20\cdot\left(\log_{10}\left(\sqrt{1+100\cdot\omega^2}\right)-\left(\log_{10}\left(\sqrt{100+\omega^2}\right)+\log_{10}\left(\sqrt{1+\omega^2}\right)\right)\right)=$$ $$20\cdot\left(\frac{\log_{10}\left(1+100\cdot\omega^2\right)}{2}-\left(\frac{\log_{10}\left(100+\omega^2\right)}{2}+\frac{\log_{10}\left(1+\omega^2\right)}{2}\right)\right)=$$ $$10\cdot\left(\log_{10}\left(1+100\cdot\omega^2\right)-\log_{10}\left(100+\omega^2\right)-\log_{10}\left(1+\omega^2\right)\right)=$$ $$\frac{10}{\ln\left(10\right)}\cdot\left(\ln\left(1+100\cdot\omega^2\right)-\ln\left(100+\omega^2\right)-\ln\left(1+\omega^2\right)\right)\tag3$$