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

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.)

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.

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: $$H(\omega)=\frac{10j\omega+1}{(j\omega+10)(j\omega+1)}\approx\frac{(10j\omega+1)(1-j\omega)}{10}\approx0.1-0.9j\omega$$

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:

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

• Well, I already had this, modulo using the factor of 2 to get rid of the square roots. This doesn't answer the question of how to plot that function $(3)$ against $\log_{10}\omega$ by hand. Apr 28, 2017 at 16:04