Given is transfer function: $G(s)=\frac{s(s^2+4s+100)}{s+1} $ I have to draw bode magnitude plot.
What I did:
$G(s)=100\frac{s(\frac{s^2}{100} +\frac{s}{25} +1)}{s+1} $
$Zeros: s_1=0, s_2=10 $
$Pole: s_3=-1$
So for zeros, slopes are: $+20dB$ for $s_1$, $+40dB$ for $s_2$.
For pole, slope is: $-20dB$ for $s_3$.
But I dont understand how to draw it now. Here is the solution. Can some explain me how to use my calculations to draw this plot?
Well, you need to use:
$$\text{s}=\omega\text{j}\tag1$$
Where $\text{j}^2=-1$
So, for the magnitude plot you need to plot:
$$20\log_{10}\left(\color{red}{\left|\frac{\omega\text{j}\left(\left(\omega\text{j}\right)^2+4\omega\text{j}+100\right)}{1+\omega\text{j}}\right|}\right)\tag2$$
Now, for the red part:
$$\left|\frac{\omega\text{j}\left(\left(\omega\text{j}\right)^2+4\omega\text{j}+100\right)}{1+\omega\text{j}}\right|=\frac{\left|\omega\text{j}\right|\left|\left(\omega\text{j}\right)^2+4\omega\text{j}+100\right|}{\left|1+\omega\text{j}\right|}=\frac{\omega\sqrt{16\omega^2+\left(100-\omega^2\right)^2}}{\sqrt{1+\omega^2}}\tag3$$
Using Mathematica to plot, it gives me:
You where on the right track. Namely you have identify all the values for $s$ for which the dominant order of the numerator or denominator changes. From now on I will call these points break points. For real valued zeros or poles then the (absolute values) will coincide with the break points. Assuming real valued coefficients in the transfer function then it is also possible to have complex conjugate pairs for zeros or poles, often represented with $s^2 + 2\,\zeta\,\omega\,s+\omega^2=0$, where $|\zeta|\leq1$. In that case the break point will lie at $|\omega|$.
You already got almost all break points correct, namely 0 and 10 for the zeros and 1 for the poles. Now you sort this list from smallest to largest and go through this list of break points and sketch asymptotes between them, starting at 0 (even if it is not in the list) till $\infty$.
In your case the list is $\{0,1,10\}$. Between 0 to 10 the numerator can be approximated by $100\,s$ and between 10 to $\infty$ with $s^3$. Between 0 to 1 the denominator can be approximated by $1$ and between 1 to $\infty$ with $s$. Combining these approximations for numerator and denominator yields that $G(s)$ can be approximated by with,
$$ G(s) \approx \left\{\begin{array}{ll} 100\, s, & \text{if}\ \phantom{1}0 \leq |s| < 1 \\ 100, & \text{if}\ \phantom{1}1 \leq |s| < 10 \\ s^2, & \text{if}\ 10 \leq |s| < \infty \end{array}\right. $$
The power of each approximation determines the slope of that approximation, namely the slope in dB/decade is equal to 20 times the power. So in your case the first approximation has a slope of 20 dB/decade, the second 0 dB/decade and the last 40 dB/decade. You will always have to plug in some value into one of the approximations to start your sketch, but once you have started, you can always continue by starting an adjacent assymptote at the end of the previous asymptote at the break point.
When sketching the complete bode magnitude plot you can use these asymptotes as guide lines. But at the break points neither of the asymptotes will be accurate in general, so I always just calculate those exactly using the original transfer function and sketch a line through that points that approaches the two neigboring asymptotes. However if you have a lot of break points close to each other, then the plot might not even approach the asymptotes in between break points. This happens more often for the phase of the bode plot. In this case you could improve the sketch by also calculating the slope at the break points.