I have the following transfer function and I need to draw the bold plot (magnitude and phase):
G = 1/(2*(s*1E-2)*(1+s*1E-2));
$$G(s)=\frac{1}{2\cdot10^{-2}s(10^{-2}s+1)}. $$
I am having a difficult time identifying the zeros and poles for this function because of the presence of the 2 constant. My guess is this function has 2 poles : one in 2*(10^2) and another one in 10^2. Is this correct ? Can someone please help me plot this function correctly ?
PS: Sorry for the bad format of the mathematic formula, this is my first question.
EDIT : I did the plots using matlab:
MATLAB ploted
I need help understanding how the magnitude plot is influenced by the 2 constant.
Best Answer
The poles are at $s=-j\omega=0$ and at $s=-j\omega=-100$ (so $0$ and $100$, not $200$ and $100$). Each pole adds a negative -20dB/decade slope to your magnitude plot and a -90 degrees to your phase plot.
Magnitude Plot: Because the first pole happens at $\omega=0$, the magnitude plot will start with the -20dB/decade slope. Then, at $\omega=100$ you will add an extra -20dB/decade for a total of -40dB/decade slope after that. You can make a rough sketch of this graph ignoring for the moment the values of the y-axis. Then, you can compute the gain of the system for some random frequency (let's say $\omega=10$) and adjust up and down the plot so that your sketch passes through the computed magnitude at $\omega=10$.
Phase Plot: The pole at 0 gives you a starting constant line at $-90^o$. The pole at 100 adds another $-90^o$. The transition from $-90^o$ to $-180^o$ roughly starts one decade earlier (at $\omega =10$) and ends one decade later (at $\omega = 1000$). Right at the pole location it goes through half the transition $-135^o$.