Bode function - poles and zeros

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Grigorie Vlad
Grigorie Vlad 2020 年 3 月 16 日
コメント済み: Star Strider 2020 年 3 月 16 日
I have the following transfer function : G = 1/(2*(s*1E-2)*(1+s*1E-2));
I am looking to draw the phase and magnitude plot for this function. I used the following script :
s = tf('s');
G = 1/(2*(s*1E-2)*(1+s*1E-2));
bode(G)
[mag,phase,wout] = bode(G);
Is my aproach correct and if it is, can someone explain to me what are the poles and the zeros for the function ?
My guess is that I have 2 poles : one in 2*E2 and one in 1*E2. Is this correct ?

採用された回答

Star Strider
Star Strider 2020 年 3 月 16 日
The pzplot function:
s = tf('s');
G = 1/(2*(s*1E-2)*(1+s*1E-2));
bode(G)
[mag,phase,wout] = bode(G);
figure
pzplot(G)
shows one pole at -100 and another pole at 0.
Note that there is an error in the coding for ‘G’ that I corrected. There is either a misplaced parenthesis or a missing multiplication operator. The poles remain unchanged regardless how I permute those corrections.
  8 件のコメント
Grigorie Vlad
Grigorie Vlad 2020 年 3 月 16 日
Ok, now it is all clear. Thank you very much for the answer and for your time !
Star Strider
Star Strider 2020 年 3 月 16 日
As always, my pleasure!

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