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I need to solve this three exercises: $$G_1(s) = \frac{1}{s}$$

$$G_2(s) = \frac{1}{s^2}$$

$$G_3(s) = \frac{1}{s^3}$$

This is what I did: enter image description here

For the single pole ($1/s$), magnitude of A is $\infty$ and phase is 0

For the single pole ($1/s$), magnitude of B is $\infty$ and phase is -45

For the single pole ($1/s$), magnitude of C is $\infty$ and phase is -90


For the 2 poles one ($1/s^2$), magnitude of A is $\infty$ and phase is 0

For the 2 poles one ($1/s^2$), magnitude of B is $\infty$ and phase is -90 ($=2*(-45)$)

For the 2 poles one ($1/s^2$), magnitude of C is $\infty$ and phase is -180 ($=2*(-90)$)


For the 3 poles one ($1/s^3$), magnitude of A is $\infty$ and phase is 0

For the 3 poles one ($1/s^3$), magnitude of B is $\infty$ and phase is -135 ($=3*(-45)$)

For the 3 poles one ($1/s^3$), magnitude of C is $\infty$ and phase is -270 ($=3*(-90)$)


MATLAB can't do these Nyquist analysis in a correct way. That's why I'm asking here.

Thanks

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Your answer is correct. Each pole gives 90 degrees phase lag. Furthermore, here can you download a file lnyquist1 which is able to give a more accurate result when you have a pole at zero.

num = 1;
den = [1 0 0 0];
lnyquist1(num,den);

In addition, WolframAlpha is also able to give an accurate result, see here.

| improve this answer | |
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  • $\begingroup$ That's a nice script, but the infinite is too small I think. If an improvement can be made, infinite has to be something like 300 instead of 80. $\endgroup$ – Euler Jul 5 '16 at 13:32

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