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I'm trying to verify whether the transfer function in this control diagram is the one stated in this document.

enter image description here enter image description here enter image description here I tried solving it and got this: Shifting the armature current $$I_a$$ feedback point from the left position all the way to the right at the angular velocity point and ignoring the torque disturbance input since I'm only concerned with the voltage input transforms the $$K_{cf}$$ feedback gain into $$\frac{0.075s}{13.2}$$. The lower feedback loop is reduced to $$\frac{2.5}{0.02s+1}\times\frac{13.2}{s}=\frac{33}{s(0.02s+1)}$$ which when combined with the $$K_e=1.49$$ gain becomes $$\frac{33}{49.17+s(0.02s+1)}$$. Now, only the upper feedback remains and upon multiplication with the power amp $$K_{pa}=80$$ gain, it becomes $$\frac{2640}{49.17+s(0.02s+1)}$$. Solving for the feedback loop yields $$\frac{34848}{198s}$$ and after multiplying with the radius $$R_p=1.25$$ and integrator $$\frac{1}{s}$$ yields $$\frac{43560}{198s^2}$$ which is very different from what is proposed in the second image in the post.

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