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I was going through some practice problems on Control System Modelling and came across this problem. It seems like the answer to this question is that the time constant is "reduced" but the result that we get from a quick tranfer function analysis is counter to this.

Let's assume that the tranfer function of the system without the tachometer feedback is $$G(s) = \frac{k}{1 + s\tau} \\\small\text{ where }\tau \text{ is the initial time constant}$$ Applying the tachometer feedback F(s) = fs in feedback loop, the overall transfer function will be $$G^{'}(s) = \frac{G(s)}{1 + G(s)F(s)} = \frac{\frac{k}{1 + s\tau}}{1 + \left ( \frac{k}{1 + s\tau}\right ) fs} = \frac{k}{1 + s\tau + kfs} = \frac{k}{1 + s(\tau + kf)} = \frac{k}{1 + s\tau_{new}}$$ Thus, $$\tau_{new} = \tau + kf$$ Hasn't the time constant increased? Or is there something wrong with this?

Couldn't really find a source which discusses this at length, but here's a relevant link: Effect of tachometer feedback in a control system

ocrdu
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erstan
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1 Answers1

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It`s not so simple. The principle called "tachometer feedback" is better known as "Pseudo-Derivative Control". I propose to use this term for starting a google search for more information. There are many articles on the subject.

After applying this method the single-loop feedback system has become a new system with a local loop (in addition to the overall loop) which realizes the tachometer principle (speed in and voltage out). That means - you must not simply add a function k*s in the feedback path to apply this principle.

LvW
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