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For example, the step response of $\frac{s+1}{s+2}$ is decreasing although its gain is +1, and this system has higher gain at high frequencies than low frequencies based on its Bode plot. But if I make the transfer function strictly proper, no matter how transient the pole I added is, such as $\frac{s+1}{(s+2)(s+1000000000000)}$, the system tends to respond "nicely" like those responses shown in textbooks and professor's examples.

  1. Mathematically, $\frac{s+1}{s+2}=1-\frac{1}{s+2}$, so its response to a step input is an exponential decay, is there a more physical/intuitive way to explain this?
  2. Why systems with higher gain at high frequencies have decreasing response to step inputs?
  3. Is there a reason that we prefer increasing(nice) step response (or do we prefer it at all)?

By nice, I mean something like this

By not nice, I mean something like this

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Normally a lead compensator is used to increase the phase of the open loop transfer function in a certain frequency range, while trying to keep the magnitude close to constant. So one would normally not encounter a lead compensator all by itself.

But to also answer the question regarding the shape of the step response you can use that

$$ \lim_{s\to0} \frac{s+1}{s+2} = \frac{1}{2} $$

$$ \lim_{s\to\infty} \frac{s+1}{s+2} = 1 $$

So low frequencies get halved, while high frequencies pass straight through. The moment the step starts the input has an infinite slope which one could also view as a "high frequency", so passes straight through the lead compensator. But the steady state response to the step would be a "low frequency", so that magnitude would get halved. So initially the step would pass straight through the lead compensator, but as time progresses its magnitude would drop to a half.

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  • $\begingroup$ Thanks for your answer, it makes more sense now. $\endgroup$ – n33 Apr 14 at 21:43
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There's something wrong with the simulation program you're using. When you add the transient pole the program realizes the main pole is -2 and makes the calculations right. Try to simulate the step responses after eliminating s+1 (the zero). Maybe the problem is that the zero is confusing the program.

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