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


2 Answers 2


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.

  • $\begingroup$ Thanks for your answer, it makes more sense now. $\endgroup$
    – n33
    Commented Apr 14, 2019 at 21:43

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.