# Why does a lead-compensator type system decrease to a step response

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 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.