circuit analysisoperational-amplifier
$$
R_s = 1k\Omega.
$$
http://i.stack.imgur.com/0iv6H.png
I don't understand why when the resistance is infinite $$v_1 = V_s$$.
It's clear that when a resistor has a very large value, the intensity is $$If = \frac{V}{ inf} = 0$$ and then $$I_s = I_1$$.
But in the second case (when the resistence f is 40kΩ) i don't know why the $$I_s=0$$ and $$v_1 = \frac{5}{6}vs$$
The usual blanket statement of no overshoot for a critically damped system makes an implicit assumption about the initial conditions. A critically damped system will overshoot at most once.
To intuitively see this, imagine a mass spring damper system where the mass is released at t=0 and finds an equilibrium condition. That response will have no overshoot (say a door is release and it closes automatically).
On the other hand, if the mass is moving rapidly at t=0 (say the door is given a mighty push just before t=0) it may well overshoot before reaching a final equilibrium.
For a detailed mathematical treatment, minus hand waving, see, for example, this.
If it's a step response from equilibrium there should be NO overshoot, and any overshoot is an indication of underdamping.
As Dave Tweed mentions, if some overshoot can be tolerated, then a faster rise time can be achieved by slightly underdamping. For example, for a tolerable overshoot of 5%, \$\zeta \$ should be about 0.69. The 10% to 90% rise time is then about 35% faster than the critically damped case. Perhaps they are illustrating such a design decision, which is pretty common.
There seems to be quite a few bad examples of formulas out there, so be careful. Here's one example from here, which is totally incorrect.
When an ideal op amp is connected with negative feedback, it obeys two rules:
In your first circuit, \$V_S\$ is only connected to the non-inverting input. By rule #2, no current flows into that input. This lets us calculate the equivalent input resistance:
$$I_S = 0\ \mathrm A$$ $$R_{in} = \frac{V_S}{I_S} = \frac{V_S}{0\ \mathrm A} = \infty \ \Omega$$
Your second circuit is drawn incorrectly. You've connected R2 to one of the op amp's power pins, but it should be connected to the output. In this circuit, current can flow from \$V_S\$ to the output, although none flows into the inverting input. Here we need rule #1, which tells us that the voltage at the inverting input is equal to the voltage at the non-inverting input -- zero volts (ground). So the circuit acts like the right side of \$R_1\$ is grounded, which makes \$R_1\$ the input resistance. We can work out the math, too:
$$I_S = \frac{V_S - 0\ \mathrm V}{R_1} = \frac{V_S}{R_1}$$
$$R_{in} = \frac{V_S}{I_S} = \frac{V_S}{\frac{V_S}{R_1}} = R_1$$
Best Answer
You have things switched up. It's exactly opposite what you say. When Rf = inf there's no feedback path, and then V1=5/6Vs and Is = V1/5k because there's no other path for the current to go.
When Rf=40k now there's a feedback path, and if you perform the math by equating Vo to Vi by tracing the currents from the right to the left side through the op-amps you'll end up with Vo=9V1. And if you determine the current through the 5k and determine the current through Rf, you'll find that they're equal and opposite which means that all of the current flowing into or out of the 5kohm resistor is actually going through the 40k resistor as well leaving no current left to go through Rs. Rf chosen to be 40k is what allows for it in this circuit.
Your two cases are actually these:
1. Rf=inf. V1=5/6*Vs. (Vs-V1)/1k = Is
2. Rf=40k. V1=Vs. Is=0
If you're still confused, let me know and I'll add in equations and numbers.