L.P and H.P circuits

[omerysmi]

In this circuit the input is step signal, I need to find an experssion of Vo(0+) and of Vo(∞).

My answer is:
Vo(0+) = (E*R2)/(R1+R2)
Vo(∞) = E

Am I right?

 

[Harald Kapp]

Welcome to our forum.
Unfortunately, your answer is not correct - at least as far as I understand it. Assuming that Vo(0+) is the output voltage for the time where Vi=0V, then how and why should it be influenced by "E" at all?
If the assumption is incorrect, then Vo(0+) makes no sense.

You need to find an expression for Vo=Vo(t)=f(E,t) where t=time with t=0 at the moment of the step change (this is for ease of calculation, you can set t=0 anywhere, but the expression becomes slightly more complicated). f(E,t) will contain terms using R1, R2 and C1 (the capacitor is missing in your equation).
You will need to use the relation I(c)=C*dv/dt or equivalently V(c)=1/C*integral(i(c)*dt).

To get a feeling for the circuit imagine the two following operating conditions:
- what happens if Vi is a DC signal? What effect has this on Vo?
- what happens if Vi is a high frequency AC signal? Consider the impedance of C1 for high frequencies and its relation to R1, R2. What is Vo (in a first approximation)?

 

[Harald Kapp]

One more hint: using I(c)=C*dv/dt you will arrive at a differential equation. There are many ways to solve a differential equation. A good starting point is to assume an exponential output like Vo=a*(1-exp(b*t) and solve for the parameters a and b.

 

[hevans1944]

Vo(0+) occurs after the input voltage, Vi, transitions from 0 to E. At time Vo(0+), Vi = E. Vo(0+) will then be Vi R1/(R1+R2), assuming C1 is fully discharged at Vo(0). So the first part of your answer is correct.

Vo will rise exponentially to toward Vi = E until at Vo(∞) = Vi = E. So the second part of your answer is correct.

Of course in a real circuit Vo never reaches Vi, but after a sufficient number of "time constants" (typically six) we assume the capacitor is fully charged to the input voltage Vi and no more current flows in R1 or R2. Note that after the capacitor is fully charged, the voltage divider consisting of R2 in series with R1 no longer has any effect on Vo. It only has an effect when the capacitor is charging or discharging. For this circuit, the RC time constant is (R1+R2) C1.

 

[Arouse1973]

Nice Harald.
Adam

 

[hevans1944]

In this circuit the input is step signal, I need to find an experssion of Vo(0+) and of Vo(∞).

My answer is:
Vo(0+) = (E*R2)/(R1+R2)
Vo(∞) = E

Am I right?

Yes.
Yes.
See my post #4 above.