Hello.
1. Low logic level (Vin=0V)
We don't consider this case because we can never have Vin=0V.
Is this correct?
No, you can actually have any voltage there. In the circuit it's connected to air, which would be a very low voltage white noise or something. We will assume that these inputs will only work with 0 or 5V.
So yes, it can be 0V, if you connect them to ground they will actually be in 0V.
Fourth diode (that is connected to emitter of input voltage) is reverse biased.
It's actually not even biased. Not forward, not reverse, because both potentials are theorically zero. Maybe in real life there's a few houndred of milivolts, but the diode won't polarize neither. Then the output transistor base is at 0V and the transistor is
OFF. Because of that, the output is "high".
Where to connect capacitor Cp and how to find fall and rise time of Vout?
What is the value of Vout in this case?
I would assume that the capacitor is connected between Vout and GND. That will make the output to rise and fall slower than without the capacitor.
This rise and fall times are how much it takes the output to reach 90% of the opposite state.
RISE transient:
The cap will charge through a 6K resistor.
Vc= Vi + Vf * e^(-t/(R*C)). --> Charging Formula
Vc = Vi + Vcc * e^[t/(6K*220uF)].
Vi is the initial voltage of the capacitor. It will be
Uces=0.2V. Vcc is 5V. "t" is what you have to calculate. Replace "Vc" by 90% of 5V (4,5V) and find "t". Apply "ln" (e base logarythm) to both terms to "eliminate" the "t" from the exponent. Do your math.
FALL transient:
The cap will discharge only through the transistor which ideally won't have any resistance (it does), but as it's not specified you should assume that the fall time is zero.
The logic function is NAND.