Electronic – Op-amp circuit analysis without input signal

I don't know what it says precisely in sedra smith but you can't do anything about input offset current. However, if you mean input bias current then that is eradicated by making the dc resistance seen by both inputs the same value. So, if there is a 100k feedback and a 100k input resistor to the inverting terminal then, ideally, you should have 50k seen by the non-inverting terminal to 0V. Thus bias currents produce an error voltage at each input terminal that is identical and thus cancel out.

Input offset current is the difference between the bias currents on each input terminal and this is not so easily got rid of.

Of course if the bias current is only 1nA and you are using 1k resistors (i.e. quite small values) then the worst case dc voltage error produced (if you don't have a resistor in the inverting input) is 1 uV - probably far less than the input offset voltage of the op-amp and therefore not worth bothering about.

See this (MT-038: Op Amp Input Bias Current - Analog Devices) for further reading on the subject.

Start by analyzing the steady state. That means all capacitors are open circuits. The first opamp then just inverts Vinput with respect to ground. The second then inverts it again, but with a gain of 1/2. Therefore Voutput = Vinput / 2.

Since both opamps are ideal and have feedback so that they operate in their linear region, the negative inputs of both will be held at 0 V.

Analyzing the dynamic behavior is more tricky, but you already know what the steady state will be after everything settles. C2 feeds current proportional to the derivative of Vinput into the node at the negative input of the second amp. This will add to the steady state signal. Voutput will therefore have a component that is proportional to -dVinput/dt.

C1 and R2 form a high pass filter, which in the feedback path results in a the second stage being a low pass filter. Whatever the Voutput would have otherwise been, it will be low pass filtered by a single pole at 1/2πR₂C₁. When R is in Ohms and C in Farads, then this expression is in Hz.