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## Homework Statement

Design an integrator to integrate a ##2 kHz## square wave with peak to peak amplitude ##0.5V##.

Choose ##R_F = 150 k##, ##C_I = 10 \mu F##.

Choose ##C_F## such that ##R_F >> \frac{1}{\omega C_F}##.

Choose ##R_I## such that ##R_IC_F << 0.5 ms## and ##\frac{1}{\omega C_I} << R_I##.

Comment on the phase shift.

## Homework Equations

## The Attempt at a Solution

I am having difficulty with this for some reason.

I derived the equation for the square wave (in volts) to be:

##v_i(t) = 0.25, \space 0 \leq t \leq \frac{1}{4000} s##

##v_i(t) = -0.25, \space \frac{1}{4000} s \leq t \leq \frac{1}{2000} s##

Now here's where I had trouble. I don't know how to meet the specifications which have been prescribed.

I tried deriving a transfer function (which was an absolute mess) using:

$$\frac{v_o}{v_i} = - \frac{Z_2}{Z_1} = - \frac{Z_{C_F} || R_F}{R_S + R_I + Z_{C_I}}$$

After massaging the expression a bit, I got:

$$\frac{v_o}{v_i} = - \frac{ \frac{R_F}{C_F} }{ (\frac{R_S}{C_F} + \frac{R_I}{C_F} + \frac{R_F}{C_I}) + j( \omega(R_F R_S + R_F R_I) - \frac{1}{\omega C_F C_I} ) }$$

Which I believe to be incorrect, because when I substitute ##\omega = \omega_0 = \frac{1}{R_F C_F}## into the gain, the magnitude of the denominator is not ##\sqrt{2}## due to the ##- \frac{R_F}{C_I}## term.

I can't see what I'm doing wrong in terms of the transfer function, and I'm not sure it's the right way.

It would be much appreciated if someone could assist me with this.

Thank you.