To obtain the transfer function of the circuit, we substitute $Z_1=R_1$ and $Z_2=R_2 \|\left(1 / s C_2\right)$. Since $Z_2$ is the parallel connection of two components, it is more convenient to work in terms of $Y_2$; that is, we use the following alternative form of the transfer function:

$$
\frac{V_o(s)}{V_i(s)}=-\frac{1}{Z_1(s) Y_2(s)}
$$

and substitute $Z_1=R_1$ and $Y_2(s)=\left(1 / R_2\right)+s C_2$ to obtain

$$
\frac{V_o(s)}{V_f(s)}=-\frac{1}{\frac{R_1}{R_2}+s C_2 R_1}
$$


This transfer function is of first order, has a finite dc gain (at $s=0, V_o / V_i=-R_2 / R_1$ ), and has zero gain at infinite frequency. Thus it is the transfer function of a low-pass STC network and can be expressed in the standard form of Table 1.2 as follows:

$$
\frac{V_o(s)}{V_i(s)}=\frac{-R_2 / R_1}{1+s C_2 R_2}
$$

from which we find the dc gain $K$ to be

$$
K=-\frac{R_2}{R_1}
$$

and the $3-\mathrm{dB}$ frequency $\omega_0$ as

$$
\omega_0=\frac{1}{C_2 R_2}
$$

We could have found all this from the circuit by inspection. Specifically, note that the capacitor behaves as an open circuit at dc; thus at dc the gain is simply $\left(-R_2 / R_1\right)$. Furthermore, because there is a virtual ground at the inverting input terminal, the resistance seen by the capacitor is $R_2$, and thus the time constant of the STC network is $C_2 R_2$.

Now to obtain a dc gain of 40 dB , that is, $100 \mathrm{~V} / \mathrm{V}$, we select $R_2 / R_1=100$. For an input resistance of $1 \mathrm{k} \Omega$, we select $R_1=1 \mathrm{k} \Omega$, and thus $R_2=100 \mathrm{k} \Omega$. Finally, for a $3-\mathrm{dB}$ frequency $f_0=$ 1 kHz , we select $C_2$ from

$$
2 \pi \times 1 \times 10^3=\frac{1}{C_2 \times 100 \times 10^3}
$$

which yields $C_2=1.59 \mathrm{nF}$.
The circuit has gain and phase Bode plots of the standard form. As the gain falls off at the rate of $-20 \mathrm{~dB} /$ decade, it will reach 0 dB in two decades, that is, at $f=100 f_0=100 \mathrm{kH}$. As Fig. indicates, at such a frequency which is much greater than $f_0$, the phase is approximately $-90^{\circ}$. To this, however, we must add the $180^{\circ}$ arising from the inverting nature of the amplifier (i.e., the negative sign in the transfer function expression). Thus at 100 kHz , the total phase shift will be $-270^{\circ}$ or, equivalently, $+90^{\circ}$.