Exercise 4.1 Signal sources, loads and overall gain

a)

An answer:

Assuming voltage-controlled voltage sources as amplifiers or voltage controlled current sources with resistor — whatever floats your goat — you get:

$$\begin{array}{rcll}& \frac{v_{o1}}{v_i}& =a_{v1}\cdot \frac{r_{\mathit{in}1}}{r_{\mathit{in}1}+R_g}& \\ & & =30\cdot \frac{6\mathrm{\text{k}}\mathrm{\Omega }}{6\mathrm{\text{k}}\mathrm{\Omega }+600\mathrm{\Omega }}\approx 27& \\ & \frac{v_{o2}}{v_i}& =a_{v2}\cdot \frac{r_{\mathit{in}2}}{r_{\mathit{in}2}+R_g}& \\ & & =60\cdot \frac{600\mathrm{\Omega }}{600\mathrm{\Omega }+600\mathrm{\Omega }}=30& \end{array}$$

b)

An answer:
The equivalent relevant part of the circuit for this question is given below.

$$\begin{array}{rcll}& H(\mathit{j\omega })& =\frac{v_o}{v_i}& \\ & & =\frac{r_{\mathit{in}}}{r_{\mathit{in}}+R_g+\frac{1}{\mathit{j\omega }{C}_C}}& \\ & & =\frac{\mathit{j\omega }{r}_{\mathit{in}}{C}_C}{1+\mathit{j\omega }(r_{\mathit{in}}+R_g)C_C}& \\ & & =\frac{r_{\mathit{in}}}{r_{\mathit{in}}+R_g}\cdot \frac{\mathit{j\omega }(r_{\mathit{in}}+R_g)C_C}{1+\mathit{j\omega }(r_{\mathit{in}}+R_g)C_C}& \end{array}$$

The cut-off frequency of this first order high pass transfer function is

$$\begin{array}{rcll}{f}_c& =\frac{1}{2\pi C_C(r_{\mathit{in}}+R_g)}& & \end{array}$$

Solving for $C_C$ gives:

$$\begin{array}{rcll}{C}_C& =6\mu \mathrm{\text{F}}& (r_i=600\mathrm{\Omega })& \\ C_C& =1.2\mu \mathrm{\text{F}}& (r_i=6\mathrm{\text{k}}\mathrm{\Omega })& \end{array}$$

c)

An answer: