A Hartley oscillator

a)

An answer:
$C_{0}$ makes sure that there is zero DC-current in $L_{2}$ , and that hence $L_{2}$ does not influence the bias point

b)

An answer:
$g_{m} = 40 \cdot I_{C} = 0.2 S$

c)

An answer:

pict

d)

An answer:

\begin{array}{l} A_{loop} & = - \frac{g_{m} \cdot Z_{A} \cdot Z_{C}}{Z_{A} + Z_{B} + Z_{C}} \equiv 1 \end{array}

with

\begin{array}{l} Z_{A} = \frac{jω L_{2} R}{jω L_{2} + R} \\ Z_{B} = \frac{1}{jωC} \\ Z_{C} = jω L_{1} \end{array}

where $R = \frac{βfe}{g_{m}}$ . This leads to

\begin{array}{rcl} - g_{m} \cdot \frac{jω L_{2} R}{jω L_{2} + R} \cdot jω L_{1} = \frac{jω L_{2} R}{jω L_{2} + R} + \frac{1}{jωC} + jω L_{1} \end{array}

Collecting the imaginary terms and equating them to zero yields

\begin{array}{l} jω L_{2} R + \frac{R}{jωC} + jω L_{1} R = 0 \to \\ - ω^{2} C (L_{1} + L_{2}) + 1 = 0 \to \\ ω_{osc}^{2} = \frac{1}{C (L_{1} + L_{2})} \end{array}

yielding the oscillation (radian) frequency $ω_{osc} = \frac{1}{\sqrt{C (L_{1} + L_{2})}}$

e)

An answer:

Equating the loop gain, at the oscillation frequency, and equating that to 1 gives

\begin{array}{l} - g_{m} \cdot j ω_{osc} L_{2} R \cdot j ω_{osc} L_{1} = \frac{L_{2}}{C} + j ω_{osc} L_{1} j ω_{osc} L 2 \to \\ (g_{m} \cdot R + 1) \cdot ω_{osc}^{2} L_{2} L_{1} = \frac{L_{2}}{C} \to \\ (βfe + 1) \cdot \frac{L_{2} L_{1}}{C (L_{1} + L_{2})} = \frac{L_{2}}{C} \to \\ (βfe + 1) \cdot \frac{L_{1}}{L_{1} + L_{2}} = 1 \to \\ L_{2} = βfe \cdot L_{1} \end{array}