Exercise 11.3 Impedances from reactances

a)

An answer:
$Z_{\mathit{cap}}(\mathit{j\omega })=\frac{1}{\mathit{j\omega C}}+\mathit{j\omega l}\cdot 2\mathit{nH}∕\mathit{mm}$ with $l$ the total wire length in mm.

b)

An answer:

$$\begin{array}{llll}\hfill Z_{\mathit{cap}}(\mathit{j\omega })& =\frac{1}{\mathit{j\omega C}}+\mathit{j\omega L}& \hfill & \\ \hfill & =-j32\mathrm{\Omega }+j25\mathrm{\Omega }=-j7\mathrm{\Omega }& \hfill & \end{array}$$

c)

An answer:
About 230 pF.

d)

e)

An answer:
About 12 pF.

f)

An answer:
This yields a negative capacitor... that would be inductive behavior at 100 MHz. Value wise it is just a little below 40 nH.

g)

An answer:

$$\begin{array}{llll}\hfill \frac{1}{\mathit{j\omega C}}& =\frac{1}{\mathit{j\omega }{C}_{\mathit{target}}}-\mathit{j\omega L}& \hfill & \\ \hfill \frac{1}{\mathit{j\omega C}}& =\frac{1+{\omega }^{2}L{C}_{\mathit{target}}}{\mathit{j\omega }{C}_{\mathit{target}}}& \hfill & \\ \hfill & \to & \hfill & \\ \hfill C& =\frac{C_{\mathit{target}}}{1+{\omega }^{2}L{C}_{\mathit{target}}}\mathrm{\text{with C=100pF and L=40nH}}& \hfill & \\ \hfill C& \approx 39\mathit{pF}& \hfill & \end{array}$$