Solution:
$
\begin{aligned}
& \frac{1}{\sqrt{1+\varepsilon^2}}=0.8 \Rightarrow \varepsilon=0.75 \\\\
& \frac{1}{\sqrt{1+\lambda^2}}=0.2 \Rightarrow \lambda=4.899\\
\end{aligned}
$
Ws $=0.6$ pi rad: $w_p=0.2$ pi rad . then assume $\mathrm{T}=1 \mathrm{sec}$
$
\frac{\omega_{\mathrm{s}}}{\omega_{\mathrm{p}}}=\frac{\Omega_{\mathrm{s}} \mathrm{T}}\\{\Omega_{\mathrm{p}} \mathrm{T}}=\frac{\Omega_{\mathrm{s}}}{\Omega_{\mathrm{p}}}=\frac{0.6 \pi}{0.2 \pi}=3\\
$
$
\mathrm{N} \geq \frac{\log \left(\frac{\lambda}{\varepsilon}\right)}{\log \left(\frac{\Omega_{\mathrm{s}}}{\Omega_{\mathrm{p}}}\right)} \geq \frac{\log \left(\frac{4.899}{0.75}\right)}{\log 3} \geq 1.71\\
$
For $\mathrm{N}=2$ the transfer …
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