An $N=3$ Chebyshev bandpass filter is to be designed with $3 d B$ passband ripple for a communication link. The center frequency is at $2.4 \mathrm{GHz}$ and filter has to meet bandwidth requirement of $20 \%$. The filter has to be inserted into $50 \Omega$ characteristic line impedance.

Solution:

center frequency of filter $=2.4 \mathrm{GH} 2$

Bandwidth requicement of filter $=20 \%$

$\%$ Bandwidth of filter $=\frac{\text { upper freq. limit - lower freq. limit }}{\text { center frequeacy }} \times 100$

$ \begin{aligned}\\ &\therefore \quad 20=\frac{\mathrm{Fu}-\mathrm{Fl}}{\mathrm{fc}} \times 100\\\\ &\therefore \frac{f_u-f_l}{f_c}=0.2\\\\ &\therefore \mathrm{fu}-\mathrm{fl}=0.2 \mathrm{fe}=0.2 \times 2.4 \mathrm{GH}=0.48 \mathrm{GH}_2\\\\ &\therefore \mathrm{F}_4-\mathrm{Fl}=0.48 \mathrm{GH}\\\\ &\therefore \quad \mathrm{Fu}+\frac{0.484 \mathrm{~Hz}}{2}=\text { uppertimit }\\\\ &\therefore \quad \mathrm{fu}=2.49 \mathrm{H}_2+\frac{0.48 \mathrm{GH}}{2}=2.64 \mathrm{GHz}\\\\ &\mathrm{Fl}=2.4 \mathrm{GHz}-\frac{0.48 \mathrm{HHz}}{2}=2.16 \mathrm{GHz}\\ \end{aligned}\\ $

$ \therefore \% B W=\frac{f 4-f l}{f c} \times 100=\frac{2.649 \mathrm{H}_2-2.164 \mathrm{HL}}{2.4 \mathrm{GH}} \times 100=20 \% \\ $

so,

lower frequency limit of $B P F=\mathrm{Fl}=2.16 \mathrm{GHz}$

upper frequency limit of $B P F=F u=2.64 G \mathrm{~Hz}$

Ripple magnitude in passband $=G R=3 d B$

$z_G=z_L=50 \Omega$

Design of Low Pass filter prototype:

order of filter $=N=3$

For Chebyshev filter of older $N=3$ with ripple of $3 d B$ value, Filter root values from table,

$ \begin{aligned} \\ &g_0=1 \\\\ &g_1=3.3487 \\\\ &g_2=0.7117 \\\\ &g_8=3.3487 \\\\ &g_4=1 \\ \end{aligned} \\ $

$g_1 \& g_3$ will provide inductor while $g_2$ will provide capacitor.

$ \begin{aligned}\\ &X_L=g_1=g_3=3.3487 \quad X C=\frac{1}{g_2}=\frac{1}{0.7117}=1.405 \\\\ &X_L=2 \pi F L \text { with } F=F_C=\text { lowel fieq.limit }=2.16 \mathrm{GH} \\\\ &\therefore L=\frac{X_L}{2 \pi F_c} \times Z_0 \\\\ &\therefore L=\frac{3.3487}{2 \pi \times 2.16 \times 10^9} \times 50=12.3 \mathrm{nH}\\ \end{aligned}\\ $

$ \begin{aligned} \\ &x_c=\frac{1}{2 \pi \mathrm{fc}} \text { with } f=f_c=10 \text { wel freq. limit }=2.16 \mathrm{GHz} \\\\ &\therefore c=\frac{1}{2 \pi \mathrm{fc}_{\mathrm{c}} \mathrm{H}_c} \times \frac{1}{20}=\frac{1}{2 \pi \times 2.16 \times 10^9 \times 1.405} \times \frac{1}{50}=1.048 \mathrm{PF}\\ \end{aligned}\\ $

$ \text { The low pass ' } T \text { ' section filter prototype is as below, }\\ $

$ \begin{aligned}\\ &\text { Design of Bandpass filter: }\\\\ &\begin{aligned}\\\\ f_0=\sqrt{f u \times f l}=\sqrt{(2.64 \times 2.16) \times 10^9 \times 10^9} &=2.387 \times 10^9 \mathrm{HH}_2 \\\\ &=2.387 \mathrm{GHL}\\ \end{aligned}\\\\ &g_L=g_1=g_3=3.3487, \quad g_c=g_2=0.7117\\ \end{aligned}\\ $

$ \begin{aligned} \\ &\omega_u=2 \pi \mathrm{Fu}_u=2 \pi \times 2.64 \mathrm{GHz}=16.59 \mathrm{GH} \\\\ &\omega_l=2 \pi \mathrm{Fl}=2 \pi \times 2.16 \mathrm{GHz}=13.57 \mathrm{GHz} \\\\ &\omega_0=\sqrt{\omega_u \times \omega_l}=\sqrt{16.59 \times 13.57} \mathrm{GHz}=15 \mathrm{GHz} \\\\ &\therefore C_{B P_1}=\frac{\omega_u-\omega l}{\omega_0^2 \times g \mathrm{~L}}=\frac{16.59 GHz-13.57GHz \mathrm{}}{\left(15 \mathrm{HH}_2\right)^2 \times 3.57 \mathrm{GHz}} \\\\ &C_{B P_1}=\frac{(16.59-13.57) \times 10^9}{\left(15 \times 10^9\right)^2 \times 3.3487}=4 \mathrm{PF}\\ \end{aligned}\\ $

$ \begin{aligned}\\ &C B P_1=\frac{(16.59-13.57) \times 10^9}{\left(15 \times 10^9\right)^2 \times 3.3487}=4 \mathrm{PF}\\\\ &L_{B P_1}=\frac{g_L}{\omega u-\omega d}=\frac{3.3487}{(16.59-13.57) \times 10^9}=1.1 \mathrm{nH}\\\\ &L_{B P_2}=\frac{w_4-w l}{w_0^2 g_c}=\frac{(16.59-13.57) \times 10^9}{\left(15 \times 10^9\right)^2 \times 0.7117}=18.85 \mathrm{PH}\\\\ &C_{B P_2}=\frac{g_c}{w_u-w_l}=\frac{0.7117}{(16.59-13.57) \times 10^9}=0.235 \mathrm{nF}\\ \end{aligned}\\ $

$ \begin{aligned} \\ &\text { Scaliug capacitors \& inductors: } \\\\ &C_{B P_1}=\frac{C B P_1}{50}=\frac{4 P F}{50}=0.08 \mathrm{PF} \\\\ &L_{B P_1}=L_{B P_1 \times 50}=1.1 \mathrm{nH} \times 50=55 \mathrm{nH} \\\\ &L B P_2=L_{B P_2} \times 50=18.85 \mathrm{PH} \times 50=942.5 \mathrm{PH}=0.94 \mathrm{nH} \\\\ &C_{B P_2}=\frac{C B P_2}{50}=\frac{0.235 \mathrm{nF}}{50}=4.7 \times 10^{-12}=4.7 \mathrm{PF} \\ \end{aligned} \\ $

Replacing series arm of low pass filter (inductor) with series combination of $L B P_1 \&{ }C B P_1$ and parallel aim of low pass filter (capacitor) with parallel combination of $L B P_2 \& C B P_2$ we get desired band pass filter.