CAPACITIVELY COUPLED SHUNT RESONATOR BANDPASS FILTER DESIGN

Design a third-order bandpass filter with a 0.5 dB equal-ripple response using capacitively coupled short-circuited shunt stub resonators. The center frequency is 2.5 GHz, and the bandwidth is 10%. The impedance is 50 Ω. What is the resulting attenuation at 3.0 GHz?

Question

CAPACITIVELY COUPLED SHUNT RESONATOR BANDPASS FILTER DESIGN

Design a third-order bandpass filter with a 0.5 dB equal-ripple response using capacitively coupled short-circuited shunt stub resonators. The center frequency is 2.5 GHz, and the bandwidth is 10%. The impedance is 50 Ω. What is the resulting attenuation at 3.0 GHz?

Accepted Answer

We first calculate the attenuation at 3.0 GHz. Using (8.71) to convert 3.0 GHz to normalized low-pass form gives

[latex]\omega \leftarrow \frac{1}{\Delta } (\frac{\omega }{\omega _{0}} - \frac{\omega _{0}}{\omega } ) =\frac{1}{0.1}(\frac{3.0}{2.5}-\frac{2.5}{3.0}) =3.667[/latex].

Then, to use Figure 8.27a, the value on the horizontal axis is

[latex]\left|\frac{\omega }{\omega _{c}} \right|-1 = \left|-3.667\right| -1=2.667,[/latex]

from which we find the attenuation as 35 dB. Next we calculate the admittance inverter constants and coupling capacitor values using (8.136) and (8.137):

[latex]Z_{0}J_{01}=\sqrt{\frac{\pi \Delta }{4g_{1}} }[/latex],

[latex]Z_{0}J_{n,n+1}=\frac{\pi \Delta }{4\sqrt{g_{n}g_{n+1}} }[/latex],

[latex]Z_{0}J_{N,N+1}=\sqrt{\frac{\pi \Delta }{4g_{N}g_{N+1}} } [/latex]

[latex]C_{01}=\frac{J_{01}}{\omega _{0}\sqrt{1-(Z_{0}J_{01})^{2}}} [/latex],

[latex]C_{n,n+1}=\frac{J_{n,n+1}}{\omega _{0}} [/latex],

[latex]C_{N,N+1}=\frac{J_{N,N+1}}{\omega _{0}\sqrt{1-(Z_{0}j_{N,N+1})^{2}} } [/latex]

n	[latex]g_{n}[/latex]	[latex]Z_{0}J_{n-1,n}[/latex]	[latex]C_{n-1,n}(pF)[/latex]
1	1.5963	[latex]Z_{0}j_{01}[/latex] = 0.2218		[latex]C_{01}[/latex] = 0.2896
2	1.0967	[latex]Z_{0}j_{12}[/latex] = 0.0594		[latex]C_{12}[/latex] = 0.0756
3	1.5963	[latex]Z_{0}j_{23}[/latex] = 0.0594		[latex]C_{23}[/latex] = 0.0756
4	1	[latex]Z_{0}j_{34}[/latex] = 0.2218		[latex]C_{34}[/latex] = 0.2896

Then we use (8.138), (8.140), and (8.141) to find the required resonator lengths:

[latex]C_{n}^{′}=C_{n}+\Delta C_{n}=C_{n}-C_{n-1,n}-C_{n,n+1}[/latex],

[latex]\Delta \ell =\frac{Z_{0}\omega _{0}C}{\beta } =(\frac{Z_{0}\omega _{0}C}{2\pi } )\lambda [/latex],

[latex]\ell _{n}=\frac{\lambda }{4}+(\frac{Z_{0}\omega _{0}\Delta C_{n}}{2\pi } )\lambda [/latex] ,

n	[latex] ΔC_{n}(pF)[/latex]	[latex] Δ\ell_{n}(λ)[/latex]	[latex]\ell[/latex](deg)
1	-0.3652	-0.04565	73.6
2	-0.1512	-0.0189	83.2
3	-0.3652	-0.04565	73.6

Note that the resonator lengths are slightly less than 90° (λ / 4 ) . The calculated amplitude response of this design is shown in Figure 8.54. The stopband rolloff at high frequencies is less than at lower frequencies, and the attenuation at 3 GHz is seen to be about 30 dB, while our calculated value for a canonical lumped-element bandpass filter was 35 dB.

Question 8.10: CAPACITIVELY COUPLED SHUNT RESONATOR BANDPASS FILTER DESIGN ...

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