Question 8.1: ANALYSIS OF A PERIODIC STRUCTURE Consider the periodic capac...
ANALYSIS OF A PERIODIC STRUCTURE Consider the periodic capacitively loaded line shown in Figure 8.5 (such a line may be implemented as in Figure 8.1, with short capacitive stubs). If \ Z_{0}=50 , d = 1.0 cm, and \ C_{0}= 2.666 pF, sketch the k-β diagram and compute the propagation constant, phase velocity, and Bloch impedance at f = 3.0 GHz. Assume \ k=k_{0}.
We can rewrite the dispersion relation of (8.9a) \ \cos \beta d=\cos\theta -\frac{b}{2} \sin\theta ,as
\ \cos \beta d=\cos k_{0}d-\left(\frac{C_{0}Z_{0}c}{2d}\right) k_{0}d \sin k_{0}d.Then
\ \frac{C_{0}Z_{0}c}{2d} =\frac{\left(2.666\times 10^{-12} \right)\left(50\right) \left(3\times 10^{8}\right) }{2\left(0.1\right) } =2.0,so we have
\ \cos \beta d=\cos k_{0}d-2k_{0}d\sin k_{0}dThe most straightforward way to proceed at this point is to numerically evaluate the right-hand side of the above equation for a set of values of \ k_{0}d starting at zero. When the magnitude of the right-hand side is unity or less, we have a passband and can solve for βd. Otherwise we have a stopband. Calculation shows that the first passband exists for \ 0\leq k_{0}d\leq 0.96. The second passband does not begin until the sin \ k_{0}d term changes sign at \ k_{0}d=\pi. As k0d increases, an infinite number of passbands are possible, but they become narrower. Figure 8.6 shows the k-β diagram for the first two passbands.
At 3.0 GHz, we have
\ k_{0}d=\frac{2\pi \left(3\times 10^{9}\right) }{3\times 10^{8}}\left(0.01\right) =0.6283=36so βd = 1.5 and the propagation constant is β = 150 rad/m. The phase velocity is
\ v_{p}=\frac{k_{0}c}{\beta } =\frac{0.6283}{1.5}c=0.42c,which is much less than the speed of light, indicating that this is a slowwave structure. To evaluate the Bloch impedance, we use (8.2) and (8.12):
\ \frac{b}{2} =\frac{\omega C_{0}Z_{0}}{2} =1.256,\ \theta =k_{0}d=36°,
\ A=\cos \theta -\frac{b}{2}\sin \theta =0.0707,
\ B=j\left(\sin \theta +\frac{b}{2}\cos \theta -\frac{b}{2} \right) =j0.3479.
Then,
\ Z_{B}=\frac{BZ_{0}}{\sqrt{A^{2}-1}} =\frac{\left(j0.3479\right)\left(50\right) }{j\sqrt{1-\left(0.0707\right)^{2} } } =17.4\Omega
