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 Z0 = 50 , d = 1.0 cm, and C0 = 2.666 pF, sketch the k-β diagram and compute the propagation constant, phase velocity

Question

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 [latex]\ Z_{0}=50[/latex] , d = 1.0 cm, and [latex]\ C_{0}= 2.666 pF[/latex], sketch the k-β diagram and compute the propagation constant, phase velocity, and Bloch impedance at f = 3.0 GHz. Assume [latex]\ k=k_{0}[/latex].

Accepted Answer

We can rewrite the dispersion relation of (8.9a) [latex]\ \cos \beta d=\cos heta -\frac{b}{2} \sin heta ,[/latex]as

[latex]\ \cos \beta d=\cos k_{0}d-\left(\frac{C_{0}Z_{0}c}{2d} ight) k_{0}d \sin k_{0}d.[/latex]

Then

[latex]\ \frac{C_{0}Z_{0}c}{2d} =\frac{\left(2.666 imes 10^{-12} ight)\left(50 ight) \left(3 imes 10^{8} ight) }{2\left(0.1 ight) } =2.0,[/latex]

so we have

[latex]\ \cos \beta d=\cos k_{0}d-2k_{0}d\sin k_{0}d[/latex]

The 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 [latex]\ k_{0}d[/latex] 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 [latex]\ 0\leq k_{0}d\leq 0.96[/latex]. The second passband does not begin until the sin [latex]\ k_{0}d[/latex] term changes sign at [latex]\ k_{0}d=\pi[/latex]. 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

[latex]\ k_{0}d=\frac{2\pi \left(3 imes 10^{9} ight) }{3 imes 10^{8}}\left(0.01 ight) =0.6283=36[/latex]

so βd = 1.5 and the propagation constant is β = 150 rad/m. The phase velocity is

[latex]\ v_{p}=\frac{k_{0}c}{\beta } =\frac{0.6283}{1.5}c=0.42c,[/latex]

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):

[latex]\ \frac{b}{2} =\frac{\omega C_{0}Z_{0}}{2} =1.256,[/latex]

[latex]\ heta =k_{0}d=36°,[/latex]

[latex]\ A=\cos heta -\frac{b}{2}\sin heta =0.0707,[/latex]

[latex]\ B=j\left(\sin heta +\frac{b}{2}\cos heta -\frac{b}{2} ight) =j0.3479.[/latex]

Then,
[latex]\ Z_{B}=\frac{BZ_{0}}{\sqrt{A^{2}-1}} =\frac{\left(j0.3479 ight)\left(50 ight) }{j\sqrt{1-\left(0.0707 ight)^{2} } } =17.4\Omega[/latex]

Question 8.1: ANALYSIS OF A PERIODIC STRUCTURE Consider the periodic capac...

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