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## Q. 9.3

FIELD DISPLACEMENT ISOLATOR DESIGN

Design a field displacement isolator in an X-band waveguide to operate at 11 GHz. The ferrite has $4\pi M_{s} = 3000 G$ and $ϵ_{r} = 13$. Ferrite loss can be ignored.

## Verified Solution

We first determine the internal bias field, H0, such that $µ_{e} < 0$. This can be found from Figure 9.8, which shows $µ_{e}/µ_{o}$ versus H0 for $4\pi M_{s} = 3000 G$ at 11 GHz.
We see that $H_{0} = 1200 O_{e}$ should be sufficient. Also note from this figure that a ferrite with a smaller saturation magnetization would require a much larger bias field.

Next we determine the slab position, c /a, by numerically solving (9.79) for the propagation constants,$β_±$, as a function of c /a. The slab thickness was set to t = 0.25 cm, which is approximately a/10. Figure 9.14a shows the resulting propagation constants, as well as the locus of points where $β_{+}$ and  c /a satisfy the condition of (9.88). The intersection of $β_{+}$ with this locus will ensure that $E_{y} = 0$ at x = c + t for the forward wave; this intersection occurs for a slab position of c /a =0.028. The resulting propagation constants are $β_{+} = 0.724k_{0} < k_{0}$  and $β_{−} = 1.607k_{0} > k_{0}.$

$\left(\frac{k_f}{\mu _e} \right)^2 + \left(\frac{\kappa \beta}{\mu \mu _e} \right)^2 – k_a \cot k_a c\left(\frac{k_f}{\mu _0\mu _e}\cot k_ft +\frac{\kappa \beta }{\mu_0 \mu \mu_e } \right) -\left(\frac{k_a}{\mu _0} \right) \times \cot k_a c \cot k_ad\left(\frac{k_f}{\mu _0\mu _e}\cot k_ft -\frac{\kappa \beta }{\mu_0 \mu \mu_e } \right)=0.$            (9.79)

$k^+_a=\frac{\pi }{d}$                    (9.88)

The electric ﬁelds are plotted in Figure 9.14b. Note that the forward wave has a null at the face of the ferrite slab, while the reverse wave has a peak (the relative amplitudes of these ﬁelds are arbitrary). A resistive sheet can be placed at this point to attenuate the reverse wave. The actual isolation will depend on the resistivity of this sheet; a value of 75 Ω per square is typical.