Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 7.4.2

TE $_{01}$ versus TE $_{11}$ and TM $_{01}$

An air-filled copper ( $σ_{c} = 5.8 \times 1 0^{7}$ S/m) circular waveguide of radius $a = 2.5$ cm is operated at $f = 15$ GHz.

Find the cutoff frequencies of TE $_{11}$ , TM $_{01}$ and TE $_{01}$ .
For TE $_{01}$ find $τ$ , $λ_{g}$ and $Z_{T E}$ , and write $H_{z}$ , $H_{ρ}$ , $E_{ϕ}$ with numbers (amplitude constant $C$ ).
Find the conductor attenuation of TE $_{11}$ , TM $_{01}$ and TE $_{01}$ in dB/km.
By what factor does $α_{c}$ of TE $_{01}$ drop if the frequency is doubled?

Try it yourself first.

Strategy

With $m = 0$ the TE field formulas Eq. (7.4.20) keep only $H_{z}$ , $H_{ρ}$ and $E_{ϕ}$ , and $J_{0}^{'} = - J_{1}$ turns them into first-order Bessel functions. The three attenuation constants Eq. (7.4.23) share the factor $R_{s} / (a η_{0})$ : compute it once, then multiply by the bracket and divide by $s$ of each mode.

Bookkeeping

Air, $a = 0.025$ m, $f = 15$ GHz: $λ = 2$ cm, $k = 100 π = 314.2$ rad/m, $ω μ_{0} = 1.184 \times 1 0^{5} Ω$ /m. Copper: $R_{s} = π f μ_{0} / σ_{c} = 31.95$ m $Ω$ .

a) Cutoffs

$c / (2 π a) = 1.910$ GHz, so $f_{c} = x \times 1.910$ GHz:

f_{c, TE_{11}} = 3.52 GHz, f_{c, TM_{01}} = 4.59 GHz, f_{c, TE_{01}} = 7.32 GHz .

b) TE₀₁

$h = 3.832/0.025 = 153.3$ rad/m and $s = 1 - (7.318/15)^{2} = 0.8729$ :

τ = 314.2 \times 0.8729 = 274.2 rad/m, λ_{g} = \frac{2 cm}{0.8729} = 2.29 cm, Z_{T E} = \frac{120 π}{0.8729} = 431.9 Ω.

With $m = 0$ , Eq. (7.4.20) gives $H_{ρ} = - \frac{j τ}{h} C J_{0}^{'} (h ρ) = \frac{j τ}{h} C J_{1} (h ρ)$ and $E_{ϕ} = \frac{j ω μ _{0}}{h} C J_{0}^{'} (h ρ) = - \frac{j ω μ _{0}}{h} C J_{1} (h ρ)$ , where $τ / h = 1.789$ and $ω μ_{0} / h = 772.7 Ω$ :

H_{z} H_{ρ} E_{ϕ} = C J_{0} (153.3 ρ) e^{- j 274.2 z}, = j 1.789 C J_{1} (153.3 ρ) e^{- j 274.2 z}, = - j 772.7 C J_{1} (153.3 ρ) e^{- j 274.2 z}

( $V/m$ for $C$ in $A/m$ ).

c) Losses

$R_{s} / (a η_{0}) = 0.03195/ (0.025 \times 377) = 3.390 \times 1 0^{- 3}$ Np/m, and

TE_{11} : s = 0.9721, (\frac{f _{c}}{f})^{2} + \frac{1}{1.84 1 ^{2} - 1} = 0.0550 + 0.4184 = 0.4734, α_{c} = \frac{3.390 \times 1 0 ^{- 3} \times 0.4734}{0.9721} = 1.651 \times 1 0^{- 3} Np/m = 14.3 dB/km, TM_{01} : s = 0.9520, α_{c} = \frac{3.390 \times 1 0 ^{- 3}}{0.9520} = 3.561 \times 1 0^{- 3} Np/m = 30.9 dB/km, TE_{01} : (\frac{f _{c}}{f})^{2} = 0.2380, α_{c} = \frac{3.390 \times 1 0 ^{- 3} \times 0.2380}{0.8729} = 9.24 \times 1 0^{- 4} Np/m = 8.0 dB/km .

d) Doubling the frequency

At $30$ GHz $R_{s}$ grows by $2$ , $(f_{c} / f)^{2}$ falls by $4$ to $0.0595$ , and $s$ rises from $0.8729$ to $0.9698$ . Hence $α_{c} = 2.56$ dB/km, smaller by the factor

\frac{4}{2} \times \frac{0.9698}{0.8729} = 2.83 \times 1.111 = 3.14.

Far above cutoff ( $s \to 1$ ) the factor would be $2^{3/2} = 2.83$ , the $f^{- 3/2}$ law.

Answer

f_{c} = 3.52, 4.59, 7.32 GHz TE_{01} : τ = 274.2 rad/m, λ_{g} = 2.29 cm Z_{T E} = 432 Ω α_{c} = 14.3 (TE_{11}), 30.9 (TM_{01}), 8.0 (TE_{01}) dB/km 30 GHz : 2.56 dB/km, factor 3.14

Check

$- E_{ϕ} / H_{ρ} = 772.7/1.789 = 431.9 Ω = Z_{T E}$ , as Eq. (7.4.14) requires for TE modes. $E_{ϕ} \propto J_{1} (153.3 ρ)$ vanishes at the wall, since $J_{1} (3.832) = 0$ : the boundary condition holds.

(a) \alpha_c(f) of the three modes. — (a) $α_{c} (f)$ of the three modes.

(b) TE_{01}, a=2.5 cm. — (a) $α_{c} (f)$ of the three modes.

Remark

The price of the low loss is mode purity. At $15$ GHz this pipe carries $17$ modes, $30$ field patterns, and every imperfection converts TE $_{01}$ into them (Application 7.4.2).

Problem 7.4.2: TE_{01} versus TE_{11} and TM_{01}