Waveguides

Exercise n. 1

Find the dimensions of a rectangular waveguide with the following specifications: . 1. it is single mode in the band between fmin= 10 GHz and fmax= 15 GHz

2. the attenuation of the first higher order mode must be at least αlim= 8 dB/cm for the whole band [fmin,fmax]

3. dispersion must be minimized

4. the waveguide must carry up to PT = 150 kW without discharge when it is matched If the waveguide designed above is connected to a load with reflection coefficient |ΓL| = 0.2, what is the maximum power that the guide can carry without discharge?

Solution

Certainly we choose b < a/2 so that the first higher order mode is T E20. Next find a.

Enforce

fmin ≥ fc10= c 2a fmax≤ fc20= c

a

from which we find c

2fmin ≤ a ≤ c fmax

that is

1.5 ≤ a ≤ 2 cm Compute the attenuation of mode T E20:

α20=

Note that the value of αlimgiven in dB must be converted in Neper, by diving it by 8.6859.

Dispersion is minimized if a = 1.919 cm is chosen. This value is inside the range 1.5 ≤ a ≤ 2 cm, hence it is acceptable. Then the maximum value for b is bmax= a/2 = 0.960 cm

Now find b on the basis of the power to be transmitted PT. The active power flow on the T E10

modal line is

Pwg=1 2

|V⁺|² Zt10

because the modal line is matched.

The maximum value of the electric field is reached at x = a/2, so that

Emax= r2

abV⁺

Substituting in the previous expression, we find that the power in the guide for a given value of Emaxis

Now, setting Emax= R = 20 kV/cm the dielectric rigidity of air, we obtain the power at the onset of discharge Pdisch:

Pdisch= 1

Note that this power is a function of frequency because Zt10is. At f = fminthe modal impedance is maximum, so Pdischis minimum: this is the most critical situation. If the condition Pdisch(fmin) ≥ PT is satisfied, Pdisch(f ) ≥ PT for all f . Hence enforce

from which we find

b ≥ 4Zt10(fmin)PT

aR² = 0.472 cm

In conclusion we find 0.472 ≤ b ≤ 0.960 cm; any value of b in this range is acceptable.

Suppose b = 0.472 cm is chosen. Then the discharge power is exactly Pdisch = PT = 150 kW. If the guide is mismatched, the discharge power is reduced by the factor 1/S,

Pdisch= 1

so the discharge power is Pdisch= 150/1.5 = 100 kW. In conclusion, the guide previously designed can carry only a maximum power of 100 kW, in mismatched conditions, if discharge is to be avoided.

¥

Exercise n. 2

The rectangular waveguide WR90 of Fig. B.8 is connected to a load that has a reflection coefficient ΓL = 0.4 exp(−j45^◦) for the fundamental mode T E10, at the frequency f = 12 GHz. Design a small piece of dielectric to be inserted at a convenient position in the guide so that matching is obtained. Then draw the voltage plot |V (z)| with V^inc= 1 V.

Solution

Compute first the parameters of the modal lines. The waveguide is WR90, with a = 0.9 in=0.9 ·

Γ

L

ε

r

A B

A B ^C

Γ

L

zd

,

d

k Z

_∞

k

_z0

, Z

_∞0

0

,

0

k

z

Z

_∞

Figure B.8. Matching of a load by means of a piece of dielectric and equivalent modal circuit for the T E10mode

2.54 cm. and fc0= c/(2a) = 6.56 GHz. Recall

kz0= k0

s 1 −

µfc0

f

¶₂

kzd= k0

s εr−

µfc0

f

¶₂

Z∞0= Z0

s 1 −

µfc0

f

¶₂ Z∞d= Z0

s εr−

µfc0

f

¶₂

which yield

kz0= 2.1043 rad/cm λg0= 2.986 cm Z∞0= 450.2789 Ω

The matching device is a λ/4 line. It must be positioned in such a way that ZB is real ZB = RB. Then the characteristic impedance of the line is computed according to

Z∞d=p RARB

where RA = Z∞0, the desired input impedance. To find RB draw on the Smith chart the circle through Γ_L with center in the origin. The intersections with the real axis are

rm= 1 − |ΓL|

1 + |ΓL| and rM =1 + |ΓL| 1 − |ΓL| = 1

rm

Which one of the two must be selected? Let r be one of the two. Then

Z∞d= Z∞0

√r

From the equations above, Z∞d≤ Z∞0, hence r ≤ 1 so the point rmmust be chosen. Now we can obtain εr≤ 1, which is not realizable.

In order to draw the voltage plot, i.e. the electric field plot in the waveguide, we mark some points on the Smith chart of Fig. B.9. From ΓL turn clockwise (Toward Generator) up to z_B⁺ = rm= 0.4286. Then as it should be.

To draw the plot, note that zA⁻ = 1 implies that |V (z)| is flat to the left of A since there is no backward wave. Moreover in A⁺ there is a maximum of voltage and in B⁻ a minimum, since the corresponding points are on the positive and negative part of the real axis, respectively. Then V_A− = 1 V. The VSWR on AB is

S = 1 + |Γ_A⁺|

1 − |Γ_A+| = 1 + |Γ_B⁻|

1 − |Γ_B−| = 1.5275

Γ

_L

z

A⁺

z

B⁺

=r

m

z

B⁻

z

A⁻

r

M

Figure B.9. Smith Chart plot relative to the design of the matching device and to the voltage plot

This number, of course, coincides with zA⁺. Then

V_B− = V_min= Vmax

S = 0.6547 Of course VB⁻ = VB⁺. To draw the part relative to BC, note that

VC= V_C⁺(1 + ΓL) = V_B⁺+e^−jkz0LBC(1 + ΓL) V_B⁺+ = V_B⁺−

1 + Γ_B−

1 + ΓB⁺

V_B⁺− = V_A⁺+e^−jkzdLAB = −jV_A⁺+

V_A⁺+ = V_A⁺−

1 + ΓA⁻

1 + Γ_A⁺

We find |VC| = 1.4333 V.

Alternatively and more rapidly, we can use the conservation of energy to compute |V (z)|. Since the structure is lossless, the net power is the same everywhere in the circuit. In general

1 2

|V^inc|² Z∞0 = 1

2|V (z)|²R{Y (z)}

The left hand side is the net power in A⁻, because the reflected power is zero. In particular

|VC|²= |V^inc|²

Z∞0R{YL} = |V^inc|² R{yL}

|VB|²= |V^inc|²

Z∞0R{YB} = |V^inc|²

R{y_B+} = |V^inc|² Z∞0Y∞dR{y_B−} The resulting plot is shown in Fig. B.10

−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5

0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

z (cm)

|V(z)|

A

B

C

Figure B.10. Voltage plot

¥

Exercise n. 3

Consider the waveguide structure of Fig. B.11. It consists of a semi-infinite rectangular waveguide, short circuited on the left and infinite on the right, excited by a dipole antenna in B. The guide is empty up to C and then it is completely filled with a dielectric with relative permittivity εr = 4.

The guide cross section has dimensions a = 5 cm, b = 2 cm. The frequency of the source is f = 4 GHz. Moreover LAB = 2.5 cm and LBC = 2 cm. The dipole can be represented on the fundamental mode equivalent line by a current generator is= 0.01 A.

Compute the power that the source radiates beyond C and select LAB so as to maximize it.

Solution

Compute first the parameters of the modal lines. The waveguide has a = 5 cm, hence the critical

frequency of the T E10 mode is fc0= c/(2a) = 3 GHz. Recall

kz0= k0

s 1 −

µfc0

f

¶₂

kzd= k0

s εr−

µfc0

f

¶₂

Z∞0= Z0

s 1 −

µfc0

f

¶₂ Z∞d= Z0

s εr−

µfc0

f

¶₂

which yield

kz0= 55.4125 rad/m λg0= 0.1134 m kzd= 155.3245 rad/m λgd= 0.0405 m Z∞0= 569.9704 Ω Z∞d= 203.3387 Ω

ε

r

A B

B C

zd

,

d

k Z

_∞

0

,

0

k

z

Z

_∞

i

s

A

C

Figure B.11. Waveguide structure with a dipole source and modal circuit for the fundamental mode

B

⁺

B

⁻

Z

B−

Z

B⁺

i

s

V

B

Figure B.12. Loads seen by the current generator

Then compute the load seen by the generator. Start with that looking to the right. The line to the right of C is infinitely long, hence ZC = Z∞d. Compute

zC⁻ =Z_∞d

Z∞0 = 0.3568

ΓC⁻ =z_C−− 1

These values have been obtained by means of a computer, but can be obtained, with less accuracy, by the Smith chart. Similarly for the load looking to the left.

←−

The original modal circuit can be lumped into the one of Fig. B.12. The voltage at the generator terminals is

V_B⁻ = V_B⁺= − is

←−

Y_B⁻+−→

Y_B⁺ = −4.3934 − j6.2010 V (B.1) In order to compute the power radiated beyond C we can apply the principle of energy conservation since the structure is lossless. Then

PC = PB⁺=1

2|VB⁺|²R{−→

YB⁺} = 0.0219 W since−→

Y_B⁺= (7.5756 − j7.4007) · 10⁻⁴ S.

If we want a more detailed computation, that yields also the phase of voltage and current, we can proceed step by step across the discontinuities:

V_B⁺+ = VB⁺ exactly as found above by the energy method.

If we compute the power radiated to the left, we should find zero, since the waveguide is shorted in A. Indeed

←− PB⁻ =1

2|V_B⁻−|R{←−

YB⁻} = 0

because←−

YB⁻ is pure imaginary.

If we want to maximize the radiated power, it is necessary to maximize the magnitude of V_B. Since the numerator of (B.1) is fixed, we must minimize the magnitude of the denominator. Note that −→

Y_B+ is fixed and that ←−

Y_B− is pure imaginary and its value depends on the length LAB. Clearly, in these conditions, the denominator has minimum amplitude if it is real. Now, −→y_B⁺ = 0.4318 − j0.4218, so that we must enforce ←−y_B⁻ = j0.4218, which requires LAB/λg0= 0.3135 and LAB= 3.5551 cm.

In ordinary applications, the waveguide is matched when looking to the right of the source, so that −→y_B+ = 1 and the optimum situation is reached if ←−y_B− = 0, which implies LAB= λg0/4.

[1] S. Orphanidis, Electromagnetic waves and antennas, available on-line from www.ece.rutgers.edu/ orphanidi/ewa.

[2] B. Thid´e, Electromagnetic field theory, available on-line from www.plasma.uu.se/CED/Book/

[3] D. M. Pozar, Microwave engineering, Reading: Addison-Wesley, 1990.

[4] R. E. Collin , Foundations for microwave engineering, New York: McGraw-Hill 1992.

[5] R. Harrington Time harmonic electromagnetic fields, McGraw-Hill, New York 1961.

[6] C. A. Balanis,Antenna theory: analysis and design, Wiley Interscience, 2005

[7] S. Ramo, J. R. Whinnery, Th. VanDuzer, Campi e onde nell’elettronica per le comunicazioni, Milano, Angeli, 1980

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In document Notes Electromagnetic Fields (Page 144-153)