Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
Question
Book Icon
Chapter 12, Problem 7P

(a)

To determine

Find the value of the dimensions of an air-filled rectangular waveguide.

(a)

Expert Solution
Check Mark

Answer to Problem 7P

The value of the dimensions a×b of an air-filled rectangular waveguide is 3cm×1.25cm.

Explanation of Solution

Calculation:

Given that the cutoff frequency fc of TE10 mode is 5GHz and TE01 mode is 12GHz.

For TE10 mode:

Write the expression to calculate the cutoff frequency for the TE10 mode.

fc=u2a        (1)

Here,

u is the phase velocity of uniform plane wave in dielectric medium and

a is the inner dimension of the waveguide.

Given waveguide is air-filled, therefore the phase velocity u is the speed of light in vacuum c, which is 3×108m/s.

Therefore the Equation (1) becomes,

fc=c2a

Rearrange the above equation.

a=c2fc

Substitute 3×108m/s for c and 5GHz for fc in above Equation.

a=3×108m/s2(5GHz)=3×108m/s2(5×1091/s)=3×102m=3cm

For TE01 mode:

Write the expression to calculate the cutoff frequency for the TE01 mode.

fc=u2b        (2)

Here,

b is the inner dimension of the waveguide.

Substitute c for u in Equation (2).

fc=c2b

Rearrange the above equation.

b=c2fc

Substitute 3×108m/s for c and 12GHz for fc in above Equation.

b=3×108m/s2(12GHz)=3×108m/s2(12×1091/s)=1.25×102m=1.25cm

Conclusion:

Thus, the value of the dimensions a×b of an air-filled rectangular waveguide is 3cm×1.25cm.

(b)

To determine

Find the value of the cutoff frequencies of the next three higher TE modes in an air-filled rectangular waveguide.

(b)

Expert Solution
Check Mark

Answer to Problem 7P

The value of the cutoff frequencies (fc) of the next three higher TE modes TE20, TE01 and TE11 modes are 10GHz, 12GHz and 13GHz respectively.

Explanation of Solution

Calculation:

From part (a), the dimension of the waveguide is in form of a>b.

The condition a>b can also be expressed as 1a<1b.

Write the general expression to calculate the cutoff frequency for the waveguide.

fc=u2(ma)2+(nb)2

Substitute c for u in above equation.

fc=c2(ma)2+(nb)2

Substitute 3×108m/s for c, 3cm for a and 1.25cm for b in above Equation.

fc=(3×108m/s)2(m3cm)2+(n1.25cm)2=(1.5×108)(m3×102)2+(n1.25×102)21/s {1c=102}

fc=(1.5×108)(m0.03)2+(n0.0125)2Hz {1Hz=11s}        (3)

For TE10 mode:

The integers are m=1 and n=0.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(10.03)2+(00.0125)2Hz=(1.5×108)(1111.11+0)Hz=5×109Hz=5GHz

For TE20 mode:

The integers are m=2 and n=0.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(20.03)2+(00.0125)2Hz=(1.5×108)(4444.44+0)Hz=10×109Hz=10GHz

For TE30 mode:

The integers are m=3 and n=0.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(30.03)2+(00.0125)2Hz=(1.5×108)(10000+0)Hz=15×109Hz=15GHz

For TE40 mode:

The integers are m=4 and n=0.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(40.03)2+(00.0125)2Hz=(1.5×108)(17777.778+0)Hz=20×109Hz=20GHz

For TE01 mode:

The integers are m=0 and n=1.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(00.03)2+(10.0125)2Hz=(1.5×108)(0+6400)Hz=12×109Hz=12GHz

For TE02 mode:

The integers are m=0 and n=2.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(00.03)2+(20.0125)2Hz=(1.5×108)(0+25600)Hz=24×109Hz=24GHz

For TE11 mode:

The integers are m=1 and n=1.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(10.03)2+(10.0125)2Hz=(1.5×108)(1111.11+6400)Hz=13×109Hz=13GHz

For TE21 mode:

The integers are m=2 and n=1.

Therefore, the Equation (3) becomes,

fc=(1.5×108)(20.03)2+(10.0125)2Hz=(1.5×108)(4444.44+6400)Hz=15.62×109Hz=15.62GHz

Therefore, the next three higher TE modes are TE20, TE01 and TE11 modes.

Conclusion:

Thus, the value of the cutoff frequencies (fc) of the next three higher TE modes TE20, TE01 and TE11 modes are 10GHz, 12GHz and 13GHz respectively.

(c)

To determine

Find the value of the cutoff frequency for the TE11 mode in an air-filled rectangular waveguide.

(c)

Expert Solution
Check Mark

Answer to Problem 7P

The value of the cutoff frequency (fc) for the TE11 mode in an air-filled rectangular waveguide is 8.67GHz.

Explanation of Solution

Calculation:

For TE11 mode, the integers are,

m=1n=1

Write the general expression to calculate the cutoff frequency for the waveguide.

fc=u2(ma)2+(nb)2        (4)

Write the expression to calculate the phase velocity of uniform plane wave in the lossless dielectric medium.

u=cεr

Here,

c is the speed of light in vacuum which is 3×108m/s and

εr is the relative permittivity of the medium.

Substitute cεr for u in Equation (4).

fc=(cεr)2(ma)2+(nb)2=c2εr(ma)2+(nb)2

Substitute 3×108m/s for c, 2.25 for εr, 1 for m and n, 3cm for a and 1.25cm for b in above Equation.

fc=3×108m/s22.25(13cm)2+(11.25cm)2=(1.5×1082.25)(13×102)2+(11.25×102)21/s {1c=102}=8.67×109Hz {1Hz=11s}=8.67GHz

Conclusion:

Thus, the value of the cutoff frequency (fc) for the TE11 mode in an air-filled rectangular waveguide is 8.67GHz.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Complete the following problems. Show your work/calculations, save as.pdf and upload to the assignment in Blackboard. missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill). 1. What are the x and y dimensions for the center position of holes 1,2, and 3 in the part shown in Figure 26.2 (below)? 6.0000 Zero reference point 7118 1.0005 1.0000 1.252 Bore 6.0000 .7118 Cbore 0.2180 deep (3 holes) 2.6563 1.9445 Figure 26.2 026022 (8lot and Drill Part) (Setup Instructions--- (UNITS: Inches (WORKPIECE NAT'L SAE 1020 STEEL (Workpiece: 3.25 x 2.00 x0.75 in. Plate (PRZ Location 054: ' XY 0.0 - Upper Left of Fixture TOP OF PART 2-0 (Tool List ( T02 0.500 IN 4 FLUTE FLAT END MILL #4 CENTER DRILL Dashed line indicates- corner of original stock ( T04 T02 3.000 diam. slot 0.3000 deep. 0.3000 wide Intended toolpath-tangent- arc entry and exit sized to programmer's judgment…
A program to make the part depicted in Figure 26.A has been created, presented in figure 26.B, but some information still needs to be filled in. Compute the tool locations, depths, and other missing information to present a completed program. (Hint: You may have to look up geometry for the center drill and standard 0.5000 in twist drill to know the required depth to drill).
We consider a laminar flow induced by an impulsively started infinite flat plate. The y-axis is normal to the plate. The x- and z-axes form a plane parallel to the plate. The plate is defined by y = 0. For time t <0, the plate and the flow are at rest. For t≥0, the velocity of the plate is parallel to the 2-coordinate; its value is constant and equal to uw. At infinity, the flow is at rest. The flow induced by the motion of the plate is independent of z. (a) From the continuity equation, show that v=0 everywhere in the flow and the resulting momentum equation is მu Ət Note that this equation has the form of a diffusion equation (the same form as the heat equation). (b) We introduce the new variables T, Y and U such that T=kt, Y=k/2y, U = u where k is an arbitrary constant. In the new system of variables, the solution is U(Y,T). The solution U(Y,T) is expressed by a function of Y and T and the solution u(y, t) is expressed by a function of y and t. Show that the functions are identical.…
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY