WiFi Lab C
Answer Keys
Instructions:
1. The test is designed for 50-minutes, with a mix of multiple choice, short answer, and calculation,
totaling 54 questions, 166 points.
2. For calculation questions, it is not required that you show your work, however partial credit will be
assigned if correct steps are shown with an incorrect answer.
3. Answers must be given with appropriate units to receive full credit.
4. Allowed materials: 3-ring binder, writing utensils + scratch papers (you may need a few), two calculators of any type.
Written by: Lei Jiang (leijianghome@gmail.com)
WiFi Lab C
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MCQ on Radio waves (24 pts)
1. [1 pt] Wireless signal’s frequency and its wavelength has the following relation:
A. For higher frequency, wavelength is shorter
B. For higher frequency, wavelength is longer
C. There is no fixed relationship between frequency and wavelength
D. The wavelength of wireless waves depends on the signal’s bandwidth
2. [1 pt] What is the propagation speed of the wireless wave in free space?
A. 300,000,000 m/s
B. 300,000 m/s
C. 300,000 miles/hour
D. 186,000 miles/hour
3. [1 pt] An electromagnetic wave with a single frequency component has a electrical field as
A. A sinusoidal wave
B. A square wave
C. A isolated pulse
D. A narrow pulse repeated at a fixed interval
4. [1 pt] A sinusoidal voltage wave with peak amplitude of 100V has an average voltage magnitude of
A. 0 V
B. 35.4 V
C. 70.7 V
D. 100 V
5. [1 pt] Radio power of 0.4 kW can be expressed as
A. 400 dBm
B. 56 dBm
C. 34 dBm
D. 34 dB
6. [1 pt] The amount of wireless wave energy in free space follows the following relationship:
A. Proportional to the square of distance (d2 ) and the square of frequency (f2 )
B. Proportional to the square of distance (d2 ) and inversely proportional to the square
of frequency (f−2 )
C. Proportional to distance (d) and the square of frequency (f2 )
D. Proportional to the square of distance (d2 ) and independent of frequency
7. [1 pt] The impedance of free space is
A. 50 ohms
B. 100 ohms
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C. 377 ohms
D. 1000 ohms
8. [1 pt] If you stand 3 meters away from a bright 100 W electric bulb, which of the following best approximates the electric field strength that you will experience as caused by the bulb?
A. 1 V/m
B. 10 V/m
C. 20 V/m
D. 10 mV/m
E. < 1 mV/m
9. [1 pt] The speed of electromagnetic wave in a coaxial cable is roughly
A. 0.95X the speed of light in vacuum
B. 0.65X the speed of light in vacuum
C. 1.05X the speed of light in vacuum
D. 1.54X the speed of light in vacuum
E. Same as the speed of light in vacuum
10. [1 pt] For a perfect matching between feeder and antenna, the VSWR, or Voltage Standing Wave Ratio,
is approximately
A. 1:1
B. 1:3
C. 3:1
D. 1.3 : 1
11. [1 pt] Antenna A has a gain of 14.5 dBd, and antenna B has a gain of 18.5 dBi. The strength of signal
at same distance is therefore
A. Antenna B is stronger by 1.85 dB
B. Antenna A is stronger by 4 dB
C. Antenna B is stronger by 4 dB
D. Antenna B is stronger by 1.276 dB
12. [1 pt] Electromagnetic exposure is a hot topic. According to IEEE, recommended limits are 600 microW/cm2
at 900 MHz, and 1000 microW/cm2 at 1800MHz. Why is it different vs. frequency?
A. Low-frequency radio waves cannot penetrate the human body.
B. Human body absorption of EM wave is frequency dependent.
C. Higher frequency radio waves are less common in consumer electronics.
D. Lower frequency radio waves carry less energy versus high frequency.
13. [1 pt] Regular microwave oven power is
times more than WiFi from laptop
A. 100
B. 1000
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C. 10,000
D. 100,000
14. [1 pt] For microwave or WiFi transmission circuit, a matched impedance in the network means
A. Reaching the maximum possible output voltage
B. Reaching the maximum possible output current
C. Reaching the maximum possible output power
D. Reaching the least possible internal resistive power loss
15. [1 pt] An antenna has an output signal that is 100 times stronger than the input signal. The gain for
this antenna is
A. 10 dB
B. 20 dB
C. 100 dB
D. 1 dB
16. [1 pt] A circuit’s output signal power is one-fifth of its input power, the gain of this circuit is
A. -7 dB
B. 3.5 dB
C. -3.5 dB
D. - 5 dB
17. [1 pt] What is the power of a 0 dBm radio signal after going through a circuit with 23 dB gain?
A. 0.2 W
B. 0.23 W
C. 2.3 W
D. 0.023 W
18. [1 pt] Which of the following can help increase the resonant frequency of a dipole antenna?
A. Increasing the arm length
B. Adding a loop at middle of the arms
C. Shortening the arm length
D. Adding capacitors at the end of the arms
19. [1 pt] A half-wave dipole antenna has two arms with each the length of
A. 1/4-th the wavelength of target frequency
B. Half the wavelength of target frequency
C. Twice the wavelength of target frequency
D. Four times the wavelength of target frequency
20. [1 pt] Antenna A has a gain of 0 dBd, and antenna B has a gain of 2 dBi. With the same driving power
and using the same setup, one can compare the received signal strength at an equal distance along each
antenna’s maximum gain direction. Which of the following is correct?
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A. Antenna A and B are similar in signal strength
B. A is similar to a half-wave dipole antenna, Antenna B is slightly worse vs. A
C. Antenna A signal is null, while antenna B has twice the signal strength.
D. A is similar to a half-wave dipole antenna, Antenna B is stronger by 2dB.
21. [1 pt] What type of antenna are Yogi, quad, and dish?
A. Omi-directional antennas
B. Non-resonant antennas
C. Directional antennas
D. Loop antennas
22. [1 pt] What is the function of a balun?
A. Radio emitter between a balanced and unbalanced circuit to cut off any unintended cross
coupling
B. Allow antenna and feedline to have matching impedance
C. Expand the working bandwidth of an antenna
D. Reducing the reflected wave amplitude for an antenna
23. [1 pt] SNR, Signal-to-Noise Ratio, refers to
A. The ratio of the signal voltage amplitude to noise floor voltage amplitude
B. The ratio of received wireless signal power versus the power of noise floor
C. The ratio of received signal power versus the sum of signal and noise power
D. The ratio of received signal electrical field amplitude versus the total E-field amplitude of signal
and noise
24. [1 pt] A standard chart to help perform impedance matching for transmission and antenna is
A. A VSWR vs. frequency chart
B. A polar Gain plot
C. A Smith chart
D. A radiation pattern plot
E. S-parameter vs. Frequency chart
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Short Answers (29 pts)
25. [2 pts] Which of the following bands contain 2.4 to 5 GHz spectrum?
√
C
⃝ K
⃝ L
⃝ P
√
S
⃝ X
26. [3 pts] Which of the following standard(s) support 54 Mbps at 2.4 GHz band?
⃝ 802.11a
√
802.11ac
√
802.11ad
⃝ 802.11b
√
802.11g
√
802.11n
27. [3 pts] List three basic forms of ground waves
space wave , surface wave , tropospheric wave
28. [3 pts] Describe how the metal screen on the microwave oven can prevent escape of microwave energy
while allowing us to see through.
Solution: Microwave is 2-4GHz, or 15 cm to 7.5cm wavelength. The screening mesh used on
microwave oven doors is sufficiently fine (much smaller than 7.5 cm) that it behaves as a continuous,
thin, metal sheet, preventing the escape of the radar energy.
29. [6 pts] Explain Circular Polarization.
Solution: In CP, the radiation is a superposition between two orthogonal LPs. The electromagnetic
field of the wave has a constant magnitude, and its direction rotates at a constant rate in a plane
perpendicular to the direction of the wave.
30. [6 pts] List three or more benefits of circular polarization, compared to linear polarization in WiFi or
5G design.
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Solution:
CP minimizes the sensitivity of orientations between the receiving and transmitting antennas. - No
need to tune the direction of antenna
It can ease multipath interference in contrast to linear polarized (LP). - Avoid phase cancellation.
CP antennas can send and receive signals in all planes having strong absorption and reflection of
radio signals.
CP is more resistant to signal degradation due to extreme weather conditions.
31. [6 pts] List three or more limitations of circular polarization, compared to linear polarization in WiFi
or 5G design.
Solution:
Need special transmitter and receiver to take advantage of benefit
Lower range due to higher power consumed
May not show performance benefit if surrounding remove one of the polarizations
LH vs. RH CP is not compatible, so both transmitter and receiver need to have the same handedness.
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EM wave and Antenna Analysis (63 pts)
32. [3 pts] A perfectly designed antenna has a measured power of 20 dB, but the power drops by 40 dB at
half meter away. What is the receiver signal power at half-meter away in a mW unit?
0.01 mW
33. [3 pts] For an electromagnetic wave with a wavelength of 9 cm, what is the photon energy?
Solution: E = hc/λ = 2.2E − 24 Joule.
34. [3 pts] For a parabolic antenna, the gain equation is expressed as
G (dBi) = 10 log(ν(πD/λ)2
where D is the diameter of the antenna (in meters), and λ is the wavelength. The efficiency ν is generally
0.6. For a 2.4GHz wave with a 15-cm antenna diameter, what is the maximum gain possible?
Solution: G = 10log(0.6 × (π15/12.5)2 ) = 9.3 dBi
35. [6 pts] Describe the main decay difference between near-field vs far-field radiation. [2 pts]
If a Science Olympiad competitor designed an antenna at the maximum allowed dimension, At least
how far is it considered outside the near field relative to the edge of his antenna for a 2.4GHz, 2mW
transmitter?
Solution:
Near field is stronger but decays much more rapidly as 1/r4 function.
The minimum distance of the near field is r = 2d2 /λ. λ = 12.5cm, so rmin = 36 cm.
36. [8 pts] [Friis equation] The received power Pr (in W) at the users is given by the Link equation below:
Pr =
Pt Gt Gr λ2 L
(4πR)2
where Pt is the access point transmit power in W, Gt and Gr are the gain of the access point antenna
and the user antenna, respectively. λ is the wavelength, and L is the loss factor to account for the
environment and system. This link equation can be used to estimate the maximum range of coverage R
for general antennas.
Assume we have a poor WiFi receiver with 0 dB Gain, and link loss of 20 dB, a threshold signal of -80
dBm. To achieve a robust connection at 100 meter, what is the gain needed for the antenna with a
transmitter of 2mW at 2.4G?
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Solution:
Pr = −80dBm =10[−80−30]/10 = 10−8 mW;
Gt =(10−8 /2) × (4π100)2 /0.1252 /(0.01) = 50.5, or 17.0 dB
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WiFi Lab C
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Answer the following four questions with the gain chart below for a 2.4 GHz, 12 dBi directional antenna
(Horizontal plane shown).
37. [2 pts] What is the gain at 55-degree direction?
Solution: 0 dBi
38. [2 pts] What is the beamwidth in degrees?
Solution: 65 degree [+/-3degree allowed]
39. [2 pts] What is the front-to-back gain ratio in dB?
Solution: 25 [25 to 27]
40. [4 pts] If the vertical-plane beam-width is 35 degrees (gain graph not shown), what is the directivity of
this antenna as measured in dBi?
Solution:
D = 4π/Φ, where Φ is the solid angle. Here the solid angle is approximated by half beam width of
the horizontal and vertical planes. D = 41253/(17.5)/(32.5) = 72.5; 10 log10(72.5) = 18.6.
41. [6 pts] A microwave transmitter operating at the carrier frequency of 3 GHz is protected by a Plexiglass
enclosure which has a permittivity three times that of free space ϵ = 3ϵ0 .
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A) What is the refraction index of the enclosure? [2 pts] B) What is the wavelength of 3GHz wave inside
the plexiglass in cm? [2 pts] C) What is the minimum thickness of the plexiglass for it to be transparent
to the wave? [2 pts]
Solution:
A) Refraction index = sqrt(e/e0) = 1.73; Wavelength = 3e8/3e9 = 10 cm;
B) Within the plexiglass, it is 10/1.73 = 5.8 cm;
C) Minimum thickness is half wavelength = half wavelength = 2.9 cm.
42. [8 pts] For a 2.5-GHz antenna,
(A) calculate the radius of the first Fresnel zone at a point of 50 meter from the source and 50 meter
from the receiving end in free space. (Show your work) [3 pts] (B) Explain briefly the meaning of Fresnel
zone, [2 pts] and (C) Describe what decision can be advised based on this dimension. [3 pts]
Solution:
p
p
(A) Rf = 17.3 1/f (D1 ∗ D2)/(D1 + D2) = 17.3 1/2.5 ∗ (0.05 ∗ 0.05/0.1)
= 1.73 meter (3pts)
(B) Fresnel zone radius describes the influence region of diffraction effect.
(C) Generally a radio-opaque object should not be more than 0.6 R tall to avoid strong diffraction
attenuation. (3 pts)
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A circular coil with radius a and resistance R is shown below. The amplitude of the incoming signal
is proportional to the induced current. The signal is a monochromatic, linearly polarized wave (much
larger wavelength than the radius a), with the wave vector pointing to the Z direction of a Cartesian
reference frame, and the electrical field in along the X axis.
Answer the following three questions (Q. 43-46)
43. [6 pts] What is the best orientation of the antenna to detect the maximum signal? Briefly explain.
Solution:
Loop antenna is magnetic field driven
Best is to place the circle in the X-Z plane, as the B field is along the Y axis. If the B field is
perpendicular to the circle surface, the electromotive force is maximum as it is proportional to time
variation of the flux of B field through the circle.
44. [6 pts] For a distance of 50 meter, and assuming the source of the signal is 100 W,
(A) What is the electric field magnitude at the circular antenna?
(B) What is the maximum magnetic field strength?
Solution:
(A) I = P/A = 100/(4π502 ) = 3.18 mW/m2 [2 pts]
2
; Emax = 1.55 V/m (ϵ = 8.854 × 10−12 ) [2 pts]
I = 21 ϵ × c × Emax
(B) Magnetic field = E/c = 5.16 × 10−9 Tesla [2 pts]
45. [6 pts] If the source has a 2.5GHz frequency, what is the maximum possible EMF induced in the radio
antenna, if the radius is 7 cm?
Solution:
EMF = d(B)/dt ×A = (2.5 × 2π × 109 ) × (5.16 × 10−9 ) × [π × (0.07)2 ] = 1.248V
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Signal and Noise Analysis (46 pts)
46. [4 pts] Assume an ideal, noiseless channel with a bandwidth of 30 MHz that is transmitting a signal
with four discrete levels.
A) What is the maximum bit rate possible?
B) What is the term usually assigned to this maximum limit?
Solution:
A) 2 x Bandwidth x log2 V = 2 x 3 x106 x 2 = 120 Mbps
B) Nyquist Bit Rate or Nyquist Rate
47. [4 pts] Assume random noise sources to be present, a channel with a bandwidth of 30 MHz with a SNR
ratio of 8 dB. What is the maximum bit rate (capacity) of transmission?
Solution: Based on Shannon’s capacity formula, and convert SNR from dB unit
2 x Bandwidth x (1+ SNR) = 2 × 30 × 106 × log2 (1 + 100.8 ) = 172.2 Mbps
48. [4 pts] What is the thermal noise power in dBm at an absolute temperature of T = 290 K with 30MHz
bandwidth?
Solution: Thermal noise = -174 dBm + 10 log (30 MHz) = -103.5 dBm
49. [4 pts] The sensitivity of a circuit can be defined by the smallest level of signal it can detect above all
the noise levels. If all systematic characteristics (antenna etc) are to remain the same, and if data rate
is increased 10 fold, how does a receiver sensitivity change?
Solution: Sensitivity reduces linearly with the data rate - so 10X reduction in sensitivity.
50. [8 pts] For a radio receiver with 20 dB noise figure accounting for internal circuit noise,
A) what is the sensitivity of the receiver with a desired SNR of 8 dB at 30 MHz bandwidth, at 290K?
[4 pts]
B) What is the sensitivity if the temperature is raised to 110 C? [4 pts]
Solution:
The noise factor is defined as the ratio of the output noise power of a device to the portion thereof
attributable to thermal noise in the input termination at standard noise temperature 290 K.
The noise figure is the noise factor expressed in decibels.
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Thermal noise power = kTB, B=Bandwidth, T is in Kelvin, k is the Boltzmann constant
A) Sensitivity (in dB form) = Thermal noise + SNR + NF = -75.5 dBm
B) At 110 C, thermal noise is increased to 10log(kT × 30 MHz)
10log(1.38 × 10−23 × 383 × 30 × 106 ) = −98 dBm
So sensitivity level = -70 dBm.
Three WiFi antennas (A to C) have been tested and show the following data:
Type
A
B
C
VSWR peak
9.0
2.0
1.2
Bandwidth
200 MHz
60 MHz
20 MHz
Answering the following three questions.
51. [9 pts] What is the amount (%) of power reflected in each case? [3 pts each]
Solution: Reflected amplitude =[VSWL-1]/[VSWL+1] = 0.8, 0.33, 0.09, respectively.
Reflected power = 0.64x, 0.1x, and 0.0081
Accept answer at 63-64% power, 10% power, and 1% power
Reference: VSWL relation with Γ, the reflection coefficient. The power percentage is 100 × Γ2 .
52. [9 pts] What is the corresponding return loss for each? [3 pts for each]
Solution: Return loss = −10log10(Pr /P ) or = −20log(Γ)
Return loss is 2 dB, 10 dB, 20 dB (accept rounding error)
53. [4 pts] Which one of the A,B,C would you recommend as best WiFi antenna?
Solution:
B is the best choice for a balance of bandwidth vs loss (Technically A is too poor a design to quote
a bandwidth). Return loss < 10 dB is very good already.
One can argue that for WiFi applications, a 20 MHz bandwidth is good enough which is also a valid
answer (C).
54. [2 pts] What is your favorite antenna design?
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