Antennas Questions and Answers – Critical Frequency

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This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Critical Frequency”.

1. When a wave is incident normally then the acceptable highest frequency at which signal can be returned is the ______
a) critical frequency
b) LUF
c) optimum frequency
d) dominating frequency
View Answer

Answer: a
Explanation: When a wave is incident normally then the acceptable highest frequency at which signal can be returned is the critical frequency. Beyond critical frequency, wave is penetrated into another region. When the frequency is greater than critical frequency, still some part is returned back by varying the angle of incidence and this frequency is called MUF.
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2. What is the critical frequency when the electron density in the F1 layer is 20×1010/cm3?
a) 40.2GHz
b) 4.02 MHz
c) 40.2 MHz
d) 40.2Hz
View Answer

Answer: b
Explanation: Critical frequency \(f_c=9\sqrt{N\, max} \)
\(f_c =9\sqrt{20×10^{10}}=4.02MHz \)

3. What is the electron density of the layer if critical frequency is 3 MHz?
a) 11×1012/cm3
b) 0.11×1012/cm3
c) 1.1×1012/cm3
d) 0.11×1010/cm3
View Answer

Answer: b
Explanation: \(f_c=9\sqrt{N\, max} \)
\(N_{max}=\frac{f_c^2}{81}=\frac{(3×10^6)^2}{81}=0.11×10^{12}/cm^3\)
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4. For a regular layer, the critical frequency is proportional to the ______ of electron density.
a) Square
b) Inverse
c) Square root
d) Inverse square root
View Answer

Answer: c
Explanation: For a regular layer, the critical frequency is proportional to the square root of Nmax electron density. \(f_c=9\sqrt{N \,max} \)

5. Find the critical frequency when the refractive index of the layer is 0.54 and MUF is 9MHz?
a) 0.84MHz
b) 7.57MHz
c) 0.75MHz
d) 8.4MHz
View Answer

Answer: b
Explanation: Refractive index \(n=\sqrt{(1-\frac{81N_{max}}{f^2})}\)
⇨ \(N_{max}=\frac{(1-n^2)f^2}{81}=\frac{(1-0.54^2)(9×10^6)^2}{81}=0.708 ×10^{12}/cm^3\)
⇨ critical frequency \(f_c=9\sqrt{N \,max} =9\sqrt{0.708 ×10^{12}}=7.57Mhz\)
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6. The value of refractive index when the MUF is equal to the critical frequency is ______
a) 1
b) 0
c) 0.5
d) 0.29
View Answer

Answer: b
Explanation: Refractive index \(n=\sqrt{(1-\frac{81N_{max}}{f^2})}\)
Critical frequency \(f_c=9\sqrt{N \,max} \)
When \(f_c=f_{MUF}, n=\sqrt{(1-\frac{81N_{max}}{f_{MUF}^2})}=1-1=0\)

7. For fMUF ≥ fc, the wave will reflects back irrespective of the angle of incidence.
a) True
b) False
View Answer

Answer: b
Explanation: For fMUF ≤ fc, the wave will reflects back irrespective of the angle of incidence. For fMUF ≥ fc, the wave will reflects back depending of the angle of incidence (which should be small).
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8. Which of the following frequency is greater than the critical frequency?
a) MUF
b) LUF
c) Optimum frequency
d) VLF
View Answer

Answer: a
Explanation: According to the secant law, fMUF=fc secθi
So MUF is greater than or equal to the critical frequency depending on the secθi

9. When the frequency decreases below _____ frequency, the signal reception becomes too weak and the Noise increases.
a) MUF
b) LUF
c) Optimum frequency
d) Critical frequency
View Answer

Answer: b
Explanation: When the frequency is lowered below the lowest usable frequency, the signal gets absorbed. At low frequencies the noise level is also more and signal reception will be also difficult.
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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter