Antennas Questions and Answers – Maximum Usable Frequency

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

1. The maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer is called as_______
a) Maximum usable frequency
b) Critical frequency
c) Resonance frequency
d) Dominating frequency
View Answer

Answer: a
Explanation: The maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer is called as maximum usable frequency (MUF). For a specified angle there will be a maximum frequency for which the wave is reflected back. If wave exceeds MUF then it’s not reflected back.
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2. If wave exceeds the MUF then it is not reflected back.
a) True
b) False
View Answer

Answer: a
Explanation: Maximum usable frequency (MUF) is defined as the maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer. If the wave exceeds MUF then it’s not reflected back and is transmitted to other upper layers and signal is lost.

3. The frequency below which the entire power gets absorbed is called as ____
a) MUF
b) LUF
c) Critical frequency
d) Optimum frequency
View Answer

Answer: b
Explanation: The frequency below which the entire power gets absorbed is called as LUF (Lowest usable frequency). The maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer is called as maximum usable frequency (MUF). The frequency at which there is optimum return of the wave is the optimum frequency.
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4. Suppose a ray is incident normally in the ionosphere region with electron density 36×1010/cm3, then the critical frequency is _____
a) 5GHz
b) 54MHz
c) 5.4MHz
d) 324GHz
View Answer

Answer: c
Explanation: \(f_c=9\sqrt{N \,max}\) where Nmax= electron density and angle of incidence is ἰ = 0 (given)
∴ \(f_c =9\sqrt{36×10^{10}} = 5.4MHz\)

5. Relation between MUF and critical frequency is ______
a) fc = fMUF secθi
b) fMUF = fc sec2 θi
c) fc = fMUF sinθi
d) fMUF = fc secθi
View Answer

Answer: d
Explanation: The fMUF in terms of critical frequency is given by fMUF = fc secθi for a given angle of incidence between two locations.
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6. Skip distance is the _____
a) Minimum distance at which wave returns back at the lowest possible frequency
b) Maximum distance at which wave returns back at the critical frequency
c) Minimum distance at which wave returns back at the critical angle
d) Maximum distance at which wave returns back at the lowest possible frequency
View Answer

Answer: c
Explanation: The minimum distance at which the wave returns back at the critical angle is called skip distance.

7. What is the value of maximum usable frequency when the incident angle is 60° and the critical frequency is 4.5MHz?
a) 4.5MHz
b) 2.25MHz
c) 9MHz
d) 18MHz
View Answer

Answer: c
Explanation: MUFfMUF = fcsecθi = 4.5MHz ×sec60=9MHz.
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8. Which of the following is true when a ray is incident normally in an Ionosphere region?
a) MUF is equal to critical frequency
b) MUF is greater than critical frequency
c) MUF is less than critical frequency
d) MUF is zero
View Answer

Answer: a
Explanation: When a ray is incident normally in an Ionosphere region, θi=0
⇨ fMUF=fc secθi=fc
Therefore, MUF is equal to critical frequency.

9. Which of the following statements is false?
a) MUF is always greater than or equal to critical frequency depending on the incident angle
b) Optimum frequency is the frequency at which optimum reflection of wave takes place
c) Beyond the MUF, the entire wave gets reflected back
d) Below LUF, the entire power of wave gets absorbed
View Answer

Answer: c
Explanation: Beyond MUF, (greater than critical frequency) the rays get penetrated into the region and no part of it is reflected back.
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10. Suppose a ray is incident normally in the ionosphere region with electron density 25×1010/cm3, then the Maximum usable frequency is _____
a) 5GHz
b) 45MHz
c) 4.5MHz
d) 125GHz
View Answer

Answer: c
Explanation: \(f_c=9\sqrt{N \,max}\) where Nmax= electron density and angle of incidence is ἰ = 0 (given)
∴ \(f_c = 9\sqrt{25×10^{10}} = 4.5MHz\)
⇨ And fMUF = fc secθi = fc = 4.5MHz

11. What is the refractive index of the region operating at 16MHz frequency with the electron density 49×1010/cm3?
a) 0.518
b) 0.919
c) 0.155
d) 0.845
View Answer

Answer: b
Explanation: Refractive index \(n=\sqrt{(1-\frac{81N}{f^2})}=\sqrt{(1-\frac{81×49×10^{10}}{(16×10^{6})^2})}=\sqrt{(1-0.155)}=\sqrt{0.845}\)
n=0.919

12. Find the MUF for the wave operating at a critical frequency 6MHz and having the skip distance d as 25km and virtual height of 100km in the ionosphere layer.
a) 6.05MHz
b) 1.25MHz
c) 1.025MHz
d) 3.25MHz
View Answer

Answer: a
Explanation: \(f_{MUF}=f_c \sqrt{((\frac{d}{2h})^2+1)}=6×10^6×\sqrt{((\frac{25}{2×100})^2+1)}=6.05MHz\)

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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