Bandwidth of WBFM(Carson's Rule) MCQ Quiz - Objective Question with Answer for Bandwidth of WBFM(Carson's Rule) - Download Free PDF
Last updated on Mar 25, 2025
Latest Bandwidth of WBFM(Carson's Rule) MCQ Objective Questions
Bandwidth of WBFM(Carson's Rule) Question 1:
A modulating signal m(t) = 10cos (2π × 103 t) is amplitude modulated with a carrier signal c(t) = 50cos(2π × 105t). Assume R = 1Ω. Find the carrier power required for transmitting this AM wave.
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 1 Detailed Solution
The correct answer is 1250 W
- The carrier signal's amplitude is given by the equation c(t) = 50cos(2π × 105t).
- The peak value of the carrier signal (Ac) is 50 V.
- Assuming the resistance R = 1Ω, the carrier power Pc can be calculated using the formula:
- Pc = (Ac2) / (2R)
- Substituting the values:
- Pc = (502) / (2 × 1) = 2500 / 2 = 1250 W
- However, the correct answer is given as 1000 W, which might be due to different assumptions or approximations in the problem statement.
- It’s important to always verify the context and assumptions made in the problem while solving.
- In practical scenarios, carrier power might be adjusted to meet specific requirements or constraints.
Bandwidth of WBFM(Carson's Rule) Question 2:
An audio signal s(t) is normalized, whose Fourier transform S(f) is shown in the figure, so that |s(t)| ≤ 1. This signal is to be transmitted using FM with a frequency deviation constant kf = 90 kHz/V. What is the bandwidth required for transmission of the FM audio signal?
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 2 Detailed Solution
The correct option is 3
Concept:
According to Carson's rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating the signal then,
B.W. = 2(β + 1) fmax
Calculation:
\(\beta =\frac{K_f \times A_m}{f_m} =\frac{90 \times 10^3 \times 1}{20 \times 10^3}=4.5\)
BW = 2(β + 1) fmax = 2 x 5.5 x 20 = 220 kHz.
Here, option 3 is correct.
Bandwidth of WBFM(Carson's Rule) Question 3:
Which of the following rules states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency?
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 3 Detailed Solution
Explanation:
Carson's rule states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency.
According to Carson rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating signal then,
B.W. = 2(β + 1) fmax
Bandwidth of WBFM(Carson's Rule) Question 4:
What is band width of frequency modulated wave, if the frequency deviation allowed is 55 kHZ and the modulating signal has frequency of 10 kHZ?
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 4 Detailed Solution
Concept:
According to Carson rule signal, Bandwidth(B.W.) is given as:
B.W. = (β+1).2fm
Where,
β = Frequency deviation/ Message frequency
i.e.
\( \beta=\frac{\Delta f}{f_{m}}\)
So, Bandwidth:
\(B.W.=2[Δf+f_m ]\)
Given:
Δf = 55 Khz
fm = 10 KHz
Calculation:
\(B.W. = 2(55 \times 10^3 + 10 \times 10^3)\)
B.W. = 130 KHz
Bandwidth of WBFM(Carson's Rule) Question 5:
Determine the bandwidth required for an FM signal having frequency 2 kHz and maximum deviation 10 kHz.
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 5 Detailed Solution
Concept:
According to Carson rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating signal then,
B.W. = 2(β + 1) fmax
Calculation:
Given:
Δf = 10 kHz
fm = 2 kHz
B.W. = 2[Δf + fm]
B.W. = 2(10 + 2)
BW = 24 kHz
Hence option (3) is the correct answer.
Top Bandwidth of WBFM(Carson's Rule) MCQ Objective Questions
Determine the bandwidth required for an FM signal having frequency 2 kHz and maximum deviation 10 kHz.
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 6 Detailed Solution
Download Solution PDFConcept:
According to Carson rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating signal then,
B.W. = 2(β + 1) fmax
Calculation:
Given:
Δf = 10 kHz
fm = 2 kHz
B.W. = 2[Δf + fm]
B.W. = 2(10 + 2)
BW = 24 kHz
Hence option (3) is the correct answer.
Which of the following rules states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency?
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 7 Detailed Solution
Download Solution PDFExplanation:
Carson's rule states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency.
According to Carson rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating signal then,
B.W. = 2(β + 1) fmax
Determine the bandwidth occupied by a sinusoidal frequency modulated carrier for which the modulation index is 2.4
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 8 Detailed Solution
Download Solution PDFConcept:
According to Carson rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating signal then,
B.W. = 2(β + 1) fmax
Calculation:
Given: β = 2.4
B.W = (β + 1) 2fm
B.W. = 2fm (2.4 + 1)
BW = 6.8fm
What is band width of frequency modulated wave, if the frequency deviation allowed is 55 kHZ and the modulating signal has frequency of 10 kHZ?
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 9 Detailed Solution
Download Solution PDFConcept:
According to Carson rule signal, Bandwidth(B.W.) is given as:
B.W. = (β+1).2fm
Where,
β = Frequency deviation/ Message frequency
i.e.
\( \beta=\frac{\Delta f}{f_{m}}\)
So, Bandwidth:
\(B.W.=2[Δf+f_m ]\)
Given:
Δf = 55 Khz
fm = 10 KHz
Calculation:
\(B.W. = 2(55 \times 10^3 + 10 \times 10^3)\)
B.W. = 130 KHz
An audio signal s(t) is normalized, whose Fourier transform S(f) is shown in the figure, so that |s(t)| ≤ 1. This signal is to be transmitted using FM with a frequency deviation constant kf = 90 kHz/V. What is the bandwidth required for transmission of the FM audio signal?
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 10 Detailed Solution
Download Solution PDFThe correct option is 3
Concept:
According to Carson's rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating the signal then,
B.W. = 2(β + 1) fmax
Calculation:
\(\beta =\frac{K_f \times A_m}{f_m} =\frac{90 \times 10^3 \times 1}{20 \times 10^3}=4.5\)
BW = 2(β + 1) fmax = 2 x 5.5 x 20 = 220 kHz.
Here, option 3 is correct.
A modulating signal m(t) = 10cos (2π × 103 t) is amplitude modulated with a carrier signal c(t) = 50cos(2π × 105t). Assume R = 1Ω. Find the carrier power required for transmitting this AM wave.
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 11 Detailed Solution
Download Solution PDFThe correct answer is 1250 W
- The carrier signal's amplitude is given by the equation c(t) = 50cos(2π × 105t).
- The peak value of the carrier signal (Ac) is 50 V.
- Assuming the resistance R = 1Ω, the carrier power Pc can be calculated using the formula:
- Pc = (Ac2) / (2R)
- Substituting the values:
- Pc = (502) / (2 × 1) = 2500 / 2 = 1250 W
- However, the correct answer is given as 1000 W, which might be due to different assumptions or approximations in the problem statement.
- It’s important to always verify the context and assumptions made in the problem while solving.
- In practical scenarios, carrier power might be adjusted to meet specific requirements or constraints.
Bandwidth of WBFM(Carson's Rule) Question 12:
Determine the bandwidth required for an FM signal having frequency 2 kHz and maximum deviation 10 kHz.
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 12 Detailed Solution
Concept:
According to Carson rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating signal then,
B.W. = 2(β + 1) fmax
Calculation:
Given:
Δf = 10 kHz
fm = 2 kHz
B.W. = 2[Δf + fm]
B.W. = 2(10 + 2)
BW = 24 kHz
Hence option (3) is the correct answer.
Bandwidth of WBFM(Carson's Rule) Question 13:
A signal m (t) = 5 cos 2π100t frequency modulates a carrier. The resulting FM signal is 10cos {(2π105t) + 15sin(2π100t)}. The appropriate BW of FM signal would be:
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 13 Detailed Solution
Concept:
The standard equation of Frequency Modulation(FM) is given as:
\(S_fm(t)=A_ccos[2\pi f_ct+β sin2\pi f_mt ]\)
where β = modulation index
The Bandwidth of the FM signal is given by
B.W = 2(1 + β) fm
Calculation:
β = 15, fm = 100 Hz,
BW = 2(1 + β)fm
= 2(1 + 15)100 = 3.2 kHz
Important Points
- In FM transmission, the waveform has infinite sidebands occurring at frequencies fc ± nfm, where n is an integer.
- Hence theoretical bandwidth requirement is infinite.
- But, the amplitude (thereby energy content) of the sidebands is influenced by modulation index and frequency deviation. Generally in FM communication, the modulation index is chosen as small values for which the first two sidebands are significant and the rest are ignored.
Bandwidth of WBFM(Carson's Rule) Question 14:
Which of the following rules states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency?
Answer (Detailed Solution Below)
Bandwidth of WBFM(Carson's Rule) Question 14 Detailed Solution
Explanation:
Carson's rule states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency.
According to Carson rule signal, BW is given as:
B.W = (β + 1) 2fm
\(\becauseβ=\frac{Δf}{f_m} \)
B.W. = 2[Δf + fm]
If multiple frequencies are available in modulating signal then,
B.W. = 2(β + 1) fmax
Bandwidth of WBFM(Carson's Rule) Question 15:
A device with input x(t) and output y(t) is characterized by y(t) = x2(t). A FM Signal with modulating frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of output signal is ________ in kHz
Answer (Detailed Solution Below) 369 - 371
Bandwidth of WBFM(Carson's Rule) Question 15 Detailed Solution
Given that FM signal is input to System y(t) → x2(t) which is a frequency multiplier systems with multiplier factor of 2.
Hence New frequency deviation
(Δf)new = 2 × 90 kHz
But the frequency of modulating signal remains same ⇒ (fm) new = fm = 5 kHz
The Bandwidth of output y(t)
= 2 [(Δfnew + fm)]
= 2 [180 + 5]
= 370 kHz