
The number of AM broadcasting stations that can be accommodated in a 100 kHz bandwidth if the highest frequency modulating a carrier is 5 kHz are
(A) 5
(B) 7
(C) 20
(D) 10
Answer
139.8k+ views
Hint: Bandwidth is basically the total space which can be allotted with carrier waves. A wave being modulated will produce the same frequency of lower and upper bands. So, we will need double space in the bandwidth to accommodate that wave.
Complete step by step solution
Given: Bandwidth (B.W.) = 100 kHz
\[{f_m}\] = Highest modulation frequency = 5 kHz
We know,
$B.W. = 2n{f_m}$ ……(i)
Where, B.W. = Bandwidth.
n = Number of AM broadcasting stations.
\[{f_m}\] = Highest modulation frequency.
Putting the value of B.W. and \[{f_m}\] in equation (i) we get
$100kHz = 2n \times 5kHz$
Cross multiplying, we get
$n = \dfrac{{100}}{{10}} = 10$ .
(Option D)
Additional information:
The need for modulating a wave are as follows-
(i) In case of direct transmission there is a possibility of signals from different stations getting lost or mixed up.
(ii) Direct transmission requires an antenna of abnormal size.
(iii) It is limited to only a short distance.
Note above formula gives the maximum number amplitude modulated broadcasting stations in a given bandwidth. There are three types of modulation
(i) Amplitude modulation: In this case the amplitude of a high frequency wave is varied in accordance with the instantaneous value of a low frequency signal. The high frequency wave is called a carrier wave and the low frequency wave is called the modulating voltage.
(ii) Frequency modulation: In this case the frequency of the carrier wave is varied in accordance with the instantaneous value of the modulating signal. This is advantageous because all man made and atmospheric disturbances cause variation in the amplitude to the carrier wave which is eliminated in case of frequency modulation.
(iii) Phase modulation: In this case the phase of the carrier wave is varied in accordance with the instantaneous value of the modulating signal.
Complete step by step solution
Given: Bandwidth (B.W.) = 100 kHz
\[{f_m}\] = Highest modulation frequency = 5 kHz
We know,
$B.W. = 2n{f_m}$ ……(i)
Where, B.W. = Bandwidth.
n = Number of AM broadcasting stations.
\[{f_m}\] = Highest modulation frequency.
Putting the value of B.W. and \[{f_m}\] in equation (i) we get
$100kHz = 2n \times 5kHz$
Cross multiplying, we get
$n = \dfrac{{100}}{{10}} = 10$ .
(Option D)
Additional information:
The need for modulating a wave are as follows-
(i) In case of direct transmission there is a possibility of signals from different stations getting lost or mixed up.
(ii) Direct transmission requires an antenna of abnormal size.
(iii) It is limited to only a short distance.
Note above formula gives the maximum number amplitude modulated broadcasting stations in a given bandwidth. There are three types of modulation
(i) Amplitude modulation: In this case the amplitude of a high frequency wave is varied in accordance with the instantaneous value of a low frequency signal. The high frequency wave is called a carrier wave and the low frequency wave is called the modulating voltage.
(ii) Frequency modulation: In this case the frequency of the carrier wave is varied in accordance with the instantaneous value of the modulating signal. This is advantageous because all man made and atmospheric disturbances cause variation in the amplitude to the carrier wave which is eliminated in case of frequency modulation.
(iii) Phase modulation: In this case the phase of the carrier wave is varied in accordance with the instantaneous value of the modulating signal.
Recently Updated Pages
Average fee range for JEE coaching in India- Complete Details

Difference Between Rows and Columns: JEE Main 2024

Difference Between Length and Height: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

Algebraic Formula

Difference Between Constants and Variables: JEE Main 2024

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
