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Question: An audio signal consists of two distinct sounds: one a human speech which is in the frequency band o...

An audio signal consists of two distinct sounds: one a human speech which is in the frequency band of 200Hz to 2700Hz, while other is a high-frequency music signal in the frequency band of 10200Hz to 15200Hz. The ratio of A.M signal bandwidth required to send both signals to the A.M signal bandwidth required to send just the human speech is:
A). 2
B). 5
C). 6
D). 3

Explanation

Solution

Hint: The bandwidth of an A.M wave is the difference between the larger maximum frequency and the minimum frequency. If we want to send both of the signals together, the minimum frequency will belong to the human speech range, and the maximum frequency will belong to the music signal. The AM bandwidth for human speech will be the difference between the maximum and the minimum frequency of human speech.

Formula Used:
The bandwidth of the carrier wave for amplitude modulation is given by,
A.M Bandwidth=fmaxfmin\text{A}\text{.M Bandwidth}={{f}_{\max }}-{{f}_{\min }}
fmax{{f}_{\max }} is the maximum frequency in the message frequency range.
fmin{{f}_{\min }} is the minimum frequency in the message frequency range.

Complete step by step answer:
In the question, it is given that the frequency band of human speech is from 200Hz to 2700Hz. So, the AM band which can be used to send these signals is the difference between the maximum and minimum frequency of the human speech range. So, we can write,
A.M Bandwidth=fmaxfmin\text{A}\text{.M Bandwidth}={{f}_{\max }}-{{f}_{\min }}
Substituting the values in the above equation we get,
A.M Bandwidth=2700Hz200Hz\text{A}\text{.M Bandwidth}=2700Hz-200Hz
A.M Bandwidth=2500Hz\text{A}\text{.M Bandwidth}=2500Hz … equation (1)
It is also mentioned that the message contains a high-frequency musical signal whose frequency ranges from 10200Hz to 15200Hz. So, if we want to send both the human speech signal and musical signal in a single bandwidth, we should design our bandwidth to encompass the frequency ranges of both these signals. So, the maximum frequency of the bandwidth will be the maximum frequency of the musical signal, and the minimum will be the minimum frequency of human speech. So, the A.M bandwidth is given by,
A.M Bandwidth=fmaxfmin\text{A}\text{.M Bandwidth}={{f}_{\max }}-{{f}_{\min }}
A.M Bandwidth=15200Hz200Hz\text{A}\text{.M Bandwidth}=15200Hz-200Hz
A.M Bandwidth=15000Hz\text{A}\text{.M Bandwidth}=15000Hz … equation (2)
So, the ratio of A.M signal bandwidth required to send both signals to the A.M signal bandwidth required to send just the human speech is
Ratio=150002500=6Ratio=\dfrac{15000}{2500}=6
So, the ratio is 6.
So, the answer to the question is option (C).

Note: Amplitude Modulation is a type of modulation that is used for radio transmission. Modulation is necessary since information in the message signal can get distorted or lost when the signal travels long distances. So, it is beneficial to have the message signal superimposed with a carrier wave, so it can be sent to a very long distance. This process of superimposing a message wave or signal with the carrier wave is called modulation. In amplitude modulation, the amplitude of the carrier wave is adjusted or modulated according to the message signal amplitude.
Some application Amplitude modulation is,
i). Broadband Transmission
ii). Airband radio
iii). Single sideband
iv). Quadrature amplitude modulation