
A carrier signal \[C(t) = 25\sin (2.51 \times {10^{10}}t)\]is an amplitude modulated by a message signal \[m(t) = 5\sin (1.57 \times {10^8}t)\]and transmitted through an antenna. What will be the bandwidth of the modulated signal?
A. 1987.5 MHz
B. 2.01 GHz
C. 50 MHz
D. 8 GHz
Answer
162k+ views
Hint:Before we proceed with the problem, it is important to know about the message signal and bandwidth. The signal which contains a message to be transmitted is called a message signal. It is a baseband signal and has to undergo the process of modulation, to get transmitted. Hence, it is also called the modulating signal. The difference between the upper and lower frequencies of a signal is known as bandwidth and it is measured in hertz.
Formula Used:
The formula to find the bandwidth of a signal we have,
\[\text{Bandwidth} = 2{f_m}\]…….. (1)
Where, \[{f_m}\] is the frequency of the message signal.
Complete step by step solution:
By data the message signal is,
\[m(t) = 5\sin (1.57 \times {10^8}t)\]………. (2)
The general formula of message signal is,
\[m(t) = A\sin (\omega t)\]……..… (3)
Compare the equations (2) and (3) we get the value of \[\omega \]thereby calculating message frequency.
\[\omega = 1.57 \times {10^8}t\]
We know that, \[\omega = 2\pi {f_m}\]
\[ \Rightarrow {f_m} = \dfrac{\omega }{{2\pi }}\]
Substituting the value of \[\omega \]in the above equation we get,
\[ \Rightarrow {f_m} = \dfrac{{1.57 \times {{10}^8}}}{{2 \times 3.142}}\]
\[ \Rightarrow {f_m} = 0.25 \times {10^8}Hz\]
\[ \Rightarrow {f_m} = 0.25 \times {10^8}Hz\]
Now substituting the value \[{f_m}\] in equation (1)
\[\text{Bandwidth} = 2{f_m}\]
\[\text{Bandwidth} = 2 \times 0.25 \times {10^8}\]
\[ \therefore \text{Bandwidth} = 50MHz\]
Therefore, the bandwidth of the modulated signal is \[50\,MHz\].
Hence, option C is the correct answer
Note:The bandwidth depends on many factors including environment, cabling, usage, and is usually less than the theoretical maximum and the wireless network connection speeds vary widely based on the conditions of the connection. The efficiency of bandwidth is defined as the effective and useful use of every Hertz of frequency available for communication. Usually, bandwidth is represented in the number of bits, kilobits, megabits, or gigabits that can be transmitted in 1 second.
Formula Used:
The formula to find the bandwidth of a signal we have,
\[\text{Bandwidth} = 2{f_m}\]…….. (1)
Where, \[{f_m}\] is the frequency of the message signal.
Complete step by step solution:
By data the message signal is,
\[m(t) = 5\sin (1.57 \times {10^8}t)\]………. (2)
The general formula of message signal is,
\[m(t) = A\sin (\omega t)\]……..… (3)
Compare the equations (2) and (3) we get the value of \[\omega \]thereby calculating message frequency.
\[\omega = 1.57 \times {10^8}t\]
We know that, \[\omega = 2\pi {f_m}\]
\[ \Rightarrow {f_m} = \dfrac{\omega }{{2\pi }}\]
Substituting the value of \[\omega \]in the above equation we get,
\[ \Rightarrow {f_m} = \dfrac{{1.57 \times {{10}^8}}}{{2 \times 3.142}}\]
\[ \Rightarrow {f_m} = 0.25 \times {10^8}Hz\]
\[ \Rightarrow {f_m} = 0.25 \times {10^8}Hz\]
Now substituting the value \[{f_m}\] in equation (1)
\[\text{Bandwidth} = 2{f_m}\]
\[\text{Bandwidth} = 2 \times 0.25 \times {10^8}\]
\[ \therefore \text{Bandwidth} = 50MHz\]
Therefore, the bandwidth of the modulated signal is \[50\,MHz\].
Hence, option C is the correct answer
Note:The bandwidth depends on many factors including environment, cabling, usage, and is usually less than the theoretical maximum and the wireless network connection speeds vary widely based on the conditions of the connection. The efficiency of bandwidth is defined as the effective and useful use of every Hertz of frequency available for communication. Usually, bandwidth is represented in the number of bits, kilobits, megabits, or gigabits that can be transmitted in 1 second.
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