
Statement I: A speech signal of 2 KHz is used to modulate a carrier signal of 1 MHz. The bandwidth required for the signal is 4 kHz. Statement II: The side band frequencies are 1002 kHz and 998 kHz. In the light of the above statements, choose the correct answer from the options given below:
A. Both statement I and statement II are false
B. Statement I is false but statement II is true
C. Statement I is true but statement II is false
D. Both statement I and statement II are true
Answer
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Hint: Upper side band is defined as the sum of carrier frequency and message signal frequency. Band width of the modulated signal is given by the difference of the upper sideband and lower sideband. The sideband frequencies can be higher or lower than carrier frequency
Formula used
Bandwidth is given as,
\[2{\omega _m}\]
Where \[{\omega _m}\] is speech or message signal.
Upper side band is given as, U.S.B \[ = {\omega _c} + {\omega _m}\]
Lower side band is given as, L.S.B\[ = {\omega _c} - {\omega _m}\]
Where \[{\omega _c}\] is carrier signal.
Complete step by step solution:
Bandwidth required for the signal is 4 kHz.
Frequency of speech signal or message signal \[{\omega _m}\]=2 KHz
Frequency of carrier signal \[{\omega _c}\] = \[1{\rm{ MHz = 1000 KHz}}\]
The side band frequencies are 1002 kHz and 998 kHz.
As we know Bandwidth = \[2{\omega _m} = 2 \times 2\,KHz = 4\,KHz\]
There the bandwidth required for the signal is 4 kHz.
Therefore, from this we conclude statement I is correct.
Now as Upper side band, U.S.B \[ = {\omega _c} + {\omega _m}\]
Lower side band, L.S.B\[ = {\omega _c} - {\omega _m}\]
Substituting the value of \[{\omega _c}\] from statement I, we get
Upper side band, U.S.B \[ = {\omega _c} + {\omega _m} = 1000 + 2 = 1002\,{\rm{ KHz}}\]
Lower side band, L.S.B \[ = {\omega _c} - {\omega _m} = 1000 - 2 = 998\,{\rm{ KHz}}\]
Also Bandwidth = U.S.B- L.S.B =1002-998 = 4 KHz
Hence, the sideband frequencies are 1002 kHz and 998 kHz. We get this result by using the statement I.
Hence option D is the correct answer.
Note:As there is a maximum amount of data that is transmitted over an internet connection in some amount of time. Bandwidth is how much information is received every second but the speed is how fast that information is received. When there is more in a computer it can faster send and receive data.
Formula used
Bandwidth is given as,
\[2{\omega _m}\]
Where \[{\omega _m}\] is speech or message signal.
Upper side band is given as, U.S.B \[ = {\omega _c} + {\omega _m}\]
Lower side band is given as, L.S.B\[ = {\omega _c} - {\omega _m}\]
Where \[{\omega _c}\] is carrier signal.
Complete step by step solution:
Bandwidth required for the signal is 4 kHz.
Frequency of speech signal or message signal \[{\omega _m}\]=2 KHz
Frequency of carrier signal \[{\omega _c}\] = \[1{\rm{ MHz = 1000 KHz}}\]
The side band frequencies are 1002 kHz and 998 kHz.
As we know Bandwidth = \[2{\omega _m} = 2 \times 2\,KHz = 4\,KHz\]
There the bandwidth required for the signal is 4 kHz.
Therefore, from this we conclude statement I is correct.
Now as Upper side band, U.S.B \[ = {\omega _c} + {\omega _m}\]
Lower side band, L.S.B\[ = {\omega _c} - {\omega _m}\]
Substituting the value of \[{\omega _c}\] from statement I, we get
Upper side band, U.S.B \[ = {\omega _c} + {\omega _m} = 1000 + 2 = 1002\,{\rm{ KHz}}\]
Lower side band, L.S.B \[ = {\omega _c} - {\omega _m} = 1000 - 2 = 998\,{\rm{ KHz}}\]
Also Bandwidth = U.S.B- L.S.B =1002-998 = 4 KHz
Hence, the sideband frequencies are 1002 kHz and 998 kHz. We get this result by using the statement I.
Hence option D is the correct answer.
Note:As there is a maximum amount of data that is transmitted over an internet connection in some amount of time. Bandwidth is how much information is received every second but the speed is how fast that information is received. When there is more in a computer it can faster send and receive data.
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