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A signal '4 t is band limited to the frequency range 0 to fn. The signal frequency is translated by multiplying it by the signal V, t —cos2πfct . The value of fc such that the bandwidth of the translated signal is 1% of the carrier frequency fc will be
- a)200 fm
- b)100 frn
- c)2 fm
- d)50 fm
Correct answer is option 'A'. Can you explain this answer?
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A signal 4 t is band limited to the frequency range 0 to fn. The signa...
BW after multiplexing =
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A signal 4 t is band limited to the frequency range 0 to fn. The signa...
To understand why the correct answer is option 'A', let's break down the problem step by step.
Given Information:
- The signal is band-limited to the frequency range 0 to fn.
- The signal frequency is translated by multiplying it by the signal V, t cos2fct.
- The bandwidth of the translated signal is 1% of the carrier frequency fc.
1. Bandwidth of the Original Signal:
The band-limited signal has a frequency range from 0 to fn. Therefore, the bandwidth of the original signal is given by B = fn.
2. Effect of Multiplying by Cosine:
When the original signal is multiplied by the cosine function, it is frequency-shifted. The multiplication introduces sidebands around the original frequency.
3. Frequency Translation:
The frequency translation is determined by the cosine function, V, t cos2fct. The cosine function has a frequency component of 2fc. So, the translated signal will have a frequency range from 2fc - fn to 2fc + fn.
4. Bandwidth of the Translated Signal:
The bandwidth of the translated signal is the difference between the upper and lower frequencies. Let's calculate it:
Bandwidth = (2fc + fn) - (2fc - fn)
= 2fn
5. Percentage of Bandwidth:
The problem states that the bandwidth of the translated signal is 1% of the carrier frequency fc. Let's express it mathematically:
2fn = 1% of fc
fn = 0.5% of fc
6. Comparing Bandwidths:
Now, let's compare the original signal bandwidth with the translated signal bandwidth:
B = fn
2fn = 0.5% of fc
Since the translated signal's bandwidth is 0.5% of fc, which is smaller than the original signal's bandwidth, the translated signal's bandwidth can never be 1% of fc.
Therefore, option 'A' (100 fn) is incorrect.
Given Information:
- The signal is band-limited to the frequency range 0 to fn.
- The signal frequency is translated by multiplying it by the signal V, t cos2fct.
- The bandwidth of the translated signal is 1% of the carrier frequency fc.
1. Bandwidth of the Original Signal:
The band-limited signal has a frequency range from 0 to fn. Therefore, the bandwidth of the original signal is given by B = fn.
2. Effect of Multiplying by Cosine:
When the original signal is multiplied by the cosine function, it is frequency-shifted. The multiplication introduces sidebands around the original frequency.
3. Frequency Translation:
The frequency translation is determined by the cosine function, V, t cos2fct. The cosine function has a frequency component of 2fc. So, the translated signal will have a frequency range from 2fc - fn to 2fc + fn.
4. Bandwidth of the Translated Signal:
The bandwidth of the translated signal is the difference between the upper and lower frequencies. Let's calculate it:
Bandwidth = (2fc + fn) - (2fc - fn)
= 2fn
5. Percentage of Bandwidth:
The problem states that the bandwidth of the translated signal is 1% of the carrier frequency fc. Let's express it mathematically:
2fn = 1% of fc
fn = 0.5% of fc
6. Comparing Bandwidths:
Now, let's compare the original signal bandwidth with the translated signal bandwidth:
B = fn
2fn = 0.5% of fc
Since the translated signal's bandwidth is 0.5% of fc, which is smaller than the original signal's bandwidth, the translated signal's bandwidth can never be 1% of fc.
Therefore, option 'A' (100 fn) is incorrect.
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