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A DS spread spectrum system transmit at a rate of 1 kbps in the presets of a tone jammer. The jammer power is 20 dB greater then the desired signal, and the required εb/J0 to achieve satisfactory performance is 10 dB.
Que: The spreading bandwidth required to meet the specifications is
- a)107 Hz
- b)103 Hz
- c)105 Hz
- d)106 Hz
Correct answer is option 'D'. Can you explain this answer?
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A DS spread spectrum system transmit at a rate of 1 kbps in the preset...
We have a system where (Jav/Pav) = 20 dB, R =1000, and (εb/J0)db = 10 dB
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A DS spread spectrum system transmit at a rate of 1 kbps in the preset...
To determine the spreading bandwidth required to meet the specifications, we need to consider the given information:
- The spread spectrum system transmits at a rate of 1 kbps.
- The power of the jammer is 20 dB greater than the desired signal.
- The required b/J0 to achieve satisfactory performance is 10 dB.
Let's break down the solution into steps:
1. Find the desired signal power (Ps):
Since the jammer power is 20 dB greater than the desired signal, we can express it as:
Pj = Ps + 20 dB
2. Convert the power difference to a linear scale:
Using the formula:
P(dB) = 10 * log10(P(linear))
We can rewrite the equation as:
20 dB = 10 * log10(Pj/Ps)
Solving for Pj/Ps:
Pj/Ps = 10^(20/10)
Pj/Ps = 100
3. Determine the processing gain (G):
The processing gain represents the ratio of the spread bandwidth to the information bandwidth. In this case, the information bandwidth is 1 kbps. The processing gain can be calculated as:
G = Spread Bandwidth / Information Bandwidth
G = Spread Bandwidth / 1000 Hz
4. Calculate the jammer power in terms of the desired signal power:
Pj = 100 * Ps
5. Determine the required SNR (Signal-to-Noise Ratio) for satisfactory performance:
Since the required b/J0 is 10 dB, we can express it as:
SNR = Signal Power / Noise Power
SNR = Ps / Pn
SNR = 10^(10/10)
SNR = 10
6. Calculate the noise power:
Pn = Ps / SNR
Pn = Ps / 10
7. Calculate the spreading bandwidth required:
The spreading bandwidth can be calculated using the formula:
Spreading Bandwidth = (Signal Power + Jammer Power) / (Noise Power + Jammer Power)
Substituting the values:
Spreading Bandwidth = (Ps + 100 * Ps) / (Ps / 10 + 100 * Ps)
Spreading Bandwidth = (101 * Ps) / (Ps / 10 + 100 * Ps)
Spreading Bandwidth = 101 / (1/10 + 100)
Spreading Bandwidth = 101 / (0.01 + 100)
Spreading Bandwidth = 101 / 100.01
Spreading Bandwidth ≈ 1.0099 Hz
Therefore, the spreading bandwidth required to meet the specifications is approximately 1.0099 Hz, which can be rounded to 1.01 Hz. Hence, the correct answer is option 'D' (106 Hz).
- The spread spectrum system transmits at a rate of 1 kbps.
- The power of the jammer is 20 dB greater than the desired signal.
- The required b/J0 to achieve satisfactory performance is 10 dB.
Let's break down the solution into steps:
1. Find the desired signal power (Ps):
Since the jammer power is 20 dB greater than the desired signal, we can express it as:
Pj = Ps + 20 dB
2. Convert the power difference to a linear scale:
Using the formula:
P(dB) = 10 * log10(P(linear))
We can rewrite the equation as:
20 dB = 10 * log10(Pj/Ps)
Solving for Pj/Ps:
Pj/Ps = 10^(20/10)
Pj/Ps = 100
3. Determine the processing gain (G):
The processing gain represents the ratio of the spread bandwidth to the information bandwidth. In this case, the information bandwidth is 1 kbps. The processing gain can be calculated as:
G = Spread Bandwidth / Information Bandwidth
G = Spread Bandwidth / 1000 Hz
4. Calculate the jammer power in terms of the desired signal power:
Pj = 100 * Ps
5. Determine the required SNR (Signal-to-Noise Ratio) for satisfactory performance:
Since the required b/J0 is 10 dB, we can express it as:
SNR = Signal Power / Noise Power
SNR = Ps / Pn
SNR = 10^(10/10)
SNR = 10
6. Calculate the noise power:
Pn = Ps / SNR
Pn = Ps / 10
7. Calculate the spreading bandwidth required:
The spreading bandwidth can be calculated using the formula:
Spreading Bandwidth = (Signal Power + Jammer Power) / (Noise Power + Jammer Power)
Substituting the values:
Spreading Bandwidth = (Ps + 100 * Ps) / (Ps / 10 + 100 * Ps)
Spreading Bandwidth = (101 * Ps) / (Ps / 10 + 100 * Ps)
Spreading Bandwidth = 101 / (1/10 + 100)
Spreading Bandwidth = 101 / (0.01 + 100)
Spreading Bandwidth = 101 / 100.01
Spreading Bandwidth ≈ 1.0099 Hz
Therefore, the spreading bandwidth required to meet the specifications is approximately 1.0099 Hz, which can be rounded to 1.01 Hz. Hence, the correct answer is option 'D' (106 Hz).
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A DS spread spectrum system transmit at a rate of 1 kbps in the presets of a tone jammer. The jammer power is 20 dB greater then the desired signal, and the required εb/J0 to achieve satisfactory performance is 10 dB.Que: The spreading bandwidth required to meet the specifications isa)107Hzb)103Hzc)105Hzd)106HzCorrect answer is option 'D'. Can you explain this answer?
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