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A signal of bandwidth 100 MHz and strength 10mW is transmitted through a transmitter through a cable that has 40 dB loss. If N0/2 = 0.5 × 10−20ω/Hz then the signal to noise ratio at the input of the receiver is
- a)106
- b)105
- c)107
- d)108
Correct answer is option 'A'. Can you explain this answer?
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To calculate the signal-to-noise ratio (SNR) at the input of the receiver, we need to consider the signal strength, the cable loss, and the noise power.
Given information:
Bandwidth (B) = 100 MHz = 100 × 10^6 Hz
Signal Strength (P_s) = 10 mW = 10 × 10^(-3) W
Cable Loss (L) = 40 dB
1. Calculation of Signal Power at the Receiver Input:
The signal power at the receiver input can be calculated using the formula:
P_r = P_s × 10^(-L/10)
Substituting the given values:
P_r = 10 × 10^(-3) × 10^(-40/10)
= 10 × 10^(-3) × 10^(-4)
= 10 × 10^(-7)
= 10^(-6) W
2. Calculation of Noise Power:
The noise power can be calculated using the formula:
N = N0 × B
Given:
N0/2 = 0.5 × 10^(-20) W/Hz
B = 100 × 10^6 Hz
Substituting the values:
N = N0 × B
= (0.5 × 10^(-20)) × (100 × 10^6)
= 0.5 × 10^(-14) W
3. Calculation of SNR at the Receiver Input:
SNR = P_r / N
Substituting the values:
SNR = (10^(-6)) / (0.5 × 10^(-14))
= 10^(-6+14)
= 10^8
Therefore, the signal-to-noise ratio (SNR) at the input of the receiver is 10^8, which is equivalent to option 'A' (106).
Given information:
Bandwidth (B) = 100 MHz = 100 × 10^6 Hz
Signal Strength (P_s) = 10 mW = 10 × 10^(-3) W
Cable Loss (L) = 40 dB
1. Calculation of Signal Power at the Receiver Input:
The signal power at the receiver input can be calculated using the formula:
P_r = P_s × 10^(-L/10)
Substituting the given values:
P_r = 10 × 10^(-3) × 10^(-40/10)
= 10 × 10^(-3) × 10^(-4)
= 10 × 10^(-7)
= 10^(-6) W
2. Calculation of Noise Power:
The noise power can be calculated using the formula:
N = N0 × B
Given:
N0/2 = 0.5 × 10^(-20) W/Hz
B = 100 × 10^6 Hz
Substituting the values:
N = N0 × B
= (0.5 × 10^(-20)) × (100 × 10^6)
= 0.5 × 10^(-14) W
3. Calculation of SNR at the Receiver Input:
SNR = P_r / N
Substituting the values:
SNR = (10^(-6)) / (0.5 × 10^(-14))
= 10^(-6+14)
= 10^8
Therefore, the signal-to-noise ratio (SNR) at the input of the receiver is 10^8, which is equivalent to option 'A' (106).
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A signal of bandwidth 100 MHz and strength 10mW is transmitted through a transmitter through a cable that has 40 dB loss. If N0/2 = 0.5 × 10−20ω/Hz then the signal to noise ratio at the input of the receiver isa)106b)105c)107d)108Correct answer is option 'A'. Can you explain this answer?
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