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A 1W video signal having bandwidth of 200 MHz is transmitted to a reciver through a cable that has 50 dB loss. If the efficitive two sided noise power spectred density at the receiver is 10-20 Watt/Hz, then the signal to noise ratio at the receiver is
  • a)
    50 x 105
  • b)
    25 x 105
  • c)
    107
  • d)
    106
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A 1W video signal having bandwidth of 200 MHz is transmitted to a reci...
PS = 1 W
Since cable loss is 50 dB (105), So
power at the input of reciever
PS = 10-5 Watts
Noise power = 2 x200 x106 x10-20
2, Because of two sided PSD
= 4 x 10-12 Watts
Therefore
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Most Upvoted Answer
A 1W video signal having bandwidth of 200 MHz is transmitted to a reci...
Given Data:
- Bandwidth of the video signal = 200 MHz
- Cable loss = 50 dB
- Effective two-sided noise power spectral density = 10-20 Watt/Hz

Calculating Signal-to-Noise Ratio (SNR):
- Signal power = 1 W
- Noise power = 10-20 Watt/Hz * 200 MHz = 2-4 W

Calculating Signal-to-Noise Ratio (SNR) in dB:
- SNR = Signal Power / Noise Power = 1 W / 2-4 W = 0.25 - 0.5
- SNR in dB = 10 * log10(SNR) = 10 * log10(0.25 - 0.5) = 25 - 50 dB

Answer:
- The signal-to-noise ratio at the receiver is 25 x 10^5. Hence, option B is correct.
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A 1W video signal having bandwidth of 200 MHz is transmitted to a reciver through a cable that has 50 dB loss. If the efficitive two sided noise power spectred density at the receiver is 10-20 Watt/Hz, then the signal tonoise ratio at the receiver isa)50 x 105b)25 x 105c)107d)106Correct answer is option 'B'. Can you explain this answer?
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