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A slow FH binary FSK system with non coherent detection operates at ε0/J0 10, with hopping bandwidth of 2 GHz, and a bit rate of 10 kbps.
Que: The processing gain of this system is
- a)23 dB
- b)43 dB
- c)43 dB
- d)53 dB
Correct answer is option 'D'. Can you explain this answer?
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A slow FH binary FSK system with non coherent detection operates at ...
We are given a hopping bandwidth of 2 GHz and a bit rate of 10 kbs.
Most Upvoted Answer
A slow FH binary FSK system with non coherent detection operates at ...
The processing gain of a frequency-hopped binary frequency shift keying (FH/BFSK) system can be calculated using the formula:
Processing Gain (G) = Bandwidth of the FH system / Bit Rate
Given:
Hopping Bandwidth (B) = 2 GHz = 2 x 10^9 Hz
Bit Rate (R) = 10 kbps = 10^4 bps
To calculate the processing gain, we need to convert both the bandwidth and the bit rate into the same units. Let's convert the Bit Rate into Hz.
Bit Rate (R) = 10 kbps = 10^4 bps = 10^4 Hz
So, now both the bandwidth and bit rate are in Hz.
Now, let's substitute the values into the formula:
Processing Gain (G) = Bandwidth of the FH system / Bit Rate
G = B / R
G = 2 x 10^9 Hz / 10^4 Hz
G = 2 x 10^5
The processing gain is 2 x 10^5.
To convert the processing gain into decibels (dB), we can use the formula:
Gain (dB) = 10 log10(G)
Substituting the value of G into the formula:
Gain (dB) = 10 log10(2 x 10^5)
Gain (dB) = 10 log10(2) + 10 log10(10^5)
Gain (dB) = 10 x 0.301 + 10 x 5
Gain (dB) = 3.01 + 50
Gain (dB) = 53.01 dB
Therefore, the processing gain of this system is 53 dB.
Processing Gain (G) = Bandwidth of the FH system / Bit Rate
Given:
Hopping Bandwidth (B) = 2 GHz = 2 x 10^9 Hz
Bit Rate (R) = 10 kbps = 10^4 bps
To calculate the processing gain, we need to convert both the bandwidth and the bit rate into the same units. Let's convert the Bit Rate into Hz.
Bit Rate (R) = 10 kbps = 10^4 bps = 10^4 Hz
So, now both the bandwidth and bit rate are in Hz.
Now, let's substitute the values into the formula:
Processing Gain (G) = Bandwidth of the FH system / Bit Rate
G = B / R
G = 2 x 10^9 Hz / 10^4 Hz
G = 2 x 10^5
The processing gain is 2 x 10^5.
To convert the processing gain into decibels (dB), we can use the formula:
Gain (dB) = 10 log10(G)
Substituting the value of G into the formula:
Gain (dB) = 10 log10(2 x 10^5)
Gain (dB) = 10 log10(2) + 10 log10(10^5)
Gain (dB) = 10 x 0.301 + 10 x 5
Gain (dB) = 3.01 + 50
Gain (dB) = 53.01 dB
Therefore, the processing gain of this system is 53 dB.
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A slow FH binary FSK system with non coherent detection operates atε0/J0 10, with hopping bandwidth of 2 GHz, and a bit rate of 10 kbps.Que: The processing gain of this system isa)23 dBb)43 dBc)43 dBd)53 dBCorrect answer is option 'D'. Can you explain this answer?
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