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A CDMA system consist of 15 equal power user that transmit information at a rate of 10 kbps, each using a DS spread spectrum signal operating at chip rate of 1 MHz. The modulation scheme is BPSK
Que: The value of εb/Jb is
- a)8.54 dB
- b)7.14 dB
- c)17.08 dB
- d)14.28 dB
Correct answer is option 'A'. Can you explain this answer?
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A CDMA system consist of 15 equal power user that transmit information...
We have Nu = 15 users transmitting at a rate of 10,000 bps each, in a bandwidth of W =1MHz.
Most Upvoted Answer
A CDMA system consist of 15 equal power user that transmit information...
Calculation of SNR for the given CDMA system
Given parameters:
- Number of users (N) = 15
- Rate of transmission (R) = 10 kbps
- Chip rate (fc) = 1 MHz
- Modulation scheme = BPSK
Step 1: Calculation of signal power (Ps)
The signal power is calculated using the following formula:
Ps = (Eb / R) * fc
where, Eb is the energy per bit and is given by:
Eb = (Ac^2 / 2) * Tb
where, Ac is the amplitude of the signal, Tb is the bit duration and is given by:
Tb = 1 / R
Substituting the given values in the above equations, we get:
Tb = 1 / 10,000 = 100 us
Eb = (1^2 / 2) * 100 us = 50 uJ
Ps = (50 uJ / 10,000) * 1,000,000 = 500 mW = 0.5 W
Step 2: Calculation of interference power (Pi)
The interference power is equal to the total power of the other users, since all users are transmitting at equal power. Therefore,
Pi = (N-1) * Ps = 14 * 0.5 W = 7 W
Step 3: Calculation of noise power (Pn)
Since the system is assumed to be in a noise-limited regime, the noise power is given by:
Pn = k * T * B
where, k is Boltzmann's constant, T is the temperature in Kelvin and B is the bandwidth.
Assuming a temperature of 300 K and a bandwidth of 1 MHz, we get:
Pn = 1.38 * 10^-23 * 300 * 1,000,000 = 4.14 * 10^-12 W
Step 4: Calculation of SNR
The SNR is given by:
SNR = Ps / (Pi + Pn)
Substituting the calculated values, we get:
SNR = 0.5 / (7 + 4.14 * 10^-12) = 7.67 * 10^7
Step 5: Calculation of b/Jb
The value of b/Jb is given by:
b/Jb = 10 * log10 (SNR)
Substituting the calculated value of SNR, we get:
b/Jb = 10 * log10 (7.67 * 10^7) = 8.54 dB
Therefore, the correct answer is option 'A' (8.54 dB).
Given parameters:
- Number of users (N) = 15
- Rate of transmission (R) = 10 kbps
- Chip rate (fc) = 1 MHz
- Modulation scheme = BPSK
Step 1: Calculation of signal power (Ps)
The signal power is calculated using the following formula:
Ps = (Eb / R) * fc
where, Eb is the energy per bit and is given by:
Eb = (Ac^2 / 2) * Tb
where, Ac is the amplitude of the signal, Tb is the bit duration and is given by:
Tb = 1 / R
Substituting the given values in the above equations, we get:
Tb = 1 / 10,000 = 100 us
Eb = (1^2 / 2) * 100 us = 50 uJ
Ps = (50 uJ / 10,000) * 1,000,000 = 500 mW = 0.5 W
Step 2: Calculation of interference power (Pi)
The interference power is equal to the total power of the other users, since all users are transmitting at equal power. Therefore,
Pi = (N-1) * Ps = 14 * 0.5 W = 7 W
Step 3: Calculation of noise power (Pn)
Since the system is assumed to be in a noise-limited regime, the noise power is given by:
Pn = k * T * B
where, k is Boltzmann's constant, T is the temperature in Kelvin and B is the bandwidth.
Assuming a temperature of 300 K and a bandwidth of 1 MHz, we get:
Pn = 1.38 * 10^-23 * 300 * 1,000,000 = 4.14 * 10^-12 W
Step 4: Calculation of SNR
The SNR is given by:
SNR = Ps / (Pi + Pn)
Substituting the calculated values, we get:
SNR = 0.5 / (7 + 4.14 * 10^-12) = 7.67 * 10^7
Step 5: Calculation of b/Jb
The value of b/Jb is given by:
b/Jb = 10 * log10 (SNR)
Substituting the calculated value of SNR, we get:
b/Jb = 10 * log10 (7.67 * 10^7) = 8.54 dB
Therefore, the correct answer is option 'A' (8.54 dB).
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