Question 1. [4 marks] Consider a wireless communication system with a transmitter power of 15microwatts. Additive White Gaussian Noise (AWGN) corrupts the signal at the receiver front-end. Thewireless communication link is operating in free-space propagation conditions with a separation distanceof d = 500m Signaling bandwidth is B = 5 MHz, the effective noise temperature of the device is Te = 290degree Kelvins. Recall free-space pathloss equation as discussed in Week 1 lecture, path loss is the ratioof transmitter power Pt to receiver power Pr and is expressed in dB scale as belowPLdB = 10log10( PtPr)PLdB = −10log10( GtGrλ2(4πd)2)Transmitter and receiver gains are assumed Gt = Gr = 10dB and carrier frequency of operation fc = 2GHz. Note: Gains are defined in dB, convert to linear scale as required for use with the above pathlossequation. You may want to expand the above equation in dB in which case you can directly use theantenna gain defined in dB (refer to Lecture 5 slide number 3).1) Calculate the transmitter power in dBm.2) Calculate the pathloss in dB.3) What is the power at the receiver in dBm?4) Calculate the receive signal power to noise power ratio (SNR) at the receiver input.5) Considering a 4-QAM communication system, write the expression for calculating the Symbol errorrate (SER) using the Union bound.(Note: it is enough if you write your answer as an expressionwith Q function and substitute appropriate values for the SNR and minimum Eculidean distance.Not necessary to give numerical value for the symbol error rate)
Question
Question 1. [4 marks] Consider a wireless communication system with a transmitter power of 15microwatts. Additive White Gaussian Noise (AWGN) corrupts the signal at the receiver front-end. Thewireless communication link is operating in free-space propagation conditions with a separation distanceof d = 500m Signaling bandwidth is B = 5 MHz, the effective noise temperature of the device is Te = 290degree Kelvins. Recall free-space pathloss equation as discussed in Week 1 lecture, path loss is the ratioof transmitter power Pt to receiver power Pr and is expressed in dB scale as belowPLdB = 10log10( PtPr)PLdB = −10log10( GtGrλ2(4πd)2)Transmitter and receiver gains are assumed Gt = Gr = 10dB and carrier frequency of operation fc = 2GHz. Note: Gains are defined in dB, convert to linear scale as required for use with the above pathlossequation. You may want to expand the above equation in dB in which case you can directly use theantenna gain defined in dB (refer to Lecture 5 slide number 3).1) Calculate the transmitter power in dBm.2) Calculate the pathloss in dB.3) What is the power at the receiver in dBm?4) Calculate the receive signal power to noise power ratio (SNR) at the receiver input.5) Considering a 4-QAM communication system, write the expression for calculating the Symbol errorrate (SER) using the Union bound.(Note: it is enough if you write your answer as an expressionwith Q function and substitute appropriate values for the SNR and minimum Eculidean distance.Not necessary to give numerical value for the symbol error rate)
Solution
- Calculate the transmitter power in dBm.
The transmitter power in dBm can be calculated using the formula:
P(dBm) = 10 * log10(P(mW))
Given that the transmitter power is 15 microwatts, we first need to convert this to milliwatts:
P(mW) = 15 microwatts = 0.015 mW
Substituting this into the formula gives:
P(dBm) = 10 * log10(0.015) = -11.76 dBm
- Calculate the pathloss in dB.
The pathloss in dB can be calculated using the formula:
PL(dB) = -10 * log10(Gt * Gr * λ^2 / (4πd)^2)
Given that the transmitter and receiver gains are 10 dB (which is equivalent to a linear gain of 10), the carrier frequency is 2 GHz (which gives a wavelength λ = c/fc = 310^8 m/s / 210^9 Hz = 0.15 m), and the separation distance is 500 m, substituting these values into the formula gives:
PL(dB) = -10 * log10(10 * 10 * (0.15)^2 / (4π*500)^2) = 92.45 dB
- What is the power at the receiver in dBm?
The power at the receiver can be calculated by subtracting the pathloss from the transmitter power:
Pr(dBm) = Pt(dBm) - PL(dB)
Substituting the previously calculated values gives:
Pr(dBm) = -11.76 dBm - 92.45 dB = -104.21 dBm
- Calculate the receive signal power to noise power ratio (SNR) at the receiver input.
The SNR can be calculated using the formula:
SNR = Pr / N
Where N is the noise power. The noise power can be calculated using the formula:
N = k * Te * B
Where k is Boltzmann's constant (1.3810^-23 J/K), Te is the effective noise temperature (290 K), and B is the bandwidth (5 MHz = 510^6 Hz). Substituting these values into the formula gives:
N = 1.3810^-23 * 290 * 510^6 = 2*10^-16 W = -174 dBm
Therefore, the SNR is:
SNR = Pr / N = -104.21 dBm / -174 dBm = 69.79 dB
- Considering a 4-QAM communication system, write the expression for calculating the Symbol error rate (SER) using the Union bound.
The symbol error rate for a 4-QAM system can be calculated using the formula:
SER = 2 * Q(sqrt(2*SNR))
Where Q(x) is the Q-function, which is a function of the Gaussian error function, and SNR is the signal-to-noise ratio. In this case, the SNR is 69.79 dB, so the SER is:
SER = 2 * Q(sqrt(2*69.79))
Similar Questions
Which of the following statements about the characteristics of wireless links are true?Group of answer choicesThe bit error rate (BER) of a wireless channel decreases as the signal-to-noise ratio (SNR) increases.The "hidden terminal problem" happens when A sends to B over a wireless channel, and an observer, C (that can be even closer to A than B), does not detect/receive A's transmission because of physical obstacles in the path between A and B.Multipath propagation occurs when portions of the electromagnetic wave reflect off objects and the ground taking paths of different lengths between the sender and a receiver, and thus arriving at the receiver at slightly different points in time.The "hidden terminal problem" happens when A sends to B over a wireless channel, and an observer, C (that is further away from A than B), does not detect/receive A's transmission because the signal strength of A's transmission has faded significantly by the time it reaches C.The bit error rate (BER) of a wireless channel increases as the signal-to-noise ratio (SNR) increases.Path loss refers to the dropping of link-layer frames that are being relayed among wireless access points due to buffer overflow, just as network-layer datagrams are dropped at routers with full buffers.The "hidden terminal problem" happens when A sends to B over a wireless channel, and an observer, C (that can be even closer to A than B), does not detect/receive A's transmission because of physical obstacles in the path between A and C.Multipath propagation occurs when a sender sends multiple copies of a frame to a receiver, which is relayed over different by base stations or other wireless devices to the receiver.Path loss refers to link-layer frames that are corrupted due to the higher bit error rates in wireless channels.Path loss refers to the decrease in the strength of a radio signal as it propagates through space.The "hidden terminal problem" refers to the fact that many people can never seem to find their mobile phones.
In free space transmission, the signal attenuation increasesa.proportionally with distanceb.proportionally with the square distancec.proportionally with distance cube
When transmitting over a perfectly reflecting, smooth, plane earth, the path loss tends to increasea.with the square of distanceb.with the fourth power of distancec.linearly with distanced.with distance cube
What is Free space path loss? Line-of-sight between two antennas with no obstacles Line-of-sight between two antennas with obstacles
Choose the best answer for the following multiple-choice: Rapid variation of the received signal level due to the effect of multiple signal paths is known as: a. Path loss b. Time variance c. Flat fading d. Shadowing For the log-distance path loss model, the reference distance (do) is properly defined based on: a. Signal frequencies b. Antenna Gains c. Propagation environments d. Signal bandwidths Multipath delay spread in the wireless channels causes: a. Doppler spread b. Time dispersion c. Frequency dispersion d. Fast fading A probability density function of the signal received in the LOS environment follows: a. Rician distribution b. Gausian distribution c. Rayleigh distribution d. Normal distribution In the MIMO wireless system, increasing of the QAM modulation levels leads to: a. Data rate ↑, Error rate ↓ b. Data rate ↑, Error rate ↑ c. Data rate ↓, Error rate ↓ d. Data rate ↓, Error rate ↑ In multiple- antenna systems, the data rates mainly increased by: a. Spatial diversity techniques b. Beamforming techniques c. Modulation techniques d. Spatial multiplexing technique
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