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A signal with a power of \(13 \mathrm{dBm}\) is input to a low-noise amplifier along with noise of \(1 \mathrm{~mW}\) and an interfering signal of \(2 \mathrm{~mW}\). (a) What is the SIR at the input to the amplifier? Express your answer in decibels. (b) QPSK modulation is used and coding results in a processing gain, \(G_{P}\), of \(13 \mathrm{~dB}\). What is the ratio of the effective energy per bit to the noise power per bit (i.e., what is \(\left.E_{b} / N_{o}\right)\) after despreading?

Short Answer

Expert verified
a) 7 dBb) 20 dB

Step by step solution

01

- Calculate SIR

The Signal-to-Interference Ratio (SIR) can be calculated by dividing the power of the signal by the power of the interference.Power of signal, P_s = 13 dBm = 10^((13-30)/10) WPower of interference, P_i = 2 mW = 2 * 10^(-3) WSIR (linear) = P_s / P_i = (10^((13-30)/10)) / (2 * 10^(-3)) =0.02 / 0.002 = 5To convert SIR into decibels, use the formula:SIR (dB) = 10 * log_10(SIR)SIR (dB) = 10 * log_10(5)SIR (dB) ≈ 10 * 0.699 = 6.99 ≈ 7 dB
02

- Processing Gain

Given the processing gain, G_P = 13 dBApplying processing gain, the effective energy per bit to noise power per bit ratio (E_b/N_o) can be calculated using the formula:E_b/N_o (dB) = SIR (dB) + G_PGiven SIR = 7 dBE_b/N_o = 7 dB + 13 dB = 20 dB

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Low-Noise Amplifier
A low-noise amplifier (LNA) is a crucial component in the reception of signals, particularly in wireless communication systems. It amplifies very weak signals without significantly increasing the noise. This is essential because noise can drown out weak signals, making them difficult to detect and process. The primary goals of an LNA are:
  • Minimize additional noise introduced during amplification.
  • Maintain a high Signal-to-Interference Ratio (SIR).
  • Enhance the quality of the received signal.
By carefully designing the LNA, engineers strive to achieve these goals and ensure the integrity and usability of the amplified signal. The performance of an LNA is often evaluated based on its noise figure, which is a measure of how much additional noise the amplifier introduces compared to an ideal noise-free amplifier.
QPSK Modulation
Quadrature Phase Shift Keying (QPSK) is a modulation scheme used in telecommunications to transmit data efficiently. It converts digital bits into a signal suitable for transmission by varying the phase of a carrier wave. Here are the core elements of QPSK:
  • QPSK encodes two bits per symbol, thus doubling the data rate within the same bandwidth compared to binary Phase Shift Keying (BPSK). This makes it more efficient.
  • The modulation scheme has four distinct symbols, each representing a unique phase shift: 45°, 135°, 225°, and 315°.
  • The use of multiple phases allows for better utilization of available bandwidth and improves data transmission rates.
QPSK is widely used in various communication systems, including satellite and cellular networks, due to its robustness and efficiency.
Processing Gain
Processing gain is a measure of how much an error-correcting technique or spread spectrum system can improve the signal-to-noise ratio (SNR) of a communication system. It is fundamental in systems using spread spectrum techniques such as Code Division Multiple Access (CDMA). The processing gain is defined as: \ \ \text{\(G_{P} = 10 \log_{10}(\frac{Bandwidth_{_ss}}{Data\text{ }Rate})\)} \ In the problem, the processing gain is given as 13 dB, which helps in improving the effective energy per bit to noise power per bit ratio (\(E_b / N_o\)). Here’s why processing gain is vital:
  • It increases the effective SNR, improving the reliability of the received signal.
  • It aids in interference rejection, allowing the system to distinguish the desired signal from noise and interference more effectively.
  • It enhances the overall system performance, especially in environments with high levels of interference.
By applying the processing gain, the system effectively spreads the signal over a wider bandwidth, then de-spreads it at the receiver, thus boosting the strength of the desired signal relative to noise and interference.

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Most popular questions from this chapter

An OFDM system with 12 data subcarriers, uses a coding rate of \(3 / 4,\) and each subcarrier uses \(16-\) QAM modulation (with a modulation efficiency of \(2.7 \mathrm{bit} / \mathrm{s} / \mathrm{Hz})\) with a bandwidth of \(250 \mathrm{kHz}\). What is the maximum data rate supported?

A new communication system is being investigated for sending data to a printer. The system will use GMSK modulation and a channel with \(25 \mathrm{MHz}\) bandwidth and an information bit rate of \(10 \mathrm{Mbit} / \mathrm{s}\). The modulation format will result in a spectrum that distributes power almost uniformly over the 25 MHz bandwidth. [Parallels Example 5.3] (a) What is the processing gain? (b) If the received RF SIR is \(6 \mathrm{~dB},\) what is the effective system SIR (or \(\left.E_{b, i} / N_{o, i}\right)\) after DSP? Express your answer in decibels.

The receiver in a digital radio system receives a \(100 \mathrm{pW}\) signal and the interference from other radios at the input of the receiver is \(20 \mathrm{pW}\). The receiver has an overall gain of \(40 \mathrm{~dB}\) and the noise added by the receiver, referred to the out- $$2$$ put of the receiver, is \(100 \mathrm{nW}\). (a) What is the RF SIR at the output of the receiver? (b) If 16-QAM modulation with a modulation efficiency of \(2.98 \mathrm{bit} / \mathrm{s} / \mathrm{Hz}\) is used and the processing gain is \(30 \mathrm{~dB}\), what is the effective SIR after despreading, i.e. what is \(E_{b, \text { eff }} / N_{o, b} ?\)

An OFDM system with 48 subcarriers carrying data uses 16-QAM modulation of each subcarrier and a coding rate of \(2 / 3\). The actual modulation efficiency for the 16-QAM system here is 2.7 bit \(/ \mathrm{s} / \mathrm{Hz}\). What is the maximum data rate supported in Mbit/s when the bandwidth of each modulated subcarrier is \(312 \mathrm{kHz}\) ?

The channel bandwidth in the GSM cellular phone system is \(200 \mathrm{kHz}\) and the GMSK modulation scheme used has a spectral efficiency of \(1.354 \mathrm{bit} / \mathrm{s} / \mathrm{Hz}\) (a) What is the data rate of one frequency channel? (b) A time slot is \(577 \mu\) s long. How many bits are there in one (i.e. a duration of 8.25 bits). How many data bits are there in a GSM time slot? (c) A GSM frame duration is \(4.615 \mathrm{~ms}\) long and has eight time slots and a voice user has one time slot every frame. How many data bits per second are available to a single user?

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