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Analyzing image transmission quality using filter and C-QAM
Published in Debatosh Guha, Badal Chakraborty, Himadri Sekhar Dutta, Computer, Communication and Electrical Technology, 2017
Md. Khaliluzzaman, Deepak Kumar Chy, Kaushik Deb
32 C-QAM is a cross-shape constellation. The constellations for odd values of n as Q = 2n have a cross shape and hence called 32 cross-constellation QAM, where the value of n is 5. The baud rate or symbol rate is T, where T denotes symbol rate or baud rate. In general, the symbol rate is equal to the bit rate divided by the total number of bits transmitted with each symbol. However, the signal bandwidth depends on symbol rate in such situation. Each symbol of 32 cross-constellation QAM carries m = log2 (k) bits of information. Therefore, 32 cross-constellation QAM carries 5 bit of information per symbol. The average energy can be calculated of 32 cross-constellation QAM, which is a revolved version of a 36-QAM square constellation, where the corner points are deleted, as they consume more power than other points in 36-QAM. If a 36-QAM is rotated, the energy is unchangeable. Therefore, the total energy of 36 QAM is first estimated. Then, the energy of the four-corner points is determined. Finally, the estimated energy of 36-QAM will be subtracted by the four-corner points’ energy. The resulting energy divided by 32 gives average energy of 32 cross-constellation.
Technologies
Published in Henry H. Perritt, Eliot O. Sprague, Domesticating Drones, 2016
Henry H. Perritt, Eliot O. Sprague
A Morse Code CW signal has a bandwidth of only 50–100 Hz, because the rate of information transfer is fairly low—less than 35 words per minute. AM voice communications signals require bandwidth on the order 3.0 KHz. High-fidelity sound requires about twice that. Broadcast television signals require about 6 megahertz, and digitally encoded signals, whatever their content, require bandwidth proportional to the bitrate, which depends on the modulation scheme—8 bits per symbol is about the best currently achievable. So 6 MHz bandwidth can handle, at most, 25 MB/s. The difference between bandwidth and data rates arises because bandwidth depends on symbol rate; one symbol can represent more than one bit in advanced modulation schemes.
From launch to transmission: satellite communication theory and SNG
Published in Jonathan Higgins, Satellite Newsgathering, 2012
The term ‘symbol’ is often used in connection with the transmitted bits, and all this means is that in a BPSK signal each bit represents a ‘symbol’, and also that in a QPSK signal, each pair of bits also represents a ‘symbol’. Therefore, a symbol in a QPSK signal carries twice the information that a BPSK symbol does in the same bandwidth. The symbol rate is measured in symbols per second, historically referred to as the ‘baud rate’.
Design and Implementation of an Enterprise Credit Risk Assessment Model Based on Improved Fuzzy Neural Network
Published in Applied Artificial Intelligence, 2023
Simulation condition: The frequency hopping bandwidth is set. According to the interval of each frequency slot , the entire frequency hopping bandwidth is divided into N = 50 frequency slots. The channel is used in the simulation, and the noise power spectral density is set to . For narrowband interference, the number of interference channels is set to , and the interference power spectrum is −30dBW. We set the narrowband interference bandwidth to be the same as the frequency slot bandwidth, and the partial frequency band interference factor is . The interference is randomly distributed in frequency slots, the modulation method is MFSK, the frequency hopping speed is , and the symbol rate is .
Maritime cognitive radio spectrum sensing based on multi-antenna cyclostationary feature detection
Published in International Journal of Electronics, 2020
Jingbo Zhang, Feng Ran, Da Liu
At present, some useful research results have been obtained for the spectrum sensing method in the land wireless communication environment. Some spectrum sensing methods have been proposed in (Y¨ucek & Arslan, 2009), such as energy detection, cyclostationary detection and matched filter detection. Energy detection is the most widely used blind detector because it requires a priori knowledge of very few signals. However, it is sensitive to noise anomalies, and detection performance depends on accurate noise power (Tandra & Sahai, 2008). Matched filter detection requires a large amount of prior knowledge of the signal, which is often difficult to satisfy. The cyclostationary feature detection does not require prior knowledge of the signal and is not affected by the noise wall problem. In addition, it can detect the modulation type, symbol rate, carrier frequency and other characteristics of the signal (Gardner, Brown, & Chen, 1987), so it has received extensive attention. The typical cyclostationary feature detection method is calculated by the Fourier cycle spectral density (Tkachenko, Cabric, & Brodersen, 2007) or its multiple Loève model (Haykin, Thomson, & Reed, 2009), and the computational complexity is relatively large. Literature (Jeon, Jeong, Han, Ko, & Song, 2008) proposed a method to reduce the computational complexity by using energy detection results to trigger cyclostationary feature detection.
VCSEL-based differential modulation technique for high-speed gigabit passive optical networks
Published in Journal of Modern Optics, 2019
G. M. Isoe, S. Wassin, E.K. Rotich Kipnoo, A. W. R. Leitch, T. B. Gibbon
Bit error ratio (BER) calculation is a key metric to asses link performance in optical communication. With sufficiently high SNR and Gaussian noise, BER can be approximated as where I is the average photodetector current, is the RMS current noise and d is the average hamming distance between adjacent symbols (14). Compared to OOK modulation fomart and assuming AWGN in the link, the optical penalty for M-PAM at the same symbol rate can be calculated as (15); When comparing M-level PAM to OOK fomart at the same bit rate, the power penalty is, therefore, calculated as; From Equation (3), it is evident that although the bandwidth requirement for 4-PAM is halved compared to that of OOK, more optical power is needed to achieve the same data rate and symbol rate.