Chapter 7
Binary Data Communication
7.1 Problem Solutions
Problem 7.1 The signal-to-noise ratio is z= A2 T A2 = N0 N0 R
Trial and error using the asymptotic expression Q (x) exp -x2 /2 / 2x shows that 2z = 10-5 for z = 9.58 dB = 9.078 PE =
Classification of Signals
Continuous time and
discrete time signals
Analog and digital signals
Periodic and aperiodic
signals
Energy and power signals
Deterministic and radaom
signals
Physical description is known
completely in either
mathematical or grap
Analysis and
Transmission of Signals
1
Content
Aperiodic Signal and Fourier Integral
Transforms of Some Useful Functions
Properties of Fourier Transform
Signal Transmission Through Linear
System
Signal Energy and Power
2
2
Aperiodic Signal:
Fourier Integr
Signals and Signal
Space
Content
Classifications of Signals
Operations of Signals
Samples of Signals
Correlation of Signals
Fourier Series
2
Signal & System Analysis
To analyze (& design) communication systems, we need
to precisely describe signals & sys
ANGLE MODULATION AND
DEMODULATION
Content
Nonlinear Modulation
Bandwidth of Angle-Modulated Waves
Generating FM Signals
Non-linear Distortion and Interference
Analog AM/FM Receivers and FM
Broadcasting System
2
Nonlinear Modulation
Frequency Modulation (F
ELCT 332
Fundamental of Wireless Communications
Pre-requisite: ELCT 321 and STAT 509
Instructor: Dr. Guoan Wang
Office: SWGN 3A12
Phone: 803-777-6303
Email: [email protected]
Lecture Hours: MWF 9:40 10:30 SWGN 2A22
Office Hours: by appointment
Textbook: Mo
Chapter 5
Random Signals and Noise
5.1 Problem Solutions
Problem 5.1 The various sample functions are as follows. Sample functions for case (a) are horizontal lines at levels A, 0, -A, each case of which occurs equally often (with probability 1/3).
Chapter 4
Probability and Random Variables
4.1 Problem Solutions
Problem 4.1 S = sample space is the collection of the 25 parts. Let Ai = event that the pointer stops on the ith part, i = 1, 2, ., 25. These events are exhaustive and mutually exclus
Chapter 3
Basic Modulation Techniques
3.1 Problems
Problem 3.1 The demodulated output, in general, is yD (t) = Lp{xc (t) 2 cos[ c t + (t)]} where Lp {} denotes the lowpass portion of the argument. With xc (t) = Ac m (t) cos [ c t + 0 ] the demodul
Chapter 9
Optimum Receivers and Signal Space Concepts
9.1 Problems
Problem 9.1 a. Given H1 , Z = N , so fZ (z|H1 ) = fN (n) |z=n = 10e-10z u (z) Given H2 , Z = S +N , where S and N are independent. Thus, the resulting pdf under H2 is the convolutio
Chapter 6
Noise in Modulation Systems
6.1 Problems
Problem 6.1 The signal power at the output of the lowpasss .lter is PT . The noise power is N0BN , where BN is the noise-equivalent bandwidth of the .lter. From (5.116), we know that the noise-equi
Chapter 8
Advanced Data Communications Topics
8.1 Problem Solutions
Problem 8.1 Use the relationship Rb = (log2 M ) Rs bps where Rb is the data rate, Rs is the symbol rate, and M is the number of possible signals per signaling interval. In this cas
Chapter 2
Signal and Linear System Theory
2.1 Problem Solutions
Problem 2.1 For the single-sided spectra, write the signal in terms of cosines: x(t) = 10 cos(4t + /8) + 6 sin(8t + 3/4) = 10 cos(4t + /8) + 6 cos(8t + 3/4 - /2) = 10 cos(4t + /8) + 6
Chapter 10
Information Theory and Coding
10.1 Problem Solutions
Problem 10.1 The information in the message is I (x) = - log2 (0.8) = 0.3219 bits I (x) = - loge (0.8) = 0.2231 nats I (x) = - log10 (0.8) = 0.0969 Hartleys
Problem 10.2 (a) I (x) = -