INTRO - QPSK.pdf

Comparison of analog and digital signals  Communication System Advantages of Digital Communication Systems 1. Privacy · Achieved through encryption 2. Error Correction and Detection Achieved through encoding 3. Ease of Multiplexing 4. Simpler to measure and evaluate Disadvantages of...

Comparison of analog and digital signals  Communication System Advantages of Digital Communication Systems 1. Privacy · Achieved through encryption 2. Error Correction and Detection Achieved through encoding 3. Ease of Multiplexing 4. Simpler to measure and evaluate Disadvantages of Digital Communication Systems 1. Larger bandwidth is needed. 2. Synchronization is needed. 3. Incompatible with older analog transmission facilities Digital Communication Ø Digital Communications covers a broad area of communications techniques, including: § Digital Transmission § Digital Radio Ø Digital Transmission – is the transmission of digital signals between the transmitter and receivers and requires physical transmission medium such as cable, fiber optic , etc Ø Digital Modulation – is the transmission of digitally modulated signals between transmitter and receivers and requires free space (air) transmission medium  Digital Communication Digital Transmission Digital Transmission Medium Information Digital Information Transmitter Receiver Wire, cable fiber optic, etc Digital Digital Information Information DAC ADC ADC – Analog to Digital Converter Analog Analog Information DAC – Digital to Analog Converter Information  Digital Communication Digital Radio Transmission Medium – free space Digital Modulation Digital Information Digital Information Transmitter Receiver Digital Digital Information Information DAC ADC ADC – Analog to Digital Converter Analog Analog Information DAC – Digital to Analog Converter Information Types of Signal Transmission  Signal Analysis Representation of Signals Ø 1. Time Domain § Amplitude - The amplitude of a signal is the height of the wave above or below a given reference point A 1 0.8 0.6 0.4 0.2 0 1 22 43 64 85 106 127 148 169 190 211 232 253 274 295 316 337 358 379 400 421 442 463 484 505 526 547 568 589 610 631 652 673 694 715 t Amplitude -0.2 -0.4 -0.6 -0.8 -1  Signal Analysis Representation of Signals Ø 2. Frequency Domain § A description of signal with respect to frequency. § Amplitude vs Frequency § A frequency analyzer is a frequency domain instrument A A T 1 f= T t Amplitude F f Time Domain Frequency Domain  Signal Analysis Representation of Signals Ø 3. Phase § The phase of a signal is the position of the waveform relative to a given moment of time or relative to time zero § Polar Diagram § I and Q axis Q R Q q I I I = R Cos q Q = R Sin q FORMS OF DIGITAL MODULATION ASK FSK PSK QAM INFORMATION THEORY It is the study of the efficient use of bandwidth to propagate information through electronic communication system INFORMATION CAPACITY – number of independent symbols that can be carried through a system in a given unit of time. HARTLEY’S LAW where: I = information capacity (bits per second) B = bandwidth (Hertz) t = transmission time (sec) SHANNON LIMIT FOR INFORMATION CAPACITY Where: I = information capacity (bps) B = bandwidth ( Hz) S/N = signal to noise power ratio Sample Problems 1. For a standard voice band communication channel with a signal to noise ratio of 1000 and a bandwidth of 2.7 kHz, determine the information capacity 2. Assuming that a PSTN has a bandwidth (B) of 3000 Hz and a typical signal-to-noise ratio (SNR) of 20 dB, determine the maximum theoretical transmission rate (C) that can be achieved. 3. What is the channel capacity for a signal power of 200 W, a noise power of 10 W and a bandwidth of 2kHz of a digital system. Find the spectrun efficiency. 1. For a standard voice band communication channel with a signal to noise ratio of 1000 and a bandwidth of 2.7 kHz, determine the information capacity 2. Assuming that a PSTN has a bandwidth (B) of 3000 Hz and a typical signal-to-noise ratio (SNR) of 20 dB, determine the maximum theoretical transmission rate (C) that can be achieved. 3. What is the channel capacity for a signal power of 200 W, a noise power of 10 W and a bandwidth of 2kHz of a digital system. Find the spectrum efficiency. BIT RATE, BAUD AND M – ARY ENCODING BIT RATE – number of bits transmitted during one second (bps) - rate of change of digital information signal BAUD - rate of change of a signal on a transmission medium after encoding and modulation have occurred. BAUD where: baud = symbol rate (baud per second) ts = time of one output signaling element (sec) M – ary Encoding Where: N = number of bits necessary M = number of conditions, levels or combination possible with N bits NYQUIST BANDWIDTH —The theoretical bandwidth necessary to propagate a signal. —MULTILEVEL SIGNALING Where: fb = channel capacity (bps) B = minimum Nyquist bandwidth (Hz) M = number of discrete signal or voltage levels Examples: 1. An analog signal carries 4 bits/signal elements. If 1000 signal elements are sent per second. Find the bit rate. fb = 4 kbps Total no. of elements = 16 2. An analog signal has a bit rate of 8000 bps and a baudrate of 1000 baud. How many data elements are carried by each signal element? How many signal elements are needed? 2.N = 8 bits/element = 8 bits/ element — m = 256  Amplitude Shift Keying ØThe simplest digital modulation techniques, where a binary information signal directly modulates the amplitude of analog carrier. ØASK is similar to standard amplitude modulation except there are only two output amplitudes possible  Amplitude Shift Keying ASK modulated wave equation Vc Vask(t) =[1+ Vm(t) ] [ 2 Cos wct] v Where: Vask(t) – amplitude shift keying waveform Vm(t) – digital information (modulating signal, 1(+1V or 0 (- 1V) Vc - amplitude of the unmodulated carrier, V wc – analog carrier angular frequency, rad/sec  1. Amplitude Shift Keying ASK – Baud Rate Ø The rate of change of the ASK waveform is the same as the rate of change of the binary input (bps) Ø Thus in ASK, the bit rate is equal to baud rate ASK – Bandwidth ØIn ASK, the bit rate is equal to baud rate and is also equal to the minimum Nyquist bandwidth BW of ASK considering the modulation and filtration value: BW = (1+d)BR d = value that depends on the modulation and filtration process; factor for modulation and filtering process [ 0 or 1] = 0 is the ideal value BW = BR = 1 is the worst case value BW = 2BR ASK BW analysis is same as AM BW analysis except that fm = fa= fb/2 in ASK  Problem: Determine the baud and minimum BW necessary to pass a 10 kbps binary signal using amplitude shift keying. FREQUENCY SHIFT KEYING FREQUENCY SHIFT KEYING FSK is a form of constant-amplitude angle modulation similar to standard frequency modulation (FM) except the modulating signal is a binary signal that varies between two discrete voltage levels rather than a continuously changing analog waveform.  FSK Characteristics and Applications ¡ FSK is a form of narrowband FM ¡ reliable in the presence of noise ¡ not very efficient in terms of bandwidth Applications ¡ Used for low data rate applications ¡ HF radio systems for radioteletype (RTTY) transmissions ¡ Voice grade lines for data rates up to 1200 bps ¡ LAN Frequency Shift Keying (FSK) Simple and low performance Similar to conventional FM, but the modulating signal is digital For logic 1: mark frequency For logic 0: space frequency FSK in the time domain (a) waveform MATHEMATICAL EXPRESSION OF FSK: !"#$ % = !' ()* +2-./' + !1 (%)∆f]t} In FSK… ¡ the bit rate equals the Voltage-controlled baud rate. oscillator (VCO) ¡ the highest modulating – the most widely used FSK frequency, fa, is equal to modulator one-half of the bit rate. Phase-locked loop (PLL) – the most common circuit 1 fa = fb for demodulating FSK signals 2 HIGHEST FUNDAMENTAL FREQUENCY FSK Formulas MODULATION INDEX, M DEVIATION RATIO Df M= also called h-factor modulation index that yields the fa widest output bandwidth fm - fs 1 Df = fa = fb Occurs when both the frequency 2 2 deviation and the modulation f -f M= m s frequency are maximum fb Df = frequency deviation fa = modulating signal frequency Df(max) BANDWIDTH deviation ratio = fa(max) B = 2(Df + fb ) B minimum Nyquist bandwidth (hertz) Example 1: Determine the peak frequency deviation, bandwidth and the baud for an FSK signal with a mark frequency of 99 kHz, a space frequency of 101 kHz and a bit rate of 10 kbps. Ans: freq deviation = 1 kHz. B = 22 kHz, S = 10 kbaud Example 2: Determine the bandwidth using the Bessel Table for the FSK signal with a mark frequency of 49 kHz, a space frequency of 51 kHz, and an input bit rate of 2 kbps. Review: BW = (1+d)BR  Spectrum and bandwidth of BFSK Total BW of BFSK = (1+d)BR + 2∆f Ex. 3 Find the bandwidth of a system which spans from 200 to 300 kHz using FSK with d = 1. Determine the carrier frequency and the bit rate. BW = 100 kHz. fc =250 kHz. fb = 25 kbps Spectrum and bandwidth of multilevel FSK Total BW of BFSK = (1+d)BR + (M-1)(2∆f) FSK transmitter INTERLEVEL FSK OUTPUT FSK RECEIVER NONCOHERENT FSK DEMODULATOR COHERENT FSK DEMODULATOR Minimum Shift Keying (MSK) A form of continuous phase FSK (CP-FSK) Similar to FSK except that the fm and fs are synchronized with the bit rate There is a smooth phase transition in the analog output signal when the frequency changes. fm - fs = 0.5 fB fb fm and fs = n ; n is any odd integer 2 MSK: Pros and Cons ¡MSK has a better bit error performance than FSK ¡Disadvantage: requires synchronizing circuits which makes the over-all circuitry complicated. MSK Waveform Gaussian Minimum Shift Keying Special case of FSK used in cellular and PCS systems GMSK uses less bandwidth than conventional FSK due to a filter. The word Gaussian refers to the shape of this filter. PHASE SHIFT KEYING PHASE SHIFT KEYING Phase Shift-Keying (PSK) The output signal has a constant amplitude and phase angle that varies with the digital modulating signal Variations of PSK: Binary PSK Quaternary PSK 8-PSK 16-PSK Differential BPSK Binary Phase Shift Keying (BPSK) There two output phases possible Logic 1 – 0° phase shift, Logic 0 – 180 ° phase shift When the input digital signal changes state, the output carrier shifts between angles that are 180 degrees out of phase. Also called phase reversal keying or biphase modulation BPSK Transmitter Analog Binary Balanced Bandpass PSK Data modulator Filter Output Reference carrier Highest fundamental frequency oscillator fa = 12 fb Bandwidth fN = fb Baud rate baud rate = fN = fb BPSK Modulator (+90o) cos ωct Binary input Output phase - sinωct sinωct (180o) (0o) Logic 0 180o Logic 1 Logic 1 0o Logic 0 (a) cos ωct cos ωct (-90o) (b) +/-180o 0o Reference Logic 0 Logic 1 (a) Truth table (b) Phasor diagram (c) Constellation diagram sometimes called signal state-space diagram, is similar to a phasor except that the entire phasor is not - cos ωct drawn. In this diagram, only the relative positions of the peaks of the phasors are (c) shown. Nyquist Bandwidth The minimum double-sided Nyquist bandwidth (fN) is 2fa. Therefore, fN = 2 ( fb/2) = fb The minimum bandwidth (fN) required to pass the worst- case BPSK output signal is equal to the input bit rate.  Problems: BPSK Modulator For a BPSK modulator with a carrier frequency of 80MHz and an input bit rate of 20Mbps, determine the maximum and minimum upper and lower side frequencies, draw the output spectrum, determine the minimum Nyquist bandwidth and calculate the baud. BPSK Receiver BPSK +/- sin ωc t Binary data out input Balanced LPF modulator sin ωc t Coherent carrier recovery Mathematical Demodulation of BPSK At positive input, sinωc t: At negative input, - sinωc t: Output = (sin ωc t) (sin ωc t) Output = (-sin ωc t) (sin ωc t) = -sin2 ωc t = sin2ωc t -sin2ωc t = - ½ (1 – cos2ωc t) sin2ωc t = ½ (1 – cos2ωc t) = - ½ + ½ cos2ωc t = ½ - ½ cos2ωc t filtered out filtered out Therefore; Therefore; Output = - ½ V = logic 0 Output = + ½ V = logic 1 Quaternary Phase Shift Keying v Also called quadrature PSK v Another form of angle-modulated, constant-amplitude digital modulation. v It is an M-ary encoding technique where M=4 (hence the name “quaternary” meaning 4). v There four possible output phases for a single carrier frequency. v The binary input data are combined into groups of 2 bits called dibits. v The rate of change at the input (baud rate) is one-half of the input bit rate M = 4; N = 2  Problem: QPSK For the QPSK Transmitter, construct the truth table , phasor diagram, and constellation diagram.  QPSK Modulator The QPSK modulator is two BPSK modulators connected in parallel. I bit – modulates the carrier in phase with the reference oscillator Q bit – modulates the carrier that in quadrature with the carrier The bit rate per channel is half the input bit rate. QPSK Output QPSK Output Waveform The four possible outputs have exactly the same magnitude. The angular separation between the output phasor is 90°. Binary QPSK Input Output Q I phase 0 0 -135 0 1 -45 1 0 135 1 1 45 Bandwidth of QPSK I Channel +/- sinωc t Balanced Fb/2 +/- 1 modulator Binary sinωc t input data fb Q I cosωc t Fb/2 +/- 1 +/- cosωc t Balanced Q Channel modulator Highest fundamental freq fa = 14 fb Input I Q I Q I Q I Q I Q I data fb 1 1 0 1 1 0 1 0 1 1 0 Bandwidth fb fN = 2 fa = I channel data fb/2 Highest 2 fundamental Baud rate frequency fb Q channel data fb/2 baud rate = fN = 2  Problem: QPSK n For the QPSK Transmitter, construct the truth table , phasor diagram, and constellation diagram.  Problem2: QPSK n Using the QPSK Transmitter block diagram as the modulator model, find the output of transmitter for a binary data input of Q = O and I= 0  Problem3: QPSK n For a QPSK modulator with an input data rate (fb) equal to 10 Mbps and a carrier frequency of 70 MHz, determine the minimum double-sided Nyquist bandwidth (fN) and the baud.  OFFSET QPSK n Modified form of QPSK n The bit waveform on I and Q channels are offset or shifted in phase each other by one-half of a bit time n Advantage: limited phase shift is needed during modulation n Disadvantage: higher bandwidth requirement n The bandwidth and baud rate are twice the conventional QPSK. QPSK vs OQPSK Applications of QPSK § Satellite Communications § WiMax – Wireless MAN § HSDPA – high-speed downlink packet access