Digital Modulation Principles - Lecture Slides | ECE 284, Study notes of Electrical and Electronics Engineering

Download Digital Modulation Principles - Lecture Slides | ECE 284 and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! 1 Digital Modulation Principles Curt Schurgers 2 Electrical waveform Electro-magnetic waveform Digital Wireless Comm. System Source coding Source coding Multiple access Modulation & baseband Wireless channel Channel coding RF Source decoding Source decoding Multiple access Demodulation & baseband Channel decoding RF 0 1 0 1 1 1 0 0 1 0 1 0 V, I HE rr , Multi- plexing Demulti- plexing Information Radio Dedicated hardware (on wireless card) Protocol processor (on wireless card) Application processor PTx PRx C ha nn el c od in g ca n be d on e in e ith er h ar dw ar e or so ftw ar e (in te gr at ed w ith th e di gi ta l ra di o ha rd w ar e or ru nn in g on th e pr ot oc ol p ro ce ss or ) 2 3 Analysis of Radio Modes: Tx Important remarks ● PS ≠ PT PA is not 100% efficient ● PTx ≠ PT Transmitter power is not the same as transmitted power Transmitted power PT = power sent out by antenna Transmitter power PTx = power consumed by transmitter (depends on PT) ● PRx ≠ PR Receiver power is not the same as received power Received power PR = power captured by antenna Receiver power PRx = power consumed by receiver (independent of PR) The wireless channel behavior (LP) has a major impact on system performance and energy/power efficiency ● Important part of transmitter power PETx PS PERx PT PR PTx PRx LPTransmit electronics Receive electronicsPA ● Transmitted power depends on power consumed by PA ● Relates transmitted power to received power ● Received power impacts wireless communication performance 4 Sending Data Modulation: mapping bits into symbols ● Symbols are essentially a set of possible waveforms ● The cardinality of this set determines how many bits of information are carried Baseband processing: some filtering, etc. 0100 1110 b = 1 bit/symbol b = 2 bits/symbol 10 b bits/symbol ⇔ M = 2b possible waveforms 1 1 0 1 1 0 1 1 Modulation & baseband RF bits symbols upconverted symbolsupconversion 5 9 Information Mapping Examples (2-D) s2 s1 s2 s1 s2 s1 s2 s1 Send either s1 or s2. Send s1, s2, both or none of them. Send ±s1 or ±s2. Send any of these combinations. 00 01 11 1 Y 0 X y1x1 y1x2 …… y3 Y x3 X 01 10 YX 01 1 0 -10 0 Y -1 X M = 4 M = 2 M = 4 M = 8 This is called the ‘signal space representation’ or ‘constellation diagram’ The signals s1 , s2 , s3 ,… are basis functions (could have more than 2!). They could be any signal as long as they satisfy the orthonormality condition (different modulation schemes could have different basis functions, but the same signal space representation, or vice versa). 10 Some Basic Modulation Schemes FSK (Frequency Shift Keying) PAM (Pulse Amplitude Modulation) Remember, the baseband signals can also be upconverted to a certain carrier frequency s2 s1 Baseband PAM M = 4 M = {2, 4, 8, 16, …} M-ary PAM s2 s1 M = 2 M = {2, 4, 8, 16, …} M-ary FSK (Note: the constellation diagram is M-dimensional, therefore hard to represent for M > 2) Note: On-Off Keying OOK, is binary FSK, where one frequency is zero, i.e. no signal) Passband PAM Easiest is here to consider the constellation diagram as relating directly to the RF frequencies s3 s2 s1 M = 3 6 11 Some Basic Modulation Schemes PSK (Phase Shift Keying) QAM (Quadrature Amplitude Modulation) The basis functions are a cos and sin waves at the carrier frequency. They are called the in-phase (I) and quadrature (Q) components. s2 s1 M = 16 M = {4, 8, 16, 32, …} M-QAM M = {4, 8, 16, 32, …} M-ary PSK )2cos(2 tf T c ⋅⋅⋅ π )2sin(2 tf T c ⋅⋅⋅ π I Q The basis functions are a cos and sin waves at the carrier frequency. They are called the in-phase (I) and quadrature (Q) components. s2 s1 M = 16 )2cos(2 tf T c ⋅⋅⋅ π )2sin(2 tf T c ⋅⋅⋅ π I Q 12 Some Basic Modulation Schemes s2 s1 PSK QAM s2 s1 )2cos(2 tf T c ⋅⋅⋅ π )2sin(2 tf T c ⋅⋅⋅ π I Q [ ])2sin()2cos(2 tfbtfa T cici ⋅⋅⋅+⋅⋅⋅⋅= ππ [ ])2sin()2cos(2)()()( 21 tfbtfaTtsbtsats ciciiii ⋅⋅⋅+⋅⋅⋅⋅=⋅+⋅= ππ ( ) ( )[ ])2sin()2cos(Re2 tfjtfbja T ccii ⋅⋅⋅−⋅⋅⋅⋅+⋅= ππ (ai, bi) Ai θi ( ) ( )[ ]tfjji ci eeAT ⋅⋅⋅−⋅ ⋅⋅⋅= πθ 2Re2 ( )[ ]ic tfji eAT θπ −⋅⋅⋅−⋅⋅= 2Re2 ( )ici tfAT θπ −⋅⋅⋅⋅= 2cos 2 )()()( 21 tsbtsats iii ⋅+⋅= φφφ sincos ⋅+= je j In PSK, only the phase is different for different constellation points 7 13 Some Basic Modulation Schemes 4-QAM 16-QAM 64-QAM 4-PSK 8-PSK 16-PSK QAM (Quadrature Amplitude Modulation) PSK (Phase Shift Keying) Q Q Q Q Q Q I I I I I I Configurations with M = {8, 32, ..} exist, but they are not as regular.. A θ Each constellation point can be viewed as a complex number (with amplitude A and phase θ). In PSK, only the phase is varies, hence the name. 14 Symbol Energy Signal i is received as The instantaneous power in that signal is The energy in that symbol over one symbol duration is (s1(t) and s2(t) are normalized and orthogonal to each other) The energy averaged over all possible symbols (assuming each one is equally likely) is the energy per symbol ES (at the receiver) Q I ai + j·bi ri θi )()()( 21 tsbtsats iii ⋅+⋅= )()(2)()()()( 21 2 2 22 1 22 tstsbatsbtsatstP iiiiii ⋅⋅⋅⋅+⋅+⋅== 222 0 2 0 )()( iii T i T ii rbadttsdttPE =+=== ∫∫ Q I2 1 21 RMS M i iS rrM E == ∑ = s1(t) s2(t))2cos( 2 tf T c ⋅⋅⋅ π )2sin(2 tf T c ⋅⋅⋅ π I Q 10 19 Bit Error Rate 000 001 011 010 110 111 101 100 0000 01001100 1000 0010 01101110 1010 0011 01111111 1011 0001 01011101 1001 )(log2 M SERBER = ● Each symbol error results in a number of bit errors. By carefully choosing the mapping from bits to symbols (e.g. Gray encoding), one symbol error typically results in just one bit error. SER SNR SER 0N Eb BER 0N Eb S b RM WSNR N E ⋅ ⋅= )(log20 )(log2 M SERBER = ● Performance is given as Symbol Error Rate (SER) or Bit error Rate (BER) versus SNR or Eb/N0 Typical waterfall curve 20 Performance Example: QAM AWGN channel 11 21 Comparisons Error rate: error performance depends on constellation size 4-QAM 16-QAM 64-QAM 4-QAM 16-QAM 64-QAM (1) Choose ES the same, i.e. same rRMS2 Look at performance for the same SNR => N the same => SER increases (2) Try to achieve the same SER with the same amount of noise N This means that constellation can have the same spacing => rRMS2 increases => ES increases => required SNR increases Q I ri ∑ = == M i iRMSS rM rE 1 22 1 N RE N PSNR SSR ⋅== Nxx ~ 22 σσ = We can use the same technique to compare different modulation schemes, e.g. QAM versus PSK 22 Data Rate Data rate or bit rate: ● Higher constellation size M results in a higher bit rate Rb (bits/s) for the same RS Define bandwidth efficiency as (bits/s/Hz) ● This is a measure for how much information we can pack within a certain bandwidth )(log1 2 MRT R S b b ⋅== T RS 1 = W Rb b =η 1 1 0 1 1 0 1 1 Modulation & baseband RF bits symbols upconverted symbolsupconversion TS Tb TS )( fS W f )( fS W f cf 12 23 ● In general, W is a function of RS Could be a complex relationship, e.g. FSK ● Some common schemes have a convenient relationship: QAM PSK SSB (single side band) PAM (i.e. with appropriate filtering such that redundant information is eliminated) The above schemes all have the same bandwidth efficiency Also in these schemes Data Rate )(log)(log 22 MW MR W R Sb b = ⋅ ==η )(log2 000 M N E N E WN RE N PSNR bSSSR ⋅== ⋅ ⋅ == SRT W =≈ 1 )( fS W f )( fS W f cf upconversion cf )( fS Wnew = W/2 f filtering Only possible for PAM (not QAM or PSK) SSB-PAM 24 Comparison SER = 10-5 Source: http://www.mhhe.com/e ngcs/electrical/proakis/s tudent/images.mhtml FSK W Rb b =η 0N Eb 15 29 Strictly speaking, the performance is determined by the SNR, ratio of power in the signal versus that in the noise, at the decision device after the LNA (i.e. the slicer) However, this SNR is the same as the one before the LNA Therefore, to evaluate the performance, we can use the power PR received by the antenna (and use the noise figure NF to take inefficiencies of the LNA and other elements into account) Effect of Tx-Rx Chain Q I Channel loss LP TP P T R L PP = PA Q ILNA+ Noise N PAG LNAG N P NG PGSNR R LNA RLNA = ⋅ ⋅ = )(SNRfBER = WN P N PSNR RR ⋅ == 0 P T R L PP = 30 For modulation schemes where the decision regions depend on the amplitude ● E.g. QAM ● The PA (with gain GPA) and LNA (with gain GLNA) together have to offset the channel loss LP ● This implies that the channel loss needs to be estimated ● Typically, the GLNA is adjusted using automatic gain control (AGC) for fine grain adjustments Other Issues with Different Schemes Q I Q I Q I Channel loss LP TP P T R L PP = PA LNA+ Noise N Q I PAG LNAG Other modulation schemes do not require amplitude information in their decision regions, only phase ● E.g. PSK 16 31 Coherent versus non-coherent detection ● Coherent: require phase information e.g. QAM, PSK ● Non-coherent: does not require phase information e.g. FSK (but could also be coherent), DPSK Other Issues with Different Schemes Q I Q I Constant versus non constant envelope modulation ● Constant envelope: can use more efficient amplifiers e.g. PSK, FSK ● Non-constant envelope: e.g. QAM, PAM t FSK t PSK t QAM Summary ● QAM: amplitude-sensitive, coherent, non-constant envelope ● PSK: non-amplitude-sensitive, coherent, constant envelope ● FSK: non-amplitude-sensitive, non-coherent, constant envelope 32 Beyond Noise: Interference Other concurrent (and uncorrelated) transmissions also degrade the performance Modeled as the signal to interference + noise ratio (SINR) ● I is the interference power ● The interference of multiple concurrent transmissions is the sum of their powers ● In all prior equations, simply replace the SNR by the SINR IN PSINR R + = PR N I SINR BER or SER Q I Channel loss LP TP P T R L PP = PA Q ILNA+ Noise N PAG LNAG Interference I 17 33 Effects of the Channel The wireless channel varies in time and space due to fading effects Impact on two levels ● For a fixed PT the PR will vary and therefore the SNR ● Amplitude-sensitive modulation (e.g. QAM) cannot be decoded if changes are not tracked To track the changes in the channel, we need channel estimation Q I Channel loss LP TP P T R L PP = PA Q ILNA+ N + I PAG LNAG 321 PPPp LLLL ⋅⋅= 34 Effects of the Channel Two ways of compensating for channel variations 1. The LNA tracks the changes through AGC Does not improve the SNR variations Is necessary for amplitude-sensitive modulation Relatively fast adjustment 2. The PA tracks the changes SNR variations are compensated for Amplitude-sensitive modulation is possible Problem: requires feedback from Rx to Tx, which is only possible if variations are slow (typically constant for multiple packets) Often GPA is already at the maximum value (hardware and FCC constraints) These two ways only have a limited range of compensation and cannot compensate deep fades Q I Channel loss LP TP P T R L PP = PA Q ILNA+ N + I PAG LNAG 20 39 Other Simplified Channel Model Channel Model useful for simulation of upper protocol layers (i.e. with a fixed PHY layer) in the presence of fading As an approximation, divide the SNR range in non-overlapping intervals ● Each interval represents a state in a Markov chain ● Transitions are only possible between neighboring states ● Assume a more or less constant SNR in each interval: model with a fixed BER in each interval ● Last level often represents a deep fade, which cannot be compensated for ⇒ BER = 0.5[Wan95] 40 Two-State Markov Chain The fading margin ρ defines the SNR transition level between good and bad. The average number of level crossing is then: This depends on the mobile speed and carrier frequency through the Doppler rate The average time spent in each state is: ρ ρ ρπ −⋅⋅⋅= efN D2 GS GB PR N t 1⋅= ρ BS BG PR N t 1⋅= ρ ρ−−= ePB 1 PG PB tGG tGB tBG tBB GBGG tt −=1 BGBB tt −=1 λ vf c vf cD =⋅= SERG SERB The transition probabilities can be approximated as: [Source: Wan95] Example: two-state Markov Chain for a Rayleigh fading channel ● Gilbert-Elliott is a special case, where the Good state is error free and the Bad state is completely corrupted (BER = 0.5) ● In General, each state has a symbol error rate (SER): SERgood and SERbad ρ−=−= ePP BG 1 21 41 Example Consider a system with a binary modulation scheme and a data rate of 1 Mbps. It is operating in the 2.4 GHz band and has a fading margin of 15 dB. Construct the 2-state Markov model for a mobile moving at 25 km/h. )15(0316.0 dB−=ρ smv /94.6= Hzfc 9104.2 ⋅= HzfD 55.55= 98.23=ρN 969.0=GP 0311.0=BP 51048.2 −⋅=GBt 41074.7 −⋅=BGt ⇒ ⇒ baudRS 610= Average time in the bad state: symbols tBG 12921 = ms Rt SBG 292.111 =⋅ 42 References [Hay] Haykin, Moher, “Modern Wireless Communications,” Prentice Hall [Pro] Proakis, “Digital Communications,” McGraw-Hill [Jak] Jakes, “Microwave Mobile Communications,” IEEE Press [Gol] Goldsmith, “Wireless Communications,” Cambridge University Press [Pot] Pottie, Kaiser, “Principles of Embedded Networked Systems Design,” Cambridge University Press [Wan95] Wang, H., Moayeri, N., “Finite-State Markov Channel-A Useful Model for Radio Communication Channels,” IEEE Trans. on Vehicular Technology, vol.44, no.1, pp.163-171, Feb. 1995. [Zor95] Zorzi, M., Rao, R., Milstein, L., “On the accuracy of a first-order Markov model for data transmission on fading channels,” Proceedings of ICUPC '95, Tokyo, Japan, pp. 211-215, 1995. [Agi01] Agilent, “Digital Modulation in Communications Systems – An Introduction,” http://cp.literature.agilent.com/litweb/pdf/5965- 7160E.pdf#search=%22introduction%20digital%20modulation%22 No required reading. However, if you want more information on modulation, you can take a look at [Agi01] or any of the textbooks mentioned.