8.1 Geometric Representation of Signal Waveforms

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1 Haberlesme Sistemlerine Giris (ELE 361) 30 Ekim 2017 TOBB Ekonomi ve Teknoloji Universitesi, GÃ 1 4 z Dr. A. Melda Yuksel Turgut & Tolga Girici Lecture Notes Chapter 8 Digital Modulation Methods in an Additive White Gaussian Noise Channel AWGN: Simplest model for channel impairments Advantages of Digital Transmission: Digital Transmission: Binary data is mapped to analog waveforms A nite number of analog waveforms are used Reason of using analog waveforms: Will rst consider baseband transmission In reality communications occur in a pass band, away from f 0 Binary data is mapped to the Phase of the carrier signal Frequency of the carrier signal Both phase and amplitude of the carrier In this chapter we concentrate on the transmission of a single symbol. 8.1 Geometric Representation of Signal Waveforms Suppose we want to send block of 6 bits. There are bit combinations In digital transmission we assign each bit sequence to a dierent analog signal.

2 So dierent analog signals are needed. What should the receiver do? Compare the received signal to these 64 signals? There is a simpler way Most of the time the 64 signals can be written as a combination of 1 or 2 base signals. Then the receiver calculates the projection of the received signal on these 2 signals. Much simpler and systematic. Now we will learn to determine minimal a set of base signals corresponding to a set of signals. In order to send k bits, we need M 2 k dierent analog signals.s m (t), 1 m M We can express these M signals as linear combinations of N orthonormal signals ψ n (t) What is orthonormal? orthogonal: and normal There are innite possible choices of orthonormal basis sets Systematic way: Gram-Schmidt Orthonormalization Gram-Schmidt Orthonormalization Procedure Begin with s 1 (t) and normalize it to nd ψ 1 (t) Take s 2 (t) and compute its projection on ψ 1 (t) compute s 2 (t) c 21 ψ 1 (t) to yield... Normalize d 2 (t) to nd ψ 2 (t) 2

3 Take s k (t) and compute its projection on ψ 1 (t),..., ψ k 1 (t) compute s k (t) k 1 i1 c kiψ i (t) to yield Normalize d k (t) to nd ψ k (t) If at any step d k (t) then no new ψ(t) Example 8.1.1: Apply Gram-Schmit procedure to these signals. Solution: Expressing s m (t) as a vector in the ψ space. Find the projections of s m (t) on each ψ n (t): So s m (t) Energy of s m (t): E m Vector: Example 8.1.2: Determine the vector representations of s m (t) in the previous example. Solution: 3

4 A set of signals and their orthogonalization Most of the time a convenient orthonormal basis set can be found without using Gram- Schmidt method. 4

5 A set of signals and their orthogonalization 5

6 8.4 M-Ary Digital Modulation Binary modulation: Transmit one bit at a time. Two signal waveforms are required Binary antipodal and binary orthogonal are two main binary modulation schemes. Others are special cases of these two. 1 bit/symbol Some examples M-ary modulation: Use M dierent waveforms. Transmit k log 2 M bits at a time. R s 1 T R b kr s T b Optimal Receiver for M-ary Signals in AWGN r(t) s m (t) + n(t), 0 t T, m 1, 2,..., M n(t) is Signal Demodulator: Receiver is divided into two such as 6

7 M-ary bit and symbol intervals 1. Demodulator 2. Detector M-ary signal waveform can be represented as s m (t) N s mk Ψ(t), 0 t T, m 1,..., M k1 Signal can be represented as a vector Demodulator: CORRELATOR. 7

8 Correlation type demodulator T 0 r(t)ψ k (t) y k y s m + n r(t) n (t) 8

9 n (t) is orthogonal to every dimension of the signal waveform (s mk ), so it is irrelevant. E[n k ] E[n k n m ] So, N noise components are uncorrelated and independent. Distribution is Correlator output: y k s mk + n k, k 1,..., N Distribution of y k is: y k is also independent of n (t) Another type of demodulator: MATCHED FILTER r(t) s m (t) + n(t), 0 t T, m 1,..., M h k (t) Ψ k (T t), 0 t T, k 1,.., N y k (t) t 0 r(τ)h k (t τ)dτ sample at tt same as correlation type 9

10 Matched Filter: Impulse Response matched to the signal waveform Signal s(t): matched lter h(t) s(t t) Matched Filter example Shetch: s(t) s(t t) t 0 s(τ)s(t t + τ) Maximizes the output SNR. Output SNR depends on the energy of the matched lter but not on the signal waveform itself. In the frequency domain: 10

11 h(t) s(t t) s(t) S(f) s( t) S (f) s(t t) Y (f) S(f)H(f) y(t) y(t ) output signal power What about the noise component? output noise power SNR Example 8.4.1: Consider 4-PAM. Determine the PDF of the received signal at the output of the correlator and sketch its PDF. Solution: 11

12 Matched Filter type demodulator Example 8.4.1: Consider M-4 orthogonal signaling. Determine the PDF of the received 12

13 signal at the output of the correlator and sketch its PDF. Examples of orthogonal signaling: 1. Pulse position modulation 2. Frequency Shift Keying Solution: Optimum Detector The problem is:: arg max m1,...,m P (signal m was transmitted r) "Find the most likely transmitted waveform, given the demodulator output" This is Maximum A Posteriori Probability (MAP) detector 13

14 Signal constellation, noise cloud and received vector (M4, N3) so the problem is P (s m y) f(y s m)p (s m ) f(y) M wheref(y) f(y s m )P (s m ) m1 max m f(y s m )P (s m ) f(y) simplification For equiprobable signals, f(y s m ) MAP rule becomes Maximum Likelihood detection. N 1 e (y k s mk ) 2 /N o πno k1 take logarithm: ln f(y s m ) max P (s m y) min m m D(y, s m ) 14

15 D(y, s m ) is the Example 8.4.3: Consider binary PAM. Signal points are s 1 s 2 ɛ b. Prior probabilities are unequal ( i.e. P (s 1 ) P (s 2 )) Determine the metric. Solution: MAP detection minimizes the probability of error a Probability of error for binary PAM Output of demodulator is Assume equiprobable signaling Assume s 1 (t) was transmitted. Then demodulator output is 15

16 P (error) Due to symmetry Standardize the Gaussian noise Write in terms of the distance A Union Bound on the Probability of Error For binary equiprobable signaling (which is ) 8.5 M-ary Pulse Amplitude Modulation M-ary PAM is used to transmit bits per symbol. Single dimension M-ary PAM s m (t) s m Ψ(t), 0 t T, m 1, 2,..., M s m (2m 1 M)d Averagesymbolenergyɛ s Averagebitenergyɛ b 16

17 Example 8.5.1: Sketch the transmitted waveform for 4-PAM and the bit sequence Solution: Carrier Modulated PAM for bandpass channels (M-ary ASK) A regular baseband PAM signal is multiplied with a cosine to make it bandpass This is like a The occupied bandwidth is u m (t) s m (t) cos(2πf c t), m 1,..., M u m (t) s m 2Ψ(t) cos(2πfc t), m 1,..., M s m (2m 1 M)d su m (f) ɛ m u 2 m(t)dt 17

18 8.5.2 Demodulation and Detection of Bandpass PAM The received signal is correlated with The result is The detector decides according to Probability of Error for M-PAM For baseband PAM, if s m (t) is transmitted, the demodulator output is: Error event is related to the instantaneous noise being greater than d. Sketch: 18

19 y s m + n P M M 2 M P ( y s m > d) + 2 M P (y s m > d) 2(M 1) M Q( 2d 2 ) N o Distance parameter d can be related to the average symbol energy ɛ av as, P M In terms of the average bit energy ɛ bav P M Example 8.5.3: Using the gure below, determine the approximate SNR/bit required to achieve a SER of P M 10 6 for M2,4,8. Phase Shift Keying Each analog waveform has equal energy. Equally-spaced on a circle around the origin of the constellation diagram. 19

20 Bit error rate for PAM: The horizontal axis is ɛ bav N o (in db). Every extra 1 bit/symbol requires extra 4dB SNR. For better spectral eciency we need more transit power. Special case: For rectangular pulse Special case : For rectangular pulse 2ɛs u m (t) T cos(2πf ct + 2πm ), 0 t T m 1,..., M M ɛ av ɛ bav log 2 M Generalcase : Pulseg T (t) u m (t) g T (t) cos(2πf c t + 2πm ), 0 t T m 1,..., M M expanding the cosineu m (t) Sketch the modulator: 20

21 Example of a 4-PSK Geometric representation of M-PSK Signals For general pulse g T (t) Sketch the constellation for M2,4,8 21

22 s m Ψ 1 (t) Ψ 2 (t) Minimum Euclidian distanced min d min plays an important role in determining the error probability Example 8.6.1: For M8 PSK determine how many db the transmitted energy must be increased to achieve the same d min as M4. Solution: 22

23 23

24 8.6.2 Demodulation and Detection of PSK Signals r(t) u m (t) + n(t)m 0, 1..., M 1 n(t) n c (t) cos(2πf c t) + n s (t) sin(2πf c t) Correlated withψ 1 (t) n c (t) cos(2πf c t) andψ 2 (t) n c (t) cos(2πf c t) Output vector y s m + n by definitionn c n s E[n 2 c] E[n 2 s] Optimal detector: Compute the phase of the detector output ( ) and nd the signal s m whose phase is closest to Probability of Error for Phase Coherent PSK Let s 0 ( ɛ s, 0) be transmitted Demodulator output is (y 1, y 2 ) ( ɛ s + n c, n s ). Then the phase of y is checked. Error probability is about the phase being θ > π/m or θ < π/m. The pdf of Θ is complicated, and exact error probability calculation requires numerical integration. 2ɛ For M2: equivalent to 2PAM (antipodal signaling) p 2 Q( b N o ) For M4: We have two 2PSK signals (one in the horizontal and the other in the vertical axis of the constellation) 24

25 P 4 1 (1 P 2 ) 2 [ 2ɛb 1 1 Q( ) N o ] 2 for high SNR For general M: An approximate approach: Sketch the constellation and calculate the minimum distance: For high SNR: Each point in the constellation can be mixed with only the nearest neighbors: 25

26 8.7 Quadrature Amplitude Modulated Digital Signals For M-PSK analog waveforms have the same energy In the constellation diagram In M-QAM we don't have this constraint. Like PSK u m (t) Amcg T (t) cos(2πf c t) A ms g T (t) sin(2πf c t), m 1,..., M u mn (t) A m g T (t) cos(2πf c t + θ n ), m 1,..., M 1, n 1,..., M 2 M 1 2 k 1 M 2 2 k 2 M M 1 + M 2 26

27 Bit error rate for PSK: The horizontal axis is ɛ bav N o (in db). Every extra 1 bit/symbol requires extra 4dB SNR. For better spectral eciency we need more transit power Geometric Representation of QAM Signals Ψ 1 (t) Ψ 2 (t) s m ɛ s ɛ av 27

28 16QAM Constellation. It's like modulating the two quadrature carriers by 4PAM Functional block diagram of QAM Example 8.7.1: Determine the average energy of the below QAM contellations Solution: 28

29 Example QAM constellations 29

30 Example Demodulation and Detection of QAM Signals r(t) Amcg T (t) cos(2πf c t) A ms g T (t) sin(2πf c t) + n(t) Cross correlated byψ 1 (t)andψ 2 (t) y s m + n D(y, s m ) Optimum detector selects the signal corresponding to the smallest value of D(y, s m ) 30

31 8.7.3 Probability of Error for QAM 1st 4QAM contellation ɛ av ɛ s 2nd 4QAM contellation ɛ av ɛ s 1st 8QAM contellation ɛ av ɛ s 2nd 8QAM contellation ɛ av ɛ s 3rd 8QAM contellation ɛ av ɛ s 4th 8QAM contellation ɛ av ɛ s Two 4-point QAM Constellations 31

32 A circular 16QAM constellation. Actually rectangular QAM constellations are preferred, as they are much easier to generate (two PAM signals). When k is evenp QAM M 1 (1 P P AM M ) When k is oddp QAM M Compare with MPSK : P P SK M Ratio of the argumentsr Comment: 32

33 QAM symbol error rate graphs. Determine the required increase in transmit energy for each extra bit/symbol End-of-chapter problems:

Lecture #11 Overview. Vector representation of signal waveforms. Two-dimensional signal waveforms. 1 ENGN3226: Digital Communications L#

Lecture #11 Overview. Vector representation of signal waveforms. Two-dimensional signal waveforms. 1 ENGN3226: Digital Communications L# Lecture #11 Overview Vector representation of signal waveforms Two-dimensional signal waveforms 1 ENGN3226: Digital Communications L#11 00101011 Geometric Representation of Signals We shall develop a geometric