Wireless Communication Systems @CS.NCTU Lecture 2: Modulation and Demodulation Reference: Chap. 5 in Goldsmith’s book Instructor: Kate Ching-Ju Lin ( 林靖茹 ) 1
Modulation From Wikipedia: The process of varying one or more properties of a periodic waveform with a modulating signal that typically contains information to be transmitted. 1 0.5 0 -0.5 -1 modulate 1 1 1 0.5 0.5 0.5 0 0 0 -0.5 -0.5 -0.5 -1 -1 -1 2
Example 1 = bit-stream? (a) 10110011 (b) 00101010 (c) 10010101 3
Example 2 = bit-stream? (b) 00101011 (a) 01001011 (c) 11110100 4
Example 3 = bit-stream? (a) 11010100 (b) 00101011 (c) 01010011 (d) 11010100 or 00101011 5
Types of Modulation Amplitude ASK Frequency FSK Phase PSK
Modulation • Map bits to signals TX 0 1 0 bit stream 1 1 modulation transmitted Signal s(t) wireless channel
Demodulation • Map signals to bits TX RX 0 1 0 bit stream 1 1 1 0 1 1 0 demodulation modulation received transmitted signal x(t) Signal s(t) wireless channel
Analog and Digital Modulation • Analog modulation ⎻ Modulation is applied continuously ⎻ Amplitude modulation (AM) ⎻ Frequency modulation (FM) • Digital modulation ⎻ An analog carrier signal is modulated by a discrete signal ⎻ Amplitude-Shift Keying (ASK) ⎻ Frequency-Shift Keying (FSK) ⎻ Phase-Shift Keying (PSK) ⎻ Quadrature Amplitude Modulation (QAM) 9
Advantages of Digital Modulation • Higher data rate (given a fixed bandwidth) • More robust to channel impairment ⎻ Advanced coding/decoding can be applied to make signals less susceptible to noise and fading ⎻ Spread spectrum techniques can be applied to deal with multipath and resist interference • Suitable to multiple access ⎻ Become possible to detect multiple users simultaneously • Better security and privacy ⎻ Easier to encrypt 10
Modulation and Demodulation AWGN Channel n(t) m ={b ,...,b } s(t) ^ ^ x(t) ^ i 1 m ={b ,...,b } K + 1 K Transmitter Receiver modulate demodulate • Modulation ⎻ Encode a bit stream of finite length to one of several possible signals • Delivery over the air ⎻ Signals experience fading and are combined with AWGN (additive white Gaussian noise) • Demodulation ⎻ Decode the received signal by mapping it to the closest one in the set of possible transmitted signals 11
Band-pass Signal Representation • General form s ( t ) = a ( t ) cos (2 π f c t + φ ( t )) frequency amplitude phase • Amplitude is always non-negative ⎻ Or we can switch the phase by 180 degrees • Called the canonical representation of a band-pass signal 𝑡 𝑢 𝑏 𝑢 2𝜌𝑔 & 𝑢 + 𝜚 𝑢 12
In-phase and Quadrature Components s ( t ) = a ( t ) cos(2 π f c t + φ ( t )) = a ( t )[cos(2 π f c t ) cos( φ ( t )) − sin(2 π f c t ) sin( φ ( t ))] = s I ( t ) cos(2 π f c t ) − s Q ( t ) sin(2 π f c t ) : In-phase component of s ( t ) • s I ( t ) = a ( t ) cos( φ ( t )) : Quadrature component of s ( t ) • s Q ( t ) = a ( t ) sin( φ ( t )) q Amplitude: a ( t ) = s 2 I ( t ) + s 2 Q ( t ) φ ( t ) = tan − 1 ( s Q ( t ) Phase: s I ( t ) ) 13
Band-Pass Signal Representation s ( t ) = s I ( t ) cos(2 π f ( t ) t ) − s Q ( t ) sin(2 π f ( t ) t ) • We can also represent s(t) as Q s ( t ) = < [ s 0 ( t ) e 2 j π f c t ] 𝑡′ 𝑢 exp ( i θ ) = cos ( θ ) +jsin ( θ ) 𝑏 𝑢 • 𝜚 𝑢 s 0 ( t ) = s I ( t ) + js Q ( t ) I • s ’( t ) is called the complex envelope of the band-pass signal • This is to remove the annoying e 2 j π f c t in the analysis
Types of Modulation s ( t ) = A cos ( 2 π f c t+ 𝜚 ) • Amplitude ⎻ M-ASK: Amplitude Shift Keying • Frequency ⎻ M-FSK: Frequency Shift Keying • Phase ⎻ M-PSK: Phase Shift Keying • Amplitude + Phase ⎻ M-QAM: Quadrature Amplitude Modulation
Amplitude Shift Keying (ASK) • A bit stream is encoded in the amplitude of the transmitted signal • Simplest form: On-Off Keying (OOK) ⎻ ‘1’ à A=1, ‘0’ à A=0 RX TX bit stream 1 0 1 1 0 0 1 0 1 1 b(t) demodulation modulation signal s(t) 16
M-ASK • M-ary amplitude-shift keying (M-ASK) � , if 0 ≤ t ≤ T A i cos(2 π f c t ) s ( t ) = , otherwise , 0 where i = 1 , 2 , · · · , M A i is the amplitude corresponding to bit pattern i 17
Example: 4-ASK • Map ‘00’, ‘01’, ‘10’, ’11’ to four different amplitudes Binary 0 0 0 1 1 0 1 1 sequence 1 Time 0 (a) m ( t ) 3 Time 4-ary 1 signal 0 -1 -3 (b) s ( t ) 3 A A 4-ASK Time 0 signal -A T T -3 A (c) 18
Pros and Cons of ASK • Pros Bandwidth is the difference between the upper and lower frequencies in a ⎻ Easy to implement continuous set of frequencies. ⎻ Energy efficient ⎻ Low bandwidth requirement • Cons ⎻ Low data rate § bit-rate = baud rate 1 baud 1 second ⎻ High error probability § Hard to pick a right threshold
Types of Modulation s ( t ) = A cos ( 2 π f c t+ 𝜚 ) • Amplitude ⎻ M-ASK: Amplitude Shift Keying • Frequency ⎻ M-FSK: Frequency Shift Keying • Phase ⎻ M-PSK: Phase Shift Keying • Amplitude + Phase ⎻ M-QAM: Quadrature Amplitude Modulation
Frequency Shift Keying (FSK) • A bit stream is encoded in the frequency of the transmitted signal • Simplest form: Binary FSK (BFSK) ⎻ ‘1’ à f=f 1 , ‘0’ à f=f 2 RX TX bit stream 1 0 1 1 0 1 0 1 1 0 demodulation modulation signal s(t) 21
M-FSK • M-ary frequency-shift keying (M-FSK) � , if 0 ≤ t ≤ T A cos(2 π f c,i t ) s ( t ) = , otherwise , 0 where i = 1 , 2 , · · · , M f c,i is the center frequency corresponding to bit pattern i • Example: Quaternary Frequency Shift Keying (QFSK) ⎻ Map ‘00’, ‘01’, ‘10’, ’11’ to four different frequencies 22
Pros and Cons of FSK • Pros ⎻ Easy to implement ⎻ Better noise immunity than ASK • Cons ⎻ Low data rate § Bit-rate = baud rate ⎻ Require higher bandwidth § BW(min) = N b + N b
Types of Modulation s ( t ) = A cos ( 2 π f c t+ 𝜚 ) • Amplitude ⎻ M-ASK: Amplitude Shift Keying • Frequency ⎻ M-FSK: Frequency Shift Keying • Phase ⎻ M-PSK: Phase Shift Keying • Amplitude + Phase ⎻ M-QAM: Quadrature Amplitude Modulation
Phase Shift Keying (PSK) • A bit stream is encoded in the phase of the transmitted signal • Simplest form: Binary PSK (BPSK) ⎻ ‘1’ à 𝜚 =0, ‘0’ à 𝜚 = π RX TX bit stream 1 0 1 1 0 1 0 1 1 0 s(t) demodulation modulation signal s(t) 25
Constellation Points for BPSK • ‘1’ à 𝜚 =0 • ‘0’ à 𝜚 = π • cos ( 2 π f c t+0 ) • cos ( 2 π f c t+ π ) = cos (0) cos( 2 π f c t )- = cos ( π ) cos( 2 π f c t )- sin (0) sin ( 2 π f c t ) sin ( π ) sin ( 2 π f c t ) = s I cos ( 2 π f c t ) – s Q sin ( 2 π f c t ) = s I cos ( 2 π f c t ) – s Q sin ( 2 π f c t ) 𝜚 =0 Q Q 𝜚 = π I I ( s I , s Q ) = ( 1 , 0 ) ( s I , s Q ) = (- 1 , 0 ) ‘1’ à 1+0i ‘0’ à - 1+0i
Demodulate BPSK • Map to the closest constellation point • Quantitative measure of the distance between the received signal s’ and any possible signal s ⎻ Find |s’-s| in the I-Q plane Q ‘0’ ‘1’ n 1 =|s’-s 1 |=|s’-(1+0i)| s’=a+bi n 0 n 1 n 0 =|s’-s 0 |=| |s’-(-1+0i)| I s 1 =1+0i s 0 =-1+0i since n 1 < n 0, map s’ to (1+0i) à ‘1’
Demodulate BPSK • Decoding error ⎻ When the received signal is mapped to an incorrect symbol (constellation point) due to a large error • Symbol error rate ⎻ P(mapping to a symbol s j , j ≠ i | s i is sent ) Q ‘0’ ‘1’ Given the transmitted symbol s 1 s’=a+bi I à incorrectly map s’ to s 0 =-1+0i s 1 =1+0i s 0 =(-1+0) à ‘0’, when the error is too large
SNR of BPSK • SNR: Signal-to-Noise Ratio Q SNR = | s | 2 | s | 2 s’ = a+bi n 2 = n | s � − s | 2 I | 1 + 0 i | 2 = | ( a + bi ) − (1 + 0 i ) | 2 SNR dB = 10 log 10 ( SNR ) • Example: ⎻ Say Tx sends (1+0i) and Rx receives (1.1 – 0.01i) ⎻ SNR?
SER/BER of BPSK • BER (Bit Error Rate) = SER (Symbol Error Rate) Minimum distance of any SER = BER = P b two cancellation points ✓ d min r ! ◆ 2 E b √ = Q = Q = Q ( 2 SNR ) √ 2 N 0 N 0 From Wikipedia: Q(x) is the probability that a normal (Gaussian) random variable will obtain a value larger than x standard deviations above the mean. 30
Constellation point for BPSK • Say we send the signal with phase delay π cos(2 j π f c t + π ) = cos(2 j π f c t ) cos( π ) − sin(2 j π f c t ) sin( π ) = − 1 ∗ cos(2 j π f c t ) − 0 ∗ sin(2 j π f c t ) Band-pass representation =( − 1 + 0 i ) e 2 j π f c t Q 𝜚 = π Illustrate this by the constellation point (-1 + 0i) in an I-Q plane I -1+0i 31
Quadrature PSK (QPSK) • Use four phase rotations 1 /4 π , 3 /4 π , 5 /4 π , 7 /4 π to represent ‘00’, ‘01’, ‘11’, 10’ A cos(2 j π f c t + π / 4) = A cos(2 j π f c t ) cos( π / 4) − A sin(2 j π f c t ) sin( π / 4) Q =1 ∗ cos(2 j π f c t ) − 1 ∗ sin(2 j π f c t ) ‘01’ ‘00’ =(1 + 1 i ) e 2 j π f c t I A cos(2 j π f c t + 3 π / 4) ‘10’ ‘11’ = A cos(2 j π f c t ) cos(3 π / 4) − A sin(2 j π f c t ) sin(3 π / 4) = − 1 ∗ cos(2 j π f c t ) − 1 ∗ sin(2 j π f c t ) =( − 1 + 1 i ) e 2 j π f c t 32
Recommend
More recommend
