Digital Communications Chapter 9 Digital Communications Through Band-Limited Channels Po-Ning Chen, Professor Institute of Communications Engineering National Chiao-Tung University, Taiwan Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 1 / 51
9.2 Signal design for band-limited channels Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 2 / 51
Motivation For the baseband waveform s l (t) = I n g(t nt ) n= Channel is band-limited to [ W, W ]. Transmitted signals outside [ W, W ] will be truncated. How to design g(t) to yield optimal performance? 1 Here we use s l (t) instead of v(t) as being used in text to clearly indicate that s l (t) is the lowpass equivalent signal of s(t). Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 3 / 51
Recall that s l (t) = I n g(t nt ) n= is a cyclostationary random process with i.i.d. discrete random process {I n }. Autocorrelation function of s l (t) is is periodic with period T. R sl (t + τ, t) = E[s l (t + τ)s l (t)] The time-average autocorrelation function R sl (τ) = = T 1 T 2 R sl (t + τ, t) dt T 2 1 R I (n) T n= g(t + τ nt )g (t) dt Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 4 / 51
R sl (τ) = 1 T n R I (n) g(t + τ nt )g (t) dt The time-average power spectral density of s l (t) is S sl (f ) = = 1 T [ R sl (τ)e ı 2πf τ dτ n= R I (n)e ı 2πfnT ] G(f ) 2 Assuming I n is zero mean and i.i.d., R I (n) = σ 2 δ n, hence S sl (f ) = σ2 T G(f ) 2 Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 5 / 51
For static channel with impulse response c l (t), band-limited to [ W, W ], i.e. C l (f ) = 0 for f > W The received signal is r l (t) = c l (t) s l (t) + n l (t) = = n= c l (τ) ( I n n= I n g(t τ nt )) dτ + n l (t) g(t nt τ)c l (τ) dτ + n l (t) whose time-average power spectral density is S rl (f ) = σ2 T G(f ) 2 C l (f ) 2 + 2N 0 rect ( f 2W ) Note that n l (t) is a band-limited white noise process. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 6 / 51
r l (t) = n= I n h l (t nt ) + n l (t) with h l (t) = g(t) c l (t) Assume channel Impulse response c l (t) known to Rx. The match-filtered received signal is y l (t) = r l (t) h l (T t) = n I n x l (t nt ) + z l (t) where x l (t) = h l (t) hl (T t) and z l (t) = n l (t) hl (T t). For simplicity, one may use hl ( t) instead of h l (T t). Sampling at t = kt, k Z, we get y k = y l (kt ) = n= where x k n = x l (kt nt ). I n x l (kt nt )+z l (kt ) = I n x k n +z k n Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 7 / 51
Quick summary Channel Input data Transmitting filter g(t) s l (t) Channel filter c l (t) nl (t) r l (t) Receiving filter h l ( t) y l(t) Sampler y k Transmitted signal s l (t) = n= I n g(t nt ) Received signal r l (t) = n= I n h l (t nt ) + n l (t) with h l (t) = g(t) c l (t). Matched filter output y l (t) = n= I n x l (t nt ) + z l (t) with x l (t) = h l (t) hl ( t) or equivalently X l (f ) = H l (f ) 2 = G(f ) 2 C l (f ) 2. Sampling at t = kt and y k = n= I n x k n + z k. The transmission power is not g(t) c l (t) 2, which the receiver filter is to match! The additive interference to the transmitted information I k is not just z k but includes n=,n k Inx k n. So match filter is good for single transmission but may not be proper for consecutive transmissions! Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 8 / 51
Eye pattern A good tool to examine inter-symbol interference is the eye pattern. Eye pattern: The synchronized superposition of all possible realizations of the signal of interest viewed within a particular signaling interval. t/t b BPSK signals Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 9 / 51
Perfect eye pattern (at Tx) Eye pattern for r l (t) for g(t) being half-cycle sine wave with duration T b, c l (t) = δ(t) and error-free BPSK transmission. t/t b Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 10 / 51
Now with c l (t) = δ(t), we have, for example, g(t) = 1, 0 t < T T 0, otherwise (Time-limited; hence, band-unlimited!) h l (t) = g(t) c l (t) = g(t) x l (t) = h l (t) hl ( t) = h l(τ)hl ( (t τ))dτ T t = g(τ)g(τ t)dτ = T, t T 0, otherwise From here, you shall know the difference of using hl (T t) and hl ( t), where the former samples at t = T, while the latter samples at t = 0. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 11 / 51
Perfect eye pattern (at Rx) y l (t) = n= I n x l (t nt ) for all possible {I n {±1}} n=. (y k = y l (kt ) = n= I n x l (kt nt ) = I k ; so no ISI!) Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 12 / 51
Example: BPSK with 1/T = 1K and W = 3K Output y l (t) with c l (t) being ideal lowpass filter of bandwidth W. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 13 / 51
Example: BPSK with 1/T = 1K and W = 1K Output y l (t) with c l (t) being ideal lowpass filter of bandwidth W. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 14 / 51
Example: BPSK with 1/T = 1K and W = 0.9K Output y l (t) with c l (t) being ideal lowpass filter of bandwidth W. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 15 / 51
Example: BPSK with 1/T = 1K and W = 0.8K Output y l (t) with c l (t) being ideal lowpass filter of bandwidth W. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 16 / 51
Example: BPSK with 1/T = 1K and W = 0.7K Output y l (t) with c l (t) being ideal lowpass filter of bandwidth W. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 17 / 51
Example: BPSK with 1/T = 1K and W = 0.6K Output y l (t) with c l (t) being ideal lowpass filter of bandwidth W. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 18 / 51
Example: 4PAM with 1/T = 1K and W = 1K Output y l (t) with c l (t) being ideal lowpass filter of bandwidth W. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 19 / 51
Conclusion: The smaller W Horizontal: Decision is more sensible to timing error. Vertical: Decision is more sensible to noise. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 20 / 51
Example (revisited) Now changing to c l (t) = δ(t) + δ(t T ), we have, for example, 1, 0 t < T g(t) = T (Time-limited; hence, band-unlimited!) 0, otherwise h l (t) = g(t) c l (t) = g(t) + g(t T ) = g(t/2) x l (t) = h l (t) hl (2T t) = l(τ)hl (2T (t τ))dτ causal = g(τ/2)g((2t + τ t)/2)dτ 2T t 2T = T, t 2T 2T 0, otherwise n= y k = y l (kt ) = I n x l (kt nt ) + z l (kt ) = 2I k 2 + I k 1 + I k 3 +z k ISI The question next to be asked is that how to remove the ISI? Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 21 / 51
9.2-1 Design of band-limited signals for no intersymbol interference - The Nyquist criterion Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 22 / 51
Intersymbol interference channel Design goal y k = n= I n x k n + z k Given W, T and c l (t), we would like to design g(t) such that x k n = δ k n and that y k = I n x k n + z k n= = I k + z k Here, δ k is the Kronecker delta function, defined as 1 k = 0 δ k = 0 k 0. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 23 / 51
Nyquist rate Theorem 1 (Nyquist criterion) Let x l (t) be a band-limited signal with band [ W, W ] and x l (0) = 1. Sample at rate 1 T such that x k if and only if = x l (kt ) = where X l (f ) = F{x l (t)}. x l (t)δ(t kt ) dt = δ k X l (f m m= T ) = T Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 24 / 51
Proof: The condition of x k = x l (kt ) = δ k is equivalent to x l (t) k= δ(t kt ) = x l (kt )δ(t kt ) = δ(t). (1) k= We however have (from Lemma 2 in the next slide) F { δ(t kt )} = 1 k= T δ (f m ) and F{δ(t)} = 1. m= T Taking Fourier transform on both sides of (1) gives X l (f ) [ 1 T and proves the theorem. m= δ (f m T )] = 1 T X l (f m m= T ) = 1, Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 25 / 51
Proof of key lemma Lemma 2 F { δ(t kt )} = 1 k= T Proof: Consider the function α(f ) = 1 T δ (f m m= T ) δ (f m m= T ) which is periodic with period 1 T. From Fourier series, we have 1 α(f ) = c n e ı 2πnf /(1/T ) = e ı 2πnfT 1/T n= n= where c n = 1 2T 1 2T α(f )e ı 2πnf /(1/T ) df = 1 2T 1 2T by following the replication property of δ(f ). 1 T δ(f )e ı 2πnf /(1/T ) df = 1 T Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 26 / 51
Next by linearity of Fourier transform, we have F { k= δ(t kt )} = = k= e ı 2πfkT. k= δ(t kt )e ı 2πft dt Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 27 / 51
Implication of Nyquist criterion x k = δ k iff X l (f m m= T ) = T (2) 1 2W < 1 T : X l (f m) T and X l (f m ) T do not overlap for m m. Hence (2) is impossible! 2 2W = 1 T : This means X l (f ) = T rect (Tf ), hence x l (t) = sinc ( t T ). Theoretically ok but physically impossible! 3 So we need 2W > 1 T Channel bandwidth must be larger than sampling rate. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 28 / 51
x k = δ k iff X l (f m m= T ) = T 2W > 1 T (from W > 1 T W ) Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 29 / 51
Raised cosine pulse Definition 1 X rc (f ) = T, 0 f 1 β 2T T 2 {1 + cos [ πt 1 β ( f β )]} 2T, 1 β 1+β 2T f 2T 0, otherwise where 0 β 1 is the roll-off factor. x rc (t) = sinc ( t T ) cos (πβt/t ) 1 4β 2 t 2 /T 2 Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 30 / 51
X rc (f ) = T, 0 f 1 β 2T T 2 {1 + cos [ πt 1 β ( f β )]} 2T, 1 β 1+β 2T f 2T 0, otherwise Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 31 / 51
X rc (f m ) with m = 1, 0, 1 T Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 32 / 51
x rc (t) = sinc ( t cos (πβt/t ) ) T 1 4β 2 t 2 /T 2 Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 33 / 51
Example. Assuming C l (f ) = 1 Let G T (f ) = F{g(t)}. Let G R (f ) = F{h l ( t)}, where h l(t) = g(t) c l (t) = g(t). This gives and G R (f ) = F{g ( t)} = G T (f ) X rc (f ) = G T (f )C l (f )G R (f ) = G T (f )G R (f ) = G T (f ) 2. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 34 / 51
Note x rc (t) 1 t for large t. So it can be truncated at 3 ±t 0 for some t 0 large. So we can set x rc (t) = sinc ( t T ) cos (πβt/t ) 1 4β 2 t 2 /T 2 G T (f ) = X rc (f )e ı 2πft 0 and G R (f ) = GT (f ). Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 35 / 51
9.2-2 Design of band-limited signals with controlled ISI: Partial response signals Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 36 / 51
Duobinary pulse We relax the no-isi condition so that x k = δ k + δ k 1 y k = I k + I k 1 + z k Following similar arguments (see the next slide), it shows 1 T X l (f m m= T ) = 1 + e ı 2πfT. Setting 2W = 1 T (with W being the bandwidth of X l(f )), we have X l (f ) = T (1 + e ı 2πfT ) rect ( f 2W ) x l (t) = sinc (2Wt) + sinc [2 (Wt 1 2 )] This is called duobinary pulse. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 37 / 51
Proof: The condition of x k = δ k + δ k 1 is equivalent to x l (t) k= δ(t kt ) = We however have F { δ(t kt )} = 1 k= T x l (kt )δ(t kt ) = δ(t)+δ(t T ). (3) k= δ (f m ) and F{δ(t)} = 1. m= T Taking Fourier transform on both sides of (3) gives X l (f ) [ 1 T m= δ (f m T )] = 1 T and proves the theorem. X l (f m m= T ) = 1+e ı 2πfT, Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 38 / 51
When W = 1 2T, X l (f ) = T (1 + e ı 2πfT ) rect ( f ) 2W, T rect ( f ) 2W, with controlled ISI without controlled ISI X l (f ) with controlled ISI The duobinary filter is apparently more physically realizable! because channel is not physically realizable. But now with controlled becomes physically realizable. Recall that with no controlled ISI, we require W > 1 2T bandwidth W = 1 2T ISI, W = 1 2T Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 39 / 51
9.2-3 Data detection for controlled ISI Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 40 / 51
Precoding for duobinary pulses Received signal for duobinary shaping is y k = I k + I k 1 + z k. We could precode the sequence {I k } to simplify detection. Example 1 (Binary case - Differential encoding) Given the binary bit (information) stream {b k } Define P k = b k P k 1 {0, 1}. Set I k = 2P k 1 {±1}. Receive y k = I k + I k 1 + z k. I k + I k 1 = { ±2, if b k = 0 0, if b k = 1. ˆb k = 0 if y k > 1 and ˆb k = 1 if y k 1 Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 41 / 51
Summary of precoding system 1 Compare T m= X l (f m T ) = 1 1 T m= X (p) l (f m T ) = 1 + e ı 2πfT. We can say X (p) l (f ) = X l (f ) (1 + e ı 2πfT ) and y l (t) = n= I n x l (t nt ) + z l (t) y (p) l (t) = n= I n x (p) l (t nt ) + z l (t) and y k = I k + z k y k 0 y (p) k = I k + I k 1 + z k y (p) k 1 Note that x l (t) 2 does not decide the transmission energy! Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 42 / 51
Summary of precoding system P e = Q ( P (p) 1 E[zk 2]) = Q ( E b,l / g T (t) 2 N 0 g R (t) 2 ), error rate for I k = Q ( 2Eb N 0 1 g T (t) 2 g R (t) 2 ) = Q( 2aγ b ) e = 3 2 Q( 2aγ b ) 1 2 Q(3 2aγ b ), error rate for b k where I k {±1} the (lowpass) transmission energy per bit E b,l = g T (t) 2 = 2E b the lowpass noise E[zk 2] = σ2 l g R(t) 2 a = with n l (t) having two-sided PSD σl 2 = N 0 (subject to g T (t) c l (t) g R (t) = x l (t)) 1 g T (t) 2 g R (t) 2 Hence, pre-coding technique provides a better spectrum efficiency at the price of performance degradation. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 43 / 51
9.2-4 Signal design for channel with distortion Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 44 / 51
In general, we have C l (f ) rect ( f ), 2W and in this case s l (t) = n= I n g T (t nt ) r l (t) = c l (t) s l (t) + n l (t) where y l (t) = r l (t) g R (t) = s l (t) c l (t) g R (t) + z l (t) y k = y l (kt ) = I k + z k (We hope there is no ISI!) g T (t) transmit filter c l (t) channel impulse response g R (t) receive filter Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 45 / 51
Hence x rc (t) = g T (t) c l (t) g R (t) X rc (f ) = G T (f )C l (f )G R (f ) S zl (f ) = N 0 G R (f ) 2 because we assume {I k } real If c l (t) is known to Tx, then we may choose to pre-equalize the channel effect at Tx: G T (f ) = Xrc (f ) C l (f ) and G R (f ) = X rc (f ) Then, ISI is avoided; also, the noise power remains E[z 2 k ] = S zl (f ) df = N 0 =N 0 G R (f ) 2 X rc (f ) df = N 0 x rc (0) = N 0. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 46 / 51
Signal power I n {±d} and X rc (f ) = G T (f )C l (f )G R (f ) P av,l = d 2 g T (t) 2 T = d 2 T X rc (f ) C l (f ) df 2 Error probability P b = Q 12 4E[zk 2] = Q d E[z 2 k ] = Q av,lt N 0 [ X rc(f ) df ] C l (f ) 2 Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 47 / 51
If c l (t) only known to Rx We can only equalize the channel effect at Rx: G T (f ) = Xrc (f ) X rc (f ) and G R (f ) =. C l (f ) Signal power P av,l = d 2 T g T (t) 2 = d 2 T X rc (f ) df = d 2 T Noise power E[zk 2] = X rc (f ) S zl (f ) df = N 0 C l (f ) df 2 Error probability P b = Q d E[z 2 k ] = Q P av,lt N 0 [ X rc(f ) df ] C l (f ) 2 Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 48 / 51
If however c l (t) known to both Tx and Rx We may design: G T (f ) = Signal power X rc(f ) C l (f ) and G R (f ) = X rc(f ) C l (f ) P av,l = d 2 T g T (t) 2 = d 2 T X rc (f ) C l (f ) df Noise power E[zk 2] = X rc (f ) S zl (f ) df = N 0 C l (f ) df Error probability P b = Q d E[z 2 k ] = Q P av,l T N 0 [ X rc(f ) C l (f ) df ]2 Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 49 / 51
Either Tx or Rx knows c l (t) P b,t = P b,r = Q P av,l T N 0 [ X rc(f ) df ] C l (f ) 2 Both Tx and Rx know c l (t) P b,tr = Q P av,l T N 0 [ X rc(f ) C l (f ) Note from Cauchy-Schwartz inequality [ 2 X rc (f ) C l (f ) df ] = X rc (f ), df ]2 Xrc (f ) C l (f ) X rc (f ) 2 2 Xrc (f ) X rc (f ) = C l (f ) C l (f ) 2 df This shows P b,tr P b,t = P b,r. = holds iff C l (f ) = 1. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 50 / 51 2
What you learn from Chapter 9 Match filter to input pulse shaping and channel impulse response (Good to know) Eye pattern to examine ISI Nyquist criterion Sampling rate < channel bandwidth for no ISI (I.e., increasing sampling rate will give more samples for perhaps better performance, but adjacent samples will be eventually interfered to each other) Since ISI is unavoidable for high sampling rate, let s accept and face it, and just use controlled ISI. A better performance is resulted when both Tx and Rx know the channel. Digital Communications: Chapter 09 Ver 2018.07.25 Po-Ning Chen 51 / 51