EE4061 Communication Systems Week 11 Intersymbol Interference Nyquist Pulse Shaping 0 c 2015, Georgia Institute of Technology (lect10 1)
Intersymbol Interference (ISI) Tx filter channel Rx filter a δ(t-nt) n x(t) g(t) c(t) {a k } a k ε { 1,+1} w(t) Σ n + h(t) AWGN An ideal channel c(t) only scales and time shifts the signal g(t), but otherwise leaves it undistorted, i.e. for an ideal channel y(t) kt c(t) = αδ(t t o ) g(t) c(t) = αg(t t o ) C(f) = αe j2πft o 0 c 2011, Georgia Institute of Technology (lect10 2)
ISI C(f) argc(f) Slope =!2 ft o f!w W!W W Phase response Amplitude response is linear is flat or constant f Amplitude and phase response for an ideal channel. For a more general, non-ideal, channel, let p(t) = g(t) c(t) h(t) P(f) = G(f)C(f)H(f) Then y(t) = na n p(t nt)+n(t), where n(t) = w(t) h(t) 0 c 2013, Georgia Institute of Technology (lect10 3)
ISI y i = y(it) = n = n a n p((i n)t)+n(it) a n p i n +n i where p i n = p((i n)t), n i = nt y i = a i p 0 + n ia n p i n +n i a i p o desired term, n ia n p i n ISI n i noise In the absence of ISI and noise, y i = a i p 0. Any pulse p(t), such that the sampled pulse satisfies the condition p i = p(it) = { p 0 i = 0 0 i 0 = p 0δ i0 yields zero ISI 0 c 2013, Georgia Institute of Technology (lect10 4)
Bandlimited Pulse Shaping What overall pulse shapes p(t), p(t) = g(t) c(t) h(t), will yield zero ISI? Suppose P(f) = 1 ) 2W rect( f 2W, where W = 1/2T = R/2, T is the baud duration, R is the baud rate P(f) 1/2W p(t) 1.0 -W 0 W p(t) = sinc(2wt) = sinc(t/t), T = 1/2W. Note that f -3T -2T -T T 2T 3T p i = p(it) = { 1 i = 0 0 i 0 = δ i0 This pulse results in zero ISI. Note that p(t) is noncausal. 0 c 2013, Georgia Institute of Technology (lect10 5)
ISI - Problems a 0=1 p(t) 1.0 a 2=1-2T -T T 2T 3T t a 1 =-1 Problems: 1. P(f) = 1 ) 2W rect( f 2W is and ideal low pass filter that is not realizable. 2. p(t) decays slowly with time. It decreases with 1/ t for large t. Therefore, it is very sensitive to sampler phase, i.e., a small error in the sampler timing phase can lead to significant ISI. We desire a pulse p(t) that is realizable and has tails that decay quickly in time. 0 c 2013, Georgia Institute of Technology (lect10 6)
Problems with sinc pulse y(t) = n a n p(t nt) y( t) = n = n = n = n a n p( t nt) sin[π( t nt)/t] a n π( t nt)/t sin(π t/t) cos(πn) cos(π t/t) sin(πn) π t/t nπ a n ( 1) n sin(π t/t) π t/t nπ = a 0 sinc( t/t)+ sin(π t/t) π Last term is not absolutely summable. n 0 a n ( 1) n t/t n We have seen y i = y(it) = a i p 0 + n ia n p i n +n i where p k = p(kt), n i = n(it). 0 c 2011, Georgia Institute of Technology (lect10 7)
Matched Filtering and Pulse Shaping g(t) unknown c(t) x(t) + h(t) 2/T g(t) ^ = g(t)*c(t) w(t) To maximize the signal-to-noise ratio at the output of the receiver filter h(t), in theory we match the receiver filter to the received pulse ĝ(t) = g(t) c(t), i.e., h(t) = g(t t). However, if c(t) is unknown, then so is h(t). Practical Solution: Choose h(t) matched to the transmitted pulse g(t), i.e., choose h(t) = g(t t), over-sample by a factor of 2, and process 2 samples per baud interval. This is optimal, similar to the case when c(t) is known, but the proof is beyond the scope of this course. 0 c 2011, Georgia Institute of Technology (lect10 8)
Matched Filtering and Pulse Shaping To design the transmit and receiver filters, we will assume an ideal channel c(t) = δ(t), so that the overall pulse (ignoring time delay) is p(t) = g(t) h(t) = g(t) g( t) Taking the Fourier transform of both sides P(f) = G(f)G (f) = G(f) 2 Hence G(f) = P(f) For many practical pulses, g(t), we will also see that g(t) = g( t), i.e., the pulse is even in t, so that h(t) = g(t). This means that the transmit and receive matched filters are identical filters. 0 c 2013, Georgia Institute of Technology (lect10 9)
Conditions for ISI free transmission The condition for ISI-free transmission is p k = δ k0 p 0 = p 0 k = 0 0 k 0 That is, p(t) must have equally spaced zero crossings, separated by T seconds. Theorem: The pulse p(t) satisfies p k = δ k0 p 0 iff P (f) = 1 T n= That is the folded spectrum P (f) is flat. P(f +n/t) = p 0 P(f) W=1/2T p(t) 1.0 P Σ (f)...... -W 0 W f -3T -2T -T T 2T 3T -3/2T 0 3/2T -1/2T 1/2T 0 c 2011, Georgia Institute of Technology (lect10 10)
ISI free transmission Proof: p k = P(f)ej2πfkT df = (2n+1)/2T n= = 1/2T n= 1/2T n= kt = 1/2T ej2πf n= (2n 1)/2T P(f)ej2πfkT df f = f n/t 1/2T P(f +n/t)e j2πk(f +n/t)t df P(f +n/t) df (1) To prove sufficiency, we assume that n= P(f +n/t) = p 0 T is true. Then, p k = p 0 T 1/2T To prove necessity, we have from (1) 1/2T ej2πf kt df = sinπk πk p 0 = δ k0 p k0 p k = T 1/2T 1/2T P Σ(f )e j2πf kt df 0 c 2011, Georgia Institute of Technology (lect10 11)
Nyquist Pulse Hence, p k and P Σ (f) are a Fourier series pair, i.e., P Σ (f) = k= p k e j2πfkt If p k = p 0 δ k0 is assumed true, then from the above equation P Σ (f) = p 0. Nyquist Pulse Shaping: A pulse p(t) that yields zero-isi is one having a folded spectrum that is flat. The pulse p(t) can be generated by choosing P(f) as shown on the following slide. 0 c 2013, Georgia Institute of Technology (lect10 12)
Nyquist Pulse Shaping 1 2WP(f) 1/2 -W 0 W f 1 2WP (f) N Ideal Nyquist pulse 0 -W W 2WPod (f) 1/2 -W 0 W Note P(f) = P N (f)+p od (f). P od (f) can be any function that has skew symmetry about f = W = 1/2T. 0 c 2013, Georgia Institute of Technology (lect10 13)
Nyquist Pulse Note that P Σ (f) is flat under this condition. 1 2WP Σ (f) W=1/2T -3W -2W -W 0 W 2W f f Example: Raised Cosine 2WP od (f) = 1 2 1 2 1 2 1 2 sin π( f W) 2f x sin π( f W) 2f x W f x f W W f W +f x f x = bandwidth expansion, f x W 100 = excess bandwidth (%), α = f x W factor 1 0 f W f [ x 1 2WP(f) = 2 1 sin π( f W) ] 2f x W f x f W +f x 0 f W +f x = roll off 0 c 2011, Georgia Institute of Technology (lect10 14)
Raised Cosine Pulse 2WP(f) w fx w w+fx 0 w fx w w+fx 2WP(f) α = 1 looks like a raised cosine 2W W 0 W 2W 0 c 2011, Georgia Institute of Technology (lect10 15)
Raised Cosine Impulse Response Impulse response - Since P(f) is even, the inverse cosine transform yields W+fx p(t) = 2 0 1 = 2. 2W P(f)cos2πftdf W fx 0 = sin2πwt. cos2πf xt 2πWt 1 (4f x t) 2 1 cos2πftdf +2. 2W 1 1 sin π f W cos2πftdf W f x 2 2f x W+fx T=1/2W α = 0 α = 1 3T 2T T T 2T 3T 0 c 2011, Georgia Institute of Technology (lect10 16)
Square Root Raised Cosine Pulse To implement a matched filter, we split the overall pulse P(f) between the transmit and receive filters, i.e., p(t) = g(t) g( t). However, P(f) = G(f)G (f) = G(f) 2, so that G(f) = P(f). With square-root raised cosine pulse shaping 2W G(f) = 1 0 f ww f x 1 2 [ 1 sin π( f W) 2f x 0 f W +f x ] W f x f W +f x The impulse response is g(t) = where α = f x /W. 1 β +4β/π, t = 0 (β/ 2)((1+2/π)sin(π/4β)+(1 2/π)cos(π/4β)), t = ±T/4β, elsewhere 4β(t/T) cos((1+β)πt/t)+sin((1 β)πt/t) π(t/t)(1 (4βt/T) 2 ) 0 c 2011, Georgia Institute of Technology (lect10 17)
Square Root Raised Cosine Pulse 1.2 1 raised cosine root raised cosine 0.8 g (t) 0.6 0.4 0.2 0 0.2 0 1 2 3 4 5 6 t/t Raised cosine and root raised cosine pulses with roll-off factor α = 0.5. The pulses are truncated to length 6T and time shifted by 3T to yield causal pulses. 0 c 2011, Georgia Institute of Technology (lect10 17)
M-ary QAM Quadrature Amplitude Modulation (QAM), the transmitted waveform in each baud interval takes on one of the following M waveforms where s m (t) = 2E 0 T g(t) ( a {c,s} m ) a c m cos(2πf ct) a s m sin(2πf ct) {±1,±3,±5,±(M 1)} and 2E 0 is the energy of the signal with the lowest amplitude, i.e., when a c m,a s m = ±1. You have seen this before for the case g(t) = u T (t); however, practical systems will use the root-raised cosine pulse for g(t). Note that we use the normalization, E g = g2 (t)dt = 1. 0 c 2011, Georgia Institute of Technology (lect10 18)
Eye Diagram with Ideal Nyquist Pulse Eye diagram when P(f) is an ideal low pass filter. 0 c 2011, Georgia Institute of Technology (lect10 18)
Eye Diagram with Raised Cosine Pulse Eye diagram when P(f) is a raised cosine filter with β = 0.35. 0 c 2011, Georgia Institute of Technology (lect10 19)