Objectives. Presentation Outline. Digital Modulation Lecture 03

Transcription

1 Digital Modulation Lecture 03 Inter-Symbol Interference Power Spectral Density Richard Harris Objectives To be able to discuss Inter-Symbol Interference (ISI), its causes and possible remedies. To be able to define spectral efficiency To define what is meant by the Power Spectral Density (PSD) function To be able to compute the PSD for different forms of digital modulation and discuss its properties. Communication Systems Digital Modulation Slide Presentation Outline Inter-symbol Interference What is ISI? How can it be avoided? Power Spectral Density function Communication Systems Digital Modulation Slide 3 1

2 Inter-Symbol Interference ISI - 1 Baseband Pulse Transmission System bsolute bandwidth of rectangular pulses is infinite They are filtered as they pass through the communication system The pulse for each symbol is spread into adjacent timeslots This may cause Intersymbol Interference (ISI) Communication Systems Digital Modulation Slide 4 Inter-Symbol Interference Examples of ISI on received pulses in a binary communication system Communication Systems Digital Modulation Slide 5 Inter-Symbol Interference 3 The overall transfer function for the system is H e (f) = H(f). H T (f). H C (f). H R (f) Where H(f) is the Fourier Transform of the rectangular pulse Communication Systems Digital Modulation Slide 6

3 voiding Inter-Symbol Interference For zero ISI, Nyquist found that H e (f) must satisfy the following condition: For a single pulse, the filtered pulse is non-zero at its own sampling time, but is always zero at other sampling times. Example: f r = f 0 B = f + f 0 = ( 1+ r) f 0 Raised cosine-rolloff Nyquist filter characteristics Communication Systems Digital Modulation Slide 7 Impulse Response 1 f0 Communication Systems Digital Modulation Slide 8 voiding ISI (Contd) For a communication system with a raised cosine-rolloff filtering characteristic, the (baud) symbol rate which can be supported by the system is: B/(1+r) Where B is the absolute bandwidth of the system and r is the roll-off factor of the filter. Note that 0 r 1 is a requirement Communication Systems Digital Modulation Slide 9 3

4 Spectral Efficiency To compare signalling schemes we can define a quantity called Spectral Efficiency: The ratio of the bit rate (in bits/sec) of the modulating signal to the transmission bandwidth (in Hz) of the modulated signal. ie, r b /B T bits/sec/hz. For systems operating at the Nyquist rate, B T = r, so that r b /B T = log M, and for values of M=,4,8,16 etc, the ideal spectral efficiencies are 1,,3,4 etc bits/sec/hz. For practical systems, the values are somewhat less than these ideal values. Communication Systems Digital Modulation Slide 10 Theoretical Bandwidth Efficiency Limits Modulation method MSK BPSK QPSK 8PSK 16 QM 3 QM 64 QM 56 QM Theoretical bandwidth format efficiency limits 1 bit/second/hz 1 bit/second/hz bits/second/hz 3 bits/second/hz 4 bits/second/hz 5 bits/second/hz 6 bits/second/hz 8 bits/second/hz Note: These figures cannot actually be achieved in practical systems as they require perfect modulators, demodulators, filter and transmission paths. Communication Systems Digital Modulation Slide 11 Power Spectral Density Definition Definition: (See Wikipedia) The power spectral density (PSD) describes how the power of a signal or time series is distributed with frequency. The PSD is the Fourier transform of the autocorrelation function of the signal if the signal can be treated as a stationary random process. The Power Spectral Density of a signal exists if and only if the signal is a wide-sense stationary process. Note: If the signal is not stationary, then the autocorrelation function must be a function of two variables, so no PSD exists, but similar techniques may be used to estimate a time-varying spectral density. Communication Systems Digital Modulation Slide 1 4

5 Random Process through a Linear Time Invariant Filter - 1 Time-Invariant Filters In plain terms, a time-invariant filter (or shift-invariant filter) is one which performs the same operation at all times. Thus, if the input signal is delayed (shifted) by, say, N samples, then the output waveform is simply delayed by N samples and unchanged otherwise. Thus Y(t), the output waveform from a time-invariant filter, merely shifts forward or backward in time as the input waveform is shifted forward or backward in time. Consider a random process X(t) applied as input to a linear timeinvariant filter having impulse response h(t) that produces a new random output process Y(t). X(t) Impulse response h(t) Y(t) Transmission in this situation is governed by a convolution integral, viz: Yt () = h( τ1) Xt ( τ1) 1 Communication Systems Digital Modulation Slide 13 Random Process through a Linear Time Invariant Filter - We can compute the mean value of Y(t) as µ () t Y = E [ Y ()] t = E h( τ1) X( t τ1) 1 Now, provided that the expectation E[X(t)] is finite for all t and the system is stable, we can interchange the order of expectation and integration to obtain: µ () t = h( τ ) E[ X( t τ )] Y = h( τ ) µ ( t τ ) 1 X 1 1 Communication Systems Digital Modulation Slide 14 Random Process through a Linear Time Invariant Filter - 3 When the input random process X(t) is stationary, the mean µ X (t) is a constant µ X, so we can further simplify our result as: µ Y = µ X h( τ1) 1 = µ X H (0) Where H(0) is the zero frequency (DC) response of the system. This says that the mean of the random process Y(t) produced at the output of a linear time-invariant system in response to X(t) acting as the input process is equal to the mean of X(t) multiplied by the DC response of the system. Communication Systems Digital Modulation Slide 15 5

6 Random Process through a Linear Time Invariant Filter - 4 Now consider the auto-correlation function of the output random process Y(t). R(, tu) = EYtYu [ () ( )] Y Where t and u are two values of the time at which the output process is observed. Using the convolution integral we have: RY ( t, u) = E h( τ1) X( t τ1) 1 h( τ) X( u τ) Communication Systems Digital Modulation Slide 16 Random Process through a Linear Time Invariant Filter - 5 Now, provided that the expectation E[X (t)] is finite for all t and the system is stable, we can interchange the order of expectation and integration with respect to both τ 1 and τ to obtain: RY (, t u) = h( τ1) 1 h( τ) E[ X( t τ1) X( u τ) ] = h( τ1) 1 h( τ) RX ( t τ1)( u τ) When the input X(t) is a stationary process, the autocorrelation function of X(t) is only a function of the difference between the observation times t- τ 1 and u- τ. Thus writing τ=t-u we can write: RY( τ ) = h( τ1) h( τ) RX( τ τ1+ τ) 1 Communication Systems Digital Modulation Slide 17 Random Process through a Linear Time Invariant Filter - 6 Combining with our earlier result we see that if X(t) is a stationary input process, then Y(t) is also a stationary process. Since R Y (0) = E[Y (t)] it follows that we can find this quantity by setting τ = 0 in order to obtain: E Y () t = h( τ ) h( τ ) R ( τ τ ) Which is a constant! 1 X 1 1 Communication Systems Digital Modulation Slide 18 6

7 Power Spectral Density - 1 Consider the characterisation of random processes in linear systems by using frequency domain ideas. We are interested in finding the frequency domain equivalent to the result on the previous slide that defines the mean square value of the filter output. Now: The impulse response of a linear time-invariant filter is equivalent to the inverse Fourier Transform of the frequency response of the system. Let H(f) be the frequency response of the system. Communication Systems Digital Modulation Slide 19 Power Spectral Density - 1 π τ1 h( τ ) = H( f)exp( j f ) df Substituting this expression into our previous result gives: E Y ( t) H( f)exp( j π fτ1) df = h( τ) RX ( τ τ1) 1 = dfh ( f ) h( τ ) R ( τ τ )exp( j π fτ ) X Define a new variable τ = τ 1 τ so that E Y ( t) = dfh( f) h( τ)exp( j π fτ) RX ( τ)exp( j π fτ) Communication Systems Digital Modulation Slide 0 Power Spectral Density - 3 = τ τ π τ τ π τ τ E Y ( t) dfh( f) d h( )exp( j f ) RX ( )exp( j f ) d The middle integral is H*(f) the complex conjugate of the frequency response of the filter, so simplifying gives: E Y ( t) = df H( f) RX ( τ )exp( j π fτ) Where H(f) is the magnitude response of the filter. Finally we note that the right-most integral is actually the Fourier transform of the autocorrelation function R X (τ) of the input random process X(t). Communication Systems Digital Modulation Slide 1 7

8 Power Spectral Density - 4 Let us now define the quantity S ( f) = R ( τ )exp( j π fτ) X X The function S X (f) is called the Power Spectral Density or Power Spectrum of the stationary process X(t). E Y () t = H( f) SX ( f) df The equation marked with says that the mean square value of a stable linear time-invariant filter in response to a stationary process is equal to the integral over all frequencies of the PSD of the input process multiplied by the squared magnitude response of the filter. Communication Systems Digital Modulation Slide Power Spectral Density - 5 Consider the following situation where a random process X(t) is passed through an ideal narrow band filter with magnitude response centred around the frequency f c (see figure below): 1, f ± fc < f / H( f) = 0, f ± fc > f / Hf ( ) 1.0 f c f 0 f c f f Communication Systems Digital Modulation Slide 3 Power Spectral Density - 6 Note that f is the bandwidth of the filter. Substituting these details into the formula for E[Y (t)] we have (for sufficiently small filter bandwidth compared to the mid-band frequency f c with S X (f) a continuous function: E[ Y ( t)] ( f) SX ( fc ) Communication Systems Digital Modulation Slide 4 8

9 Power Spectral Density - 7 When a digital waveform x(t) is modelled by n= x ( t) = a w( t nt ) n where {a n } represent data, n is an integer, w(t) is the signalling pulse shape and T is the duration of one data symbol. The power spectral density function PSD of x(t) can be shown to be given by σ a µ a k k P = + x ( f ) W ( f ) W δ f T T k = T T where k is an integer, σ a and µ a are the variance and mean of the data {a n } and W(f) is the Fourier transform of w(t). Communication Systems Digital Modulation Slide 5 Properties of the PSD - 1 The Power Spectral Density Function S X (f) and the autocorrelation function R X (f) of a stationary process X(t) form a Fourier-transform pair with τ and f as the variables satisfying the following: S ( f) = R ( τ )exp( jπ fτ) X R ( τ) = S ( f)exp( jπ fτ) df X X X (Einstein-Wiener-Khintchine relations) Communication Systems Digital Modulation Slide 6 Properties of the PSD - Property 1 The zero value of the PSD of a stationary process is equal to the total area under the graph of the autocorrelation function. Property The mean square value of a stationary process is equal to the total area under the graph of the PSD Property 3 The PSD of a stationary process is always non-negative Property 4 The PSD of a real valued random process is an even function of frequency; ie S X (-f) = S X (f) Property 5 The PSD, appropriately normalised has the properties associated with a Probability Density Function ttempt these as an exercise in the Tutorial Session. Communication Systems Digital Modulation Slide 7 9

10 Example PSD Sine wave with random phase - 1 Consider the following random process cos(πf c t+θ) where θ is a uniformly distributed random variable over the interval [-π, π] First we determine the autocorrelation function for this random process: RX ( τ ) = EXt [ ( + τ ) Xt ( )] = E [ cos(π ft c + π fcτ + θ)cos( π ft c + θ)] = E[cos(4π fct+ π fcτ + θ)] + E[cos( π fcτ)] π 1 = cos(4 fct fc ) d cos( fc ) π + π τ + θ θ + π τ π = 0+ cos( π fcτ) = cos( π fcτ) Communication Systems Digital Modulation Slide 8 Example PSD Sine wave with random phase - We now take Fourier transforms of both sides of the resulting expression: RX( τ) = cos( π fcτ) RX( τ )exp( j π fτ) = cos( π fcτ)exp( j π fτ) SX( f) = cos( fc )exp( j f ) d π τ π τ τ = [ δ( f fc) + δ( f + fc) ] 4 (See table of Fourier transform values on next slide) Communication Systems Digital Modulation Slide 9 Example PSD Sine wave with random phase - 3 We see that the solution to this problem is a pair of delta functions. Communication Systems Digital Modulation Slide 30 10

11 Example PSD Sine wave with random phase - 4 Note that the total area under a delta function is unity. Thus, the total area under S X (f) is equal to / as expected. S X (f) δ ( f + fc) 4 δ ( f fc ) 4 f -f c 0 f c Communication Systems Digital Modulation Slide 31 PSD of a Random Binary Wave - 1 Consider the following random sequence of 1s and 0s: + It can be shown that the autocorrelation function for this process X(t) is given by: τ 1, τ < T RX ( τ ) = T 0, τ T t d - T RX ( τ ) -T 0 +T Communication Systems Digital Modulation Slide 3 PSD of a Random Binary Wave - Taking Fourier transforms to obtain the PSD once again we find: T τ SX ( f) = 1 exp( j π fτ) T T Conversion back from tables of Fourier transforms we find the sinc (ft) function: SX ( f) = Tsinc ( ft) Where 1 if x = 0 sinc x = sin x otherwise x Communication Systems Digital Modulation Slide 33 11

12 PSD of a Random Binary Wave - 3 Plotting the Power Spectral Density function in this case gives: sinc (x) sinc(x) Communication Systems Digital Modulation Slide 34 1