Mobile Computing CS6242

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University of Manchester School of Computer
Science CS4242 Mobile Computing Section
2 Matched filtering, pulse shaping equalisation
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2.1 Intro A digital transmission system
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• Transmission using binary signalling.
• Received symbols s(t) s1(t) or s0(t).
• Affected by LP filtered AWGN n(t)
• r(t) s(t) n(t)
• Assume n(t) has 1-sided PSD N0 Watts/Hz.
• If bandwidth of noise n(t) is B,
• its
  power variance is B No
• Unipolar' or bipolar signalling with 'rect'
  pulse shape.
• For unipolar, r(t) will be as shown for '1'
  '0

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Bipolar signalling with AWGN
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A detection strategy
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• For unipolar signalling, strategy for detecting
  1 or 0 would be to sample
  r(t) at tT/2, compare it with a 'threshold'
  V/2.
• If r(T/2) gt V/2, likely s1(t) otherwise s0(t).
• For rect pulses, no matter whether we sample in
  middle,
  at beginning or
  at end .
• Wherever we sample, risk that noise n(t) will
  cause s1(t) to be mistaken
• 
  for s0(t) or
  vice-versa.
• Choice of threshold half way between zero V
  is good when
• probabilities of s0(t)
  s1(t) are known to be equal, i.e. 0.5.

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2.2 Improved detectn process for unipolar rect
symbols
Response of averager to si(t) is ai(t),
response to n(t) is n0(t). Averager (or smoother)
could ideally produce an output
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• Force integrator output to start at zero volts
  at t0.
• (A switch dumps charge
  on capacitor.)
• Integrate and dump' circuit.

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• Noise power is ?02 0.5N0/T.
• Standard deviation ?0 is square root of this
  value.
• Therefore for unipolar pulse of voltage A,

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Example 2.1A Receive 1 volt 0 volt rect
binary symbols at 100 Baud. Transmission
distorted by AWGN with zero mean PSD
N00.00025 Watts/Hz
Estimate bit-error probability with an ID ..
Assume equal occurrence of 1s 0s
appropriate threshold Solution Consider output
from averager or MF . Noise has variance.

Therefore, ?00.112. Pulses
unaffected by averager,. Decision threshold
0.5Volts.
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Bit-error probability is
Bit error-rate is one bit in 200,000. (About one
character wrong in a 6-page document) Interesting
to compare this with what would be obtained
without id averager. But under assumption
that n(t) is AWGN, its power is infinite.
Therefore Pb Q(0) 0.5, the worst we can get.
Not 1??
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• We get more sensible result when we realise that
  receiver 'front- end' has filter to restrict
  bandwidth of received signal.
• Assume that noise band-width is restricted to ?B
  Hz, for some value of B, by a 'front-end'
  filter
• Noise power now restricted to N0B Watts.
• So variance at decision stage of detector would
  also be N0B.
• Now consider the following example

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Example 2.1B Receive 1 volt 0 volt rect
binary symbols at 100 Baud distorted by zero mean
AWGN with N00.00025 Watts/Hz. Estimate
bit-error probability without an averager. Ideal
low-pass filter H((f)) employed with bandwidth
?500 Hz. Assume equal occurrence of 1 0, with
appropriate threshold. Compare with previous
example. Solution Filtered noise has power
0.00025 x 500 0.125 Watt. Standard deviation
of noise is therefore 0.35 Decision on basis of
0 1 Volt pulses threshold 0.5 Volts.
Bit-error probability is prob of noise sample
exceeding 0.5 Volts with 0 or being less than
0.5 Volts with 1. i.e. Q ( 0.5 /
0.35) Q(1.42) 8x10-2 One bit in 12.5
(About one character wrong every one or two.)
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• Correlation Detector Matched filter
• Consider signalling with p(t) for 1 q(t) for
  0.
• Received with zero mean AWGN of 2-sided PSD N0/2
  Watts/Hz.
• Received signal, r(t), passed thro averager to
  produce z(t).
• Response of filter to p(t), q(t) n(t) is p0(t),
  q0(t) n0(t).
• Detector chooses sampling point, tT, sets
  threshold
• ? ( p0(T) q0(T) ) /2.
• Assume that each pulse starts at t0 ends at t
  T.
• Assuming p0(T) gt q0(T) when z(T) ? ? detector
  delivers 1
• 
  otherwise 0.
• Correct unless n0(T) gt ? for q(t)
• or n0(T) lt?? for p(t)
• where ? is the headroom between ?
  p0(T) or q0(T).

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• ? p0(T) - ? ? - q0(T)
  (p0(T) - q0(T) )/2
• Assuming equal prob for 1 0,
• PB 0.5Q(?/?0) 0.5Q(-?/?0)
  Q(?/?0)
• where ?0 is standard deviation of filtered noise
  n0(t).
• To generalise to pulses of any shape, id tuned
  to difference in shape between p(t) q(t).

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Correlation detector

• Tuned to p(t) q(t).
• ?/?0 is maximized
• Q( ?/?0 ), i.e. the bit-error probability,
  minimized.

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• Matched filter
• Broadly equivalent approach
• Has impulse response equal to p(t)-q(t)
  modified in 3 ways
• (i) reversed in time,
• (ii) delayed by T seconds
• (iii) multiplied by any constant k.
• Matched filter for rect pulse has rect impulse
  resp. has impulse-response.
• We can design matched filters for other symbol
  shapes.

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For corelation detector
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Ed is difference energy i.e. energy of
p(t) q(t) in Joules. Ed p0(t) q0(t)
2? Bit-error probability is equal to
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Parsevals theorem
For noise
Make H(f) P(f) - Q(f) where P(f) Q(f) are
Fourier transforms of p(t) q(t) respectively
0.5 N0 Ed
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??02 0.5N0 Ed it follows that error prob is
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• To summarise this section so far,
• Receive p(t) or q(t) corrupted by AWGN n(t) of
  2-sided PSD N0 /2 Watts/Hz.
• Need to maximise ?/?0 when ? (p0(t) -
  q0(t) / 2.
• p0(t), q0(t) n0(t) are responses to p(t),
  q(t) n(t).
• ?0 is standard deviation of n0(t) symbol rate
  is 1/T.
• With binary transmission with p(t) q(t)
  defining a threshold
• ? (p0(T)q0(T))/2 (equi-prob 1 0) bit-error
  prob Q(?/?0).
• Q(z/?0) is prob of Gaussian variable of zero mean
  variance ?0 2 being greater than z.
• If Ed denotes energy of difference between p(t)
  q(t), bit-error probability with a matched
  correlation detector is

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• This formula requires only Ed and N0.
• Actual shapes of p(t) q(t) do not matter as
  long as we use a matched filter or corr detector
• When p(t) is rectangular of height A duration
  T, q(t)0, this formula gives us

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Ex2.2 A binary signal with NRZ 1 0 volt rect
symbols p(t) q(t) a symbol rate of 1/T Baud
is corrupted by AWGN with 2-sided PSD N0 /2
1x10-3 Watts/ Hz. If received signal detected
with matched filter, what is max bit-rate that
can be sent with a bit-error rate less than 1 bit
in 1000? Estimate error-rate with no MF if noise
bandwidth B10kHz? Solution
With no MF, PB Q(1/(2?)) with ?2 0.2.
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• 2.5 Average Energy per bit
• The higher the power of the transmitted signal
  conveying a sequence of symbols, the higher the
  voltages of the symbols, the better will they be
  seen above the noise therefore the lower the
  bit-error probablty.
• If average power of signal is P Watts bit-rate
  is 1/T b/s,
• average energy per
  bit, Eb, is P divided by (1/T).
• Units are Joules/bit.
• Watts (Joules/second) divided by bits/second,
  become Joules/bit.
• Useful to know how much energy each bit will cost
  us to send.
• Different for different signalling techniques.
• Required power of transmission is Eb times the
  bit-rate.
• Do not confuse Eb with Ed (energy of difference
  signal).

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• Unipolar signalling
• Let p(t) be shaped pulse q(t) 0 for 0 ? t ?
  T. This is unipolar signalling. We have shown
  that when a matched filter is employed,

with

• Assuming equi-prob 1 0, for unipolar
  signalling, Eb Ed/2.

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• Bipolar signalling
• Let p(t) s(t) q(t) -s(t) for 0 ? t ? T.
• Binary signalling with anti-podal signals
• i.e. two signals which are
  negative of each other.
• Signal s(t) may have any shape. When a matched
  filter is employed with impulse-response 2s(T-t)
  i.e. matched to p(t)-q(t)

with

• since energy of p(t) q(t) are equal.
• In terms of Eb N0

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Exercise Let p(t) A and q(t) -A for 0 ? t ?
T. This is bipolar signalling with antipodal
rectangular NRZ pulses. Calculate PB assuming
equal numbers of 1s and 0s. Solution Average
energy per bit, Eb, A2T because energy of a
rect symbol of height ?A duration T seconds is
A2T.
To do this another way, note that
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• 2.8 Comments about uni-polar bipolar signaling
• Consider which is more efficient in terms of
  having lower bit-error
• 
  probability for a given average energy per bit.
• Consider graph of PB against 10 log10 (Eb/N0) for
  matched filter
• reception of (i) unipolar
  (ii) bipolar base-band signalling.
• Horizontal axis tells us how much greater, in
  dBs, the average energy
• 
  per bit is than the PSD N0 of the noise.
• It is ratio of Eb in Joules/bit to N0 in
  Watts/Hz.
• If noise bandwidth is 0 to B Hz, its power is
  ?02, then N0?02 / B.
• You can draw this graph with the aid of a graph
  of Q(z) against z.
• When Eb N0, (0 dB),
• PB ? 0.8x10-1 for bipolar
  0.2 (BER 1 in 5) for unipolar.
• When Eb 10N0, (10 dB), PB ? 0.3x10-5 for bipolar
  10-3 for unipolar.
• Bit-error probability of 10-3 means a BER of 1 in
  1000.

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• For given average energy / bit, bipolar better
  than unipolar.
• For same rect symbol height A, Eb A2 for
  bipolar
• 
  A2/2 for unipolar ( i.e less).
• For same Eb, reduce height for bipolar to A /
  1.414.
• For same BER reduce height for bipolar to A/2.
• (reduces power by 3dB in
  comparison to unipolar).
• If PB10-4, i.e. BER of 1 in 1000,
• ratio Eb/N0 must be 8.2dB for bipolar
  11.2dB for unipolar.
• If N0 is same, unipolar must have Eb 3dB higher
  than bipolar to achieve same BER.
• To have the same error probability

• which means that Eb(unipolar)
  2Eb(bipolar).
• This factor of two is an energy difference of 3dB.

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• These results will stand us in good stead when we
  look at
• modulated data
  transmission over a radio channel.
• Unipolar signalling is base-band orthogonal
  signalling.
• Bipolar signalling is base-band antipodal
  signalling.
• Results obtained for matched filter (or
  correlation) detection of
• bipolar unipolar signalling will be same for
  
• coherently detected single carrier modulated
  antipodal
• 
  signalling (e.g. binary PSK)
• orthogonal signalling (e.g. binary ASK),
• respectively.

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Investigate effect of choice of threshold on
unipolar signalling when probabilities of ones
and zeros are not equal. prob10.9 prob00.1
Voltages are 1 and 0 nstdev 1/4
Standard dev of AWGN nstdev 1/16 pemin 1
Becomes minimum bit-error
probability bestgamma0 Becomes best
thrshold between 0 1 for i1090
gammai/100 pe prob00.5erfc((gamma/nstdev)
/1.414) pe pe prob10.5erfc(((1-gamma)/n
stdev)/1.414) disp(sprintf('gammaf
peg',gamma,pe)) if peltpemin pemin pe
bestgammagamma end end disp(sprintf('bestgamma
f',bestgamma)) disp(sprintf('peming',pemin))
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• 2.9 ISI pulse shaping
• Rect symbols not suitable for band-limited
  channels.
• Time-limited pulse requires infinite bandwidth.
• Symbol with finite bandwidth must have infinite
  time duration.
• Although using symbols which exist from t -?
  to ? may seem
• impossible, this is the ideal we must
  produce approximations.
• Pulses used in practice may not go on back in
  time for ever.
• Must be non-zero for more than T s when
  signaling rate is 1/T .
• One symbol will run into previous next
  symbols.
• Result could be inter-symbol interference
  (ISI).
• Cannot avoid overlap of symbols in time-domain,
• Find ways of making sure that data carried by
  symbols is not
• 
  affected by this overlap.
• Solution to this challenge lies in pulse
  shaping

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• Generate an impulse of the appropriate strength
  for each symbol pass it thro a shaping
  filter.
• Impulse-response of filter is symbol shape we
  wish to launch.
• FIR digital filter followed by a DAC will do this
  job nicely.
• Channel will affect shape of symbol noise will
  be added.
• At receiver, to optimize detection, filtering
  tasks required.

Samples at intervals T
Matchd filter
Shaping Filter
Equal- iser
Channel
T T
n(t)
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• ISI can occur due to ringing of one symbol into
  next.
• ISI avoided if transmitter shapes symbols so that
  zero-crossings
• at detector occur T, 2T, ... after (
  before) centre of symbol.
• If we sample at t0, T, 2T, etc, only see centre
  of one symbol.
• All other symbols are zero at those instants.
• Nice in theory possible to a fair degree in
  practice.
• Let HN((?)) be freq-response of transmitting
  filter, channel
• 
  receiving filters all combined.
• Goal achieved if HN((?)) is Nyquist freq-resp,
  band-limiting to ?1/T Hz, with a form of
  odd-symmetry about 1/(2T ) in that

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t
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• Two purely real freq-respnses satisfying this
  property shown below.
• Guarantees that zero crossings occur at t ?T,
  ?2T, ?3T, .

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The brick-wall filter

• Has Nyquist freq-resp we know that its
  impulse-response is a
• sinc function with zero
  crossings at t?T, ?2T, etc.
• It is band-limiting to minimum possible
  bandwidth,
• Difficult to deal with because side-lobes of its
  "sinc" impulse
• 
  response do not die away too quickly.
• To get reasonable approximation to this filter,
  need very long
• impulse response
  hence high order FIR filter.
• Also, with such a pulse shape, if there is
  slightest error in timing
• of sampling
  point at detector, ISI will occur.

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• 2.10 Raised cosine frequency response
• Commonly used family of Nyquist freq-responses is
  raised cosine family parameterised by r (or ?)

• Impulse-response of such a filter may be shown to
  be

• When r0, this becomes brick-wall from -1/(2T)
  to 1/(2T) Hz
• When r 0.5, Hrc(f) is as shown below when
  T0.001 s.
• From general formula above, its spectrum is

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• Note "odd-symmetry" about f500 Hz for f gt0
• about f -500
  Hz - 1/(2T) for f lt0.
• It may be shown that Hrc((f))Hrc(( 1000 - f ) )
  1.
• When r1, Hrc((?)) is pure raised cosine shaped
  response.
• No flat-top widest bandwidth of family ?1/T
  Hz.
• With 1/T1 kHz, bandwidth would be -1000 to 1000
  Hz.

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• Impulse-responses hrc(t) corresponding to
  Hrc((?)) with r0 1
• 
  shown below taking T 10.
• Note reduction of side-lobe ripples when r1, at
  expense of
• 
  doubling the bandwidth.
• Raised cosine spectrally shaped pulse for given
  value of r is
• "100r RC symbol" has
  bandwidth ?(1r)/T Hz.
• When r0.5, we have a 50 RC spectrally shaped
  symbol with bandwidth from -750 Hz to 750 Hz if
  1/T 1 kHz..
• This is 50 more than the absolute minimum
  band-width needed to avoid ISI with the
  techniques discussed up to now.
• Minimum bandwidth would be achieved with a 0 RC
  symbol which has a "brick-wall" spectrum from
  -1/(2T) to 1/(2T) Hz..

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Bandwidth efficiency at baseband 0 RC
(brick-wall) 1/T b/s in 1/2T Hz 2b/s per Hz 50
RC 1/T b/s in 3/(4T) Hz 1.333 b/s per
Hz 100 RC 1/T b/s in 1/T Hz
1 b/s/Hz
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• Apart from the practical difficulties of
  generating a pulse with such a spectrum, the
  disadvantage of the brick-wall spectrum is that
  its time-domain shape is a "sinc" pulse which
  does not die away quickly enough for our liking.
• The 100 RC spectrum (r1) produces a time-domain
  pulse which dies away much faster.
• So by increasing r from 0 to 1 we improve the
  rate of dying away at the expense of extra
  bandwidth

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• Shaping filter is FIR digital filter with coeffs
  equal to samples of h(t).
• Succession of shaped pulses viewed on
  oscilloscope as eye-diagram.
• Scope triggered at beginning of each pulse.
• Open eye as clean new pulse superimposed on
  previous pulses.
• With noise, eye closes, making threshold
  detection difficult.
• HN((?)) is required overall response of
• transmitting
  filter, channel receiving filters.
• Receiving filters are
• Matched filter to minimise effect of AWGN
• Equalising filter to cancel out filtering effect
  of channel.

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• Pulse shape as seen at detector must have RC
  spectrum.
• Receiving filters include a matched filter.
• Not correct for transmitter to send symbols with
  RC spectra.
• If we did, matched filter would have
  freq-response equal to
• 
  complx conj of the RC spectrum.
• Resulting output would have squared magnitude of
  this spectrum.
• No longer be raised cosine but raised cosine
  squared.
• Not a problem as far as minimising effect of
  noise.
• But no good for eliminating ISI .
• Detector must see, not a 'raised cosine squared'
  spectrally shaped
• pulse but a raised
  cosine spectrally shaped pulse.

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• 2.11 Root raised cosine pulse shaping
• Distribute RC freq-response equally between
  transmitter
• 
  receiving matched filter.
• Make each 100r root raised cosine (RRC)
  freq-response.
• Transmitter sends RRC spectrally shaped symbols.
  
• In time-domain, RRC symbols not too dissimilar
  from RC ones.
• They exist for all time are symmetric about
  t0.
• But they do not have zero crossings at t ?T,
  ?2T, ?3T,

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• Chose value of r in range 0 to 1.
• Apply inverse FT to calculate corresponding
  impulse resp hrrc(t).
• Normally done in sampled data form.
• hrrc(t) sampled, windowed delayed (involves
  approximation)
• Gives impulse-response ( hence coeffs) of an FIR
  digital filter
• 
  of manageable order.
• Graph of Hrrc(f) against f when 1/T1000 Hz
  r0.5 next.

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• Note lack of "odd symmetry" about f ?500.
• Matched filter at receiver performs two roles at
  same time.
• First role is to minimise error probability PB
  as usual.
• Its impulse-response must equal s(t)
  time-reversed
• 
  suitably delayed.
• As s(t) s(-t) for RRC symbol, it also squares
  spectrum of s(t) to make it RC rather than RRC.
• As well as minimising PB, matched filter also
  completes the job of generating RC Nyquist
  freq-resp as required to eliminate ISI.
• Matched filter has impulse-response s(T-t) i.e.
  time-reversed symbol delayed by T.
• Have to delay this by several more intervals of
  T to allow some of s(-t) to be included without
  making filter non-causal.
• Must also delay decision until several intervals
  beyond t0 .
• Each symbol now extends beyond single interval of
  T s.

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Exercise Does a symbol whose spectrum is
'100r RC squared' have zero-crossings at t ?T,
?2T, ?3T, as required for zero ISI? Answer
Nope (except when r0). This is the problem.
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• Exercise Without calculating hrrc(t), i.e.
  symbol generated by exciting 100r RRC filter
  with an impulse, would you expect its
  zero-crossings to occur at t ?T, ?2T, ?3T, ?
• Answer Nope.
• Strange that this RRC symbol does not have
  zero-crossings required for zero ISI.
• Transmission along channel has ISI.
• When symbols received passed thro matched
  filter thus squaring RRC spectrum, condition for
  zero ISI is satisfied.

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• 2.12 Equalisation
• Channel distorts symbol-shape because of its
  non-ideal frequency-response.
• Noise added to the received version of symbol.
• Matched filter at receiver minimises effect of
  noise.
• Equaliser filter at receiver cancels out
  symbol-shape distortion due to frequency-response
  of channel.
• Frequency-response of a radio channel not ideal
  because of multi-path propagation.
• With wired channel problem is capacitance
  inductance of line.

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• To minimise ISI, product of freq-responses of
• pulse shaping filter at transmitter
• channel
• matched filter
• equaliser
• must be Nyquist.

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• To make product of freq-responses of all filters
  channel equal to HN((?)), generally RC, take
  its square root construct two RRC filters.
• First RRC filter acts as pulse-shaper
  band-limiting signal launched into channel.
• Receivers RRC filter completes RC shaping as
  well as being MF.
• Equaliser cancels frequency response of channel.

HN((?))


Detector
RRC filter
RRC Filter
Equaliser
Channel
T T
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Equalisation of channel may be achieved using
adaptive FIR transversal filter, (or tapped
delay line) 4th order example shown below
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• Such a filter can be made zero forcing
  transversal equaliser.
• It is FIR filter but note that delay boxes are
  not z-1 but z-T
• Looks at symbols received from RRC/channel/RRC
  combination.
• Adapts its coefficients C0, C1, C2, etc. such
  that, in response to an input waveform x(t)
  centred on t0, output y(t) has exact
  zero-crossings at t0, ?T, ?2T, ... .
• An Nth order transversal filter must delay centre
  of symbol by (N/2)T to do its job.
• Transversal filter shown above is of order N 4,
  with five taps labelled C0, C1, C2, C3, C4.
• Output is

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Given symbol x(t) from RRC/channel/RRC filtering
with centre at t 0. Assume effect of channel
makes x(?T) x(?2T) non-zero. To take an example
assume x(-2T) -0.04, x(-T) 0.2, x(0) 1.0,
x(T)0.2 and x(2T)-0.04.
x(t)
t
T
-T
-2T
2T
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• To simplify calculation, use 3-tap (2nd order)
  transversal filter.
• Force zero-crossings at t T -T.
• Centre of symbol is delayed occurs at tT.
• Output of 3-tap transversal filter is

• Output at t0, T, and 2T is as follows

• We wish to make y(0)0, y(T)1, y(2T)0.
  Therefore

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• Set of 3 linear simultaneous eqns in 3 unknowns.
  In matrix form

i.e. Ac b with solution c
A-1b Use MATLAB to find A-1 as follows A 1 .2
-0.04 0.2 1 0.2 -0.04 0.2 1 inv(A) 1.0490
-0.2273 0.0874 -0.2273 1.0909
-0.2273 0.0874 -0.2273 1.0490
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C0 1.049 -0.227 0.087
0 C1 -0.227 1.091 -0.227
1 C2 0.087 -0.227 1.049
0
Therefore C0 -0.227, C1 1.091. C2 -0.227