1 1 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen Data Communication, Lecture6 Digital Baseband Transmission 2 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen Digital Baseband Transmission Why to apply digital transmission? Symbols and bits Baseband transmission – Binary error probabilities in baseband transmission Pulse shaping – minimizing ISI and making bandwidth adaptation signaling – maximizing SNR at the instant of sampling - matched filtering – optimal terminal filters Determination of transmission bandwidth as a function of pulse shape – Spectral density of Pulse Amplitude Modulation (PAM) Equalization - removing residual ISI - eye diagram 3 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen Why to Apply Digital Transmission? Digital communication withstands channel noise, interference and distortion better than analog system. For instance in PSTN inter-exchange STP-links NEXT (Near-End Cross-Talk) produces several interference. For analog systems interference must be below 50 dB whereas in digital system 20 dB is enough. With this respect digital systems can utilize lower quality cabling than analog systems Regenerative repeaters are efficient. Note that cleaning of analog-signals by repeaters does not work as well Digital HW/SW implementation is straightforward Circuits can be easily reconfigured and preprogrammed by DSP techniques (an application: software radio) Digital signals can be coded to yield very low error rates Digital communication enables efficient exchanging of SNR to BW-> easy adaptation into different channels The cost of digital HW continues to halve every two or three years STP: Shielded twisted pair 4 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen Symbols and Bits 1 1 0 0 1 1 1 1 1 0 1 0 bi ( 1/ ) ts/s b b b T r T = bitrate ( 1/ ) D r D = symbol rate baud 2 n M = : : : number of bits : number of levels Symbol duration Bit duaration ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ b n M D T 2 log = n M () ( ) ∑ = − k k s t apt kD For M=2 (binary signalling): For non-Inter-Symbolic Interference (ISI), p(t) must satisfy: This means that at the instant of decision () ( ) ∑ = − k b k s t apt kT 1, 0 () 0, , 2 ... = ⎧ = ⎨ =± ± ⎩ t pt t D D () ( ) ∑ = − = k k k s t apt kD a Generally: (a PAM* signal) () s t *Pulse Amplitude Modulation unipolar, 2-level pulses
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1 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Data Communication, Lecture6
Digital Baseband Transmission
2 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Digital Baseband TransmissionWhy to apply digital transmission?Symbols and bitsBaseband transmission– Binary error probabilities in baseband transmission
Pulse shaping– minimizing ISI and making bandwidth adaptation signaling– maximizing SNR at the instant of sampling - matched filtering– optimal terminal filters
Determination of transmission bandwidth as a function of pulse shape– Spectral density of Pulse Amplitude Modulation (PAM)
3 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Why to Apply Digital Transmission?Digital communication withstands channel noise, interference and distortion better than analog system. For instance in PSTN inter-exchange STP-links NEXT (Near-End Cross-Talk) produces several interference. For analog systems interference must be below 50 dB whereas in digital system 20 dB is enough. With this respect digital systems can utilize lower quality cabling than analog systemsRegenerative repeaters are efficient. Note that cleaning of analog-signals by repeaters does not work as wellDigital HW/SW implementation is straightforwardCircuits can be easily reconfigured and preprogrammed by DSP techniques (an application: software radio)Digital signals can be coded to yield very low error ratesDigital communication enables efficient exchanging of SNR to BW-> easy adaptation into different channelsThe cost of digital HW continues to halve every two or three years
STP: Shielded twisted pair4 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Symbols and Bits
1 1 00 1 11 110 1 0
bi( 1/ ) ts/sb b bT r T=bitrate( 1/ )D r D=symbol rate baud
2nM =
:
::
number of bits: number of levelsSymbol durationBit duaration
⎧⎪⎪⎨⎪⎪⎩ b
nMDT
2log=n M
( ) ( )∑= −kk
s t a p t kDFor M=2 (binary signalling):
For non-Inter-Symbolic Interference (ISI), p(t) mustsatisfy:
This means that at the instant of decision
( ) ( )∑= −k bk
s t a p t kT
1, 0( )
0, , 2 ...=⎧
= ⎨ = ± ±⎩
tp t
t D D
( ) ( )∑= − =k kk
s t a p t kD a
Generally: (a PAM* signal)
( )s t
*Pulse Amplitude Modulation
unipolar, 2-level pulses
2
5 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
DigitalTransmission
‘Baseband’ means that no carrier wave modulation is used for transmission
6 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Baseband Digital Transmission Link
( ) ( ) ( )∑= − − +k dk
y t a p t t kD n t
( ) ( ) ( )≠∑= + − +%K k kk K
y t a a p KD kD n t
message reconstruction at yields= +K dt KD t
message ISI Gaussian bandpass noise
Uni
pola
r PAM original message bits
decision instances
received wave y(t)
Dt
7 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Baseband Unipolar Binary Error Probability
r.v. : ( ) ( )= +k k kY y t a n tThe sample-and-hold circuit yields:
0
0
: 0,( | ) ( )
= ==
k
Y N
H a Y np y H p y
Establish H0 and H1 hypothesis:
1
1
: 1,( | ) ( )
= = += −
k
Y N
H a Y A np y H p y A
and
pN(y): Noise spectral density
Assume binary & unipolar x(t)
8 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Determining Decision Threshold0
0
: 0,( | ) ( )
= ==
k
Y N
H a Y np y H p y
1
1
: 1,( | ) ( )
= = += −
k
Y N
H a Y A np y H p y A
Choose Ho (ak=0) if Y<VChoose H1 (ak=1) if Y>V
The comparator implements decision rule:
1 1 1
0 0
( | ) ( | )( | ) ( | )
V
e Y
Veo Y
p P Y V H p y H dyp P Y V H p y H dy
−∞
∞
∫
∫
≡ < =
≡ > =
Average error error probability: 0 0 1 1= +e e eP P P PP1
20 1 0 11/ 2 ( )= = ⇒ = +e e eP P P P PAssume Gaussian noise: 2
2
1( ) exp22N
xp xσσ π
⎛ ⎞= −⎜ ⎟⎝ ⎠
Transmitted ‘0’but detected as ‘1’
3
9 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Determining Error Rate
0 σ⎛ ⎞= ⎜ ⎟⎝ ⎠e
Vp Q
2
0 2
1 exp22 Ve
xp dxσσ π
∞
∫⎛ ⎞= −⎜ ⎟⎝ ⎠
21( ) exp22 k dQ k λλ
π∞ ⎛ ⎞= − ⇒⎜ ⎟
⎝ ⎠∫
2
0 2
1 exp22
σ σσ σπ
∞
∫⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Ve
x Vp dx Q
that can be expressed by using the Q-function, defined by
and therefore
and also
0 ( )∞
∫= Ve Np p y dy
1 ( )σ−∞∫−⎛ ⎞= − = ⎜ ⎟
⎝ ⎠V
e N
A VP p y A dy Q
10 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
11 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen 12 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Baseband Binary Error Rate in Terms of Pulse Shape and γ
0 0 / 2R N bN N B N r= ≥Note that (lower limit with sinc-pulses (see later))
4
13 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Pulse Shaping and Band-limited Transmission
In digital transmission signaling pulse shape is chosen to satisfy the following requirements:– yields maximum SNR at the time instance of decision
(matched filtering)– accommodates signal to channel bandwidth:
• rapid decrease of pulse energy outside the main lobe in frequency domain alleviates filter design
• lowers cross-talk in multiplexed systems
14 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Signaling With Cosine Roll-off Signals
Maximum transmission rate can be obtained with sinc-pulses
However, they are not time-limited. A more practical choice is the cosine roll-off signaling:
( ) ( )( ) sinc sinc /1( ) [ ( )]
p t rt t DfP f F p t
r r
= =⎧⎪⎨ ⎛ ⎞= = Π⎜ ⎟⎪ ⎝ ⎠⎩
2
2( ) sinc1 (4 )cos tp t rt
tπββ
=−
2
/ 2
1( ) cos ( / 2 )2β
π=
= Πr
fP f f rr r
for raised cos-pulses β=r/2
15 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
ExampleBy using and polar signaling, the following waveform is obtained:
Note that the zero crossing are spaced by D at
(this could be seen easily also in eye-diagram)The zero crossing are easy to detect for clock recovery.Note that unipolar baseband signaling involves performance penalty of 3 dB compared to polar signaling:
/ 2β = r
0.5 , 1.5 , 2.5 ,....t D D D= ± ± ±
( ), unipolar [0 /1]
( 2 ), polar[ 1]b
e
b
Qp
Q
γ
γ
⎧⎪= ⎨±⎪⎩
16 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Matched Filtering
0
0
( ) ( )( ) ( )exp( )
R R
R R
x t A p t tX f A P f j tω
= −⎧⎨ = −⎩
2 22( ) ( )∞ ∞
−∞ −∞∫ ∫= =R R RE X f df A P f df
( )0
1[ ( ) ( )]
( ) ( )exp
dR
R d
t t tA F H f X f
H f P f j t dfA ω
−
∞
−∞
= +
∫
=
=
2 22 ( ) ( ) ( )2nH f G f df H f dfησ
∞ ∞
−∞ −∞∫ ∫= =
( )2
2
2
2
( ) ( )exp
( )2
d
R
H f P f j t dfA AH f df
ω
ησ
∞
−∞
∞
−∞
∫
∫
⎛ ⎞ =⎜ ⎟⎝ ⎠
H(f)H(f)++( )Rx t
( )nG f( )Dy t
Should be maximized
Post filter noise
Peak amplitude to be maximized
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17 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Matched Filtering SNR and Transfer Function
( )
2
2
2
2
2
2
2
( ) *( )( )
( )
( ) ( )exp
( )2
d
R
V f W f dfW f df
V f df
H f P f j t dfA AH f df
ω
ησ
∞
∞−∞
∞−∞
−∞
∞
−∞
∞
−∞
∫∫
∫
∫
∫
⎫⎪≤ ⎪⎪⎪⎬⎪⎛ ⎞ ⎪=⎜ ⎟
⎝ ⎠ ⎪⎪⎭
222 22
2 ( )2 ( )R
R
MAX
A P f dfA A W f dfσ η η
∞
∞−∞
−∞
∫∫
⎛ ⎞ = =⎜ ⎟⎝ ⎠
Schwartz’s inequalityapplies when
SNR at the moment ofsampling
Considering righthand side yields max SNR
( ) *( )V f KW f=
impulse response is:
( ) ( )*( ) ( )exp( )
( ) ( )exp( )( ) ( )
d
d
d
V f H fW f P f j t
H f KP f j th t Kp t t
ωω
= ⎫⎬= ⎭
⇒ =⇒ = −
pulseenergy
18 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
raised cos-spectra CN(f)20 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Determining Transmission Bandwidth for an Arbitrary Baseband Signaling Waveform
Determine the relation between r and B when p(t)=sinc2 atFirst note from time domain that
hence this waveform is suitable for signalingThere exists a Fourier transform pair
From the spectra we note thatand hence it must be that for baseband
21, 0
sinc0, 1/ , 2 / ...
=⎧= ⇒ =⎨ = ± ±⎩
tat r a
t a a
2 1sinc fata a
⎛ ⎞↔ Λ⎜ ⎟⎝ ⎠
−a a
1/ a
TB a≥ f
TB r⇒ ≥
6
21 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
PAM Power Spectral Density (PSD)PSD for PAM can be determined by using a general expression
For uncorrelated message bits
and therefore
on the other hand and
21( ) ( ) ( )exp( 2 )π∞
=−∞∑= −x a
nG f P f R n j nfD
D
Amplitude autocorrelation
2 2
2
, 0( )
, 0σ + =⎧
= ⎨ ≠⎩a a
a
a
m nR n
m n
2 2( )exp( 2 ) exp( 2 )π σ π∞ ∞
=−∞ =−∞∑ ∑− = + −a n a
n nR n nfD m j nfD
1exp( 2 )n n
nj nfD fD D
π δ∞ ∞
=−∞ =−∞∑ ∑ ⎛ ⎞− = −⎜ ⎟
⎝ ⎠2 22 2 2( ) ( ) ( ) ( )x a a
nG f r P f m r P nr f nrσ δ
∞
=−∞∑= + −
1/=r D
Total power
DC power
22 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
ExampleFor unipolar binary RZ signal:
Assume source bits are equally alike and independent, thus
1( ) sinc2 2
=b b
fP fr r
/ 22 2 2 2 2
0(1/ 2 ) / 4,
bT
a b a aT A dt A mσ σ∫= = =2 2
2 2( ) sinc ( )sinc16 2 16 2x b
nb b
A f A nG f f nrr r
δ∞
=−∞∑⇒ = + −
22
22 2
( ) ( )
( ) ( )x a
an
G f r P f
m r P nr f nr
σ
δ∞
=−∞∑
=
+ −
22 1
4 2b
b
A rr
⎛ ⎞⎜ ⎟⎝ ⎠
23 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Equalization: Removing Residual ISIConsider a tapped delay line equalizer with
Search for the tap gains cN such that the output equals zero at sample intervals D except at the decision instant when it should be unity. The output is (think for instance paths c-N, cN or c0)
that is sampled at yielding
( ) ( )=−∑= − −%N
eq nn N
p t c p t nD ND
[ ]( ) ( ) ( )N N
eq n nn N n N
p kD ND c p kD nD c p D k n=− =−∑ ∑+ = − = −% %
= +kt kD ND
( ) ( 2 )N Np t c p t ND= −%
( ) ( )N Np t c p t− −
= %
0 ( ) ( )Np t c p t ND−
= −%
24 Helsinki University of Technology,Communications Laboratory, Timo O. Korhonen
Tapped Delay Line: Matrix Representation
At the instant of decision:
That leads into (2N+1)x(2N+1) matrix where (2N+1) tap coefficients can be solved: