digital communications lecture 6new · delta modulation problems • there are two problems in dm...
TRANSCRIPT
EE3220, Digital CommunicationsLecture 6
Digital Representation of Analog Signals, Part 3
Dr. Essam Sourour
2
Delta Modulation
• We have seen that Pulse Code Modulation typically generates bit rate of 64 kpbs– Sampling at 8 ksps– Each sample is converted to 8 bits
• PCM does not take advantage of the high correlation between successive samples in analog audio signal
• Another method of converting the analog signal to digital signal is Delta Modulation– Sampling at higher rate, about 32 ksps– Samples are highly correlated, i.e., small change from sample to next sample– Compare each sample to the previous sample– Generate 1 if higher and generate 0 if lower– Hence, each sample is converted to 1 bit– Bit rate is 32 kbps (50% saving over PCM in this example)
3
Delta Modulation
• DM works well if the changes between successive samples is small
4
Delta Modulation
5
Delta Demodulation
• To get back the analog signal we accumulate the delta steps• After accumulating the delta steps a low pass filter is needed
to smooth the signal and get back the analog signal
6
Delta Modulation problems
• There are two problems in DM due to the value of the step 1. Slope overload distortion:
– Occurs when Δ is smaller than the maximum slope of analog signal
– Mathematically, it occurs when: ∆
2. Granular noise:– Occurs when the analog signal amplitude is not changing– The DM output is going up and down– If Δ is too large noise is higher
7
Delta Modulation problems, cont.
8
Baseband pulses and matched filter
• In the previous lecture we have seen the pulse representation of the binary bits
• Each bit (0 or 1) is represented by a pulse • The received signal x(t) is given by:
, 0 t T• The term w(t) is the additive white Gaussian noise (AWGN)
with flat (i.e., white) power spectral density 2ڥ The receiver consists of a filter h(t) followed by sampling
every bit at t = nT
9
Matched filter
• Our goal is to find the receiver filter h(t) that maximizes the signal to noise ratio (SNR) at its output:
∗ ∗
• After sampling at time t=T:– The power of the useful signal part is – The power of the noise part is , where E[ ] is expectation
• The SNR at the output of the filter is
10
Properties of the Matched filter
• The matched filter provides maximum SNR• In time domain: Time reversed delayed version of the transmitted pulse• In frequency domain: conjugate of the Fourier Transform of g(t), multiplied
by 2
• The output SNR=2E/No , where E is the energy in the pulse g(t). The SNR does not depend on the specific function g(t).
• SNR depends only on the energy of g(t).
* exp 2
opt
opt
h t g T t
H f G f j fT
11
Simple examples
12
Matched filter for the rectangular pulse
• Consider the special case of a rectangular pulse• Using the matched filter hopt(t)
opt
opt
y t x t h t
x u h t u du
0
opt
T
y T x u h T u du
A x u du
u
x(u)
hopt(T‐u)
u
T
T
A
A
13
Matched filter for the rectangular pulse
• Hence, the output of the matched filter at time T of the received signal x(t) is given by:
• This is equivalent to integrating the received signal x(t) from 0 to T • This is called “integrate and dump” receiver• Hence, in case of a rectangular pulse we can replace the matched filter by
integrate and dump• We can ignore the scale factor A since it affects both signal and noise
0
T
y T A x u du
AWGN w(t)
signal g(t) x(t)Integrator
Sample at time t=nT
y(t) y(nT)
14
Probability of error
• We want to calculate the BER at the output of the matched filter• We assume a NRZ signaling with square pulse• Hence, we use an integrate and dump receiver• The received signal x(t) is given by:
• Assume symbols 0 or 1 can be sent with equal probability.• Since symbol 1 and symbol 0 are sent with equal power (power=A2) they
have the same probability of error• Hence, assume symbol 1 is sent:
if symbol 1 is sentif symbol 0 is sent
A w tx t
A w t
AWGN w(t)
signal g(t) x(t)Integrator
Sample at time t=nT
y(t) y(T)Decision
Say 1 if y>0
Say 0 if y<0
15
Probability of error, cont.
• The integrate and dump output is:
• Since w(t) is Gaussian with zero mean, then the noise part v is Gaussian with zero mean
• We need to find the variance of v :
• Then, y is a Gaussian random variable with mean AT and variance NoT/2
0 0
T T
signal
noise v
y x t dt AT w t dt
2
0 0
0 0
0 0 2
2
T T
T T
T To
o
E v E w t w u dt du
E w t w u dt du
N t u dt du
N T
16
Probability of error, cont.
• Then
• The noise v is Gaussian with zero mean and variance 2⁄• The probability of error is the probability that y < 0
– (i.e., we say 0 is transmitted while actually 1 is transmitted)
• Error happens when y = AT + v < 0• Then, error happens when v < ‐AT• This probability of error is Pr(v < ‐AT) = Pr(v > AT)• Note that for any Gaussian random variable X we have
• The energy per bit is Eb=A2T 2 2be bit
o
EP Q QN
X
X
xP X x Q
Pr Pr
2o
ATerror v AT QN T
22Pro
A Terror QN
17
BER of the Matched Filter
18
Intersymbol Interference
• So far we assumed the channel effect is to add AWGN only• However, some channels have filtering effect Hc(f), with
impulse response hc(t), that distorts the transmitted pulse shape g(t)
• This causes intersymbol interference (ISI)• The overall pulse shape ∗ ∗ may
overlap at the sampling time nT, which causes ISI
19
Intersymbol Interference, cont.
• At the sampling time at the receiver (T, 2T, 3T, …) there is overlap between successive pulse shapes
• This causes ISI and many errors may happen• ISI can be observed using the eye diagram
20
Examples without ISI
t
+V +V +V +V
‐V ‐V
T 2T 3T 4T 5T 6T
No ISI
t
+V +V +V +V
‐V ‐V
T 2T 3T 4T 5T 6T
No ISI
T 2T 3T 4T 5T 6T
t
No ISI
21
The eye diagram without ISI
• The eye pattern is the superposition of all possible realization of the signal for a particular interval
• The sampling time nT should be open• This is eye diagram without ISI and “raised cosine pulse” (explained next)
22
Example with ISI
• At the sampling time nT there is pulse overlap from previous and next pulses
• This causes large ISI which can cause many errors in detecting the transmitted bits
t
T 2T 3T 4T 5T 6T
Large ISI
23
The eye diagram with ISI
• With ISI the eye diagram starts to close
24
Nyquist criteria
• For ISI free transmission the pulse shape must have zero value at the nT, , 2 , 3 , …
• The overall pulse shape must satisfy the Nyquist criteria:
1 00 0
np nT
n
25
Nyquist filter
• In many cases we know the channel impulse response hc(t)• We can design the overall pulse shape ∗ ∗ such
that no ISI occurs• This means that at the sampling time (T, 2T, 3T, …) there is no ISI• This overall pulse shape p(t) is called Nyquist filter or Nyquist pulse• The most important and commonly used is the raised cosine pulse shape
26
Raised Cosine Pulse Shape
• The raised cosine pulse shape satisfies Nyquist requirements that no ISI occurs
• Note that at the sampling instant (t=t, 2T, 3T, …) this pulse =0• Hence, if the overall channel impulse response p(t) is a raised cosine
shape, it does not cause ISI • The raised cosine pulse shape has a parameter called rolloff factor• The rolloff factor r decides the BW of the pulse shape
27
Raised Cosine Pulse Shape, cont.
• The minimum bandwidth is when the rolloff factor r=0– Minimum BW = 1/2T
• The maximum bandwidth is when the rolloff factor r=1– Maximum BW = 1/T
• In general the bandwidth is given by12 1
28
Raised Cosine Pulse Shape, cont.
• The bandwidth is 1
• Wo=1/2T• The Raised cosine frequency response is:
• The impulse response is:
• The rolloff factor:
2
1 for 2
2cos for 2
40 for
o
oo
o
f W W
f W WP f W W f W
W Wf W
2
cos 22 sinc 2
1 4o
o o
o
W W tp t W W t
W W t
, 0 1o or W W W r sin
sinc( )x
xx