Transcript Slide 1

Error rate due to noise
In this section, an expression for the
probability of error will be derived
The analysis technique, will be
demonstrated on a binary PCM based on
polar NRZ signaling
The channel noise is modeled as additive
white Gaussian noise 𝑤(𝑡) of zero mean
𝑁0
and power spectral density
2
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Bit error rate
The receiver of the communication system
can be modeled as shown below
x(t)
y(t)
Y
The signaling interval will be 0 ≤ 𝑡 ≤ 𝑇𝑏 ,
where 𝑇𝑏 is the bit duration
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Bit error rate
The received signal can be given by
+𝐴 + 𝑤 𝑡 , 𝑠𝑦𝑚𝑏𝑜𝑙 1 𝑤𝑎𝑠 𝑠𝑒𝑛𝑡
𝑥 𝑡 =
−𝐴 + 𝑤 𝑡 , 𝑠𝑦𝑚𝑏𝑜𝑙 0 𝑤𝑎𝑠 𝑠𝑒𝑛𝑡
It is assumed that the receiver has
acquired knowledge of the starting and
ending times of each transmitted pulse
The receiver has a prior knowledge of the
pulse shape, but not the pulse polarity
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Receiver model
The structure of the receiver used to
perform this decision making process is
shown below
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Possible errors
When the receiver detects a bit there are
two possible kinds of error
1.
2.
Symbol 1 is chosen when a 0 was actually
transmitted; we refer to this error as an error
of the first kind
Symbol 0 is chosen when a 1 was actually
transmitted; we refer to this error as an error
of the second kind
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Average probability of error
calculations
To determine the average probability of
error, each of the above two situations will
be considered separately
If symbol 0 was sent then, the received
signal is
𝑥 𝑡 = −𝐴 + 𝑤 𝑡 , 0 ≤ 𝑡 ≤ 𝑇𝑏
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Average probability of error
calculations
The matched filter output will be given by
1 𝑇𝑏
𝑦=
𝑥 𝑡 𝑑𝑡
𝑇𝑏 0
1 𝑇𝑏
𝑦 = −𝐴 +
𝑥 𝑡 𝑑𝑡 = −A
𝑇𝑏 0
The matched filter output represents a
random variable 𝑌
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Average probability of error
calculations
This random variable is a Gaussian
distributed variable with a mean of −𝐴
The variance of 𝑌 is
𝜎2
𝑌
=
𝑁0
2𝑇𝑏
The conditional probability density function
the random variable 𝑌, given that symbol 0
was sent, is therefore
𝑓𝑌 𝑦 0 =
1
𝜋𝑁0 /𝑇𝑏
𝑒
𝑦+𝐴 2
−
𝑁0 𝑇𝑏
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Average probability of error
calculations
The probability density function is plotted
as shown below
To compute the conditional probability,
𝑝10 , of error that symbol 0 was sent, we
need to find the area between 𝜆 and ∞
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In mathematical notations
𝑝10 = 𝑃 𝑦 > 𝜆 𝑠𝑦𝑚𝑏𝑜𝑙 0 𝑤𝑎𝑠 𝑠𝑒𝑛𝑡
∞
𝑝10 =
𝑓𝑦 𝑦 0 𝑑𝑦
𝜆
𝑝10 =
1
𝜋𝑁0 /𝑇𝑏
∞
𝑒
𝑦+𝐴 2
−
𝑁0 /𝑇𝑏
𝑑𝑦
𝜆
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The integral
2
erfc(𝑧) =
𝜋
∞
𝑒
−𝑧2
𝑑𝑧
𝜆
Is known as the complementary error
function which can be solved by numerical
integration methods
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By letting 𝑧 =
𝑦+𝐴
𝑁0 /𝑇𝑏
, 𝑑𝑦 =
became
𝑝10
1
=
𝜋
𝑝10
𝑁0 /𝑇𝑏 𝑑𝑧, 𝑝10
∞
𝑒
−𝑧2
𝑑𝑧
𝐴+𝜆
𝑁0 /𝑇𝑏
1
= 𝑒𝑟𝑓𝑐
2
𝐴+𝜆
𝑁0 /𝑇𝑏
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By using the same procedure, if symbol 1
is transmitted, then the conditional
probability density function of Y, given that
symbol 1 was sent is given
𝑓𝑌 𝑦 1 =
1
𝜋𝑁0 /𝑇𝑏
𝑒
𝑦−𝐴 2
−
𝑁0 𝑇𝑏
Which is plotted in the next slide
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The probability density function 𝑓(𝑌|1) is
plotted as shown below
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The probability of error that symbol 1 is
sent and received bit is mistakenly read as
0 is given by
∞
𝑝01 =
𝑓𝑦 𝑦 1 𝑑𝑦
𝜆
𝑝10 =
1
𝜋𝑁0 /𝑇𝑏
∞
𝑒
𝑦−𝐴 2
−
𝑁0 /𝑇𝑏
𝑑𝑦
𝜆
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Expression for the average
probability of error
𝑝10
1
= 𝑒𝑟𝑓𝑐
2
𝐴−𝜆
𝑁0 /𝑇𝑏
The average probability of error 𝑃𝑒 is given
by
𝑃𝑒 = 𝑃0 𝑃10 + 𝑃1 𝑃01
𝑃0
𝐴+𝜆
𝑃1
𝐴−𝜆
𝑃𝑒 = 𝑒𝑟𝑓𝑐
+ 𝑒𝑟𝑓𝑐
2
2
𝑁0 /𝑇𝑏
𝑁0 /𝑇𝑏
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In order to find the value of 𝜆 which
minimizes the average probability of error
we need to derive 𝑃𝑒 with respect to 𝜆 and
then equate to zero as
𝜕𝑃𝑒
=0
𝜕𝜆
From math's point of view (Leibniz’s rule)
𝜕erfc(𝑢)
1 −𝑢2
=−
𝑒
𝜕𝑢
𝜋
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Optimum value of the threshold 𝜆
value
If we apply the previous rule to 𝑃𝑒 , then we
may have the following equations
2
(𝐴+𝜆)
𝑝0 − 𝑁 𝑇
𝑒 0 𝑏
𝜕𝑃𝑒
1
=−
𝜕𝜆
𝜋 𝑁0 𝑇𝑏 2
+
1
2
(𝐴−𝜆)
𝑝1 − 𝑁 𝑇
𝑒 0 𝑏
𝜋 𝑁0 𝑇𝑏 2
=0
By solving the previous equation we may
have
𝑝0
=
𝑝1
(𝐴−𝜆)2
−
𝑒 𝑁0 𝑇𝑏
(𝐴+𝜆)2
−
𝑒 𝑁0 𝑇𝑏
𝑓𝑦 (𝑦|1)
=
𝑓𝑦 (𝑦|0)
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From the previous equation, we may have
the following expression for the optimum
threshold point
𝑁0
𝑝0
𝜆𝑜𝑝𝑡 =
𝑙𝑛
4𝑇𝑏
𝑝1
If both symbols 0 and 1 occurs equally in
1
the bit stream, then 𝑝0 = 𝑝1 = , then
1. 𝜆𝑜𝑝𝑡 = 0
2
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1
2. 𝑃𝑒 = 2 𝑒𝑟𝑓𝑐
𝐴
𝑁0 /𝑇𝑏
1
= 𝑒𝑟𝑓𝑐
2
𝐴2
𝑁0 /𝑇𝑏
1
= 𝑒𝑟𝑓𝑐
2
𝐴2 𝑇𝑏
𝑁0
1
= 𝑒𝑟𝑓𝑐
2
𝐸𝑏
𝑁0
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Average probability of error vs
𝐸𝑏
𝑁0
A plot of the average probability of error as
a function of the signal to noise energy is
shown below
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As it can be seen from the previous figure,
the average probability of error decreases
exponential as the signal to noise energy
is increased
For large values of the signal to noise
energy the complementary error function
can be approximated as
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Example1
Solution
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Example1 solution
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Example 2
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Example 2 solution
a) The average probability of error is given by 𝑃𝑒 =
1
2
erfc
𝐸𝑏
𝑁0
where 𝐸𝑏 = 𝐴2 𝑇𝑏 ,
1
2
We can rewrite 𝑃𝑒 = erfc
1
2
erfc
2𝐴
𝜎
, where 𝜎 =
𝐴2 𝑇𝑏
𝑁0
1
2
= erfc
2𝐴2
𝑁
2𝑇 0
𝑏
=
𝑁0
2𝑇𝑏
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Example 2 solution
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Example 2 Solution
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Example 2 solution
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Example 2 solution
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Example 2 solution
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Example 3
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Example 3 solution
Signaling with NRZ pulses represents an
example of antipodal signaling, we can use
1 × 10
1
𝑃𝑒 = 𝑒𝑟𝑓𝑐
2
𝐸𝑏
𝑁0
1
= 𝑒𝑟𝑓𝑐
2
𝐴2 (
−3
1
)
56000
10−6
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Example 3 solution
Using the erfc(𝑥) table we find
1
2
𝐴
56000 = 2.18
10−6
𝐴2 = 0.266
With 3-dB power loss the power at the
transmitting end of the cable would be
twice the power at the receiving terminal,
this means that 𝑃𝑡𝑥 = 2𝑃𝑟𝑥 = 0.532 𝑤𝑎𝑡𝑡
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