The paper presents a tutorial on using Excel functions q(z) and erfc(x) to accurately estimate error probabilities in communication systems, addressing confusions often faced by students. It emphasizes the advantages of using Excel for these calculations over traditional methods, providing several mathematical examples to illustrate the proper application of these functions. By comparing results from examples in existing textbooks, the paper highlights the effectiveness of the proposed methods in achieving accurate estimations.

1. > PAPER IDENTIFICATION NUMBER TBD< 1
Abstract—Using several mathematical examples from
three different authors in texts from different courses this
paper illustrates the easier way to avoid confusions and
always get the correct results with the least effort was to
use the proposed Excel Gamma function explained in
detail for the proper use of the Q(z) and ercf(x) functions
in most communication courses. The paper serves as a
tutorial and introduction for such functions.
Index Terms—Bit error rate, BPSK, clear sky, erfc(x), M-
ary ASK, M-ary PSK, QPSK, Q(z).
I. INTRODUCTION
THE use of two different Excel functions will be explained in
order to properly address the possible confusion of the two
main approaches to estimate error probabilities.
Excel provides an excellent way to understand the
differences between the Q(z) and the erfc(x) functions. In the
main textbooks the correct calculation of this error probability
is either dismissed as an easy task or just plainly ignored,
authors mostly referring to old tables in their appendices and
resolving to interpolation to get the estimate or just telling the
student to use the function available in any software being
used in class.
I have found this could be highly confusing for the new
communication systems student just learning a lot of technical
material. It could be also potentially disastrous or costly as the
incorrect estimation could lead to overdesign a power
requirement or incorrect selection of the proper S/N figure.
II. PAPER ORGANIZATION
The paper will first discuss the two functions mathematical
formulas: Q(z) and erfc(x). There, the two approaches using
Excel formulas will be introduced. Afterwards, several
mathematical examples form main textbooks will be detailed
to illustrate the proper use of both approaches and they will be
compared against textbook’s estimates as well as interpolation
estimates, when applicable. The use of Excel in this paper as
the main estimation tool in this paper is just because of its
wide availability and simplicity as well as its ease handling of
extended significant figures.
A. The Q(z) function
Lathi (1998) defines the Q(z) as the Cumulative Density
Submitted for review.
x
Function (CDF) upper tail of the famous standardized
Gaussian function, F(y):
𝐹(𝑧) =
1
√2𝜋
∫ 𝑒
−𝑧2
2 𝑑𝑧
𝑧
−∞
(1)
Thus, Q(z) would be calculated as:
𝑄(𝑧) = 1 − 𝐹(𝑧) = 1 −
1
√2𝜋
∫ 𝑒
−𝑧2
2 𝑑𝑧 =
𝑧
−∞
𝑃𝑒 (2)
F(*) is standardized in the sense its mean, =0 and its
standard deviation, =1 (this, of course implies its variance,
2
= 1 also). This function must be numerically evaluated to
estimate its resulting error probability (𝑃𝑒) as it does not have
an analytical closed form. Until a few decades ago, this was a
major task that was mostly tackled by interpolating tables in
the back of textbooks or mathematical handbooks, by using
statistical software or coding. Nowadays, it could be easily
estimated with the Excel functions:
𝐹(𝑧) = 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(𝑧, 1) = 𝑃′ 𝑒 (3)
and,
𝐹(𝑃′ 𝑒)−1
= 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(𝑃′ 𝑒) = 𝑧 (4)
Where the Excel function 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(𝑧, 1) requires its
second argument to be either 0 or 1. We will not discuss the
cases where the second argument is zero here as we are only
concerned with the estimation of the CDF and, in order to
obtain such CDF estimation the second argument must be
always be 1. That is, the use of 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(𝑧, 1) will
estimate the normal standardize CDF for the value z. This is,
of course, 𝑃′ 𝑒; the cumulative probability from (-∞, z). Its
inverse function 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(𝑃′ 𝑒) provides us with the z
value when the known quantity is 𝑃′ 𝑒. Note that 𝑃𝑒 + 𝑃′ 𝑒 = 1.
It stands to logic that we will need to use (2) in Excel by
just simple subtracting the result from 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(𝑧, 1)
from 1.
B. The erfc(x) function
Lathi (1998) defines the erfc(*) function in terms of the
Q(z) one:
𝑒𝑟𝑓𝑐(𝑥) =
2
√𝜋
∫ 𝑒−𝑦^2
𝑑𝑦 = 2 ∗ 𝑄(𝑥√2)
∞
𝑥
(5)
For reasons that will become apparent in the examples,
instead of performing the change of variables and use the
Q(*), this paper will use the Excel function to define:
Understanding the Differences between the erfc(x) and the Q(z) functions: A Simple Approach using Excel
X

2. > PAPER IDENTIFICATION NUMBER TBD< 2
𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2
(6)
Where the second argument, (shape parameter)  = 0.5 and
the third argument, (scale parameter) 𝛽 = 1 were chosen to
approximate the normal standard distribution and the fourth
parameter is always set to 1 to estimate the CDF. It is quite
clear that:
𝑄(𝑧) = 𝑒𝑟𝑓𝑐(𝑥) (7)
In case we have the error probability, 𝑃′ 𝑒, and need the x
value, we must use, in general:
𝑥 = 𝐺𝐴𝑀𝑀𝐴. 𝐼𝑁𝑉((1 − 𝑃′ 𝑒)^2,0.5,1) (8)
But some adjustments must be made to fit the particular
case, as we shall see. The second argument, (shape parameter)
 = 0.5 and the third argument, (scale parameter) 𝛽 = 1 were
chosen to approximate the normal standard distribution as
before.
III. MATHEMATICAL EXAMPLES
Textbooks examples were selected from Lathi (1998),
Roddy (2006) and Pratt (2003) for their different applications
of the mathematical formulas. Lathi is more concerned with
communications systems in general but Roddy and Pratt are
more interested in satellites applications.
A. MATH EXAMPLE 1: PRATT EXAMPLE 5.4.1, PAGE 195
The author is interested in obtaining the system BER (bit
error rate) for different situations (clear sky/air, rain
attenuation) with a given C/N (carrier to noise). That is, we
need to estimate the 𝑃′ 𝑒 for a given C/N - which will be the z
or x value, depending on which Excel formula is applied.
i) BER in clear sky/air condition for BPSK modulation with
a given C/N= 14 dB implying a power ratio 10
14
10
25.11886432.
Our first step will be determining what exactly would be the
values of z or x to use.
For BPSK modulation:
𝑧 = √2 ∗
𝐶
𝑁
(9)
√2 ∗ (25.11886432)√50.2377286 7.087857831
and,
𝑥 = 𝑧/√2 (10)

7.087857831
√2

7.087857831
1.414213562
5.011872336
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(7.087857831)
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(7.087857831,1) 𝟔. 𝟖𝟏𝟎𝟏𝟎𝟖𝐄 − 𝟏𝟑
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(5.0118723362,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝟐𝟓.𝟏𝟏𝟖𝟖𝟔𝟒𝟑,0.5,1,1)
2
 𝟔. 𝟖𝟏𝟎𝟏𝟎𝟖𝐄 − 𝟏𝟑
It must be noted that, in this case, the use of (6) is
straightforward as the input variable is the power ratio
calculated. It should also be noted that both approaches
provide the same result for 𝑃𝑒.
Standard linear interpolation using Pratt’s Table of Q
function (page 505) for the textbook solution of Q(7.07) gives
𝑃𝑒 8.21E-13. Text must have a typo as it states the result as
7.8E-11.
ii) BER in clean air conditions, QPSK with a given C/N= 14
dB implying a power ratio 10
14
10 25.11886432.
Our first step will be determining what exactly would be the
values of z or x to use.
For QPSK modulation:
𝑧 = √
𝐶
𝑁
(11)
√(25.11886432)5.011872336
and,
𝑥 = 𝑧/√2 (10)

5.0118723361
√2

5.011872336
1.414213562
3.543928915
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(5.011872336 )
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(5.011872336 ,1) 2.695148E-07
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(3.5439289152,0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(12.55943216,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(
𝟐𝟓.𝟏𝟏𝟖𝟖𝟔𝟒𝟑
2
,0.5,1,1)
2
 2.695148E-07
It must be noted that, in this case, the use of (6) is, again,
straightforward as the input variable is the power ratio
calculated divided by 2. It should also be noted that both
approaches provide the same result for 𝑃𝑒.
Standard linear interpolation using Pratt’s Table of Q

3. > PAPER IDENTIFICATION NUMBER TBD< 3
function (page 505) for the textbook solution of Q(5) from
Pratt table page 505 = 2.872E-07.
iii) BER with rain attenuation of 3 dB, BPSK with a given
now of C/N= 11 dB implying a power ratio 10
11
10
12.58925412.
Our first step will be determining what exactly would be the
values of z or x to use.
For BPSK modulation:
𝑧 = √2 ∗
𝐶
𝑁
(9)
√2 ∗ (12.58925412)√25.17850824 5.017819072
and,
𝑥 = 𝑧/√2 (10)

5.017819072
√2

5.017819072
1.414213562
3.548133892
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(5.017819072)
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(5.017819072,1) 𝟐. 𝟔𝟏𝟑𝟎𝟔𝟖𝐄 − 𝟎𝟕
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(3.5481338922,0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(𝟏𝟐. 𝟓𝟖𝟗𝟐𝟓𝟒𝟏𝟐, 0.5,1,1)
2

𝟐. 𝟔𝟏𝟑𝟎𝟔𝟖𝐄 − 𝟎𝟕
It must be noted that, again for this case, the use of (6) is
straightforward as the input variable is the power ratio
calculated. It should also be noted that both approaches
provide the same result for 𝑃𝑒.
Standard linear interpolation using Pratt’s Table of Q
function (page 505) for the textbook solution of Q(5.02) gives
𝑃𝑒 2.8E-07.
iv) BER with rain attenuation of 3 dB, QPSK with now a
given C/N= 11 dB implying a power ratio 10
11
10 12.58925412.
Our first step will be determining what exactly would be the
values of z or x to use.
For QPSK modulation:
𝑧 = √
𝐶
𝑁
(11)
√(12.58925412)3.548133892
and,
𝑥 = 𝑧/√2 (10)

3.5481338921
√2

3.5481338926
1.414213562
2.508909536
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(3.548133892 )
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(3.548133892 ,1) 1.939855E-04
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(2.5089095362,0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(6.294627059,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(
𝟏𝟐.𝟓𝟖𝟗𝟐𝟓𝟒𝟏𝟐
2
,0.5,1,1)
2
 1.939855E-04
It must be noted that, in this case, the use of (6) is, again,
straightforward as the input variable is the power ratio
calculated divided by 2. It should also be noted that both
approaches provide the same result for 𝑃𝑒.
Standard linear interpolation using Pratt’s Table of Q
function (page 505) for the textbook solution of Q(3.55) from
Pratt table page 505 = 2.2E-04.
From these simple examples it should be easily inferred that
when we are dealing with C/N quantities, the preferred
methodology should be using (6) with the proper modulation
adjustment.
B. MATH EXAMPLE 2: PRATT EXAMPLE 5.4.2, PAGE 197
The author is now interested in obtaining the system BER
(bit error rate) for different situations (clear sky/air, 0.1%
average annual worst-case condition) with a given C/N
(carrier to noise). That is, we need to estimate the 𝑃′ 𝑒 for a
given C/N - which will be the z or x value, depending on
which Excel formula is applied.
i) BER in clear sky/air condition for BPSK modulation with
a given C/N= 15.2 dB implying a power ratio 10
15.2
10 
33.11311215.
Our first step will be determining what exactly would be the
values of z or x to use. This situation is the same as before
with the C/N changed.
For BPSK modulation:
𝑧 = √2 ∗
𝐶
𝑁
(9)
√2 ∗ (33.11311215)√66.2262243 8.137949637
and,
𝑥 = 𝑧/√2 (10)

8.137949637
√2

8.137949637
1.414213562
5.754399373

4. > PAPER IDENTIFICATION NUMBER TBD< 4
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(8.137949637)
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(8.137949637,1) 𝟎
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(5.7543993732,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝟑𝟑.𝟏𝟏𝟑𝟏𝟏𝟐𝟏𝟓,0.5,1,1)
2
 𝟐. 𝟐𝟐𝟎𝟒𝟒𝟔𝟎𝟓𝐄 − 𝟏𝟔
It must be noted that, in this case, the use of (6) is
straightforward as the input variable is the power ratio
calculated. It should also be noted that the Q(z) approach was
unable to provide a precise estimate for 𝑃𝑒.
Standard linear interpolation using Pratt’s Table of Q
function (page 505) for the textbook solution of erfc(5.75)/2
gives 𝑃𝑒 2.5E-15.
From the example, this sat link has an 8,000,000 bits/second
symbol rate; thus, with an estimated Pe= 2.22044605E-16 an
error will occur, in average, every 5.63E+08 seconds or 17.85
years. Pratt's example states one error would occur every 1.5
years on average.
The lack of precise estimation could cause a false alarm
very often.
ii) BER 0.1% of the year fall in clear sky conditions for
BPSK modulation with a given C/N= 12.2 dB implying a
power ratio 10
12.2
10  16.59586907.
Our first step will be determining what exactly would be the
values of z or x to use. This situation is the same as before
with the C/N changed.
For BPSK modulation:
𝑧 = √2 ∗
𝐶
𝑁
(9)
√2 ∗ (16.59586907)√33.19173815 5.761227139
and,
𝑥 = 𝑧/√2 (10)

5.761227139
√2

5.761227139
1.414213562
4.073802778
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(5.761227139)
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(5.761227139,1)
𝟒. 𝟏𝟕𝟓𝟐𝟐𝟖𝟕𝟏𝟖𝟑𝐄 − 𝟎𝟗
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(4.0738027782,0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(𝟏𝟔. 𝟓𝟗𝟓𝟖𝟔𝟗𝟎𝟕, 0.5,1,1)
2

𝟒. 𝟏𝟕𝟓𝟐𝟐𝟖𝟕𝟏𝟖𝟑𝐄 − 𝟎𝟗
It must be noted that, in this case, the use of (6) is
straightforward as the input variable is the power ratio
calculated. It should also be noted that both approaches
provide the same result for Pe.
Standard linear interpolation using Pratt’s Table of erfc
function (page 506) for the textbook solution of erfc(4.75)/2
gives 𝑃𝑒 5E-09.
From the example, this sat link has an 8,000,000 bits/second
symbol rate; thus, with an estimated Pe= 4.17522872E-09 an
error will occur, in average, every 29.94 seconds. Pratt's
example states one error would occur every 25 seconds on
average.
The imprecision in the calculations are not affecting as
much this time; we can conclude the BPSK modulation is
severely affected in the worst-case situation.
iii) BER in clean air conditions, QPSK with a given C/N=
14.8 dB implying a power ratio 10
14.8
10  30.1995172.
Our first step will be determining what exactly would be the
values of z or x to use.
For QPSK modulation:
𝑧 = √
𝐶
𝑁
(11)
√(30.1995172)5.495408739
and,
𝑥 = 𝑧/√2 (10)

5.4954087391
√2

5.495408739
1.414213562
3.885840784
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(5.495408739 )
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(5.495408739 ,1) 1.949032E-08
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(3.8858407842,0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(15.0997586,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(
𝟑𝟎.𝟏𝟗𝟗𝟓𝟏𝟕𝟐
2
,0.5,1,1)
2
 1.949032E-08
It must be noted that, in this case, the use of (6) is, again,
straightforward as the input variable is the power ratio
calculated divided by 2. It should also be noted that both
approaches provide the same result for 𝑃𝑒.

5. > PAPER IDENTIFICATION NUMBER TBD< 5
Standard linear interpolation using Pratt’s Table of erfc
function (page 506) for the textbook solution of erfc(3.89)/2
gives Pe = 2E-08.
From the example, this sat link has an 16,000,000
bits/second symbol rate; thus, with an estimated Pe=
1.949032E-08 an error will occur, in average, every 3.21
seconds. Pratt's example states one error would occur every
3.12 seconds on average.
The imprecision in the calculations are not affecting as
much this time; we can conclude the QPSK modulation should
not be used in this application.
iv) BER in 0.1% of the year fall in clear sky conditions,
QPSK with a given C/N= 11.8 dB implying a power ratio
10
11.8
10  15.13561248.
Our first step will be determining what exactly would be the
values of z or x to use.
For QPSK modulation:
𝑧 = √
𝐶
𝑁
(11)
√(15.13561248)3.89045145
and,
𝑥 = 𝑧/√2 (10)

3.89045145
√2

3.89045145
1.414213562
2.750964602
Now, we use (3) with (2) to estimate 𝑃𝑒:
𝑃𝑒 = 1 − 𝐹(3.89045145 )
1 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(3.89045145 ,1) 5.002895E-05
Using (6) we obtain for 𝑃𝑒:
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(𝑥2,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(2.7509646022,0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(7.567806242,0.5,1,1)
2

1−𝐺𝐴𝑀𝑀𝐴.𝐷𝐼𝑆𝑇(
𝟏𝟓.𝟏𝟑𝟓𝟔𝟏𝟐𝟒𝟖
2
,0.5,1,1)
2
 5.002895E-05
It must be noted that, in this case, the use of (6) is, again,
straightforward as the input variable is the power ratio
calculated divided by 2. It should also be noted that both
approaches provide the same result for 𝑃𝑒.
Standard linear interpolation using Pratt’s Table of erfc
function (page 506) for the textbook solution of erfc(15.14)/2
gives Pe = 5E-05.
From the example, this sat link has an 16,000,000
bits/second symbol rate; thus, with an estimated Pe=
5.00289478E-05 an error will occur, in average, every
0.00125 seconds. Pratt's example states one error would occur
every 0.00125 seconds on average.
The imprecision in the calculations are not affecting as
much this time; we can conclude the QPSK modulation should
not be used in this application. Worst-case condition there will
be 800 errors per second.
In summary, when dealing with C/N the best way to
estimate the Pe is by using (6) properly adjusted for
modulation: Use the C/N as input directly for BPSK or use
(C/N)/2 for QPSK.
We will now use Lathis’ examples to illustrate calculations
for both approaches when the input is the Pe and a z or x
values is required to be determined.
C. MATH EXAMPLE 3: LATHI EXAMPLE 13.1 page 620
The author interest is to compare power requirements for
different transmissions schemes under a Pe constraint.
Given:
a) Required transmissions rate, 𝑅 𝑏: 2.08E06 bps
b) Channel noise PSD, N: 2E-08
c) Pe ≤ 1E-06
i) Binary, polar signaling; Required Power=𝐸 𝑏 ∗ 𝑅 𝑏 ∗ 𝑁
𝑃𝑒 = 𝑄 (√
2𝐸 𝑏
𝑁
) = 1E − 06
Using (4),
𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(𝑃𝑒) = 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(1𝐸 − 06)
−4.753424309 = √
2𝐸 𝑏
𝑁
; −4.7534243092

2𝐸 𝑏
𝑁
22.59504266
2𝐸 𝑏
𝑁
 𝟏𝟏. 𝟐𝟗𝟕𝟓𝟐𝟏
𝐸 𝑏
𝑁
𝑃𝑜𝑤𝑒𝑟 𝟏𝟏. 𝟐𝟗𝟕𝟓𝟐𝟏 ∗ 2E − 08 ∗ 2.08E06 𝟎. 𝟒𝟔𝟗𝟗𝟕𝟔𝟖𝟗 𝐖
Using (8),
𝐺𝐴𝑀𝑀𝐴. 𝐼𝑁𝑉((1 − 𝑃′ 𝑒)^2,0.5,1) = 𝐺𝐴𝑀𝑀𝐴. 𝐼𝑁𝑉((1 −
1𝐸 − 6)^2,0.5,1)11.297522.
𝑃𝑜𝑤𝑒𝑟 𝟏𝟏. 𝟐𝟗𝟕𝟓𝟐𝟐 ∗ 2E − 08 ∗ 2.08E06𝟎. 𝟒𝟔𝟗𝟗𝟕𝟔𝟗𝟏𝐖
The text estimates
𝐸 𝑏
𝑁
= 11.35; thus,
𝑃𝑜𝑤𝑒𝑟 𝟏𝟏. 𝟑𝟓 ∗ 2E − 08 ∗ 2.08E06𝟎. 0.47216000 W
ii) 16-ary ASK, M=16; Required Power=𝐸 𝑏 ∗ 𝑅 𝑏 ∗ 𝑁 ∗
(𝑙𝑜𝑔2(𝑀)
First, we made the proper adjustments for this scheme,
𝑃𝑒𝑀 = 𝑙𝑜𝑔2(16) ∗ 𝑃𝑒 = 4 ∗ 1𝐸(−06)
𝑃𝑒 = 2(
𝑀 − 1)
𝑀
)𝑄 (√
6𝑙𝑜𝑔2(16)𝐸 𝑏
𝑁(𝑀2 − 1)
) = 4E − 06
= 2(
16 − 1)
16
)𝑄 (√
6 ∗ 4 ∗ 𝐸 𝑏
𝑁(162 − 1)
) = 4E − 06

6. > PAPER IDENTIFICATION NUMBER TBD< 6
= 2(
15)
16
)𝑄 (√
24𝐸 𝑏
𝑁(256 − 1)
) = 4E − 06
= (
15)
8
)𝑄 (√
24𝐸 𝑏
𝑁(255)
) = 4E − 06
= (
15)
8
)𝑄 (√
24𝐸 𝑏
𝑁(255)
) = 4E − 06
= 𝑄 (√
24𝐸 𝑏
𝑁(255)
) = 4E − 06 ∗ (
8
15
)
= 𝑄 (√
24𝐸 𝑏
𝑁(255)
) = 4E − 06 ∗ (
8
15
)
= 𝑄 (√
24𝐸 𝑏
𝑁(255)
) = 2.13𝐸 − 06
Using (4),
𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(𝑃𝑒) = 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(2.13𝐸 − 06)
−4.597950823 = √
24𝐸 𝑏
𝑁(255)
; −4.5979508232

24𝐸 𝑏
𝑁(255)
21.14115177
24𝐸 𝑏
𝑁(255)
21.1411517 ∗ (
255
24
)
𝐸 𝑏
𝑁
224.6247
𝐸 𝑏
𝑁
𝑃𝑜𝑤𝑒𝑟224.6247 ∗ 2E − 08 ∗
2.08E06
𝑙𝑜𝑔2(16)
∗ 𝑙𝑜𝑔2(16)
224.6247 ∗ 2E − 08 ∗ 2.08E06𝟗. 𝟑𝟒𝟒𝟑𝟖𝟗 𝐖
Using (8)
𝐺𝐴𝑀𝑀𝐴. 𝐼𝑁𝑉((1 − 𝑃′ 𝑒)^2,0.5,1) = 𝐺𝐴𝑀𝑀𝐴. 𝐼𝑁𝑉((1 −
2.13𝐸 − 6)^2,0.5,1) 10.57058
𝑃𝑜𝑤𝑒𝑟10.57058 ∗ 2E − 08 ∗ 2.08E06 ∗ 2
∗ (
255
24
)𝟗. 𝟑𝟒𝟒𝟑𝟗𝟎 𝐖
The text estimates 𝐸 𝑏 = 0.499𝐸 − 05,
𝑡ℎ𝑢𝑠, 𝑔𝑖𝑣𝑒𝑠 𝑡ℎ𝑒 𝑃𝑜𝑤𝑒𝑟 𝟗. 𝟑𝟒 𝑾. Our estimates of
𝐸 𝑏 =4.49249E-06 for the Gaussian formula and 𝐸 𝑏 =
4.4925𝐸 − 06 for the Gamma methodology. Using 𝐸 𝑏 =
0.499𝐸 − 05 results in a 𝑃𝑜𝑤𝑒𝑟 𝟏𝟎𝟑. 𝟕𝟗𝟐 𝑾;therefore, the
text must have a typo. If it meant 𝐸 𝑏 = 0.499𝐸 − 06, this
result in a 𝑃𝑜𝑤𝑒𝑟 𝟏𝟎. 𝟑𝟕𝟗𝟐 𝑾 which is not the stated power
value.
Note that the power estimates are congruent until the sixth
significant figure between the two approaches. Again, the use
of the Gamma function is the most straightforward method.
iii) 16-ary PSK, M=16; Required Power=𝐸 𝑏 ∗ 𝑅 𝑏 ∗ 𝑁 ∗
(𝑙𝑜𝑔2(𝑀)
First, we made the proper adjustments for this scheme,
𝑃𝑒𝑀 = 𝑙𝑜𝑔2(16) ∗ 𝑃𝑒 = 4 ∗ 1𝐸(−06)
𝑃𝑒 2𝑄 (√
2(𝜋2)𝑙𝑜𝑔2(16)𝐸 𝑏
256𝑁
)  4E − 06
= 𝑄 (√
(𝜋2)(2)(4)𝐸 𝑏
256𝑁
)  2E − 06
= 𝑄 (√
(𝜋2)(8)𝐸 𝑏
256𝑁
)  2E − 06
= 𝑄 (√
(𝜋2)𝐸 𝑏
32𝑁
) 2E − 06
= 𝑄 (√
(𝜋2)𝐸 𝑏
32𝑁
) 2E − 06
Using (4),
𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(𝑃𝑒) = 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(2. 𝐸 − 06)
−4.611382362 = √
(𝜋2)𝐸 𝑏
32𝑁
; −4.6113823622

(𝜋2)𝐸 𝑏
32𝑁
21.26484729
(𝜋2
)𝐸 𝑏
32𝑁
21.1411517 ∗ (
32
𝜋2
)
𝐸 𝑏
𝑁
68.9465
𝐸 𝑏
𝑁
𝑃𝑜𝑤𝑒𝑟68.9465 ∗ 2E − 08 ∗
2.08E06
𝑙𝑜𝑔2(16)
∗ 𝑙𝑜𝑔2(16)
68.9465 ∗ 2E − 08 ∗ 2.08E06𝟐. 𝟖𝟔𝟖𝟏𝟕𝟔𝟐 𝐖
Using (8)
𝐺𝐴𝑀𝑀𝐴. 𝐼𝑁𝑉((1 − 𝑃′ 𝑒)^2,0.5,1) = 𝐺𝐴𝑀𝑀𝐴. 𝐼𝑁𝑉((1 −
2. 𝐸 − 6)^2,0.5,1) 10.63242
𝑃𝑜𝑤𝑒𝑟10.63242 ∗ 2E − 08 ∗ 2.08E06 ∗ 2
∗ (
32
𝜋2
) 𝟐. 𝟖𝟔𝟖𝟏𝟕𝟔𝟓 𝐖
The text estimates 𝐸 𝑏 = 137.8𝐸 − 08 = 1.378𝐸 − 06,
𝑡ℎ𝑢𝑠, 𝑔𝑖𝑣𝑒𝑠 𝑡ℎ𝑒 𝑃𝑜𝑤𝑒𝑟 𝟐. 𝟖𝟔𝟔𝟐𝟒𝟎𝟎 𝑾. Our estimates of
𝐸 𝑏 =1.3789309E-06 for the Gaussian formula and 𝐸 𝑏 =
1.3789310𝐸 − 06 for the Gamma methodology.
Note that the power estimates are congruent until the
seventh significant figure between the two approaches. Again,
the use of the Gamma function is the most straightforward
method.

7. > PAPER IDENTIFICATION NUMBER TBD< 7
D. MATH EXAMPLE 4: RODDY EXAMPLE 10.1 page
304
The author interest is to estimate BER for binary polar
transmission; given 𝐸 𝑏 = 1𝐸 − 06 & 𝑁 = 1𝐸 − 07. This
example is surprisingly confusing as simple as it can look. As
known,
𝐸 𝑏
𝑁
=
1𝐸 − 06
1𝐸 − 07
= 10
Now, the author uses erf instead of erfc. The relationship
between these two functions is that their sum =1. That is,
erf(𝑥) + 𝑒𝑟𝑓𝑐(𝑥) = 1. (12)
Now, in order to properly solve the problem, you must
know that for polar signaling the correct formulas are the ones
we used with the above example:
𝑃𝑒 = 𝑄 (√
2𝐸 𝑏
𝑁
) (13)
Solving,
𝑃𝑒 = 𝑄 (√
2𝐸 𝑏
𝑁
) = 𝑄 (√(2 ∗ 10)) =
𝑄(√20) 𝑄(4.472135955)3.872108𝐸 − 06
It is important to understand that 𝑧 = √20 4.472135955
here; thus, 𝑥 =
𝑧
√2
=
√20
√2
= √103.16227766
Using erfc,
𝑃𝑒 = 𝑒𝑟𝑓𝑐(𝑥) =
1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(𝑥2
, 0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(3.162277662
, 0.5,1,1)
2

1 − 𝐺𝐴𝑀𝑀𝐴. 𝐷𝐼𝑆𝑇(𝟏𝟎, 0.5,1,1)
2
 3.872108𝐸 − 06
The author’s solution using a graphical approach is
surprisingly precise, it gives 𝑃𝑒 = 3.9𝐸 − 06.
And again, the Gamma approach is the easier and
straightforward methodology to find the result.
A student learning from different texts could get very
confused as to what value to input in which function in order
to correctly estimate the BER. This text directs the student to
look for erf(√10)  erf(3.16227766). There are no details on
how to do this.
IV. CONCLUSION
It could be appreciated that using examples from three
different authors in texts with different topics the easier way to
avoid confusions and get the correct results with the least
effort was to use the proposed Excel Gamma function.
It is a mistake not to explain in detail the use of the Q(z)
and ercf(x) functions in most communication courses as it is
looked as an easy task not worthy of the time. This paper
illustrates the correct way to code with Excel in order to
always obtain the correct result.
V. APPENDIX
Appendices not needed in this work.
REFERENCES
[1] B.P. Lathi Modern Digital and Analog Communications
Systems 3rd
Ed., Oxford University Press NY, 1998.
[2] T. Pratt, C. W. Bostian & J. E. Allnut Satellite
Communications 2nd
Ed. John Wiley & Sons, MA,
2003.
[3] D. Roddy Satellite Communications 4th
Ed. Mc-Graw
Hill, NY 2006.
XXXXXXXXX.