QUESTION: At the 1. Calculate the small group\'s mean income and sample standard deviation of income. 2. Test the hypothesis that the small group\'s mean income is equal to the congregation\'s mean. 3. Test the hypothesis that the small group\'s income variance is equal to the congregation\'s income variance. The following is the information given from 3: A different congregation, Wealthy Street Church And Rowing Club, suspects that their members are not randomly drawn from the national population. Its 125 members have an annual average income of $57,000. Test the hypothesis that this group is randomly drawn from the national population. Solution (1) mean=(30,000+40,000+..+120,000)/5=60000 standard deviation= sqrt(((30000-60000)^2+...+(120000-60000)^2)/(5-1)) =35355.34 (2) Ho: mu=57,000 Ha:mu not equal to 57,000 The test statistic is t=(xbar-mu)/(s/vn) =(60000-57000)/(35355.34/sqrt(5)) =0.19 Assume a=0.05, the critical values are t(0.025, df=n-1=4) =-2.78 or 2.78 (from student t table) Since t=0.19 is between -2.78 and 2.78, we do not reject HO. So we can conclude that the small group\'s mean income is equal to the congregation\'s mean (3) Ho: o^2= 6000^2 Ha: o^2 not equal to 6000^2 The test statisitc is chisquare =(n-1)*s^2/o^2 =4*35355.34^2/6000^2 =138.89 Assume a=0.05, the critical values are chisquare with 0.025, df=n-1=4 is 0.48 and chisquare with 0.975, df=4 is 11.14 (from chisquare table) Since 138.89 is larger than 11.14, we reject Ho. So we cannot conclude that the small group\'s income variance is equal to the congregation\'s income variance..

1. Question:
Based on the constellation , find out the baseband modulation symbols for the binary sequence
001001110101
(a) BPSK
(b) QPSK
(c) 8PSK
(d) 16QAM
Solution
The baseband modulation symbols for the binary sequence 001001110101
(a) BPSK
The simplest form of PSK is binary phase-shift keying (BPSK), where N = 1 and M = 2.
Therefore, with BPSK, two phases (21 = 2) are possible for the carrier. One phase represents a
logic 1, and the other phase represents a logic 0. As the input digital signal changes state (i.e.,
from a 1 to a 0 or from a 0 to a 1), the phase of the output carrier shifts between two angles that
are separated by 180°. Hence, other names for BPSK are phase reversal keying (PRK) and bi
phase modulation. BPSK is a form of square-wave modulation of a continuous wave (CW)
signal.
(b) QPSK
QPSK is an M-ary encoding scheme where N = 2 and M= 4 (hence, the name "quaternary"
meaning "4"). A QPSK modulator is a binary (base 2) signal, to produce four different input
combinations,: 00, 01, 10, and 11. Therefore, with QPSK, the binary input data are combined
into groups of two bits, called dibits. In the modulator, each dibit code generates one of the four
possible output phases (+45°, +135°, -45°, and -135°).
(c) 8PSK
The idea can be extended to have 8-PSK. Here the phase is shifted by 450
(d) 16QAM
Any rectangular QAM constellation is equivalent to superimposing two ASK signals on
quadrature carriers (I and Q components). For 4-QAM modulation, each symbol is of size k =
log2(M) = log2(4) = 2 bits. For 16-QAM modulation, the symbol size is k = log2(16) = 4 bits.
There exist other constellations that are more efficient (in terms of energy required to achieve
same error probability) than the standard rectangular constellation. But due to its simplicity in
modulation and demodulation rectangular constellations are preferred.
In any M-QAM constellation, in order to restrict the erroneous symbol decisions to single bit

2. error, the adjacent symbols in the transmitter constellation should not differ more than one bit.
This is usually achieved by converting the input symbols to Gray coded symbols and then
mapping it to the desired QAM constellation. But this intermediate step can be skipped
altogether by using a Look-Up-Table (LUT) approach which properly translates the input symbol
to appropriate position in the constellation. We will exploit the inherent property of Karnaugh
Maps to generate the gray coded 16-QAM constellation
Karnaugh Maps:
To generate a gray coded 16-QAM constellation we will use a 4 variable (since for 16-QAM, the
symbol size is 4 bits) K-Map. A 16 QAM constellation can be constructed using two 4-ASK
signals. For a 4-ASK signal there are 4 amplitude levels. For our case the 4 amplitude levels are
{-3,-1,+1,+3}.
Let’s draw the constellation to verify the above mentioned approach. Plot the in-phase and
quadrature phase components (QAM output) and each point with the equivalent binary
representation of the array index gives the 16-QAM constellation where the adjacent symbols
differ by only one bit.