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(ELEC314)[2009](f)midterm~ee_jmx^_10322.pdf
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ELEC314 Midterm Solution
Question 1 solution
(a)
symbol rate=1/4. bit rate=(log24)/4=1/2. Average energy per symbol, Es=(E1+E2+E3+E4)/4=(4+4+8+8)/4=6 Average energy per bit, Eb=Es/log24=3

(b)
14


12 =
st()st()dt=. 1

EE01 2
12

1
13 =
2 11

(c) Step1: .1()t=
E1 s1( ) t=2 s1()t
Step2:
st() = . st (). ()tdt=. 2
21 21
() = () .. t= 0
ft sts ()
2 2 21 1
..2()t= 0
31() = s3( ) t.1()tdt= 2Step3: st
.
st32() = 0

f()ts() ts . () = () . t
=. t sts()
3 331 1 31
f()ts()ts. ()t
3 31
..3()t=
= 2
2
ft()dt
. 3
Step4:
st() = st (). () tdt=. 2

41 41
.
st42() = 0


st() = . st (). () tdt= 2
43 43
() = () .. ts . ()
ft sts () . t= 0
4 4 41 1 43 3
..4()t= 0 (d) st() =2.()t




11
st() =.2.() t
21
st() =2.() t+2. ()t
3 13
st() =.2.() t+2. () t
4 13
(e)

(f)




The average energy per symbol for the new constellation is 5. So it will reduce the energy by (6-5)/6=1/6
Question 2 solution
(a)
Each matched filter has an equivalent lowpass impulse response: hi(t)=si(T-t).

(b)
The output of the matched filters are ht()* s ()| tT



t = 0
12 =
ht t 2
()* s ( )| tT = AT
22 =
(c) The receiver structure is: The output of the correlators are:

s1() s2()d= 0
0 T
Ts2() s2() d= AT 2
0
(d) The outputs of the matched filters are different from the outputs of the correlators. The two sets of outputs agree at the sampling time t=T. The 2 structure are equivalent.
Question 3 solution
a) (Can be found in the notes and tutorial, for convenience, the derivation is also put here again) By considering the nearest neighbor union bound,

Lets label the constellation diagram as follows:

S6 S5

S7 S8

S10
S9

S11
S12

S2 S1
d

S3 S4

S14 S13

S15 S16
In the constellation diagram of 16-QAM, consideration of symbol error can be divided into 3 types:

the constellation,

Thus PS 3Q ...
...
b) As metioned in the question paper, we have this set of basis signals:

( ft), cos 2 f + ) , cos 2 f + 2 ) f
cos 2 (( ft ) (( ft ) , where has to 1/T to ensure
ccc
orthogonality, and
Consider the sketch of the constellation diagram in 3 dimension 16-QAM (or called it 64-3d-QAM) (number of constellation point M = 64),

We have the following observations:
The symbol error can be divided into 4 types:
Type 1 (Corner)

-It has 3 nearest neighbors.
-8 of constellation points are of this type.

Type 2 (The one near the corner) -It has 4 nearest neighbors. -24 of constellation points are of this type.
Type 3 (The one on the face, but not at the boundary of the constellation diagram) -It has 5 nearest neighbors. -24 of constellation points are of this type.
Type 4 (The interior) -It has 6 nearest neighbors.
-8 of constellation points are of this type.
.
2 ..
2 ..
2 ..
2 .. d .
. d .
. d .
. d .

8 3Q . + 24