# ECE-T501: Midterm Examination Solution

```ECE-T501: Midterm Examination Solution
1. The constellation graph:
The average symbol energy Ē = 18 (4 &times; 1 + 4 &times; 9) = 5. The minimum symbol distance dmin = 2.
For the 8-PSK modulation with an average symbol energy 5 (the same as the given modulation scheme),
its minimum symbol distance is 2.75 &gt; 2. Therefore, 8-PSK is more efficient than the given modulation
scheme.
The given modulation scheme can be viewed as two QPSK systems, so it will not be more efficient than
QPSK. Actually, since the two equivalent QPSK systems are not indepdent, the interference between
them will increase the error probability of each of them. Consequenly, the given modulatioin scheme
is inefficient compared to the QPSK.
2. Directly use the correlation receiver structure to obtain:
Z
T
0
T
1
r(t)s0 (t)dt − ks0 (t)k2 ≷01
2
Z T
1
A
r(t)s1 (t)dt − ks1 (t)k2
r(t) cos(2πfc t)dt ≷01
2
2
0
Z
0
A block diagram of this receiver is as below
r(t)
+ −&gt; 0
sign( )
− −&gt; 1
cos( )
−A/2
3. A QPSK system can be equivalent to two indepedent BPSK systems; so it performs the same as BPSK.
1
4. The PSD of x(t) is
Φx (f ) =
1
|G(f )|2 ΦI (f ),
T
where G(f ) is the Fourier transform of g(t),
G(f ) =
Z
T
A sin
0
πt
T
e−j2πf t dt =
T cos(πf T ) −jπf T
e
,
π[1 − (2f T )2 ]
and ΦI (f ) is the PSD of the i.i.d sequence In , ΦI (f ) = 1. Therefore,
Φx (f ) =
cos2 (πf T )
π 2 [1 − (2f T )2 ]2
2
```