Communication Device Theory and Design

Communication Device Theory and Design 1
COMMUNICATION DEVICE THEORY AND DESIGN
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Communication Device Theory and Design 2
Question 1
a) ALU
1. The OR operation.
A3 A2 A1 A0+ B3 B2 B1 B0 =0111
¿ 0111
0111
2. Subtracting A from B
B3 B2 B1 B0− A3 A2 A1 A0=1011
−0100
0111
Carry I bit is set to 1.
3. XOR operation has no carry in or carry out bits since the two number sequences are
compared bit-wise;
A3 A2 A1 A0+ B3 B2 B1 B0 =1011
XOR 0111
1100
4. Preset operation sets all the bits to 1
5. A and B
A3 A2 A1 A0 . B3 B2 B1 B0=0111
¿ 0111
011 1
There are no overflow and carry out bits.
6. The clear operation resets all the bits to 0
7. Subtracting B from A
A3 A2 A1 A0−B3 B2 B1 B0=0101
−0010
0011
There is a carry in bit in the operation

Communication Device Theory and Design 3
8. The addition operation makes the ALU work like a 4-bit adder (Morris 2017).
A3 A2 A1 A0+ B3 B2 B1 B0 =0111
+00 11
10 10
There is no overflow and no carry-out bit at the end of the operation.
The results obtained above are illustrated in the table below.
Table Q1
b)
Four-sensor alarm system.
The alarm operates when any of the four conditions is met hence we use OR operations (Morris
2017).
Y = AB C' D'+ AB C' D+ ABC D' + ABCD
Karnaugh map:
Minterms Y =1100+ 1101+1110+ 1111
Y =m12 +m13+ m14 +m15=∑ m(12,13,14,15)
A3 A2 A1 A0 B3 B2 B1 B0 S2 S1 S0 Operatio
n
Cn F3 F2 F1 F0 Cn+4 OV
R
1 0 1 1 1 0 1 1 1 1 0 1 A+B 0 0 1 1 1 0 0
2 0 1 0 0 1 0 1 1 0 0 1 B minus
A
1 0 1 1 1 0 0
3 1 0 1 1 0 1 1 1 1 0 0 A ⊕ B 0 1 1 0 0 0 0
4 0 0 0 0 1 0 0 0 1 1 1 PRESET 1 1 1 1 1 1 1
5 1 1 1 1 1 1 1 1 1 1 0 A∙B 0 0 1 1 1 0 0
6 1 0 1 0 1 0 1 0 0 0 0 CLEAR 0 0 0 0 0 0 0
7 0 1 0 1 0 0 1 0 0 1 0 A minus
B
1 0 0 1 1 0 0
8 0 1 1 1 0 0 1 1 0 1 1 A plus B 0 1 0 1 0 0 0

Communication Device Theory and Design 4
AB
CD
00 01 11 10
00 0 0 1 0
01 0 0 1 0
11 0 0 1 0
10 0 0 1 0
Grouping the 1’s in the Karnaugh map, the equation is simplified to:
Y = AB
That is sensors A and B must be active for the alarm to operate.
Question 2
a) Fourier representation
x (t )= 4
π {sin 4000 πt+ 1
3 sin 12000 πt + 1
5 sin 20000 πt + 1
7 sin 28000 πt +...+ ... }
This the Fourier series of a square waveform of the form:
f ( t )= 4
π ∑
n=1,3,5, ..
∞ 1
n sin ( nπt
L )
The frequency of the signal can be determined from the fundamental harmonic which has the
argument of sine function as:
nπt
L =2 πfnt
Therefore for n=1, 2 πft =4000 πt
Giving; f = 4000 πt
2 πt =2000 Hz=2 kHz
The duty cycle of this square wave is 50% since it has only odd function has only odd
harmonics, symmetrical and is active for half of its period.
b) Lower cut-off, f L=1 kHz and higher cut-off is f H =1 kHz frequencies.
Only odd harmonics exist.
Using fundamental harmonic frequency; 2 πf =4000 π, f 0=2 kHz