Physics 227/623, Assignment #3 (Analog and Digital Communications)
1. Problem 2.1. Do parts (a) and (c), skipping (b). Sketch the waveform for the first 1 msec, rather than 3 msec. Use 3 bits plus a sign bit, i.e., the 4th bit is 1 when the function is positive and 0 when the function is negative.
Question 2.1
The given wave has
$x=A* sin(2πF*t)=A*sin(6283.19*t)$
Sampling time is
$T= 1/3000=0.333 ms$
For the first 1 ms of the wave there are 4 samples taken at
t=0, t=0.333, t=0.666 and t=0.999 ms
The amplitudes of the wave at these times are
x(0)=0
x(0.33 ms)=0.866*A
x(0.66 ms)=-0.863*A
x(0.9999 ms)=0
Using 3 bits there are $2^3=8$ voltage levels on each side±of the wave at
000 at 0
001 at A/8=0.125A
010 at A/4=0.25A
011 at 3A/8=0.375A
100 at A/2=0.5A
101 at 5A/8=0.625A
110 at 6A/8=0.75A
111 at 7A/8=0.87A
Using as the 4-th bit (MSB bit-most semnificative bit)
0 for V>0
1 for V<0
the numbers transmitted are
x=0 → 0000
x=0.866A → 0110
x=-0.863A → 1110
x=0 → 0000
- The NRZ-L format is the simple format that transmit each bit value consecutively so that the numbers transmitted will be (The 0 Volts level is at the middle of the transmitted signal NRZ=non return to zero)
(The 0 Volts level is at the middle of the transmitted signal NRZ=non return to zero).
c)In the BIPHASE-L encoding a 0 bit is represented by an transition In the middle of a clock, while “1” bit is a transition. Therefore
(CLK=0@DATA=0)→transmit 0
(CLK=1@DATA=1)→transmit 0
(CLK=0@DATA=1)→transmit 1
(CLK=1@DATA=0)→transmit 1
where the length of the CLK is half the length of the DATA. Therefore
(01010101
0 0 0 0) →01010101
(01010101
0 1 1 0) →01101001
Therefore the transmitted data for BIPHASE-L is
01010101|01101001|10101001|01010101