# 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