Homework 1 (W. Stallings Lan&Wan)

**2.4 **For the bit stream 0101110, scetch the waveforms for NRZ,
NRZI, Manchester, and Differential Manchester as well as for Miller coding
and E-NRZ.

- Nonreturn-to-Zero-Level (NRZ-L)
- 0 = high level
- 1 = low level
- Nonreturn-to-Zero Inversted (NRZI)
- 0 = no transition at beginning of interval (one bit time)
- 1 = transition at beginning of interval
- Manchester
- 0 = transition from high to low in middle of interval
- 1 = transition from low to high in middle of interval
- Differential Manchester
- Always a transition in middle of interval
- 0 = transition at beginning of interval
- 1 = no transition at beginning of interval
- Enhanced Nonreturn-to-Zero (E-NRZ)
- Separate NRZ-L into 7-bit words
- Invert bits 2,3,6,7
- Add one partiy bit to each word to make the total number of 1's in the 8-bit word and odd count
- Miller coding
- 0 = transition at the end of the bit period if the next bit is 0, no transition if the next bit is 1
- 1 = transition in the middle of the interval

**3.3 **Consider the tranfser of a file containing one million
characters from one section to another. What is the total elasped time
and effective throughput for the following cases?
**a. **A circuit-switched, star topology network. Call setup time
is negligible, and the data rate of the medium is 64 kbps

**b. **A bus topology local network with two stations a distance
D apart, a data rate of B bps, and a packet size P with 80 bits of overhead.
Each packet is acknowledged with an 88-bit packet before the text is sent.
The propagation speed on the bus is 200m/microseconds. Solve for

- D=1km, B=1 Mbps, P=256 bits
- D=1km, B=10 Mbps, P=256 bits
- D=10km, B=1 Mbps, P=256 bits
- D=1km, B=50 Mbps, P=10,000 bits

**A.**
**a.** T=(8x10**6 bits)/(64x10**3 bps) = 125 seconds
**b.** Transfer consists of a sequence of cycles. On cycle consists
of:

Data Packet = Data Packet Transmission Time + Propagation Time

ACK Packet = ACK Packet Transmission Time + Propagation Time

Define:

c = Cycle time

Q = data biits per packet

T = total time required

Td = Data Packet Transmit Time

Ta = ACK Packet Tramit Time

Tp = Propagation Time

Then

Tp = D/(200x10**6m/sec)

Ta = (88 bits)/(B bps)

Td= (P bits)/(B bps)

T = (8x10**6 bits x C sec/cyle)/(Q bits/cycle)

C= Ta + Td + 2 Tp

Q = P-80

**b1**

C= 354x10**(-6)

Q=176

T = 16 sec
**b2**

C=44.4x10**(-6)

Q=176

T=2sec
**b3**

C=444*10**(-6)

Q=176

T=20sec
**b4**

C=211.6x10**(-6)

Q=9,920

T=0.17 seconds

**c. **Define

Tr = Total repeater Delay = N/B

Then

C = Td+2Ttp+Tr

Q=P-80

T=(8x10**6 x C)/Q

We show results for N=100
**c1.**

C=366x10**(-6)

T=16.6 sec
**c2.**

C=45.6x10**(-6)

T=2.1 sec
**c3.**

C=456x10**(-6)

T=20.7 sec
**c4.**

C=212x10**(-6)

T=0.17 sec