Question

Calculate the link parameter and channel utilization efficiency (in error-free channels) for a system with the following parameters, if the simplex stop and wait protocol is used. a) Bitrate: R=12 Mbps, Modulation: BPSK, distance between the transmitter and receiver: d-6 miles, propagation velocity:v3 x 108 m/s and the length of each frame: L-500 bits. b) Symbolrate: Rs.10 Msps, Modulation: QPSK, distance between the transmitter and receiver: d=15 km, propagation velocity: v 3 x 108 m/s and the length of each frame: L- 256 bits.

Answer 1

Answer: a) 0.77 and 0.39 b) 3.9 and 0.20

Explanation:

We have to find two things here i.e link parameter and channel utilization efficiency.

a)

We are given

Bitrate= 12 Mbps,

Distance= 6 miles

Propagation Velocity= 3 x 10⁸ m/s

Length of frame= 500 bits

We know that The link parameter is related to propagation time and frame time

Where Propagation time= Bitrate x Distance= 12 Mbps x 6 miles

and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 500

So,

Link parameter= $\frac{12*10^6*6*1609.34}{3*10^8*500}$

Link parameter= 0.77

Channel utilization efficiency= $\frac{1}{1+2* link parameter}$

Channel utilization efficiency= 0.39

b)

We are given

Bitrate= 10 Mbps,

Distance= 15 km

Propagation Velocity= 3 x 10⁸ m/s

Length of frame= 256 bits

We know that The link parameter is related to propagation time and frame time, Here QPSK is used so bitrate is multiplied by 2,

Where Propagation time= Bitrate x Distance= 2 x 10 Mbps x 15 km

and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 256

So,

Link parameter= $\frac{2*10*10^6*15*1000}{3*10^8*256}$

Link parameter= 3.9

Channel utilization efficiency= $\frac{1}{1+2* link parameter}$

Channel utilization efficiency= 0.20