THE UNIVERSITY OF WESTERN ONTARIOFACULTY OF ENGINEERING

DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING

SE 410a – WIRELESS LANs AND WANs

Assignment # 1(Solutions)

1. What is the received power (in dB) in free space of a signal whose transmit power is 1W and carrier frequency is 2.4 GHz, if the receiver is at a distance of 1 mile (1.6 km) from the transmitter? What is the path loss in dB?

= Path Loss in dB= 10 ; = 1.6km=1600 m;

Substituting values of and we obtain dB Loss or -52.0642 is the path loss.

10 =-52.0642 W

2. A voice-grade channel of the telephone network has a bandwidth of 3.4 kHz. (a) Calculate the information capacity of the telephone channel for a signal-to-noise ratio of 35dB; (b) Calculate the minimum signal-to-noise ratio required to support information transmission through the telephone channel at the rate of 9600 bits per second.(a) bits/sec; 35dB=10 . Therefore,

kbps(b) 9600=3.4 2.8235=log

. Therefore, dB

3. (a) Find the average capacity in bits/sec that would be required to transmit a high resolution B/W TV signal at the rate of 32 pictures per second, if the picture is made up of 2x10 picture elements and 16 different brightness levels. All picture elements are assumed to be independent and all levels have equal likelihood occurrence. (b) For color TV, this system additionally provides for 64 different shades of color. How much more system capacity is required for a color system compared to B/W system?

(a) High Resolution B/W TV requires: Mbps=The system capacity

(b) For Colour TV the system capacity is: Mbps=System Capacity

4. Given a wireless bridge with 200mW of output power connected through a 100 foot cable with 6 dB loss to an antenna with 9 dBi of gain, what is the EIRP at the antenna in dBm?6 dB loss (Cable Loss)9 dBi gain +3dB+3dB+3dB (Antenna Gain)200mW is the input. Therefore, 200mW when comes out of the cable will become 200mW/4= 50mW. 50mW through the antenna will yield 50mW8=400mW = 26.0206dBm. Therefore, EIRP=26.0206dBm

5. It is desired to send a sequence of computer screen images over an optical fiber. The screen is 480x640 pixels, each pixel is 24 bits. There are 60 screen images per second. How much bandwidth is required?Data Rate= =442.368 Mbps. Assuming 1 bps/Hz, the bandwidth required is 442.368 MHz

6. How long does it take to transmit an 8 inch by 10 inch image by facsimile over an ISDN B channel (64 kbps)? The facsimile digitizes the image into 300 pixels per inch and assigns 4 bits per pixel. Current FAX machines go faster than this over ordinary telephone lines. How do you think they do it?The paper has 80 sq. inches, for a total of 7.2 pixels. Transmitting this image at 4 bits/pixel requires 2.88 bits. At 64 kbps (ISDN B Channel), transmission takes 450sec. Current FAX machines use 1 bit/pixel, which saves a factor of 4, but these FAX modems operate at 1.44 kbps, not 64 k bps, so transmitting a full page still takes 450 sec. The reason they appear to go faster is that most pages are 95% white spaces and runlength encoding used removes it all.

7. Compare the delay in sending x-bit message over a k-hop path in a circuit-switched network and in a lightly loaded packet switched network. The circuit setup time is‘s’ seconds, the propagation delay is‘d’ seconds per hop, the packet size is ‘p’ bits, and data rate is ‘b’ bits per second. Figure out the conditions under which a packet network has a lower delay.

For circuit switching:

For packet switching: The last bit is sent at . To get to the final destination, the last packet must be retransmitted times by intermediate nodes, each retransmission taking sec, so the total delay is given by:

Comparing the two delays, packet switching is faster if

8. A TDMA system operates at 100Mbps with a 2ms frame time. Assume that all the slots are of equal length and that a guard time of 1 s is required between slots. (a) Compute the efficiency of the system for the case of 1, 2, 5, 10, 20, 50, and 100 slots per frame; (b) Repeat (a) assuming a 100-bit preamble is required at the start of each slot. Compute the efficiency of the system; (c) Graph the results of (a) and (b).

(a) Let = number of slotsFor , there is no guard time, and therefore the efficiency ( ) is 100%For the efficiency is given by:

2 5 10 20 50 1000.999 0.9975 0.995 0.990 0.975 0.950

(b) Preamble time= , For , the efficiency ( ) is

For the efficiency is given by:

2 5 10 20 50 1000.998 0.995 0.990 0.980 0.950 0.90

Graph the values yourself.