1. 1) By using hamming code (even – parity), Show the correct binary number that transmitted by the sender if the receiver received 1111101 binary number.

2. 1) Answer: BCDX= 1111 (0) No error ACDY= 1110 (1) Error ABDZ= 1111 (0) No error So the correct binary number transmitted by the sender is 1111111 The binary number 1111

3. 2) By using hamming code (even – parity), Show the correct 4-bits number in denary number that transmitted by the sender if the receiver received 1111101binary number.

4. 2) Answer: BCDX= 1111 (0) No error ACDY= 1110 (1) Error ABDZ= 1111 (0) No error So the correct binary number transmitted by the sender is 1111111 The binary number 1111 is equivalent to 15 denary number

5. 3) By using Run-length coding as a compression algorithms for a stream of 300 bits, the first 120 are 0s and the 100 that follows are 1s and the third 60 are again 0s, and the remaining are 1s. What is the compressed version of the message?

6. 3) Answer: In binary (01111000)0(01100100)1(00111100)0(00010100)1 In decimal 120(0) 100(1) 60(0)20(1)

7. 4) By using Run-length coding as a compression algorithms for a stream of 300 bits, the first 120 are 0s and the 100 that follows are 1s and the third 60 are again 0s, and the remaining are 1s What is the compression ratio in this transmission?

8. 4) Answer: In binary (01111000)0(01100100)1(00111100)0(00010100)1, In decimal 120(0) 100(1) 60(0)20(1) The compression ratio is the number of bits in the original message divided by the number in the compressed message which is 300/36= 8.33

9. 5) Given that radio waves propagate at 3 × 10 8 m/sec, estimate the propagation time for a path length of 60 km

10. 5) Answer: Velocity (speed of light) = distance (traveled by the signal) ÷ time (propagation time) for 60 km (that is, 60000 m or 6 * 10 3 m) the propagation time is 10 3 ÷ (3 × 10 8 ) s which is approximately 0.0006

11. 6) Suppose that we have an analogue signal covers the frequency range from 5.5 kHz to 16.5 kHz What is the bandwidth of the signal?

12. 6) Answer: (2B)Bandwidth = 16.5 – 5.5 = 11 KHz = 11000 Hz

13. 7) Suppose that we have an analogue signal covers the frequency range from 5.5 kHz to 16.5 kHz What is the minimum sampling rate required for a signal?

14. 7) Answer: Bandwidth = 16.5 – 5.5 = 11 KHz = 11000 Hz Sampling rate = 2 x Bandwidth = 2 x 11000 = 22000 Hz = 22 KHz

15. 8) Suppose that the sampling frequency is 10 KHz, and it is used along with 10 bits representation for each sample. Then what data rate will be generated after digitization.

16. 8) Answer: Data rate = 10 (bit/sample) * 10 (Ksamples/sec) = 100 Kbps

17. 9) Suppose FDMA system with 800 KHz bandwidth, each user use 45 kHz, and there is 5 kHz is allocated as a guard band so the maximum number of user in the system can be

18. 9) Answer: Number of users = 800 / (45+5) = 16 users

19. 10) Calculate the frequency if the period of the waveform (periodic time (T)) = 0.01 seconds

20. 10) Answer frequency (f) = 1 / 0.01 = 100 Hz

21. 11) Calculate the Period of the waveform if the frequency (f) = 200 Hz

22. 11) Answer Period of the waveform(T) = 1 / 200 = 0.005

23. 12) The following are all multiples of the standard unit hertz (abbreviated Hz), put them into decreasing order. 1)MHz 2)GHz 3)KHz

24. 12) Answer: 1) GHz 2)MHz 3)KHz

25. 13) The following are all fractions of the standard unit the meter (abbreviated m). Put them into increasing order. 1) mm 2) nm 3) µm

26. 13) Answer: 1) nm 2) µm 3) mm

27. 14) Convert (D3) 16 hexadecimal system to the equivalent number in Binary system

28. 14) Answer D3 16 = 11010011 2

29. 15) Convert (100011011) 2 binary system to the equivalent number in Denary system.

30. 15) Answer 100011011 2 =283 10