1. Digital Transmission Dr. Hassan Yousif Department of Electrical Engineering College of Engineering at Wadi Aldwasser Slman bin Abdulaziz University Class 11 May. 2 nd, 2015

2. Dr Hassan Yousif- EE-CE- SAU

3. Block vs. Stream Information Block Information that occurs in a single block – Text message – Data file – JPEG image – MPEG file Size = Bits / block or bytes/block – 1 kbyte = 2 10 bytes – 1 Mbyte = 2 20 bytes – 1 Gbyte = 2 30 bytes Stream Information that is produced & transmitted continuously – Real-time voice – Streaming video Bit rate = bits / second – 1 kbps = 10 3 bps – 1 Mbps = 10 6 bps – 1 Gbps =10 9 bps

4. Transmission Delay Use data compression to reduce L Use higher speed modem to increase R Place server closer to reduce d L number of bits in message R bps speed of digital transmission system L/R time to transmit the information t prop time for signal to propagate across medium d distance in meters c speed of light (3x10 8 m/s in vacuum) Delay = t prop + L/R = d/c + L/R seconds

5. Dr Hassan Yousif- EE-CE- SAU

9. 1. Quantization error Every signal is analog in nature Since analog to Digital conversion follows the method described below Analog->sampling->quantization->encoding->Digital signal Quantization error is introduced while quantization process and this error cannot be removed anyhow during its transmission, hence this error poses a problem for Digital signals. 2. High power consumption Analog signals need to be converted it to analog and vice versa which requires extra circuitry and hence consumes more power. 3. Infinite bandwidth Bandwidth is inversely proportional to the time period of the pulses As pulse width reduces Bandwidth requirement increases. Therefore, if we want to have fast switching i.e. fast speed of communication we will have to reduce the time period of pulses or increase the frequency of pulses, this calls for higher and higher Bandwidth, moreover if we wish to reduce the time period of pulses to almost zero,we want an infinite Bandwidth to accomplish our task, which is not possible in any practical communication system. 4. Difficult transmission Owing to the infinite Bandwidth requirement,transmission of Digital signals becomes very difficult for communication engineers.

10. Dr Hassan Yousif- EE-CE- SAU

12. Digitization of Analog Signal Sample analog signal in time and amplitude Find closest approximation         Original signal Sample value Approximation R s = Bit rate = # bits/sample x # samples/second 3 bits / sample

13. Dr Hassan Yousif- EE-CE- SAU

25. Compression Information usually not represented efficiently Data compression algorithms – Represent the information using fewer bits – Noiseless: original information recovered exactly E.g. zip, compress, GIF, fax – Noisy: recover information approximately JPEG Tradeoff: # bits vs. quality Compression Ratio #bits (original file) / #bits (compressed file)

26. H W =++ H W H W H W Color image Red compone nt image Green compone nt image Blue compone nt image Total bits = 3  H  W pixels  B bits/pixel = 3HWB bits Example: 8  10 inch picture at 400  400 pixels per inch 2 400  400  8  10 = 12.8 million pixels 8 bits/pixel/color 12.8 megapixels  3 bytes/pixel = 38.4 megabytes Color Image

27. TypeMethodFormatOriginalCompressed (Ratio) TextZip, compress ASCIIKbytes- Mbytes (2-6) FaxCCITT Group 3 A4 page 200x100 pixels/in 2 256 kbytes 5-54 kbytes (5-50) Color Image JPEG8x10 in 2 photo 400 2 pixels/in 2 38.4 Mbytes 1-8 Mbytes (5-30) Examples of Block Information

28. Bit Rate of Digitized Signal Bandwidth W s Hertz: how fast the signal changes – Higher bandwidth → more frequent samples – Minimum sampling rate = 2 x W s Representation accuracy: range of approximation error – Higher accuracy → smaller spacing between approximation values → more bits per sample

29. Example: Voice & Audio Telephone voice W s = 4 kHz → 8000 samples/sec 8 bits/sample R s =8 x 8000 = 64 kbps Cellular phones use more powerful compression algorithms: 8-12 kbps CD Audio W s = 22 kHertz → 44000 samples/sec 16 bits/sample R s =16 x 44000= 704 kbps per audio channel MP3 uses more powerful compression algorithms: 50 kbps per audio channel

30. Video Signal Sequence of picture frames – Each picture digitized & compressed Frame repetition rate – 10-30-60 frames/second depending on quality Frame resolution – Small frames for videoconferencing – Standard frames for conventional broadcast TV – HDTV frames 30 fps Rate = M bits/pixel x (WxH) pixels/frame x F frames/second

31. Video Frames Broadcast TV at 30 frames/sec = 10.4 x 10 6 pixels/sec 720 480 HDTV at 30 frames/sec = 67 x 10 6 pixels/sec 1080 1920 QCIF videoconferencing at 30 frames/sec = 760,000 pixels/sec 144 176

32. Dr Hassan Yousif- EE-CE- SAU

33. Time-division Multiplexing (TDM) Time-division multiplexing (TDM) is a digital process that can be applied when the data rate capacity of the transmission medium is greater than the data rate required by the sending and receiving devices.

34. TDM TDM is a digital multiplexing technique to combine data. Time Division Multiplexing TDM Frequency Power User# 1 User# 2 User# 3 Time

35. Time-division Multiplexing (TDM) TDM can be implemented in two ways: synchronous TDM and asynchronous TDM. In synchronous time-division multiplexing, the term synchronous means that the multiplexer allocates exactly the same time slot to each device at all times, whether or not a device has anything to transmit. Frames Time slots are grouped into frames. A frame consists of a one complete cycle of time slots, including one or more slots dedicated to each sending device.

36. Time Division Multiplexing TDM Frequency Power User# 1 User# 2 User# 3 Time

37. Dr Hassan Yousif- EE-CE- SAU Digital Modulation

38. Bandpass Channels Bandpass channels pass a range of frequencies around some center frequency f c – Radio channels, telephone & DSL modems Digital modulators embed information into waveform with frequencies passed by bandpass channel Sinusoid of frequency f c is centered in middle of bandpass channel Modulators embed information into a sinusoid f c – W c /2 f c 0 f c + W c /2

39. Information 111100 +1 0 T 2T2T 3T3T 4T4T5T5T 6T6T Amplitude Shift Keying +1 Frequency Shift Keying 0 T 2T2T 3T3T 4T4T5T5T 6T6T t t Amplitude Modulation and Frequency Modulation Map bits into amplitude of sinusoid: “1” send sinusoid; “0” no sinusoid Demodulator looks for signal vs. no signal Map bits into frequency: “1” send frequency f c +  ; “0” send frequency f c -  Demodulator looks for power around f c +  or f c - 

40. Phase Modulation Map bits into phase of sinusoid: – “1” send A cos(2  ft), i.e. phase is 0 – “0” send A cos(2  ft+  ), i.e. phase is  Equivalent to multiplying cos(2  ft) by +A or -A – “1” send A cos(2  ft), i.e. multiply by 1 – “0” send A cos(2  ft+  ) = - A cos(2  ft), i.e. multiply by -1 We will focus on phase modulation +1 Phase Shift Keying 0 T 2T2T 3T3T 4T4T5T5T 6T6T t Information 111100

41. Modulate cos(2  f c t) by multiplying by A k for T seconds: AkAk x cos(2  f c t) Y i (t) = A k cos(2  f c t) Transmitted signal during kth interval Demodulate (recover A k ) by multiplying by 2cos(2  f c t) for T seconds and lowpass filtering (smoothing): x 2cos(2  f c t) 2A k cos 2 (2  f c t) = A k {1 + cos(2  2f c t)} Lowpass Filter (Smoother) X i (t) Y i (t) = A k cos(2  f c t) Received signal during kth interval Modulator & Demodulator

42. 111100 +A -A 0 T 2T 3T 4T5T 6T Information Baseband Signal Modulated Signal x(t) +A -A 0 T 2T 3T 4T5T 6T Example of Modulation A cos(2  ft)-A cos(2  ft)

43. 111100 Recovered Information Baseband signal discernable after smoothing After multiplication at receiver x(t) cos(2  f c t) +A -A 0 T 2T 3T 4T5T 6T +A -A 0 T 2T 3T 4T5T 6T Example of Demodulation A {1 + cos(4  ft)} -A {1 + cos(4  ft)}

44. Dr Hassan Yousif- EE-CE- SAU Convert the Sin wave shown in below figure to digital signal by quantizing it into 5 levels ranging in the following levels of volt. -0.5 v, 0 v, 0.5 v, 1 v, 1.5 v

45. What is the size of 8  10 inch picture at 400  400 pixels per inch 2 Dr Hassan Yousif- EE-CE- SAU