1 What is Multimedia? Multimedia can have a many definitions Multimedia means that computer information can be represented through media types: – Text – Audio and speech – Images – Graphics – Video

2 What is Multimedia systems? A Multimedia System is a system capable of processing multimedia data and applications A Multimedia System is characterised by the processing, storage, generation, manipulation and rendition of Multimedia information.

3 Multimedia is Multidisciplinary Image, audio, speech processing Computer vision, pattern recognition Computer networks, operating system Human computer interaction Computer graphics Multimedia computing

4 Multimedia Computing Multimedia systems involve some basic enabling techniques: – Multimedia data representation and compression. – Multimedia data processing and analysis. – Transmitting multimedia data through communication networks. – Multimedia database, indexing and retrieval.

Digital Audio, Image and Video

Digital Media In computers, audio, image and video are stored as files just like other text files. For images, these files can have an extension like – BMP, JPG, GIF, TIF, PNG, PPM, … For audios, the file extensions include – WAV, MP3, … The videos files usually have extensions: – AVI, MOV, …

Digital Media Capturing To get a digital image, an audio or a video clip, we need some media capturing device such as – a digital camera or a scanner, – a digital audio recorder, – or a digital camcorder. All these devices have to complete tasks: – Sampling: To convert a continuous media into discrete formats. – Digitization: To convert continuous samples into finite number of digital numbers. – There are probably some further compression process.

Sampling

Sampling period Ts, fs =1/Ts Signal Period T, f = 1/T Intuitively T should >= 2Ts

Nyquist Theorem Analog signal is sampled every Ts seconds Ts is referred to as the sampling interval Fs =1/Ts is called the sampling rate or sampling frequency Nyquist theorem – The necessary condition of reconstructing a continuous signal from the sampling version is that the sampling frequency fs > 2f max f max is the highest frequency component in the signal. – If a signal’s frequency components are restricted in [f1, f2], we need fs >2 (f2-f1).

Example An input max signal frequency equals 100 HZ and Fs=300HZ can we make a sampling for this signal ?! Fs > 2 Fmax 2Fmax =2*100=200 Fs > 2 Fmax Sampling can be done

Example An input max signal frequency equals 1 KHZ and Fs=500HZ can we make a sampling for this signal ?! Fs > 2 Fmax 2Fmax =2*1000=2000 Fs < 2 Fmax Sampling can not be done

Example Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. The sampling rate therefore is 8000 samples per second. According to Nyquist Theorem Fs > 2 Fmax

Example 2 For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: f s = 4f (2 times the Nyquist rate), f s = 2f (Nyquist rate), and f s = f (one-half the Nyquist rate). Figure 4.24 shows the sampling and the subsequent recovery of the signal. It can be seen that sampling at the Nyquist rate can create a good approximation of the original sine wave (part a). Oversampling in part b can also create the same approximation, but it is redundant and unnecessary. Sampling below the Nyquist rate (part c) does not produce a signal that looks like the original sine wave.

Example 2

4.16 Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals

Example 3 A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Solution Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. The bandwidth of a low-pass signal is between 0 and f, where f is the maximum frequency in the signal. Therefore, we can sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is therefore 400,000 samples per second.

Example 4 A complex band pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Solution Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. We cannot find the minimum sampling rate in this case because we do not know where the bandwidth starts or ends. We do not know the maximum frequency in the signal.

4.19 Quantization Sampling results in a series of pulses of varying amplitude values ranging between two limits: a min and a max. The amplitude values are infinite between the two limits. We need to map the infinite amplitude values onto a finite set of known values. This is achieved by dividing the distance between min and max into L zones, each of height   = (max - min)/L

4.20 Quantization Levels The midpoint of each zone is assigned a value from 0 to L-1 (resulting in L values) Each sample falling in a zone is then approximated to the value of the midpoint.

4.21 Quantization Zones Assume we have a voltage signal with amplitutes V min =-20V and V max =+20V. We want to use L=8 quantization levels. Zone width  = ( )/8 = 5 The 8 zones are: -20 to -15, -15 to -10, -10 to -5, -5 to 0, 0 to +5, +5 to +10, +10 to +15, +15 to +20

4.22 Assigning Codes to Zones Each zone is then assigned a binary code. The number of bits required to encode the zones, or the number of bits per sample as it is commonly referred to, is obtained as follows: n b = log 2 L Given our example, n b = 3 The 8 zone (or level) codes are therefore: 000, 001, 010, 011, 100, 101, 110, and 111 Assigning codes to zones: – 000 will refer to zone -20 to -15 – 001 to zone -15 to -10, etc.

4.23 Bit rate The bit rate of a PCM signal can be calculated form the number of bits per sample x the sampling rate Bit rate = n b x f s We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows: