Important QnA on Multimedia Computing - Comprehensive

Text Books

T1 Ze-Nian Li & Mark S. Drew, "Fundamentals of Multimedia", Pearson Education, 2004

Question.1 

Explain the difference between Temporal and Frequency masking.

Answer:

Frequency Masking: When an audio signal consists of multiple frequencies the sensitivity of the ear changes with the relative amplitude of the signals. If the frequencies are close and the amplitude of one is less than the other close frequency then the second frequency may not be heard. 

Temporal masking: 

• After the ear hears a loud sound, consisting of multiple frequencies, it takes a further short while before it can hear a quieter sound close in frequency.
• Temporal Masking accomplished 50% overlap between consecutive transform windows and then relating frequency masking.
• MPEG Audio encrypts frequency masking by quantizing each filter bank with adaptive values from nearby bands, defined by a look up table.

Question. 2

Consider the following set of protocols (SIP, RTSP, RSVP, RTCP, RTP on top of UDP. If you want to design a protocol stack (control and data plane) for Video‐On‐Demand (VOD) service between client and server, (a) which protocols would you use and why, and (b) in which order would you apply your selected protocols? Explain how the protocol stack of selected protocols would be used ?

Answer:

To design the Video‐On‐Demand service,

RTP protocol : for transmission of the video . RTP/UDP enables real‐time transmission. 

RTCP : The accompanying control protocol for RTP is the RTCP that would allow the receiver to provide feedback to the sender if some parameters need to be adjusted during the streaming session. 

RTSP : Since VOD will use commands such as play, stop, pause, fast forward, rewind, so anyone should use the RTSP protocol because it specifies signaling for completing these control tasks between the client and the server.

RSVP: If an IP level also requires the built-in reservation capabilities of VOD traffic, it is important to enable the RSVP protocol to start integrated Guaranteed Services using the RSVP Reservation Protocol.

 

RSVP to reserve bandwidth for the VOD session. 

Once the resources are reserved, VOD traffic should be transmitted via RTP which is on top of UDP/IP. 

Along with this, RTP RTCP and RTSP should provide (1) RTP (Control) to provide control feedback about the status of traffic and receiver, and (2) controlling the channel through signaling (RTSP) control , Stop, and otherwise control the VOD playback.

Question. 3

Compress the string “DEAF!DEFEATED” using LZW algorithm. Find compression ratio by

supposing that originally the characters are represented in 8-bit (B=66, A = 65, ,., Z = 90, Y = 89, , ! = 33).

Answer:

Input: DEAF!DEFEATED

Previous	Current	Output	Code	String
			065	A
			068	D
			069	E
			070	F
			084	T
			033	!
D	E	068	256	DE
E	A	069	257	EA
A	F	065	258	AF
F	!	070	259	F!
!	D	033	260	!D

D	E
DE	F	256	261	DEF
F	E	070	262	FE
E	A
EA	T	257	263	EAT
T	E	084	264	TE
E	D	069	265	ED
D	End Of File	068

The compressed output is 068 069 065 070 033 256 070 257 084 069 068

Compression ratio = 13/11 = 1.18181

Question:
Compress the string “PROPER_PROPERTY” using LZW algorithm by using codeword of 10 bit. Find compression ratio by assuming that originally the characters are represented in 8-bit (A = 65, B=66... Y = 89, Z = 90,  _ = 95).

Answer: 

Input: PROPER_PROPERTY
                                               

S	C	Output	Code	String
			069	E
			079	O
			080	P
			082	R
			084	T
			089	Y
			_	95
P	R	080	256	PR
R	O	082	257	RO
O	P	079	258	OP
P	E	080	259	PE
E	R	069	260	ER
R	_	082	261	R_
_	P	095	262	_P
P	R
PR	O	256	263	PRO
O	P
OP	E	258	264	OPE
E	R
ER	T	260	265	ERT
T	Y	084	266	TY
Y	NULL	089

The compressed output is 080 082 079 080 069 082 095 256 258 260 084 089


Question:
Compress the string “ROAD_ROADIES” using LZW algorithm by using codeword of 10 bit. Find compression ratio by assuming that originally the characters are represented in 8-bit (A = 65, B=66,.., Y = 89, Z = 90, ! = 33).

Answer: 

8-bit set has 0 to 255 ASCII characters. initially table contains 256 entries from 0 to 255

Input: ROAD!ROADIES

Previous	Current	Output	Code	String
			065	A
			068	D
			069	E
			073	I
			079	O
			082	R
			083	S
			033	!
R	O	082	256	RO
O	A	079	257	OA
A	D	065	258	AD
D	!	068	259	D!
!	R	033	260	!R
R	O
RO	A	256	261	ROA
A	D
AD	I	258	262	ADI
I	E	073	263	IE
E	S	069	264	ES
S	End Of File	083

The compressed output is 082 079 065 068 033 256 258 073 069 083
Compression ratio = 12/10 = 1.2

Question. 4

Suppose we use a predictor as follows: 

Also, suppose we adopt the quantizer equation

If the input signal has values as follows 20  32  42  54   78   97 115  129  150  173   205
Then find the following using DPCM
a. Predicted signal
b. Error signal
c. Quantized error signal
d. Reconstructed signal

Answer:

Question. 5

Complete the following table:

	# of frequency bands	Min # of Samples	Max # of Samples
MPEG Layer 1
MPEG Layer 2
MPEG Layer 3

Answer:

	# of frequency bands	Min # of Samples	Max # of Samples
MPEG Layer 1	32	12	12
MPEG Layer 2	32	36	36
MPEG Layer 3	32	12	36

Question. 6

You have to design a DVR server which may serve up to 100 simultaneous users over a 1 gbps connection. One of the key requirements is to support trick modes for video, including aspects like 16X rewind. Typical video will be stored in high definition approx. 20 mbps, 24 frames per second.  Each I frames take 8x data, while P frame takes 2x and B frames take 1x data.  What would be the average size for each I, P and B frames in bytes. 

Answer:

24 fps. It has IBBPBBPBBPBB

In this 1-I, 3-P, 8-B

Each of these has

I- 8X, P-2X and B-1X data

1*8X+3*2X+8*1X=22X

22X = 0.5 sec

1 sec -> 20 mb

0.5 sec -> ?  => 10 mb

22X = 10 mb

X= 106 bits/ 22 = 106 /(22*8) bytes = 5682 bytes

Question. 7

Using the following frequency table compress the string “DEED” (originally 8-bit representation) & find the compression ratio using arthimetic compression(64-bit arthimetic).

Character	A	B	C	D	E	F	G	H
frequency	10	20	10	20	50	70	90	30

Answer:

Characters	Frequency	Range
A	10	[0, 0.03)
B	20	[0.03, 0.1)
C	10	[0.1, 0.13)
D	20	[0.13, 0.2)
E	50	[0.2, 0.36]
F	70	[0.36, 0.6]
G	90	[0.6, 0.9)
H	30	[0.9, 1)

Compression

Symbol	Low	High	Range
	0	1.0	1.0
D	0.13	0.2	0.07
E	0.144	0.1552	0.0112
E	0.14624	0.148032	0.001792
D	0.14647296	0.1465984	0.00012544

After compression consider the lower value 0.14647296

Compression ratio = 64/18 =3.555

Question. 8

Ten channels, each with a 150-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 20 kHz between the channels to prevent interference?

Answer:

For ten channels, we need at least nine guard bands. This means that the required bandwidth is at least
10*150KHz + 9*20 KHz =  1680 KHz

Question. 9

Integrating RTP / RTCP type protocols works better for applications like Skype as opposed to relying on RSVP. Mention True or False and Briefly explain the reason.

Answer:

RSVP will not provide QoS across multiple autonomous systems, especially because the user base is world-wide. While RTP can be used with/without RSVP (i.e. explicit reservation)for Skype, the feedback uses RTCP like mechanism to find delay/jitter/loss-rates and adjust its approach accordingly. In other works rather than expecting quality from network, it modifies application behaviour to align with observed network characteristics.

Question. 10

The LZ compression is used to compress a text document. The dictionary table contains 512 entries. The average character per word in the document is 5. Calculate the compression achieved.

Answer:

The dictionary table contains 512 entries. Hence, it is represented by a 10 – bit code word (2"9 = 512). Thus, the index of each word in the document is a 9-bit number. If the compression scheme was not applied, each character would have been represented by a 7-bit ASCII code. Since there is an average of 5 characters per word, each word would need to be represented by a 35-bit number. However, after compression each word is represented by a 9-bit number. Hence, the compression ratio is 35:9

Question. 11
Calculate the storage capacity per image for SVGA given its resolution as 1920x1080 pixels and the color format is 24bits/pixel.

Answer:
Image Resolution=1920x1080 pixels, Colour Depth=24bits/pixel
Storage capacity = Resolution X (no. of bits per pixel / 8 bits per byte)
    = 1920x1080 x(24/8) bytes = 6220800 bytes

Question:
Image Analysis plays a very vital role for extracting descriptions from images. - Explain. 

Answer:

• Image Analysis is concerned with techniques for extracting descriptions from images that are necessary for higher-level scene analysis method. By itself knowledge of the position and value of any particular pixel almost conveys no information related to the recognition of an object, the description of an object’s shape, it’s position or orientation, the measurement of any distance on the object or whether the object is defective. Hence, image analysis techniques include computation of perceived brightness and colour, partial or complete recovery of three-dimensional data in the scene, location of discontinuities corresponding to objects in the scene and characterization of the properties of uniform regions in the image.
• Image Analysis is important in many arenas: aerial surveillance photographs, slow-scan television images of the moon or of planets gathered from space probes, television images taken from an industrial robot’s visual sensor, x-ray images and computerized axial tomography (CAT) scans. Sub areas of image processing include image enhancement, pattern detection and recognition and scene analysis and computer vision.
• Image Enhancement deals with improving image quality by eliminating noise (extraneous or missing pixels) or by enhancing contrast.
• Pattern Detection and Recognition deal with detecting and clarifying standard patterns and finding distortions from these patterns.
• Scene Analysis and Computer Vision deal with recognizing and re-constructing 3D models of a scene from several 2D images.

Question:
A resource manager provides components for the different phases of the allocation and management process.” – What are the necessary components? Explain.

Answer:
A resource manager provides components for the different phases of the allocation and management process:

i) Schedulability Test
The resource manager checks with the given QoS parameters (e.g., throughput and reliability) to determine if there is enough remaining resource capacity available to handle this additional request.

ii) Quality of Service Calculation
After the Schedulability Test, the resource manager calculates the best possible performance (e.g., delay) the resource can guaranty for the new request.

iii) Resource Reservation
The resource manager allocates the required capacity to meet the QoS guaranties for each request.

iv) Resource Scheduling
Incoming messages from connections are scheduled according to the given QoS guaranties. For process management, for instance, the allocation of the resource is done by the scheduler at the moment the data arrive for process.

Question: