Commonly, medical digital ultrasound studies consist of about 25 images extracte

Question

Commonly, medical digital ultrasound studies consist of about 25 images extracted from a full-motion ultrasound examination. Each image consists of 512 by 512 pixels, each with 8 intensity information. How many bits are there in the 25 images? Ideally however doctor would like 512 times 512 8bit facto what is (frames per second). Ignoring possible compression and overhead ultrasound? the minimum channel capacity required to sustain this full-motion ultrasound? How many bytes of storage would be needed to store a single study in uncompressed from? Suppose that a digitized TV picture is to be transmitted from a source that uses a matrix of 480 times 500 picture elements (pixels), where each pixel can take of 32 intensity values. Assume that 30 pictures are sent per second. This digital source is roughly equivalent to broadcast TV standards that have been adopted.) Find the source rate R (bps). Assume that the TV picture is to be transmitted over a channel with 4.5-MHz band width and a 35 dB signal-to-noise ratio. Find the capacity of the channel (bps). Discuss how the parameters given in part (a) could be modified to allow transmission of color TV signals without increasing the required value for R(bps) Compare the white noise density levels in closed rooms in London and Sydney in January, with average temperatures of C and computed as line has a of 2800 Hz and its signal-to-noise ratio has been the highest bit rate of this line Data needs to be sent over a noiseless channel of 30-kHz bandwidth a. Find the number of signal levels M needed the bit rate descried is approximately 400 kbps. What maximum bit rate will you get with M signal levels found part (a) in 50 MH. What are the values of SNR and o and the of noise power? What is the a telephone transmission fad lit." nominal SNR of 56 dB (400,000), and a certain level of distortion What is the theoretical maximum channel capacity (kbps) of What can we say about the actual maximum channel capacity? Study the works of Shanna 20. on the bit rate of a channel on channel places Consider a based on two different approaches. What is the upper limit to the data rate that the channel can carry? the result of part (a) is the upper limit. However, as a practical matter, better performance will be achieved at a lower data rate. Assume we choose a data rate of 2/3 as the maximum theoretical limit. How many signal levels are needed to achieve this data rate?

Explanation / Answer

3.12)

a) 512*512 = 2,62,144

1 pixel = 8 bits

2,62,144*8 = 2,097,152 bits is one image

25 images = 2,097,152*25 = 52,428,800

b) 512*512*8 = 2,097,152 bits

2,097,152*30 = 62,914,560 bits

c) It take 512*512*8 = 2,097,152

25 frames would take = 25*(512*512*8)

= 25*2,097,152

= 52,428,800

For suppose if it ask how many such studies could fit on a 600 MB CD-ROM

1 MB = 1,048,5776 bytes

600 MB = 1,048,5776*600

= 629,145,600 bytes

1 byte = 8 bits

629,145,600 bytes = 629,145,600*8 = 5,033,164,800 bits

629,145,600*25 = 1,572,864,000 bits

5,033,164,800/1,572,864,000 = 3.2 times

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