Commonly, medical digital radiology ultrasound studies consist of about 25 image

Question

Commonly, medical digital radiology ultrasound studies consist of about 25 images extracted from a full-motion ultrasound examination. Each image consists of by 512 pixels, each with 8 b intensity information. How many bits are there in the 25 images? Ideally, however, doctors would like to use 512 times 512 8-bit frames at 30 fps (frames per second). Ignoring possible compression and overhead factors, what is the minimum channel capacity required to sustain this full-motion ultrasound? Suppose each full-motion study consists of 25 s of frames. How many bytes of storage would be needed to store a single study in uncompressed form?

Explanation / Answer

Ans a. Total no of bits = No. Of Images x Size of Image x Pixel info size

=25×512×512×8

=52428800 bits.

B ans. Minimum channel capacity

= Speed of franes x size of frames x pixel info size

=30×512×512×8

=62914560 bits per second.

C ans. Required storage memory

= No. Of frames in each study x size of frane x pixel info size

=30×25×512×512×8

=1572864000 bits

=196608000 Bytes.

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