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 512 by 512 pixels, each with 8 b of intensity information. a. How many bits are there in the 25 images? b. 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? c. 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

a)Bits in the 25 images:

No of pixels *bits per pixel *no of images=   512*512*8*25

                                                                             =52428800 bytes

b)Minimum channel capacity:

Channel capacity is usually given in mega bits per second.

One frame per second would take (512 x 512) x 8 nits so 30 fps =

=(512 x 512) x 8 x 30 bytes

=62914560bytes

c) It take (512 x 512 x8) to store one frame =2097152

for 25 frames would take 25 x (512 x 512 x 8)= 2097152*25=52428800 bytes

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