Typical ultrasound consists of about 25 images extracted from a fullmotion ultra
ID: 3868521 • Letter: T
Question
Typical ultrasound consists of about 25 images extracted from a fullmotion ultrasound examination. Each of these images consists of 512 by 512 pixels, each with 8 bit intensity information.
a) Ideally the doctors would like to playback the ultrasounds at 30 frames per second. Assuming that when playing back each ultrasound frame is compressed by a factor of 0.8, what is the data rate required to play the ultrasound?
b) Consider now that this ultrasound needs to be viewed in real-time at a remote location using wireless communication. If the communication system has a signals to noise ration of 30dB and an overhead of 15%, what is the minimum bandwidth required for this communication.
c) Assume a practical error correction method is able to achieve a 75% of the above data rate. How many signal levels are required to achieve this data rate?
Explanation / Answer
Bits in the 25 images:
No of pixels *bits per pixel *no of images= 512*512*8*25
=52428800 bytes
a) Minimum channel capacity:
Channel capacity is usually given in mega bits per second.
One frame per second would take (512 x 512) x 8 x 30 frames per second.
=(512 x 512) x 8 x 30 bytes
=62914560 bytes
Compressed factor = 0.8
After compression, required data rate is = 62914560 * 0.8 = 50331648 bits/sec
b) Data Rate = 50331648
Signal to Noise Ratio = 30 dB
Overhead = 15%
Bandwidth = x
Bandwidth =
= = =283115520 Hz
c) Channel Capacity = Bandwidth * log2n
Here, 2n = Signal Level
log2n = Channel Capacity / Bandwidth = 62914560 / 283115520 =0.222
Taking log both sides
log(log2n) = log(0.222)
2n = 1
Signal Level = 1
Thus, to achieve this data rate, it required only one data signal.
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