In: Electrical Engineering
1 (20 pts) You are working for NASA/JPL on the Juno mission.
Juno is a large spacecraft orbiting Jupiter.
It's current distance from Earth is 536 million miles. Assume the
speed of light is c = 2:0 108 m/s. The
transfer rate from Juno to Earth is 50 Mbps. The frame is made up
of a 24 byte header plus a variable size
payload of 1 - 1000 bytes. Answer the following:
(a) The total transmission time to send 128 GB of pictures to
Earth. Assume only transmission time and
propagation delay.
(b) The amount of time required to transfer the pictures taking in
to account propagation delay, transmis-
sion delay, queuing delay, and processing delay. Queuing delay is 1
ms per frame and processing delay
is 0.5 ms per frame
a) given data
Transmitting speed is 50 Mbps.
total frame size 24+1 to 1000 bytes
we have a picture of 128 GB
which means 128*1024*1024*1024 bytes ( since 1 GB = 1024*1024*1024 bytes)
total transmission time = transmission time + propagation delay.
total header bytes required = (since we have 24 byte header file for every 1000 bytes of payload)
= 131 Mbytes
total data size of image is = 128*1024 + 131 Mbytes
= 131203 Mbytes
transmission time = 131203/50 sec
= 2624 sec.
since
propagation time = Distance / Velocity
Distance is given that (d) = 536 million miles.
since 1mile = 1609 meters
therfore d = 536*1609*10^6 meters
propagation time = sec
= 4312 sec
Hence total transmission time = 2624 + 4312 sec
= 6936 sec
b) by introducing queing delay and processing delay total transmession delay is increases
queuing delay means frame to frame delay
processing delay means time require to process the frame
Hence total trasmission time = actual transmission time + total frames*queuing delay + tot frames * processing del
total frames = 128*1024*1024*1024/1000
= 137438953
total trasmission time = 1693 sec + 137438953 * ( 1.5) msec
= 1693 + 206158 sec
= 207851 sec