4. Magnetic field detector Two instruments each reading 1 Magnetic field reading

Question

4. Magnetic field detector

Two instruments each reading 1 Magnetic field reading + 3 components (X, Y, Z) @8 bits

Instrument 1: covering +/- 8.190 nT 0.004 nT and +/- 65.52 nT 0.032 nT
Instrument 2: covering +/- 8 nT 0.0039 nT, +/- 64 nT 0.0312 nT, +/- 2048 nT 1.0000 nT, +/- 44000 nT 21.50 nT

Timing for both: 2 vectors/second

Data rate in KB/sec?

5. A polar weather satellite like NPOESS using the AVHRR instrument sends a high resolution data stream that consists of three separate channels, each a line of 2048 10-bit single pixels (that have a resolution of 1km/pixel, not needed for this problem), and transmits at 6 lines per second. It is allocated 40 kbit/sec of bandwidth to transmit this. By how much must this raw data be compressed to fit into its bandwidth allocation (i.e. not at all, by 2x, by 10x, etc).

Explanation / Answer

Ans:-Generally we called magnetometer or magnetic field detector . it instrument that measures the magnetization of magnetic material (ferromagnetic) al well as the find the direction and strength of a magnetic field in location. The Tesla is denoted as T is a unit. if we use IS units so that  kgs2A1.

Lorentz force law is the main principal of the magnetic field. its defined as in terms of force on moving charge.

Now instrument 1 - covering +/- 8.190 nT 0.004 nT and +/- 65.52 nT 0.032 nT ( T = tesla , n = no. of vector)

suppose we choose MDI 10242(michelson doppler image detector is minimum limit +/- 8.190 nT 0.004 nT and maximum limit +/- 65.52 nT 0.032 nT . so that time coverage (h/d) 24 , temporal cadence 60 sec. so 3 component XYZ is @8 bits. it means the rate of Data increase 8 bits/sec

so if MDI 10242 detector = time coverage (h/d) / (min - max limit ) 60

                                 = 24 / ((8.190*0.004) - (65.53*0.032)) x 60

                                 = 24/ ( 0.0328           -       2.096) x 60

                                = 24 /( -2.063 x 60) = - 24 / 123.794

                                = 0.1939 X 8 X 2               (given 8 bits 3 component in XYZ in 2 vectors per sec)

                                 = 3 .1 kbps.

Now instrument 2- covering +/- 8 nT 0.0039 nT, +/- 64 nT 0.0312 nT, +/- 2048 nT 1.0000 nT, +/- 44000 nT 21.50 nT

suppose we choose LASCO( large angel & Spectrometric cronographic) it base on different intensity detector is +/- 8 nT 0.0039 nT, +/- 64 nT 0.0312 nT, +/- 2048 nT 1.0000 nT, +/- 44000 nT 21.50 nT . so that time coverage (h/d) 24 , temporal cadence 60 sec. so 3 component XYZ is @8 bits. it means the rate of Data increase 8 bits/sec.

In this method we are taken the maximum value for magnetic field detector   +/- 44000 nT 21.50 nT . is getting maximum date rate. 21.50 nT   in series upper value 44000 nT.

21.50 x 44000 X / vector in probability in sec x 3 component bits

=21.50 x 44000 / 8i X2i

= 946000/ 80640 = 11.73 kbps

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.