In the MS-assisted case, it is the GPS receiver which sends the measurements for the SLP to calculate and revert. I understand the measurements include the Ephemeris, Iono, DGPS etc + Doppler shift that are sent. Please let me know if my understanding is right.
Does the SET send the code (the entire data transmitted by satellites as is) that it receives as is or splits it into the above components and send?
All the assistance information in SUPL is encapsulated using RRLP protocol (Radio resource location services (LCS) protocol for GSM), RRC (Radio Resource Control for UMTS) or TIA 801 (for CDMA 2000) or LPP (LTE Positioning Protocol for LTE). I'm just looking at RRLP standard ETSI TS 101 527. The following part sounds interesting:
A.3.2.5 GPS Measurement Information Element
The purpose of the GPS Measurement Information element is to provide
GPS measurement information from the MS to the SMLC. This information
includes the measurements of code phase and Doppler, which enables the
network-based GPS method where position is computed in the SMLC. The
proposed contents are shown in table A.5 below, and the individual
fields are described subsequently.
In subsequent section it is defined as:
reference frame - optional, 16 bits - the frame number of the last measured burst from the reference BTS modulo 42432
GPS TOW (time of week) - mandatory, 24 bits, unit of 1ms
number of satellites - mandatory, 4 bits
Then for each satellite the following set of data is transmitted:
satellite ID - 6 bits
C/No - 6 bits
Doppler shift - 16 bits, 0.2Hz unit
Whole Chips - 10 bits
Fractional Chips - 10 bits
Multipath Indicator - 2 bits
Pseudorange Multipath Error - 3+3 bits (mantissa/exponent)
I'm not familiar that much with GPS operation to understand all the parameters, but as far as I understand:
C/No is simply a signal(carrier) to noise ratio
Doppler shift - gives the frequency shift for a given satellite, obviously
Whole/Fractional Chips together give the phase (and thus satellite distance)
My understanding is that things like almanac, ephemeris, Iono, DGPS are all known on the network side. As far as I know those things are transferred from network to MS in MS-based mode.
Hope that helps.
Measurements collected from MS-assisted location requests include:
Satellite ID
code phase - whole chips
code phase - fractional chips
Doppler
Signal strength
Multipath indicator
pseudorange RMS indicator
In addition, the GPS time of measurements is also provided as one value (in milliseconds) for the time which all measurements are valid.
In practice, the required fields that need to be accurate and correct are:
Satellite ID
code phase - whole chips
code phase - fractional chips
Doppler
The code phase values for each satellite are almost always used for the most accurate location calculation. Doppler values can be used to estimate a rough location but aren't usually accurate enough to really contribute to the final solution.
The other values for signal strength, multipath indication, and RMS indicator usually vary in meaning so much between vendors that they don't really provide much benefit for the position calculation. They would normally be used for things like weighting other values so that good satellites count more in the final position.
The network already knows (or should know) the ephemeris and ionospheric model. They are not measurements collected by the handset.
Related
I need to monitor an AC Voltage waveform and record the RMS value when the breakdown happens. I roughly know how to acquire data from videos I have watched, however, it is difficult for me to produce a solution that reads the Breakdown Voltage Value. Ideally, I would also take a screenshot along with the breakdown voltage value,
In case you are not familiar with this topic, When a breakdown happens the voltage will drop immediately to zero. So what I need is to measure the voltage just before it falls to zero, and if possible take a screenshot. This is an image of a normal waveform (black) with a breakdown one (red).
Naive solution*:
Take the data and get the Y values (this would depend on the datatype you have, which would depend on how you acquire the data).
Find the breakdown point by iterating over the values and maintaining a couple of flags (I would probably say track "got higher than X" and once that's true, track "got lower than Y").
From that, I would just say take the last N points (Get Array Subset) and get the array max. Or just track the maximum value as you run.
Assuming you have the graph in a control, you can just right click and select Create>>Invoke Node>>Export Image.
I would suggest trying playing with that with a VI with static data which you can repeatedly run to check how your code behaves.
*I don't know the problem domain and an not overly familiar with the various analysis VIs that ship with LV, so there are quite possibly more efficient ways of doing this.
I have two identical gps modules running the same program with antennas side by side. I get different results. I don't understand why
One example is 4211.41545 from the first unit and 4211.41481 from the second the timestamp is the same.
You don't say what the values you quote are so I will assume that the value you are showing is the Lat or Long value from the NMEA output data. If this is the case then the difference between the values is 0.00064 minutes of arc. The maximum physical distance that this will represent is around 1.1 metres (along a great circle). At 60 degrees north this would correspond to around 55cm in an E/W direction.
You do not say how far apart the antennae are or whether there are any obstructions above the level of their horizons which will introduce varying multipath signals that will be different for the two antennae. The two receivers will not necessarily be sampling the satellite signals at the same instants and so can have marginally different signal timings resulting in a different position result.
A typical consumer grade GPS will have a CEP figure of 2.5 metres meaning that 50% of the values for position in an unobstructed sky view position will lie within a 2.5 metre radius circle of the true position.
Taking this into account you should not expect any two adjacent GPS devices to give identical results.
I'm currently working on a school project to design a network, and we're asked to assess traffic on the network. In our solution (dealing with taxi drivers), each driver will have a smartphone that can be used to track its position to assign him the best ride possible (through Google Maps, for instance).
What would be the size of data sent and received by a single app during one day? (I need a rough estimate, no real need for a precise answer to the closest bit)
Thanks
Gps Positions compactly stored, but not compressed needs this number of bytes:
time : 8 (4 bytes is possible too)
latitude: 4 (if used as integer or float) or 8
longitude 4 or 8
speed: 2-4 (short: 2: integer 4)
course (2-4)
So binary stored in main memory, one location including the most important attributes, will need 20 - 24 bytes.
If you store them in main memory as single location object, additonal 16 bytes per object are needed in a simple (java) solution.
The maximum recording frequence is usually once per second (1/s): Per hour this need: 3600s * 40 byte = 144k. So a smartphone easily stores that even in main memory.
Not sure if you want to transmit the data:
When transimitting this to a server data usually will raise, depending of the transmit protocoll used.
But it mainly depends how you transmit the data and how often.
If you transimit every 5 minutes a position, you dont't have to care, even
when you use a simple solution that transmits 100 times more bytes than neccessary.
For your school project, try to transmit not more than every 5 or better 10 minutes.
Encryption adds an huge overhead.
To save bytes:
- Collect as long as feasible, then transmit at once.
- Favor binary protocolls to text based. (BSON better than JSON), (This might be out of scope for your school project)
We've been trying to figure out why we only achieve writing speed of ~53MBps on UHS104 cards that claim 90MBps.
Due to hardware constraints, clock frequency supplied to the card is only 148.5 MHz (instead of 208MHz).
Does that mean that we should achieve speed of (148.5 * 4)/8 = 74.25MBps?
Or is our caclulation wrong since it assumes that if card guarantees speed of 90MBps on frequency of 208MHz, then it should guarantee speed of 74.25MBps on frequency of 148.5?
The simplified physical layer spec states that for maximum performance you need to write full AU blocks - usually 2 or 4 MByte, otherwise the card will have to copy data around internally when writing across block boundaries. Unfortunately, most of the Speed Class Specification is missing in the 4.13 chapter.
The first AUs may have a different wear level strategy, as they are normally used for the FATs. This could make them slower to write to.
I'm trying to measure ambient brightness values with a CdS light-dependent resistor (R2 in the below diagram.) R1 is a known value, and the resistance of R2 has an inverse log relationship with the current ambient brightness.
I can use the voltage in the middle of the voltage divider to calculate the value of R2, and therefore the brightness, in software. I intend to use the A-D capabilities of my PIC chip to measure that voltage. Although CdS cells are inherently a bit imprecise, I'd like to get the best precision possible. Is the input current flowing into pin AN0 negligible?
Vdd ---
|
R1
| -|--|-
+------ AN0-|IC|-
| -|--|-
R2
|
Vss ---
Bonus question - anyone know how I can linearize the inverse log response of my LDR? At 1 lux it's 1 megaohm, at 10 lux it's 100k, at 100 lux it's 10k etc. I need to calculate values as accurately as possible between 10 and 600 lux, so if I just plug it into my A-D I'm going to have to pick one end that has much crappier resolution.
It is negligible. Figure 9-4 of the datasheet specifies the equivalent currents and capacitances.
Linearization - good question. It would be quite hard for you to design a circuit that would perform phenomenally better than your A/D software approach.
In your schematic, A/D "sees" your LDR as source resistance. PIC datasheet specifies that source impedance shouldn't be more than 2.5k. If your range of light levels begins at 10lux, then the max source resistance you'll see will be 100k, which is well above required source impedance.
You can add an OpAmp buffer with high impedance inputs. The output impedance of most OpAmps is below 2.5k.
You can add a capacitor (0.1uF or so) from A/D input to ground. It will lower source impedance that sample and hold in the A/D sees.
To bring up your circuit and firmware, you could use a fixed 100k resistor in place of LDR.
No, if you will use operation amplifier betwen resistors ladder and MCPU input in any other case yes it is negligible. Read also the datasheed which will tell you what is the minimum recommended source resistance of your resistors ladder to reach 10 bit ADC resolution.
Second answer is yes it is possible to reach 12 bit resolution but not with one ADC measurment and only if you know what and haw are you measuring!