Exam solutions FYS3240/

Transcription

1 Exam slutins FYS3240/ Prblem 1 a) Explain hw the accuracy (cnstant frequency utput) f quartz crystal scillatrs is imprved. The accuracy is imprved using temperature cmpensatin (temperature cmpensated crystal scillatr - TCXO), r temperature cntrl using an ven cntrlled crystal scillatr (OCXO), where the scillatr is enclsed in a temperature cntrlled ven. b) Explain the IEEE 1588 and the NTP prtcls. IEEE 1588: Gives sub-micrsecnd synchrnizatin in distributed systems, and prvides a standard prtcl fr synchrnizing clcks cnnected via a multicast capable netwrk, such as Ethernet. All participating clcks in the netwrk are synchrnized t the highest quality clck in the netwrk. The highest ranking clck is called the grandmaster clck, and synchrnizes all ther slave clcks. IEEE 1588 uses a prtcl knwn as the precisin time prtcl (PTP), and the level f precisin achievable using PTP depends heavily n the jitter (the variatin in latency) present in the underlying netwrk tplgy. NTP: A prtcl designed t synchrnize the clcks f cmputers ver a netwrk, and can prvide accuracies f better than 10 ms ver Ethernet. The accuracy depends n the netwrk, and pint-t-pint cnnectins prvide the highest precisin. The User Datagram Prtcl (UDP) is used. (A time request is sent t a time server r t a GPS with an NTP server). c) Explain the limitatins f sftware timing based n the cmputer clck when yu try t measure the time between events in yur sftware cde. The Real Time Clck (RTC) is an integrated circuit n the mtherbard, with a typical reslutin f 1 millisecnd (1 khz). The sftware clck, called the cmputer clck, is maintained by the perating system based n the RTC interrupts. At cmputer start-up (pwer turned n) the cmputer clck (system time) is set t a value based n the RTC and then regularly updated based n interrupts frm the RTC chip. Therefre, the time reslutin f the cmputer clck is limited t 1 ms, and the accuracy f the cmputer clck is hardware and perating system dependent. If yu use a sftware timer functins (which use the cmputer clck) t cntrl a lp, then yu must expect differences in the time interval (perid) between each iteratin f the lp (this is called jitter). (In additin, the cmputer clck drifts away frm the crrect time. At the time f synchrnizatin with a time server the clck is reset t the crrect time, but with a small ffset).

2 p time (ms) T_avg jitter d) Hw can yu make mre accurate relative time measurements in yur sftware then what yu can btain using the cmputer clck? The questin is abut relative time measurements; e.g. number f millisecnds elapsed t calculate a result. That means that synchrnizatin t UTC time r GPS time is nt imprtant. What is imprtant is t have a clck with high time reslutin (millisecnd reslutin is ften nt sufficient), and the clck must have high accuracy (lw drift). Since we are nly interested in accurate relative time, nt the abslute time, we shuld use a high precisin cunter/timer Available cmputer hardware timers are: High Precisin Event Timer (HPET) 64-bit up-cunter with a frequency higher than 10 MHz. Time Stamp Cunter (TSC) 64-bit register in the CPU (cres) that increment each prcessr clck cycle. Hwever, can be unreliable n a mdern multicre cmputer due t: multicre cmputers can have different values in their timekeeping registers. variability f the CPU frequency due t pwer management technlgies r perfrmance technlgies such as Intel Turb Bst Technlgy. T get μs reslutin: Use the functins QueryPerfrmanceCunter and QueryPerfrmanceFrequency Synchrnizing the cmputer clck t GPS will nt be sufficient, since we still get the same reslutin prblem (1 ms) frm the cmputer clck (and the cmputer clck will

3 als drift after synchrnizatin). S, an alternative t use HPET and TSC is t add a timing/cunter card t yur cmputer. These cunter/timing cards are available with high precisin scillatrs (OCXO). S instead f reading the cmputer clck yu read a cunter value frm this card, frm yur sftware. Since mst DAQ-cards als have cunters an alternative is t read a cunter value frm a DAQ-card, but a DAQ card has a less accurate scillatr cmpared t a dedicated timing card. (With a knwn scillatr frequency yu can calculate the number f micrsecnds frm the number f cunts). e) Yu are prvided tw separate text files with measurements recrded frm tw different instruments. The first clumn in bth files shuld cntain a GPS timestamp fr each data sample. Hwever, cmparing the measurement data in the tw files yu discver that it must be an ffset in time between the timestamps in the tw files. What culd be the reasn fr this time ffset? Crdinated Universal Time (UTC) is the wrld time standard based n atmic clcks, but with leap secnds added at irregular intervals t cmpensate fr the slwing f the Earth's rtatin. Remember that GPS time des nt include leaps secnds, and as f July 2012 GPS time is 16 secnds ahead f UTC because f the leap secnds added t UTC. S, if yu see at 16 secnds difference in the time clumns (when cmparing crrespnding measurement data) this is due t that ne instrument has cnverted t UTC time, while the ther has nt. Anther errr surce (if nt the 16 sec difference) culd be in the time stamping prcess; causing that the time stamp is nt assciated t the crrect data sample (e.g. due t buffered data acquisitin). This can happen in a cmputer-based DAQ system unless yu d time stamping directly in hardware in the instrument/sensr. (Fr instance many high-speed cameras timestamp each image in hardware, inside the camera, befre the image cntaining the timestamp as part f the image data is transmitted / recrded) There are f curse several ther pssible explanatins. The text describes a time ffset, and with ffset we usually think f a cnstant (time invariant) deviatin frm the true/crrect value. Sme GPS errrs culd cause an ffset time drift between tw instruments, such as ne instrument lsing GPS signals, and with nly the instrument s internal quartz clck available the time will drift until the next GPS time fix. But the questin was mre abut a cnstant time ffset, and nt a time drift. It is f curse als pssible that the tw instruments used a different GPS, and if they had different cnstant errrs this wuld give a cnstant time ffset. But this errr is likely t be s small that yu wuld nt discver it given the prblem described. Remember that the GPS system depends n very accurate clcks in rder nt t get very large psitin errrs. Given a psitin errr f 100 meter, which is very large, this crrespnds t a time errr f abut 33 μs. (S = v *t, where v = 3*10 8 m/s). S this is nt a likely explanatin. If the instruments where placed at different gegraphic lcatins with large separatin each GPS culd experience different effects n the GPS signals thrugh the transmissin path frm the satellites. This culd als cause a time ffset. But again, this is nt s likely t be the explanatin, because the errr is

4 very likely t be t small t be discvered her. (Als remember that 1 μs sample interval f an instruments crrespnd t a sample frequency f 1 MHz). Prblem 2 a) Explain why UDP is used instead f TCP in applicatins such as vide streaming. UDP is nt reliable (packets can be lst) N flw cntrl N saturatin cntrl N retransmissin f data Hwever, UDP is a fast prtcl with small verhead (cmpared t TCP), and shrter delay (reactin time dwn t abut 10 ms). Smetimes lsing a few data packets is nt a prblem, and a shrt delay is mre imprtant than receiving 100 % f the data packets, such as in vide steaming. Therefre UDP is used fr bradcast and multicast f data, such as vide. Als it is used in timing prtcls such as NTP. Smetimes there is als nt sufficient time t retransmitt the data if data are missing (and still get a real-time data stream), such as when vide signals are distributed frm satellites. b) What is a jumb frame, and why is it used? A typical netwrk data package (frame) used t be abut 1500 bytes. A jumb frame is a larger data packet, and a cmmn jumb frame size is abut 9 kb. IPv4 supprts jumb packets up t 64 kb. By using jumb packets yu can transmit the same amunt f data with fewer packets. Thugh yu save a small amunt f bandwidth (by using fewer headers), yu dramatically reduce CPU usage because the PC spends less time analyzing packets. c) What is required in rder t run data acquisitin sftware n a mdern cmputer running Windws with prtected mde? Explain. Need a device driver in rder t allw hardware I/O peratins frm applicatin prgrams. In prtected mde, there are fur privilege levels r rings, numbered frm 0 t 3, with ring 0 being the mst privileged and 3 being the least. The perating system and sme device drivers run in ring 0 and applicatins run in ring 3 d) What are the advantages and disadvantages f selecting RAID-10 when ding data acquisitin? RAID-10 = RAID0 + RAID1 = Striping and mirrring Advantages: Bth increased speed and redundancy cmpared t a single drive Can sustain multiple drive failures Fast rebuild as data is cpied blck fr blck frm the surce t the target. (N parity calculatins are required as in RAID-5 and RAID-6).

5 Disadvantages: Cnfiguratin requires twice the number f hard drives (cmpared t RAID-0) e) What is the definitin f a real time system? A real-time system gives yu determinism, meaning that yu can predict when a sectin f yur prgram will execute. Hard real-time: system where it is abslutely necessary that respnses ccur within the required deadline Sft real-time: allws fr sme deadlines t be missed with nly a slight degradatin in perfrmance but nt a cmplete failure. f) Many cmputer-based data acquisitin systems are running n nn-real-time hardware using perating systems such as Windws 7. Explain the main hardware and sftware slutins used t make this pssible. We need t add data buffers in ur system, and t make the data transfer and ur sftware as efficient as pssible (t avid verwrite and verflw errrs). Hardware slutins: Direct Memry Access (DMA) transfer mechanism (the fastest way t mve data, and allws the CPU t d ther wrk), using the DMA-cntrller.

6 Buffered acquisitin - FIFO buffer n the DAQ card RAID-hardware and/r SSD drives (t increase write speed) Sftware slutins: Since it is nt pssible t ensure that a sftware cde will finish within a specified deadline n a nn-real-time cmputer and perating system the fllwing techniques are used in sftware: Buffered acquisitin - using a circular PC-buffer in RAM Prducer-cnsumer structure using multiple threads ( independent acquisitin and write-t-file speed). Data queues t send data between prducer and cnsumer lps Queues cntain a buffer (it is a FIFO) t make sure that n data are lst. g) Hw can yu t increase the signal-t-nise-rati (SNR) in a data acquisitin system? Using an amplifier (as clse t the sensr as pssible, t amplify the signal befre the nise enter e.g. the transmissin cable), fr instance with autmatic gain cntrl (AGC), t use the entire analg-t-digital cnverter (ADC) dynamic range. Use an ADC with mre bits per sample (The SNR f an ideal N-bit ADC is SNR(dB) = 6.02*N Use versampling (fllwed by digital P-filtering and dwn-sampling). Filtering (t remve nise and limit the signal bandwidth), in hardware r sftware (Remember frm basic electrnics that thermal nise (Jhnsn nise) in a resistrs is prprtinal t the square rt f the signal bandwidth) Averaging (when cnstant r peridic signals and randm nise); Apply general nise reductin techniques: Psitin nise surces away frm data acquisitin device, cable, and sensr if pssible Place data acquisitin device as clse t sensr as pssible t prevent nise frm entering the system

7 Twisted pairs, cax cable, shielding Use differential signals Avid grund lps. Tpics nt cvered in lectures: Smthing ck-in amplifier h) The equivalent circuit fr a sensr and the DAQ frnt-end electrnics is shwn in Figure 1. The sensr is mdelled as an ideal vltage generatr with utput vltage V in fllwed by a parallel cnnectin f a resistr R s and a capacitance C s n the utput. The DAQ frnt-end electrnics is mdelled as an ideal amplifier with gain 1, with a parallel cnnectin f a resistr R and a capacitr C n the input. The signal cable between the sensr utput and the frnt-end electrnics input is assuemd ideal, with zer resistance and zer capacitance. Figure 1: Equivalent circuit fr a sensr and the DAQ frnt-end electrnics (the signal cable between the sensr utput and the DAQ frnt-end electrnics input is assuemd ideal, with zer resistance and zer capacitance). k at the fllwing tw cases: 1) very lw frequency f the input signal frm the sensr (f 0) 2) very high frequency f the input signal (f ) Shw that the cnditns R >> R s and C << C s are required in rder t get Vut Vin Hints: All calculatins can be based n V = * I (Ohm s law with a cmplex impedance). k at the circuit as a vltge divider between the sensr impedance and the lad impedance. The fllwing relatins are prvided t assist in the calculatins: Vltage divider: Vut Vin s

8 Impedance f a resistr: R R Impedance f a capacitr : C 1, where ω = 2πf jc Impedance f a parallel cnnectin f a resistr and a capacitr: p R C C C R R V in 1) w frequency case where f 0: The abslute value f the capacitr impedance ( C ) is very high, and the impedance f the parallel cnnectin can be apprximated as: p C R C R The vltge divider between the sensr impedance and the lad impedance therefre gives: V ut R R R S V in This requires Vut Vin R R in rder t get a denminatr that is apprximately equal t R, and s 2) High frequency case where f : The abslute value f the capacitr impedance is very lw, and the impedance f the parallel cnnectin can be apprximated as: p C The vltge divider between the sensr impedance and the lad impedance therefre gives: V ut C, S C, C, V in 1 C 1 1 C C S V in

9 This requires C C in rder t get the denminatr apprximately equal t 1/C, and Vut Vin S