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Fast VRM Controller Architecture with Estimated Load CurrentFeedforwardI.I NTRODUCTIONT HE specifications for modern microprocessor voltage regulation modules(VRM’s)require that the microprocessor supply voltage follows a load line[1],(1) where is a reference voltage,is the desired load line slope(or regulator output impedance),and is the current supplied to the microprocessor load.This approach is also known as adaptive or optimal voltage positioning[2]. Table I gives sample microprocessor VRM specifications reflecting current industry trends.One of the main challanges of VRM design is handling the large,high-slew-rate load current steps,at low microprocessor supply voltages,and tight regulation tolerance.For fast load current transients,the regulator output impedance depends on both the size of the output capacitors and the delay of the VRM controller[3].In fact,given afinite load current slew rate and a controller response fast compared to the load slew time,the converter closed-loop output impedance can be made smaller than the output capacitor effective series resistance(ESR).Thus,it is beneficial to develop controllers with very fast response in order to avoid the need for very large output capacitors.II.C ONTROLLER A RCHITECTUREFig.1shows the block diagram of a typical microprocessor multi-phase VRM.In the analysis below the multi-phase converter is modelled as a single phase converter for simplicity,while in Section III a full four-phase converter is simulated.A.Fast Response with Output Current FeedforwardThe problem of following accurately a load line can be approached as a reference tracking problem,where has to track the right-hand side of(1).A common approach in tracking problems is to use feedforward from the reference signal to the output to handle the bulk of the regulation action,and use the feedback only to compensate for the imperfections of the feedforward[4].This approach can generally handle better fast reference signal changes compared to pure feedback regulation,since the gain and bandwidth of the feedforward are not limited by stability considerations.Indeed,with ideal load current feedforward,a switching converter can have zero output impedance[5].B.VRM modellingA control block diagram of the VRM with load current feedforward1is shown in Fig.2,whereinput voltage12Vreference output voltage1Vmax load current100Amax load slew rate50A/sclosed-loop output impedance 1.25output tolerance band50Fig.1.Four-phase VRM block diagram.Fig.2.VRM control block diagram with load current feedforward.is the transfer function between the controller command and the output voltage,with parameters defined in Table II, ,and(3) modelling respectively the total power train inductance and series resistance.The open-loop output impedance isand(6)C.Feedback Control LawThe feedback control uses a standard PID law with an additional high-frequency pole,(7)The derivative term zero is placed slightly above the cutoff frequency to provide dB/dec rolloff.The high frequency pole is designed to approximately cancel the output capacitor ESR zero,thus maintaining the dB/dec slope.The feedback control law parameters for the present design are given in Table II.TABLE IIP ROTOTYPE VRM DESIGND.Feedforward Control LawThe feedforward control law can be derived by setting the closed-loop output impedance(6)equal to the desired value,(9) Thus,the design of the feedforward control law requires knowledge only of the converter inductance and output capacitor time constant.E.Estimating the Load CurrentThe control strategy discussed above assumes that the load current is measured.Sensing the load current directly is not practical since it will require inserting a sensing resistor after the output capacitor,thus increasing the output impedance,or using an expensive Hall-effect current sensor.Alternatively,the load current can be reconstructed from estimates of the inductor and capacitor currents[3],since.The inductor current can be estimated with an network connected in parallel to the inductor.This approach has been used successfully in commercial products[6].In the VRM implementation block diagram in Fig.3the four phases have been modeled as a single phase with quarter of the inductance and inductor and switch resistance.An estimator consisting of and is connected across the inductor.If the time constant of the estimator matches the time constant of the inductor,,the voltage across is equal to the voltage across .The inductor current can then be estimated by dividing the voltage across by an estimate of the inductor resistance.For good matching the temperature depencence of has to be compensated for in the sensing amplifier [6].Further,in an actual four phase converter the phase currents have to be summed,which can be achieved by splittingin four resistors connected to the four switching nodes,and terminating on a single.Analogously,the capacitor current can be estimated from the output voltage with an network matching the time constant of the output capacitor[3].In Fig.3,and are chosen such that.The capacitor current is derived by dividing the voltage drop across by an estimate of the output capacitor ESR,. In the case of perfect matching of the estimator and power train parameters,,the injection of in the controller does not affect the closed-loop poles and zeros of the system.In practice,there typically is some mismatch between the estimator and power train parameters,resulting in becoming a function of the converter state variablesFig.3.VRM implementation block diagram.and hence altering the system pole and zero locations.For small mismatches this effect is small,and can be tolerated in a robustly designed controller.However,this assumption should be verified at design time based on expected component and circuit tolerances.F.Critical InductanceThe above discussion assumes that the converter duty ratio does not saturate,i.e.,in Fig.3does not reachor ground.In practice satisfying this assumption is challenging due to the small inductor voltage drop(1V) available during unloading transients.For the duty ratio to remain unsaturated during large unloading transients,the total converter inductance has to be below certain critical value[3],[7].The critical inductance calculation has previously been done assuming an infinite-slew-rate load current step[3].Taking into account the finite slew rate of the load current,a less conservative value for can be derived.(Derivation will be shown in the complete paper).For the specifications in Table I,23nH,hence phase inductors were designed for nH each.G.PWM ModulatorThe implementation of a switch modulation scheme having a very short delay is essential for achieving a very fast controller response.A standard PWM modulation scheme without latching appears a suitable choice in this respect.It compares the control signal to a triangular or sawtooth modulating waveform at the switching frequency(Fig.3). Multi-phase operation is achieved by using phase-shifted modulating waveforms for the different phases.The control signal fed into the PWM modulator is the sum of the outputs of the feedforward and feedback control laws.The load current feedforward signal ideally has no ripple related to the converter switching since it is derived from an exogenous variable.Thus,the output of the feedforward control law can be fed directly into a non-latched PWM modulator without causing undesirable high-frequency behavior.The feedback signal,however,can have a substantial ripple resulting from the switching action and the derivative term in.Switching frequency ripple in the control signal can lead to undesirable limit cycling and chaotic behavior[8].To prevent this from happening,a sample-and-hold(S/H)operating at the effective switching frequency is introduced in the feedback path,thus eliminating switching ripple from the control signal.The sample-and-hold is preceded by a resettable integrator()05101520253035400.850.90.9511.05V o (V)t (µs)V c (V)t (µs)(a)nominal power train(b)and power train variationsFig.4.Simulated response of VRM design from Table II to a 100A,50A/s loading and unloading transient.averaging the feedback signal over each effective switching period (),and thus providing good DC accuracy of the feedback control.The sample-and-hold and the resettable integrator introduce some additional delay in the feedback path,however this is not critical to the overall speed of response since fast load changes are handled by the feedforward path.As pointed out in Section II-A,the feedback path only compensates for imperfections in the feedforward control,and ensures DC accuracy.III.S IMULATION R ESULTSFig.4shows a simulation of the converter design in Table II.A coupled-inductor structure is used in order to reduce the inductor current ripple [9].The feedback and feedforward control laws follow (7)and (9),respectively.The top plots show the output voltage response to a A,A/s loading and unloading transient,while the bottom plots give the controller output going to the PWM modulator.Part (a)corresponds to simulation with nominal power train parameters,while part (b)shows four waveforms corresponding to variation in the capacitor ESR,combinedwithvariation in the total inductance.Notice that due to the inductance value selection below the critical inductance,the controller output does not go negative,i.e.,the duty ratio does not saturate.Finally,experimental prototype design is underway to corroborate the proposed architecture functionality.R EFERENCES[1]Intel Corp.,“V oltage regulator–down (VRD)10.0,”[Online].Available:/design/Pentium4/guides/25288501.pdf,April2003.[2]R.Redl,B.P.Erisman,and Z.Zansky,“Optimizing the load transient response of the buck converter,”in Proc.IEEE Applied PowerElectron.Conf.,1999,vol.1,pp.170–176.[3] A.V .Peterchev,Jinwen Xiao,and S.R.Sanders,“Architecture and IC implementation of a digital VRM controller,”IEEE Trans.on PowerElectron.,vol.18,no.1,pp.356–364,Jan.2003.[4]J.-J.E.Slotine and W.Li,Applied Nonlinear Control ,New Jersey:Prentice-Hall,1991.[5]R.Redl and N.O.Sokal,“Near-optimum dynamic regulation of DC–DC converters using feed-forward of output current and input voltagewith current-mode control,”IEEE Trans.on Power Electron.,vol.PE-1,no.3,pp.181–192,July 1986.[6]International Rectifier Corp.,“IR3081:XPHASE VR 10.0control IC,”Data Sheet.[Online].Available:/product-info/datasheets/data/ir3081.pdf,April 2003.[7] A.V .Peterchev and S.R.Sanders,“Low conversion ratio VRM design,”in Proc.IEEE Power Electron.Spec.Conf.,2002.[8]S.Banerjee and G.C.Verghese (Editors),Nonlinear Phenomena in Power Electronics:Attractors,Bifurcations,Chaos,and NonlinearControl ,New York:IEEE Press,2001.[9]Jieli Li,C.R.Sullivan,and A.Schultz,“Coupled-inductor design optimization for fast-response low-voltage DC–DC converters,”in Proc.IEEE Applied Power Electron.Conf.,2002,vol.2,pp.817–823.。
Controllers FEC,Standard•Sturdy control rack requiring a minimum of space•Analogue inputs/outputs and Ethernet optional•Quick installation using the SAC sensor/actuator connector system•User-oriented software–programming the way you think or according to standard ElectroniccontrolsystemsFrontEndControllers 7.1Controllers FEC,StandardKeyfeaturesThe installation-saving controller The FEC Standard is not just a new mini controller.It shows that there is still room for innovation in mini controllers at the start of the new millennium.With its robust extruded aluminium housing,it demonstrates thatcompact design and toughness can go hand in hand.Its connector system is accessible from the front,ensuring no wastage of space within control cabinets.And the sensor/actuator connector system SAC,making its world premiere in this product,very largely replaces terminal strips in the I/O area.This means that control cabinets with FEC Standard have a decisive advantage:Up to 50%less space required,and up to 40%less time.Thanks to the integration of a high-speed counter into every CPU,this mini controller is well able to carry out counting and simple positioning operations.Additionally,the optional analogue inputs/outputs turn a smart mini controller into a smart process controller.The two serial interfaces in every CPU make the FEC Standard into a talented communicator which allows program-ming via one interface and operation and monitoring via the other,at the same time.The leading concept in communication today is Ethernet,the “network of networks”.This can of course be integrated intoFEC Standard as an option.After all,smart automation technologydemands smart network technology.With Ethernet and a web server,the FEC Standard paves the way for the visualisation technology of tomorrow:Controller surfing.E l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,StandardKey featuresHardwareThe FEC Standard has a clip for a top-hat rail and corner holes for bolt-mounting using a mounting plate.All connections are accessible from the front;there is no need for additional space for connections from above or below.Power supplyThe FEC Standard is poweredexclusively via 24V DC as per modern control cabinet technology.24V DC (+25%/-15%)power supply for the controller itself,24V DC (+/–25%)power supply for the input signals,positive switching,24V DC output signals 400mA,proof against short-circuits and low-resistance loads.The analogue inputs/outputs are 0(4)...20mA I/Os,12bitresolution.Serial interfacesEvery FEC Standard is equipped with two serial interfaces –COM and EXT.These are universal TTL interfaces with a maximum data transmission rate of 115kbits/s.Depending on requirements,the interfaces can be used as RS232c (SM14or SM15)or RS485(SM35)interfaces.Adapters should be ordered separately.The COM interface is generally used together with the SM14for program-ming,while the EXT interface can be used for an MMI device,a modem or other devices with a serial interface.Ethernet interfaceThe FEC Standard versions with an Ethernet interface incorporate an Ethernet 10BaseT interface with an RJ45connection and a data transmission rate of 10Mbits/s.A combined “Link/Active”LEDindicates the connection status.The FEC Standard supports datacommunication and programming/troubleshooting via the Ethernetinterface.ProgrammingThe FEC Standard is programmed using FST.FST is a unique programming language rich in tradition and very easy to use,allowing “programming the way you think”:IF...THEN...ELSEFST also supports STEP operation for sequence programming.FST can be used for programming via Ethernet;a web server is alsoavailable.E l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,StandardKey featuresThe sensor/actuatorconnectorTogether with the FEC Standard,we are introducing an innovative new installation concept,the sensor/actuator connector SAC.Thisconnector combines three functions in a very compact design:•Connection of inputs,outputs and power supply•Status signal by means of an LED •Replaces terminal strip for sensors andactuatorsThe three-wire version of theconnector has internally connected straps for 0V and 24V DC.This allows any sensor (up to 3wires)or actuator (up to the maximumpermissible output current)to be feddirectly to the connector.There is no need for a terminal strip for sensors and actuators.This allows space savings in control cabinets of up to 50%.The SAC uses a tension-spring contact system.This means no need for screw connections.Solid wires can simply be pushed into the connector,while in the case of finely-stranded wire,all that is necessary is to open the contact by pressing on the relevant pin and then introduce the wire.Cable end sleeves can be used if desired but are not essential.The tension-spring system and the fact that no terminal strip between the controller and sensors/actuator is required means that a time saving of up to 40%can be achieved during installation.The pin assignment for the I/O panel is simple and is always the same:Pin 1+24V DC Pin 2Bit 0Pin 3Bit 1Pin 4Bit 2Pin 5Bit 3Pin 6Bit 4Pin 7Bit 5Pin 8Bit 6Pin 9Bit 7Pin 100VThe power supply for the LEDs istaken from the signal pins in the connector.This means that the entire input assignment can be checked without acontroller.E l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,StandardKey featuresProgramming withFSTProgramming the way you think How do we describe a machine?“When a workpiece reaches here,this cylinder should advance.”How does the software interpretthis?Or does your machine work through a sequence step by step?“First,this cylinder must advance and stop the workpiece,and then the workpiece must be clamped,and thenfinally...”How,for example,can we sub-divide a task?Program 0:Organisation Program 1:Set-up program Program 2:AutomationprogramProgram 3:Fault monitoring Program 4:Manual operation ...Program 63:TroubleshootingprogramHow does one controller communicate with another?Every controller with Ethernet can send and receive data from every other controller within a network –no matter whether this data relates to inputs,outputs,flags or registers.Central programming of distributed controllersEvery controller within a network can be programmed from any desired network interface.A controller on the World Wide Web FST incorporates a web server –the Internet and the world of automationmeet.Programming just couldn’t beeasier.E l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,StandardProduct range overview The FECStandardE l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,StandardProduct range overviewThe principle of the FECStandard11235461In each case 16digital inputs,24V DC,positive-switching 2Optionally:3analogue inputs/1analogue output3In each case 8digital outputs 4Power supply5Rotary RUN/STOP switch 62serial interfaces,option ofEthernetE l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,StandardTechnical data General technical dataFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660Max.operating temperature0...55°C Max.transport and storage temperature –25...+70°CRel.humidity 0...95%(non condensing)Operating voltage 24V DC +25%/–15%Power consumption <5W Degree of protection IP20Degree of protection Degree of protection III.Power pack in accordance with IEC 742/EN60742/VDE0551/PELV with at least 4kV insulation resistance or switched-mode power supplies with safety isolation as defined by EN 60950/VDE 0805are required Certification C-TickI/O connection Tension spring connector EMC EN 61000-6-4Digital inputsFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660Number1632Number of above usable as high-speed inputs (max.2kHz)2Minimum pulse length for TRUE:250µs,Minimum pause length for FALSE:250µs Input voltage/current 24V DC,typical 5mA Nominal value for TRUE 15V DC min.Nominal value for FALSE 5V DC max.Input signal delay Typical 5msElectrical isolationYes,via optocoupler Permissible length of connecting cable Max.30mStatus display via LED Optional,in connectorAnalogue inputsFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660Number 03Signal range 0(4)...20mA Resolution12bit,±3LSB Conversion time10ms Permissible length of connecting cable Max.30mDigital outputsFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660Number 816ContactsTransistorCurrent/voltage 24V DC,max.400mA Short circuit proofYesProof against low-resistance loads Yes,up to 5W Overload-proof YesElectrical isolation Yes,via optocoupler Switching speedMax.1kHzElectrical isolation in groups Yes,in each case 1byte Maximum group current 3.2A Switching cyclesUnlimitedStatus display via LED Optional,in connectorAnalogue outputsFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660Number 01Signal range 0(4)...20mA Resolution12bit Conversion time 10ms Max.load resistance700ΩE l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,StandardTechnical dataRotary switchFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660 Number1Positions16STOP/RUN0=Stop1...F=RUNSerial interfaceFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660 Number2Connection RJ12plug socketFeatures Serial,asynchronous,TTL level,no electrical isolationUse as RS232c PS1-SM14or PS1-SM15requiredTerminal assignment SM14/15Transmit,receive,RTS,CTSUse as RS485PS1-SM35requiredUse as programming interface9600bits/s,8/N/1Use as universal interface:COM300...9600bits/s,7N1,7E1,7O1,8N1,8E1,8O1Use as universal interface:EXT300...115,000bits/s,7N1,7E1,7O1,8N1,8E1,8O1SAC connectorFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660 Number of connectors required44778 Insulating material PBT,colour blackTemperature range PS1-SAC10/SAC30:–20...+100°Cp gPS1-SAC11/SAC31:–20...+75°CFlammability class V-0Grid dimension 3.5mmConnector system Spring connectionInsulation-stripping length9...10mmClamping range0.05...1.5mm2Single-conductor H05(07)V-U0.20...1.5mm2Multi-stranded without cable end sleeves0.5...1.5mm2Multi-stranded with cable end sleeves inaccordance with DIN46228/10.5...1.5mm2Multi-stranded hot-dip galvanized0.05...0.2mm2Current rating for strap contacts16ACurrent rating for individual contacts2A(max.6A per contact,please note the admissible loads for distributor board and supply contacts)EthernetFEC-FC400FEC-FC440FEC-FC600FEC-FC640FEC-FC660 Number01011Bus interface IEEE802.3(10BaseT)Data transmission speed10Mbits/sConnector RJ45Supported protocols TCP/IP,EasyIP,httpOPC server upon requestDDE server Yes,for EasyIP ElectroniccontrolsystemsFrontEndControllers 7.1Controllers FEC,StandardTechnical data ProgrammingFSTProgramming languages Version 4.02:statement list(with version 3.2:statement list and ladder diagram in German and English)Working languageGerman and English Number of programs and tasks per project64(0...63)Permissible input addresses0 (255)addressable as bits or words Permissible output addresses 0 (255)addressable as bits or words Number of flags10,000(0...9999),addressable as bits or wordsNumber of timers and counters 256(0...255)in each case,with 1status bit,1setpoint and 1actual value Number of registers (words)0 (255)addressable as words Programming interfaceRS232or Ethernet Number of different operations >28Subroutine Up to 200different subroutines per project C/C++Yes,for modules and drivers File handling Yes RS232c Yes ABG Yes FEDYesWeb server Yes (FST from version 4)RemanenceFlag words 0...255Register 0 (126)Timer and counter preselects and counter words 0...127PasswordPerformance1.6ms/1k instructions approx.E l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1Controllers FEC,Standard Technical dataType L1L2FEC-FC4…132.1114.2FEC-FC6…174.7156.8Ordering data–The FEC Standard with FST programmingDesignation Features Part No.TypeIPC controller16I/8O183862FEC-FC400-FST 16I/8O,Ethernet185205FEC-FC440-FST32I/16O191449FEC-FC600-FST32I/16O,Ethernet191450FEC-FC640-FST32I/16O,3/1analogue I/Os,Ethernet197157FEC-FC660-FST ElectroniccontrolsystemsFrontEndControllers 7.12006/0920064/7.1-2320062006/094/7.1-24Controllers FEC,StandardTechnical dataOrdering data –Connectors for the FEC Standard Designation Features Part No.TypePlug 1-row,no LED,tension-spring system 197159PS1-SAC10-10POL Plug 1-row,with LED,tension-spring system 197160PS1-SAC11-10POL+LED Plug 3-row,no LED,tension-spring system 197161PS1-SAC30-30POL Plug3-row,with LED,tension-spring system197162PS1-SAC31-30POL+LEDOrdering data –Cables for the FEC Standard Designation Features Part No.TypeProgramming cable RS232adapter for programming from PC,complete with neutral modem cable 188935PS1-SM14-RS232Converter RS232adapter for connection of any desired devices with a serial interface,with top-hat-rail clip,no neutral modem or RS232cable 192681PS1-SM15-RS232Converter RS485adapter,with top-hat-rail clip 193390PS1-SM35-RS485CableNeutral modem cable160786PS1-ZK11-NULLMODEM-1,5M Earthing setEarthing set for earthing of cable screening via the H-rail526683FEC-ZE30Ordering data –Display and operating units Designation Features Part No.Type Operator unitDisplay and operating unit,LCD with 4lines,20characters each,illuminatedbackground,4function keys,real-time clock and expansion interface,e.g.Ethernet533531FED-50Operator unit Display and operating unit,LCD with 4lines,20characters each,illuminated background,12function keys,numeric keypad,real-time clock and expansion interface,e.g.Ethernet533532FED-90Fieldbus interface Ethernet interface module for FED 533533FEDZ-IET Programming cable Programming cable for FED533534FEDZ-PC Cable Connecting cable FEC (RJ12,COM and EXT)to FED189432FEC-KBG6Ordering data –Software and manuals for the FEC Standard Designation Features Part No.TypeProgramming softwareFST software version4.X on CD,manuals on CD191440PS1-FST2-CD-WIN g g FST software version 4.1on CD with manual DIN A5in German 537927P .SW-FST4-CD-DE FST software version 4.1on CD with manual DIN A5in English 537928P .SW-FST4-CD-EN ManualSystem manual FEC Standard,German 525368P .BE-FEC-S-SYS-DE System manual FEC Standard,English525369P .BE-FEC-S-SYS-ENE l e c t r o n i c c o n t r o l s y s t e m sF r o n t E n d C o n t r o l l e r s7.1。
EWRF 5.8G FPV VTX OSD SettingEWRF 5.8G Audio/Video transmitters work well with several other commonly used Betaflight/Cleanflight FCs when using OSD setting. Almost all F3 and F4 targets (except for those with integrated VTx) are supported.Refer below on how to use OSD setting to change your VTx parameters. EWRF VTx’s settings of bands, channels, power and Pitmode through OSD is dependent on the flight controller firmware.1 Connection and configuration1.1 F3/F4 flight controllers with integrated BetaFlight OSD∙Wire three pins PWM/OSD, VIDEO IN and GND of the VTX to the∙PWM/OSD OUT to the Tx pin of an free Uart∙VIDEO OUT and GND to the Flight Controller accordingly.∙Ensure the Flight Controller is running Betaflight firmware Release 3.1 or later.∙Then connect the FC to Betaflight configurator.∙Make sure “VTX” button in the configuration tab is enabled1.2 Port ConfigurationOpen Ports tab, on UART Port that the VTx is connected to, select TBS SmartAudio from Peripherals drop down menu2 Accessing Betaflight OSD2.1 Getting into Betaflight OSD CMS (Configuration Menu System）When powering up your craft, the below screen will appear on your FPV receiver screen receiver. To access the OSD setting page: Throttle middle, Yaw left and Pitchup3 BAND and Channel SettingUse Pitch stick to move the cursor and select FEATURESUse Roll stick to enter FEATURES Menu shown below:Select VTX SA and enter its menu as shown in figure below. Within the VTX SA menu, use Roll stick to choose the band.Notes: EWRF AV transmitters support 48 channels, however they are selectable only through the button on the VTX (NOTE only if the VTx’s OSD wire is NOT connected to the FC). Currently channel and band selection through Betaflight OSD only supports 40 channels.After choosing the band and channel, navigate to SET and select YES to save your setting.4 Transmitting Power SettingFollow the same procedure as in the previous section navigate to POWER.EWRF e708/709TM3 support only three transmitting powers of 25mw, 200mw and 600mw. There are four power options of 25mw, 200mw, 600mw and 800mw in the menu of OSD setting. When you make power selection, please note that the selecting of 500mw or 800mw the VTx transmitting power will be 600mw.EWRF 7082TM supports transmitting power of 25mw, 100mw and 200mw, note that selecting 25mw in the OSD the VTx will be set to 25mw, the 200mw selection it will be 100mw and with 600mw or 800mw selection it will be 200mw.Notes: Power setting is effective immediately which is different from the setting of the band and channel which requires SAVE to be executed.5PitMode SettingPitMode is a feature that set the VTx to a very low amplitude of power output, it allows the user to power up the video transmitter for settings changes or perform checks on the craft without interfering with others. When in PitMode, the VTx transmitting distance is less than 2 meters.To set PITMODE from the OSD, navigate to CONFIG, then select PitMode setting shown below.Using the Pitch stick move the cursor to PIT FMODE, use the Roll stick to toggle:∙left to exit the PitMode (PIR)∙right to enter the PitMode (POR)Notes: After powering up the VTx and FC again, PitMode status will be saved even if the default PIR will display on the screen upon powering up. Ignore “POR or PIR” if it appears on the screen. The indication of PITMODE status is from the LEDs mode on the VTx (see specific VTx manual for information).。
Output regulation of a class of continuous-time Markovianjumping systemsShuping He a,b,c,Zhengtao Ding c,n,Fei Liu ba College of Electrical Engineering and Automation,Anhui University,Hefei230601,Chinab Key Laboratory of Advanced Process Control for Light Industry(Ministry of Education),Institute of Automation,Jiangnan University,Wuxi214122,Chinac Control Systems Centre,School of Electrical and Electronic Engineering,University of Manchester,Sackville Street Building,Manchester M139PL,UKa r t i c l e i n f oArticle history:Received24February2012Received in revised form20June2012Accepted3August2012Available online24August2012Keywords:Output regulationMarkovian jumping systemState feedbackError feedbacka b s t r a c tThis paper studies output regulation for continuous-time Markovian jumping systems(MJSs),for which mode-dependent regulation equations are investigated.With theextension of regulation scheme to MJSs by stochastic Lyaponov–Krasovskii functionalframework,sufficient conditions are,respectively,obtained for state feedback and errorfeedback.The resulting closed-loop system is guaranteed to be stochastically stable andthe output regulation error almost asymptotically converges to zero.A semi-definiteoptimization approach via disciplined convex programming is adopted to ensurerelaxed solutions of the regulation equations.Finally,two numerical simulations aregiven to illustrate the performance of the proposed approach.Crown Copyright&2012Published by Elsevier B.V.All rights reserved.1.IntroductionMarkovian jumping systems(MJSs)have receivedconsiderable attention in systems,circuit and controlcommunity for many years,see,for example,[1–25].MJSsoften arise in practical control systems which may experi-ence abrupt changes in structures and parameters due to,for example,sudden environment changes,subsystemswitching,system noises,failures occurred in componentsor interconnections and executor faults,etc.In this classof stochastic systems,the dynamics of jumping modesand continuous states are,respectively,modeled byfinitestate Markov chains and differential equations.In recentyears,the stability,controller design andfiltering pro-blems for MJSs have regained increasing interest andsome results are also available.On another research front,the asymptotic regulationproblem of the output of a dynamical system is one of thecentral problems in control theory.An important variantof this problem is the output regulation problem and hasbeen studied since the seventies[26,27].Basically,theoutput regulation problem is either a disturbance rejec-tion problem or a tracking problem or a combination ofthese two problems.Different with the conventionaldisturbance rejection and tracking problem,the systemdisturbances or reference signals are always infinite-energy ones and generated by an external system namedexosystem.The key feature of the output regulationproblem is tofind a measured output feedback controllersuch that the closed-loop system is asymptotically stableand the regulated output asymptotically tends to zeroregardless of the exosignals affecting the system.It can beused in areas such as set point control,reference signalstracking,disturbances rejection,and observer design forautonomous systems.However,very few reports in the literature considerthe output regulation of stochastic MJSs though there areplenty of results on output regulation of linear systemsContents lists available at SciVerse ScienceDirectjournal homepage:/locate/sigproSignal Processing0165-1684/$-see front matter Crown Copyright&2012Published by Elsevier B.V.All rights reserved.n Corresponding author.E-mail addresses:shuping.he@(S.He),zhengtao.ding@(Z.Ding),fliu@(F.Liu).Signal Processing93(2013)411–419[28–30]and nonlinear systems[31–38].For linear sys-tems,the design of the linear regulator was given in terms of certain matrix equations,for example,Francis equa-tions[27].The solution depends on the property of the exosystem signals to be observable for the system output. The extension to the nonlinearfield was given by[31], [32].It has been shown that the regulation problem is solvable by means of a set of partial differential equations henceforth named the Francis–Isidori–Byrnes equations[31].In this paper,we consider the output regulation of continuous-time MJSs.Different with the main results in [29,30],the suffcient conditions of this paper are identi-fied to guarantee solutions to output regulation via state feedback and error feedback for such stochastic systems based on stochastic Lyapunov–Krasovskii functional.The design criterions are presented in the form of linear matrix inequality(LMI)[39],which can be easily checked. And the relevant regulator problem is described as a semi-definite optimization(SDP)one via disciplined con-vex programming[40].Finally,two numerical simulations are included to illustrate the effectiveness of the devel-oped techniques.Throughout this paper,we use the following notations: R n and R nÂm stand for an n-dimensional Euclidean space and the set of all nÂm real matrices,respectively;A T and AÀ1denote the matrix transpose and matrix inverse; diag f A B g represents the block-diagonal matrix of A and B;s max(p)and s min(p)denote the maximal and minimal eigenvalues of a positive-define matrix P;99*99 denotes the Euclidean norm of vectors;E{*}denotes the mathematics statistical expectation of the stochastic pro-cess or vector;L n2ð01Þis the space of n dimensional square integrable function vector overð01Þ;;99x(t)992,E denotes the mean square norm of x(t)on time-interval½0t ,where99xðtÞ992,E ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiE f x TðtÞxðtÞgp;P o0or P40means matrix P is negative-definite or positive-define;I and0are,respectively,the unit and the zero matrices with appropriate dimensions;and‘‘*’’means the sym-metric terms in a symmetric matrix.2.Problem formulationGiven a probability space(O,G,P r)where O is the sample space,F is the algebra of events and P r is the probability measure defined on G.Let us consider a class of continuous-time MJSs defined in the probability space (O,G,P r)and described by the following differential equations,_xðtÞ¼Aðr tÞxðtÞþBðr tÞuðtÞþEðr tÞdðtÞeðtÞ¼Cðr tÞxðtÞþDðr tÞdðtÞxðtÞ¼x t0,rðtÞ¼r t,t¼t08><>:ð1Þwhere x(t)A R n is the state,u(t)A R m is the controlled input,d(t)A R p is the disturbance to be rejected,e(t)A R q is the error to be regulated x tis a vector-valued initial continuous function and r0is the initial mode. A(r t)A R nÂn,B(r t)A R nÂm,C(r t)A R qÂn,D(r t)A R qÂp, E(r t)A R nÂp are the mode-dependent matrices with from an exosystem,_dðtÞ¼SðrtÞdðtÞð2ÞThe jumping parameter r t in MJSs(1)and(2)represents a continuous-time discrete state Markov stochastic process taking values on afinite set L¼f1,2,ÁÁÁ,N g with transition rate matrix P¼{p ij},i,j A L and has the following transition probability from mode i at time t to mode j at time tþD t as P r¼P r f r tþD t¼j9r t¼i g¼p ij D tþoðD tÞ,if i a j1þp ii D tþoðD tÞ,if i¼j(ð3Þwhere limD t-0oðD tÞD-0as D t40.In this relation,p ij Z0is the transition probability rate and for i,j A L,i a j,we haveX Nj¼1,j a ip ij¼Àp ii:ð4ÞRemark1.To simplify the study,we take the initial time t0¼0and let the initial values x0and r0befixed.At each mode,we assume that the continuous-time MJSs have the same dimension.The coefficient matrices in system(1) and(2)are known mode-dependent constant ones with appropriate dimensions.For convenience,we denote A(r t), B(r t),C(r t),D(r t),E(r t),S(r t)as A i,B i,C i,D i,E i,S i,respec-tively,with r t¼i,i A L.Assume that:A1.The eigenvalues of S i,i A L are with non-negative real parts;A2.MJSs(1)are Lyapunov stochastically stable;A3.The pairA i E i0S i!ðC i D iÞ2435is detectable.Wefirst consider the following full-order state feedback controller,uðtÞ¼K i xðtÞþF i dðtÞð5Þwhere K i and F i are the state feedback controller para-meters to be designed.We can get the following closed-loop MJSs(6)by substituting(5)into MJSs(1)and(2),_xðtÞ¼ðAiþB i K iÞxðtÞþðE iþB i F iÞdðtÞeðtÞ¼C i xðtÞþD i dðtÞ_dðtÞ¼SidðtÞ:8><>:ð6ÞUnder the assumptions of A1and A2,the objective of this part is to design a feedback control law,such that: B1.The feedback system(6)formed by the feedback control law u(t)¼K i x(t)þF i d(t)is stochastically stable,i.e., almost asymptotically stable([3],[4],[7],[11]);B2.For any given initial states,the controlled state x(t) of the closed-loop system can track the desired reference signal Q i d(t),wherein Q i is the coefficient matrix,i.e.,theregulated output x(t)¼x(t)ÀQ i d(t)satisfies99xðtÞ992,E¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiE f x TðtÞxðtÞgq-0over time;B3.The regulated error tends almost asymptotically toS.He et al./Signal Processing93(2013)411–419 412Remark 2.In fact,when the eigenvalues of S i ,i A L are with negative real parts,the rejected disturbances d (t )-0as t -N .and the regulated error e (t )and the regulated output x (t )of the presented MJSs (1)and (2)almost asymptotically tend to zero obviously.In this case,it just needs to design a controller u (t )¼K i x (t )that makes the closed-loop MJSs (6)stochastically stable.Thus,this paper pays more attention to almost asymptotically tracking and almost asymptotic regulation problem when d (t )a 0as time goes on.The so-called output regulation problem of linear systems was first introduced by Smith and Davison [26]and Francis and Wonham [27].For more results of this topic,we refer readers to [28–30]and the references therein.For the linear dynamic system without Marko-vian jumping,the problem of output regulation is solvable via state feedback if and only if there exists matrices Q and R that satisfy the following linear matrix equations QS ¼AQ þBR þECQ þD ¼0:(ð7ÞIn this paper,we study the output regulation problems of continuous-time MJSs.Our aim is to design a feedback controller satisfying B1–B3under the assumptions A1-A2.Before proceeding with the study,the following defini-tions for some given initial conditions can be formalized.Definition 1.The MJSs (1)and (2)(setting u (t )¼0,d (t )¼0)are said to be stochastically stable if,for any initial x (t )¼x 0and initial mode r t ¼r 0,thenlim T -1E Z T 099x ðt ,x 0,r 0Þ992dtr 0,x ðt Þ¼x 0&'o 1:ð8ÞDefinition 2.The MJSs (1)and (2)are said to be stochas-tically stabilizable if there exists a feedback control law of form (5),then the closed-loop MJSs (6)are stochastically stable.Definition 3.Define the time differential of the regula-tion output x (t )¼x (t )ÀQ i d (t )as _xðt Þ¼_x ðt ÞÀQ i _d ðt ÞÀX N j ¼1p ij Q j d ðt Þ:ð9ÞRemark 3.It should be noted that the added term P Nj ¼1p ij Q j d ðt Þin (9)is needed based on the time-dependent modes r t ¼i in MJSs (1)and (2).When we take the time differential of regulation output function x (t ),the time-dependent modes r t ¼i will be impossible to ignore.For more details about this topic,we refer readers to the results in [1],[3–5]and [11].3.Main resultsTheorem 3.1.Under the assumptions A1-A2,the problemof output regulation is solvable via state feedback if the (a1).There exist a set of matrices Q i and R i ,satisfying thefollowing regulator equations ,Q i S i ¼A i Q i þB i R i þE i ÀP N j ¼1p ij Q jn C i Q i þD i ¼0:ð10Þ(a2).The following LMI (11)holds for a set of positive-definite and mode-dependent matrices X i and mode-dependent matrices Y i .G i M ðX i Þn N ðX i Þ"#o 0ð11Þwhere G i ¼A i X i þX i A T i þB i Y i þY T i B T i þp ii X i ,M ðX i Þ¼ffiffiffiffiffiffiffip i 1p X ÁÁÁffiffiffiffiffiffiffiffiffiffiffiffiffip i i À1ðÞp X i ffiffiffiffiffiffiffiffiffiffiffiffiffiffip i i þ1ðÞp X iÁÁÁffiffiffiffiffiffiffip iN p X i h i,N ðX i Þ¼Àdiag f X 1ÁÁÁX i À1X i þ1ÁÁÁX N g :Moreover,the state feedback controller gain matrices areK i ¼Y i X À1i ,F i ¼R i ÀK i Q i :ð12ÞProof.Let x (t )¼x (t )ÀQ i d (t ).By Definition 3and consider-ing relation (10),we have_x ðt Þ¼_x ðt ÞÀXN j ¼1p ij Q j d ðt ÞþQ i _dðt Þ2435¼ðA i þB i K i Þx ðt ÞþðE i þB i F i Þd ðt ÞÀA i Q i þB i R i þE i ÀXN j ¼1p ij Q j 0@1A d ðt ÞÀX N j ¼1p ij Q j d ðt Þ¼ðA i þB i K i Þ½x ðt ÞÀQ i d ðt Þ þB i ðK i Q i þF i ÀR i Þd ðt Þ¼ðA i þB i K i Þx ðt Þð13ÞLet the mode at time t be i ;that is r t ¼i A L .Takethe stochastic Lyapunov–Krasovskii functional V (x (t ),i ,t Z 0):R n ÂR ÂR þ-R þto be V (x (t ),i )¼x T (t )P i x (t ),wherein P i 40is a positive-definite matrix for each mode i A L .The weak infinitesimal operator I [U ]of the process (x (t ),i ,t Z 0)for closed-loop MJSs (13)at the point {t ,x (t ),i }is given by [1],[3–5]and [11]I V ðx ðt Þ,i Þ¼lim D t -01D tE V ðx ðt þD t Þ,r t þD t ,t þD t Þ9x ðt Þ,r t ¼iÈÉÂÀV ðx ðt Þ,i Þ :ð14ÞTake the time differential of V (x (t ),i )along the trajec-tories of the closed-loop MJSs (13),and it yields,I V ðx ðt Þ,i Þ¼x Tðt ÞðA i þB i K i ÞT P i þP i ðA i þB i K i ÞþX N j ¼1p ij P j 2435x ðt Þ:Thus,it concludes that I V (x (t ),i )o 0can be guaranteed byðA i þB i K i ÞTP i þP i ðA i þB i K i ÞþX N j ¼1p ij P j o 0:ð15ÞPre-and post-multiplying inequality (15)by block-diagonal matrices P À1i ,applying Schur complementformula and letting X i ¼P À1i and Y i ¼K i X i ,inequality (15)S.He et al./Signal Processing 93(2013)411–419413And if matrix inequality (15)holds,there will exist matrix X i 40,such that I V ðx ðt Þ,i Þ¼ÀE f x Tðt ÞX i x ðt Þg :Since I V (x (t ),i )o 0,we can get E f V ðx ðt Þ,i Þg o E f V ðx 0,r 0Þg ¼E f x T0X i P ðr 0Þx 0g :Then,the following relation holds,I V ðx ðt Þ,i Þx o ÀE f x Tðt ÞX i x ðt Þgx 00:Define M 1¼inf 0r d r tE f 99x ðd Þ992g ,M 2¼E {99x 0992},s 1¼min i 2Ls min ðX i Þ,s 2¼max i 2Ls max ðP ðr 0ÞÞ.Therefore,there exists a positive number s 40satisfying the following relation,I V ðx ðt Þ,i ÞE f V ððt Þ,i Þg o ÀE f x Tðt ÞX i x ðt Þg E f V ð0,r 0Þg r ÀM 1s 1M 2s 2¼Às :Since M 140,M 240,s 140,s 240and s 40,we have I V ðx ðt Þ,i Þo s E f V ðx ðt Þ,i Þg :That is,E f V ðx ðt Þ,i Þg o e Às t E f V ðx 0,r 0Þg :By letting r ¼M 2s 2,for a given small positive scalar l 40,we can getl E f x T ðt Þx ðt Þg r E f V ðx ðt Þ,i Þg o r e Às t :ð16ÞLetting t go to infinity implies thatlim t -1E 99x ðt Þ992-0:ð17Þwhich also implies that 99x (t )992,E -0as t -N by the mainresults in [1],[3–6]and [11].Taking the limit as t -N ,it follows from relation (16)thatlim t -1E Z t 099x ðt Þ992&'dt o lim t -1r 1Àe Às t Âür o 1:Recalling Definition 1,we know that MJSs (13)arestochastically stable.On the other hand,e ðt Þ¼C i x ðt ÞþD i d ðt Þ¼C i x ðt ÞþðC i Q i þD i Þd ðt Þ¼C i x ðt Þð18ÞFrom (10)and (18),we know that lim t -199e ðt Þ992,E -0.This completes the proof.Remark 4.Notice that the output regulator design in this part is under the complete access to the system states.According to the definitions in [3],[5]and [11],we know that the closed-loop MJSs (13)are mean square stable when the regulated output x (t )satisfies relation (17).By the main proof of Theorem 3.1,we can see that mean square stable implies stochastically stable.In order to obtain the output regulation condition for MJSs,the coefficient matrix Q i is selected as a mode-dependent one.By the time-differential of the selected stochastic Lyapunov–Krasovskii functional V (x (t ),i )with the defini-tion in relation (14),we can get the main results in Theorem 3.1.If the coefficient matrix Q i is selected as a mode-independent one,that is Q i ¼Q ,then we have the follow-Corollary 3.1.Under the assumptions A1and A2,theproblem of output regulation is solvable via state feedback if the following relations hold for all i A L .(b1).There exist matrices Q and R,satisfying the follow-ing regulator equations ,QS i ¼A i Q þB i R þE iC i Q þD i ¼0:(ð19Þ(b2).LMI (11)with a set of positive-definite and mode-dependent matrices X i and mode-dependent matrices Y i holds .Moreover,the state feedback controller gain matrices areK i ¼Y i X À1i ,F i ¼R i ÀK i Q :ð20ÞPractically the complete access to the states is not the fact for many reasons such as the unavailability of the sensors to measure some of the state variables,and conse-quently the previous control approach will not be feasible.To overcome such problem,we first consider the jumping observer with the following state space representation,_^x ðt Þ¼A i ^x ðt ÞþB i u ðt ÞþE i ^d ðt ÞþH 1i ½e ðt ÞÀ^e ðt Þ _^d ðt Þ¼S i ^d ðt ÞþH 2i ½e ðt ÞÀ^e ðt Þ ^e ðt Þ¼C i^xðt ÞþD i^d ðt Þ8>><>>:ð21Þwhere H 1i and H 2i are observer gains to be designed.Definethe estimation error ~xðt Þ¼x ðt ÞÀ^x ðt Þ,~d ðt Þ¼d ðt ÞÀ^d ðt Þ,then we can get the following observer error dynamic MJS (22)by substituting (21)into (1),_~x ðt Þ¼ðA i þH 1i C i Þ~x ðt ÞþðE i þH 1i D i Þ~dðt Þ_~d ðt Þ¼H 2iC i~x ðt ÞþðS iþH 2iD iÞ~dðt Þ8<:ð22ÞAnd,it can be rewritten as _z ðt Þ¼A z iz ðt Þð23Þwhere z ðt Þ¼~x ðt Þ~d ðt Þ"#,A z i ¼A iE i 0S i!þH 1i H 2i!ðC iD i Þ.It shows from assumption A3that the eigenvalues of the matrix A z i can be specified in the left-half complex plane.But it does not mean the stability of the jumping observer error dynamics in (22).For these,we can invoke the stochastic stability conditions as described in (11)in the main proof of Theorem 3.1.Stated by the following Theorem,we can get the stochastic stability results of jumping observer error dynamic MJSs (22).Theorem 3.2.Under the assumption A3,the jumping observer error dynamic MJSs (22)are stochastically stable if there exist a set of positive-definite and mode-dependent matrices P 1i and P 2i and a set of mode-dependent matrices L 1i and L 2i satisfying the following LMI :S 1S 3nS 2"#o 0ð24Þwhere S 1¼P 1i A i þA T i P 1i þL 1i C i þC T i L T1i þP Nj ¼1p ij P 1j ,S 2¼P 2i S i þS T i P 2i þL 2i D i þD T i L T 2i þP N j ¼1p ij P 2j ,S.He et al./Signal Processing 93(2013)411–419414Moreover,the jumping observer gain is given byH1i¼PÀ11i L1i,H2i¼PÀ12iL2i:ð25ÞProof.For the jumping observer error dynamic MJSs(22), we take the stochastic Lyapunov–Krasovskii functional Vð~xðtÞ,~dðtÞ,iÞas follows,Vð~xðtÞ,~dðtÞ,iÞ¼~xðtÞP1i~xðtÞþ~dðtÞP2i~dðtÞ:Then following the similar proof in Theorem3.1,we can get the main results of Theorem3.2and this completes the proof.For MJSs(22),we consider the following feedback controller by estimated error feedbackuðtÞ¼K i~xðtÞþF i~dðtÞð26Þwhere K i and F i are the error feedback controller para-meters to be designed.Then,one has the following closed-loop MJSs(27)by substituting(26)into(1)and(2), _xðtÞ¼ðAiþB i K iÞxðtÞþðE iþB i F iÞdðtÞÀB i K i~xðtÞÀB i F i~dðtÞ^eðtÞ¼Ci xðtÞþD i dðtÞ_dðtÞ¼Si dðtÞ:8><>:ð27ÞTheorem3.3.Under the Assumption A1-A3,the problem of output regulation is solvable via error feedback if the following conditions hold for all i A L.(c1).There exist a set of mode-dependent matrices Q i and R i satisfying the regulator equations in(10).(c2).The mode-dependent LMI(11)with a set of positive-definite and mode-dependent matrices X i and mode-dependent matrices Y i holds.(c3).The mode-dependent LMI(24)with a set of positive-definite and mode-dependent matrices P1i and P2i and mode-dependent matrices L1i and L2i holds.Proof.Similar to the proof in Theorem 3.1,we define x(t)¼x(t)ÀQ i d(t)and take the time-differential of x(t), then it yields,_xðtÞ¼_xðtÞÀX Nj¼1p ij Q j dðtÞþQ i_dðtÞ2 43 5:¼ðA iþB i K iÞxðtÞþðE iþB i F iÞdðtÞÀB i K i~xðtÞÀB i F i~dðtÞÀQ i S i dðtÞÀX Nj¼1p ij Q j dðtÞ¼ðA iþB i K iÞ½xðtÞÀQ i dðtÞ ÀB i K i~xðtÞÀB i F i~dðtÞþA i Q iþB iðK i Q iþF iÞþE iÀQ i S iÀX Nj¼1p ij Q j2 435dðtÞ:ð28ÞBy the relation(10)and(12),we have_xðtÞ¼ðAiþB i K iÞxðtÞÀB i K i~xðtÞÀB i F i~dðtÞ:ð29ÞBy combining(29)and the observer error dynamic MJSs(22),it follows that_xðtÞ¼ðAiþB i K iÞxðtÞÀB i K i~xðtÞÀB i F i~dðtÞ_~xðtÞ¼ðAi þH1i C iÞ~xðtÞþðE iþH1i D iÞ~dðtÞ:8>><>ð30ÞIt should be pointed out that,if both the conditions (c2)and(c3)in Theorem 3.2are satisfied,the above system is stochastically stable for99~xðtÞ992,E-0and 99~dðtÞ992,E-0as t-N.Then we have99x(t)992,E-0as t-N.Moreover,we can get the following relation by condition(c1)eðtÞ¼C i xðtÞþD i dðtÞ¼C i xðtÞþðC i Q iþD iÞdðtÞ¼C i xðtÞ:ð31ÞIt means limt-199eðtÞ992,E-0.This completes the proof. Remark5.The separation principles are used to obtain the controller design and the observer design via error feedback here.Taking into account LMI(24),one has thefact that99~xðtÞ992,E-0and99~dðtÞ992,E-0as t-N.Then wehave99xðtÞ992,E-0as t-N.Following this,the controller and the observer can be,respectively,designed by LMI (11)and(24).When there are difficulties of solving(10),we cantransform(10)into the following SDP problems viadisciplined convex programming[40],min Zsubject toZ I A i Q iþB i R iþE iÀX Nj¼1p ij Q jÀQ i S in Z I26643775Z0,Z I C i Q iþD in Z I"#Z0:ð32ÞIn fact,to make the relative terms approximate with a satisfactory precision,we can alsofirstly select a suffi-ciently small scalar Z40to meet(32).Remark6.The solutions of Theorems3.1and3.3can be obtained by solving a SDP problem via disciplined convex programming with(32)and solving LMI(11).By using the relevant Matlab Toolbox,it is straightforward to check the feasibility of the disciplined convex programming and LMI.Remark7.Indeed,the applications of output regulation are comprehensive in industrial control processes,for example,set point control,reference signals tracking, disturbances rejection,and observer design for autono-mous systems,etc.It should be observed that the con-tributions of this paper are mainly theoretical aspects. As a widely used stochastic system,the proposed methods can be considered in future research.In order to illustrate the effectiveness of the developed techniques,we will give two numerical examples in the following Section4. Remark8.As one of the central problems in control theory, output regulation control was widely investigated in theo-retical and practical aspects.In this paper,we succeeded in designing the output regulator of continuous-time stochas-tic MJSs in the cases that the system states are accessible or not completely paring with the output regulation study for piecewise-linear systems[29,30],our researches are more focused on the fact that how to simplify the output regulator design procedure of stochastic MJSs byS.He et al./Signal Processing93(2013)411–419415output regulator design schemes also adapt to the systems in which the states are not completely accessible.4.Numeral examplesExample 4.1.We consider the following continuous-time stochastic MJSs with two jumping operation modes described asA 1¼À12À2À3!,A 2¼À0:51À2À3!,B 1¼0132!,B 2¼00:532 !,C 1¼C 2¼0:1À0:1ÂÃ,D 1¼À0:20:2ÂÃ,D 1¼0:1À0:1ÂÃ,E 1¼0:10:2À0:10:1 !,E 2¼0:1À0:10:20:1 !,S 1¼S 2¼03À3!:Selecting the transition rate matrix that relates the twooperation modes as P ¼À0:50:50:3À0:3!and solving theSDP optimization problem in (32)and LMI (11),we can get the optimal Z ¼1.8274Â10À11and the solutions as follows:K 1¼À0:33470:83050:50000:0042 !,F 1¼0:58141:0126À2:8895À0:2789 !,K 2¼À0:29010:71980:43330:0036!,F 2¼3:92891:7293À5:5978À0:3418!:For the initial conditions x 1(0)¼0.8and x 2(0)¼À1.0,we can get the simulation results of the jumping modes,the tracking response of state x (t )and Q i d (t ),and the regulated error e (t )in Figs.1–3.By the simulation results,the output e (t )to be regulated almost asymptotically tends to zero and the output tracking performance is quite satisfactory although there exists obvious transient tracking errors.In fact,when the states are not available,we can also study the output regulation problems for MJSs.In this situation,it just needs to design a jumping observer system to estimate the states and disturbances of the original model.Example 4.2.We consider the following two-mode con-tinuous-time stochastic MJSsA 1¼À12À2À3 !,A 2¼À2À33À2!,B 1¼B 2¼0:2000:2 !,C 1¼0:1À0:1ÂÃ,C 2¼À0:10:1ÂÃ,D 1¼À0:20:2ÂÃ,D 1¼0:1À0:1ÂÃ,E 1¼0:10:2À0:10:1!,E 2¼0:1À0:10:20:1 !,S 1¼S 2¼03À30 !:051015202530354045500.511.522.5t/sJ u m p i n g m o d e sFig.1.Jumpingmodes.S.He et al./Signal Processing 93(2013)411–419416。
管理学原理第⼗⼀章⾃测练习题Chapter Eleven: Foundations of ControlMultiple Choice Questions1. The type of control system that focuses on such criteria as organizational values, beliefs, and traditions isa. market controlb. organizational controlc. bureaucratic controld. clan control2. Which of the following is not a reason for the importance of control?a. It ensures that the organization is moving toward established goalsb. It is related to planningc. It allows managers to delegated. It helps establish hierarchical relationships3. A disadvantage of personal observation as a method for measuring performance isa. The level of coverage of performance activities is limitedb. Infrequencyc. The time involvedd. Others’ biases4. All of the following are characteristics of an effective control system excepta. accuracyb. reasonable criteria.c. low costd. timeliness5. Once actual result have been measured, the next step in the control process isa. determination of goals consistent with organizational objectivesb. comparisons between the plan and resultsc. checking measurements against established legal standardsd. taking action to correct unfavorable variations from the plan6. Which of the following statements is most accurate?a. The best-known form of feedforward control is direct observation.b. Feedforward controls have the greatest impact on motivation.c. The type of control most widely used is feedback control.d. Feedback controls are designed to detect problems before they occur.7. The biggest problem with feedback controls is thata. It is difficult, if not impossible, to obtain the necessary informationb. They are of little use as a motivational toolc. They do not provide the type of information needed to judge the accuracy of the planning processd. When significant deviations are detected, the damage has already been done8. The MAJOR problem with feedforward controls is that theya. are costly to implementb. are time-consuming to implementc. require information that is often difficult or impossible to obtaind. indicate a problem only after it has occurred9. The BEST-KNOWN form of concurrent control isa. feedbackb. decentralizationc. internal selectiond. direct supervision10. Reviewing reports is a form ofa. feedback controlb. concurrent controlc. feedforward controld. consecutive control11. Control should be placeda. on all activitiesb. where there are problem areasc. on the single most important factord. where there are cost-effective12. Informal control are recommended whena. Organizational size is largeb. Importance of an activity is lowc. Decentralization is highd. Organizational culture is threatening13. The last step in the control process is toa. establish standards.b. determine the need for corrective action.c. compare performance against standards.d. measure performance.14. The _____, the more managers will be informal, personal management by walking around.a. greater the degree of decentralizationb. greater the degree of centralizationc. smaller the organizationd. smaller the span of control15. Very large organizations have ___________ controls.a. highly formal, impersonal feedbackb. highly informal, impersonal feedbackc. highly formal, personal feedbackd. highly informal, personal feedback16. Controls in organizations with more open and supportive the cultures area. formal, externally imposed, impersonal.b. informal, self controlled.c. informal, personal, management by walking around.d. elaborate, formal.17. The more important the activity and the more centralized the organization, thea. more elaborate, comprehensive, formal rules and regulations.b. more loose, informal controls.c. more informal, management by walking around.d. less elaborate, comprehensive controls, formal rules and regulations.18. Which of the following is not a feedforward control measure for reducing employee theft and fraud?a. Pre-hire screeningb. Anti-theft education and trainingc. Treating employees with respectd. Involving employees in developing theft and fraud deterrent policies19. _____ is the process of monitoring activities and correcting any significant deviations.a. Planningb. Organizingc. Leadingd. Controlling20. An effective control system ensures that activities are completed in ways thata. motivate employees.b. maximize profits.c. lead to attainment of the organization's goals.d. minimize conflict.21. There are three different approaches to designing control systems. Which of the following is not one of these approaches?a. market controlb. economic controlc. bureaucratic controld. clan control22. _____ is a control approach that emphasizes authority and relies heavily on administrative rules, policies, and procedures.a. Market controlb. Bureaucratic controlc. Clan controld. Price control23. _____ is a control approach often used in organizations in which teams are common and technology is changing rapidly.a. Market controlb. Clan controlc. Bureaucratic controld. Price control24. Alta Corp. has turned each of its divisions into profit centers and evaluates each division by the percentage of total corporate profits that it generates.Alta uses _____ control.a. accountingb. bureaucraticc. marketd. sector analysis25. Beta Corp. managers are allowed a great deal of autonomy and freedom, butthey must stay within their budgets and follow corporate guidelines. Beta uses _____ control.a. accountingb. bureaucraticc. marketd. sector analysis26. The key to an effective control system is to design one thata. monitors everything happening in the organization.b. monitors only the important activities in the organization.c. helps the organization to effectively and efficiently reach its goals.d. requires a minimum of management involvement.27. The standards that management uses as specific performance indicators are developed in the _____ function.a. planningb. organizingc. leadingd. controlling28. Planning _____ control.a. followb. occur at the same timec. preceded. has no relationship to29. Which of the following is not a step in the control process?a. Measure actual performance.b. Compare actual performance against the standard.c. Develop the standard.d. Take appropriate action to correct any deviations.30. Which of the following methods of measurement reports only on key areasand may ignore important facts?a. personal observationb. statistical reportsc. oral reportsd. written reports31. When a manager discovers a variation between actual performance and the standard, managers shoulda. reevaluate the standard.b. take action to control the actual performance.c. determine the acceptable range.d. do nothing and see if performance improves.32. If the source of variation has been deficient performance, the manager willwant to take corrective action. Examples of corrective action include all of the following excepta. changes in structure.b. changes in the standard.c. redesign of jobs.d. replacement of personnel.33. When problems are corrected at once to get performance back on track,_______ has been used.a. mechanistic controlb. immediate corrective actionc. basic corrective actiond. remedial corrective action34. If a manager asks how and why performance has deviated, and then correctsthe source of the deviation, she is usinga. mechanistic control.b. immediate corrective action.c. basic corrective action.d. remedial corrective action.35. The key to _____ control is taking managerial action before a problem occurs.a. standardsb. feedbackc. concurrentd. feedforward36. Control action that is taken to correct a problem that has already occurred is:a. feedforward control.b. concurrent control.c. feedback control.d. management control.37. Which of the following statements is not true?a. Very large organizations will typically have highly formalized andimpersonal feedforward and feedback controls.b. The higher one moves in the organization's hierarchy, the greater the need for a multiple set of criteria.c. The type and extent of controls should be consistent with the organization's culture.d. If control is costly, and the repercussions from the error are small, the control system must be elaborate.38. _____ is the final link in the functional chain of management.a. Planningb. Controlc. Leadershipd. Goal setting39. Controls are directed at several areas. Which of the following is usually not one of those areas?a. informationb. financesc. technology usaged. people40. Which is probably most cost effective for a small business?a. feedforward controlb. concurrent controlc. feedback controld. empirical control。
around 抛异常controller处理-回复在程序开发中，一种常见的情况是当程序遇到错误或异常时，需要抛出异常来提醒开发者。
异常处理是保证程序正确运行的重要环节，而在控制器层中，处理异常的能力尤为重要。
本文将围绕着在控制器中处理抛出的异常展开，详细介绍异常的概念、抛出异常的方式以及在控制器中处理异常的具体步骤。
1. 异常的定义与概念异常是程序运行过程中发生的错误或特殊情况，可以干扰正常程序流程的事件。
异常分为可检查异常（Checked Exception）和不可检查异常（Unchecked Exception）。
可检查异常通常是由外部因素引起的，如输入错误或外部资源不可用；而不可检查异常则是由程序内部错误引起的，如数组越界、空指针引用等。
2. 抛出异常的方式在控制器中，抛出异常可以使用`throw`关键字。
通过`throw`将异常抛给调用者，调用者可以是控制器层的其他方法或其上一级的调用链。
# 2.1. 创建异常类在抛出异常之前，需要先定义异常类。
通常情况下，一个程序会有多个自定义异常类，用于处理不同类型的异常。
自定义异常类需要继承自`Exception`或`RuntimeException`，并可以添加额外的自定义属性和方法。
# 2.2. 抛出异常实例当遇到异常情况时，使用`throw`关键字创建并抛出对应的异常实例。
可以通过异常类的构造函数传递错误信息，以便在异常处理中进行验证或展示。
3. 在控制器中处理抛出的异常控制器是MVC模式中的关键组件之一，负责接收用户请求并处理业务逻辑。
在处理过程中，不可避免地会遇到异常情况。
下面将从以下几个步骤分析如何在控制器中处理抛出的异常。
# 3.1. 捕获异常为了在执行过程中捕获异常，可以使用`try-catch`代码块。
将可能抛出异常的代码放在`try`块中，然后通过`catch`块捕获异常并进行处理。
# 3.2. 处理异常在`catch`块中，可以根据不同的异常类型进行不同的处理。
Feedforwardnet是一种常见的神经网络模型，它具有前馈传递函数，用于对输入数据进行处理并生成输出。
本文将介绍feedforwardnet 的默认传递函数，包括其定义、特点和应用。
1. 传递函数的定义在feedforwardnet中，传递函数是指在网络中一个神经元的输出如何传递给下一个神经元的规则。
默认传递函数通常指的是线性传递函数，它表示输入信号通过神经元的加权和再加上偏置项的结果。
数学表达式为:\[y = 信信 + b\]其中，y表示神经元的输出，w表示权重，x表示输入信号，b表示偏置项。
默认传递函数是一种简单的线性函数，其特点是计算简单，易于理解和实现。
2. 特点默认传递函数具有以下几个特点：- 线性计算: 默认传递函数是一种线性计算，即输入信号和权重的线性组合再加上偏置项，不涉及非线性运算，因此计算速度较快。
- 易于调整: 由于默认传递函数的简单性，可以通过调整权重和偏置项来改变网络的输出，使其具有较强的灵活性。
- 可解释性: 默认传递函数的计算过程简单明了，易于理解，可以直观地解释神经元的输出是怎样由输入信号和权重决定的。
3. 应用默认传递函数在神经网络中有着广泛的应用，主要体现在以下几个方面：- 神经网络初始化: 在构建神经网络模型时，可以使用默认传递函数作为初始传递函数，然后根据实际情况进行调整。
- 线性模型: 默认传递函数适用于需要进行线性计算的模型，例如线性回归模型、线性分类模型等。
- 网络层连接: 默认传递函数可以作为神经网络中神经元之间连接的一种方式，构建起网络的拓扑结构。
4. 总结默认传递函数是feedforwardnet中常用的传递函数之一，它具有计算简单、易于调整和可解释性的特点，适合于一些简单的神经网络模型和线性计算模型。
在实际应用中，需要根据具体的问题和数据特点选择合适的传递函数，并根据实际需求进行调整和优化。
5. 参考文献- Haykin, S. (1994). Neural networks: Aprehensive foundation. Prentice Hall.- Bishop, C. M. (2006). Pattern recognition and machine learning. Springer.默认传递函数作为神经网络中常用的传递函数之一，拥有一定的优势和特点，因此在神经网络模型的构建和应用中具有广泛的适用性。
Temperature ControlTuning a PID (Three Mode) ControllerTuning a temperature controller involves setting the proportional, integral, and derivative values to get the best possible control for a particular process. If the controller does not include anautotune algorithm, or if the autotune algorithm does not provide adequate control for the particular application, then the unit must be tuned using trial and error.The following is a tuning procedure for the OMEGA CN2000 controller. It can be applied to other controllers as well. There are other tuning procedures which can also be used, but they all use a similar trial and error method. Note that if the controller uses a mechanical relay (rather than a solid state relay), a longer cycle time (20 seconds) should be used when starting out.The following definitions may be needed:1) Cycle time - Also known as duty cycle; the total length of time for the controller to complete one on/off cycle. Example: with a 20 second cycle time, an on time of 10 seconds and an off time of 10 seconds represents a 50 percent power output. The controller will cycle on and off while within the proportional band.2) Proportional band - A temperature band expressed in % of full scale or degrees within which the controller‘s proportioning ac-tion takes place. The wider the proportional band, the greater the area around the setpoint in which the proportional action takes place. This is sometimes referred to as gain, which is the reciprocal of proportional band.3) Integral, also known as reset, is a function which adjusts the proportional bandwidth with respect to the setpoint to com-pensate for offset (droop) from setpoint; that is, it adjusts the controlled temperature to setpoint after the system stabilizes.4) Derivative, also known as rate, senses the rate of rise or fall of system temperature and automatically adjusts the proportional band to minimize overshoot or undershoot.A PID (three mode) controller is capable of exceptional con-trol stability when properly tuned and used. The operator can achieve the fastest response time and smallest overshoot by following these instructions carefully. The information for tuning this three mode controller may be different from other controller tuning procedures. Normally a SELF TUNE feature will eliminate the need to use this manual tuning procedure for the primary output; however, adjustments to the SELF TUNE values may be made if desired.After the controller is installed and wired:1. Apply power to the controller.2. Disable the control outputs if possible.3. For time proportional primary output, set the cycle time. Enter the following value:CYCLE TIME 15 SEC (Only appears if output is a time proportional output. A smaller cycle time may be required for systems with an extreme-ly fast response time.)Then select the following parameters:PR BAND 1 ______5% (PB)RESET 1 ________0 R/M (TURNS OFF RESET FUNCTION)RESET 2 ________0 R/MRATE 1 _________0 MIN (TURNS OFF RATE FUNCTION)RATE 2 _________0 MINNOTEOn units with dual three mode outputs, the primary and second-ary tuning parameters are independently set and must be tuned separately. The procedure used in this section is for a HEATING primary output. A similar procedure may be used for a primary COOLING output or a secondary COOLING output.A. TUNING OUTPUTS FOR HEATING CONTROL 1. Enable the OUTPUT(S) and start the process.2. The process should be run at a setpoint that will allow the temperature to stabilize with heat input required.3. With RATE and RESET turned OFF, the temperature will stabilize with a steady state deviation, or droop, between the setpoint and the actual temperature. Carefully note whether or not there are regular cycles or oscillations in this temperature by observing the measurement on the display. (An oscillation may be as long as 30 minutes.) The tuning procedure is easier to follow if you use a recorder to monitor the process temperature.Figure 1. Temperature OscillationsPRIMARY TIMEDivide PB by 2 if you observe this. TIMEThis is close to perfect tuning.TIMEMultiply PB by 2 if you observe this.4. If there are no regular oscillations in the temperature, divide the PB by 2 (see Figure 1). Allow the process to stabilize and check for temperature oscillations. If there are still no oscillations, divide the PB by 2 again. Repeat until cycles or oscillations are obtained. Proceed to Step5.If oscillations are observed immediately, multiply the PB by 2. Observe the resulting temperature for several minutes. If the oscillations continue, increase the PB by factors of 2 until the oscillations stop.5. The PB is now very near its critical setting. Carefully in-crease or decrease the PB setting until cycles or oscillations just appear in the temperature recording.If no oscillations occur in the process temperature even at the minimum PB setting of 1%, skip Steps 6 through 11 below and proceed to paragraph B.6. Read the steady-state deviation, or droop, between setpoint and actual temperature with the “critical” PB setting you have achieved. (Because the temperature is cycling a bit, use the average temperature.)7 Measure the oscillation time, in minutes, between neigh-boring peaks or valleys (see Figure 2). This is most easily accomplished with a chart recorder, but a measurement can be read at one minute intervals to obtain the timing.8. Now, increase the PB setting until the temperature devia-tion, or droop, increases 65%.The desired final temperature deviation can be calculated by multiplying the initial temperature deviation achieved with the CRITICAL PB setting by 1.65 (see Figure 3) or by use of the convenient Nomogram I (see Figure 4). Try several trial-and-error settings of the PB control until the desired final temperature deviation is achieved.9. You have now completed all the measurements necessary to obtain optimum performance from the Controller. Only two more adjustments are required - RATE and RESET.10. U sing the oscillation time measured in Step 7, calculate the value for RESET in repeats per minutes as follows: RESET = 8 x 1 __ __ 5 T OWhere T O = Oscillation Time in Minutes. OR Use Nomogram II (see Figure 5):Enter the value for RESET 1.11. A gain using the oscillation time measured in Step 7, calcu-late the value for RATE in minutes as follows: RESET = T O __ 10Where T O = Oscillation TimeOR Use Nomogram III (see Figure 6)Enter this value for Rate 1.12. I f overshoot occurred, it can be eliminated by decreasing the RESET time. When changes are made in the RESET value, a corresponding change should also be made in the RATE adjustment so that the RATE value is equal to: RATE = 1 ______________ 6 x Reset Valuei.e., if reset = 2 R/M, the RATE = 0.08 min.13. S everal setpoint changes and consequent RESET and RATE time adjustments may be required to obtain the proper balance between “RESPONSE TIME” to a system upset and “SETTLING TIME.” In general, fast response is accompanied by larger overshoot and consequently shorter time for the process to “SETTLE OUT.” Conversely, if the response is slower, the process tends to slide into the final value with little or no overshoot. The requirements of the system dictate which action is desired.14. W hen satisfactory tuning has been achieved, the cycle time should be increased to save contactor life (applies to units with time proportioning outputs only (TPRI)). Increase the cycle time as much as possible without causing oscillations in the measurement due to load cycling.15. P roceed to Section C.TEMPERA TURE CYCLE TIME IN MINUTESCORRECT RESET SETTING IN REPEA TS PER MINUTE0.120100.20.353211230.500.300.2010200.100.05300.030.02100TRMPERA TURE CYCLE TIME IN MINUTESCORRECT RA TE SETTING IN MINUTES5440503032201100.330.220.110.030.30.020.20.010.1Figure 2. Oscillation TimeFigure 3. Calculating Final Temperature Deviation3ϒFigure 4. Nomogram IFigure 6. Nomogram IIIFigure 5. Nomogram IIB. TUNING PROCEDURE WHEN NO OSCILLATIONSARE OBSERVED1. Measure the steady-state deviation, or droop, betweensetpoint and actual temperature with minimum PB setting.2. Increase the PB setting until the temperature deviation(droop) increases 65%. Nomogram I (see Figure 4)provides a convenient method of calculating the desired final temperature deviation.3. Set the RESET 1 to a high value (10 R/M). Set the RATE1 to a corresponding value (0.02 MIN). At this point, themeasurement should stabilize at the setpoint tempera-ture due to reset action.4. Since we were not able to determine a critical oscilla-tion time, the optimum settings of the reset and rateadjustments must be determined by trial and error. After the temperature has stabilized at setpoint, increase the setpoint temperature setting by 10 degrees. Observe the overshoot associated with the rise in actual temperature.Then return the setpoint setting to its original value and again observe the overshoot associated with the actual temperature change.Excessive overshoot implies that the RESET and/orRATE values are set too high. Overdamped response(no overshoot) implies that the RESET and/or RATE val-ues are set too low. Refer to Figure 7. Where improved performance is required, change one tuning parameter at a time and observe its effect on performance whenthe setpoint is changed. Make incremental changes inthe parameters until the performance is optimized.5. When satisfactory tuning has been achieved, the cycletime should be increased to save contactor life (applies to units with time proportioning outputs only (TPRI)). Increase the cycle time as much as possible without causing oscilla-tions in the measurement due to load cycling.C. TUNING THE PRIMARY OUTPUT FOR COOLINGCONTROLThe same procedure is used as for heating. The process should be run at a setpoint that requires cooling control before the temperature will stabilize.D. SIMPLIFIED TUNING PROCEDURE FOR PID CON-TROLLERSThe following procedure is a graphical technique of analyz-ing a process response curve to a step input. It is much easier with a strip chart recorder reading the process vari-able (PV).1. Starting from a cold start (PV at ambient), apply fullpower to the process without the controller in the loop,i.e., with an open loop. Record this starting time.2. After some delay (for heat to reach the sensor), the PVwill start to rise. After more delay, the PV will reach amaximum rate of change (slope). Record the time atwhich this maximum slope occurs and the PV at which it occurs. Record the maximum slope in degrees perminute. Turn off system power.3. Draw a line from the point of maximum slope back to theambient temperature axis to obtain the lumped system time delay Td (see Figure 8). The time delay may alsobe obtained by the equation:Td = time to max. slope-(PV at max. slope - Ambient)/max. slope 4. Apply the following equations to yield the PID param-eters:Pr. Band = Td x max. slope x 100/span = % of spanReset= 0.4 / Td = resets/minuteRate = 0.4 x Td = minutes5. Restart the system and bring the process to setpoint withthe controller in the loop and observe response. If theresponse has too much overshoot, or is oscillating, then the PID parameters can be changed (slightly, one at atime, and observing process response) in the following directions:Widen the proportional band, lower the Reset value, and increase the Rate value.Example: The chart recording in Figure 8 was obtained by applying full power to an oven. The chart scales are 10°F/ cm, and 5 min/cm. The controller range is 100 to 600°F, or a span of 500°F.Maximum slope = 18°F/5 minutes= 3.6˚F/minuteTime delay = Td = approximately 7 minutes.Proportional Band = 7 minutes x3.6°F/minutes x 100/500°F = 5%.Reset = 0.4/7 minutes = 0.06 resets/minuteRate = 0.4 x 7 minutes = 2.8 minuteTuning a PID Controller Cont’dORESET OR RA TE T OO HIGH RESET OR RA TE T OO LOWFigure 7. Setting RESET and/or RATEFigure 8. System Time Delay。
feedfordnet用法-回复Feedforward是一种反馈工具，旨在提供积极的建议和指导，以改善个人和团队的绩效。
它与传统的反馈方法不同，传统的反馈注重过去的表现，而feedforward注重的是未来的改进。
本文将一步一步回答"[feedfordnet用法]"这个问题，介绍Feedforward的基本概念、使用步骤以及应用场景。
首先，我们需要了解Feedforward的基本概念。
Feedforward，即正向反馈，是一种通过给予建设性指导和建议，帮助个人或团队改进的反馈方式。
与传统的反馈方法不同，Feedforward不重视过去的表现，而是关注如何在未来中做得更好。
为了正确运用Feedforward，以下是一些使用Feedforward的具体步骤：第一步，明确目标：在给予Feedforward之前，你需要明确自己或团队的目标。
这可以帮助你更有针对性地给出建议和指导。
例如，如果你目标是提高演讲技巧，那么你可以围绕这个目标给出相关的Feedforward。
第二步，创造安全环境：Feedforward只有在安全和信任的环境下才能发挥最大的作用。
确保你的反馈接收者知道你的意图是为了帮助他们成长而不是批评他们的能力。
创造一个开放的对话氛围，让每个人都能够积极参与和分享意见。
第三步，具体而建设性：在给出Feedforward时，确保你的建议具有明确性和可操作性。
避免模糊和抽象的描述，而是提供具体的行动步骤和实用的建议。
例如，如果你要指导他人如何改进演讲技巧，不要只是说"你需要更自信"，而是给出具体的技巧和练习方法。
第四步，积极鼓励：鼓励和赞赏在Feedforward中起着重要的作用。
当你给予建议时，不要只关注你所看到的问题，也要关注对方的优点和潜力。
肯定对方的努力和进步，给予鼓励和支持，有助于建立积极的反馈循环和提高自信心。
接下来，我们来探讨一些Feedforward的应用场景。
