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Actualizar Firmware de iPhone/iPod Touch al 3.0 No es recomendable

with 74 comments

wifi-ihpone

A diferencia de otros Firmware, la liberación del 3.0 ha generado unos percances que en comparación a sus predecesores, el 3.0 claro esta, presenta unas mejoras, tales como adjuntar fotos/imagenes a los mensaje de texto (realmente Apple se tardo mucho en anexar esto) y el bluetooth nunca ha funcionado bien en los iPhone. No he podido realizar pruebas sobre el bluetooth para dar opinión si existen correcciones o no sobre esta función. Pero cuando sus deficiencias en algunas funciones importantes para el usuario no cumplen con su servicio, esto ocasiona la decisión de esperar a que Apple solventa las cosas y no instalarlo hasta su debido momento.

Ya existen diversas opiniones sobre los problemas que posee el Firmware 3.0, tales como un consumo mayor de la bateria que al apagar algunas funciones tales como el Wifi, el bluetooth, ahorran el consumo de la bateria (principalmente el apagar el Wifi).

Pero de estos problemas, el que ha sido de mayor critica, es precisamente del Wi-Fi, el cual en el Firmware 3.0 No funciona. Existen opiniones que reportan una mala conexión, una desconexión continua e incluso, una total ausencia del servicio.

Por costumbre y en vista de no haber tenido problemas con Firmware previos, actualice al 3.0 recordando que ante cualquier problema realizamos un Restore volviendo al Firmware anterior. Pero que sorpresa me llevo que al realizar la Restauración, sigo teniendo el 3.0.

iTunes siempre cuando actualizamos el Firmware (el sistema operativo del iphone/ipod touch) él mismo hace un backup o restore de la versión previa. Desconozco el porqué este restore no se mantuvo. En el unico momento en el que el Wifi se activo fue cuando realice desde el telefono un borrado de contenido y de ajustes. Pero esto obviamente no era la solución ya que tendria el Wifi activo pero sin contactos, imagenes, canciones, un iphone virgen.

Mis consejos referente a mi experiencia con el Firmware 3.0:

  • Averigua primero las mejoras “y sobre todo” los problemas, que ya se conocen del Firmware que Apple ha liberado. Incluso hasta versiones del mismo iTunes, ya que es el puente de comunicación entre tu producto y su software.
  • Si te haz encontrado con este articulo (post de este web-blog) y no haz actualizado tu Firmware, te aconsejo que No Actualices al Firmware 3.0. Hasta que Apple realmente haya liberado una actualización del Firmware que corrija estas fallas.

Estos problemas son comentados de la actualización de un iPhone 3G original (no jailbreakdo), y es un problema presente tanto para iphone como para ipod touch. Aunque Apple ya se ha pronunciado con unas soluciones para esto, debo indicar algo sobre ello.

Apple aconseja en Ajustes > General > Información el verificar si existe una dirección Wifi presente e indican Restaurar los ajustes de red para obtener nuevamente una dirección, la cual no es como una IP sino como una dirección MAC o parecida. En mi caso, siempre he tenido dirección de Wifi de este tipo, pero aunque tenga el router a mi lado, el iPhone no consigue senal alguna, ni siquiera de otras redes del área, simplemente no se sincroniza con ningun dispositivo inalámbrico. El unico Internet que poseo actualmente es el de mi operador movil (movistar).

Sin tener mucho tiempo de vida, el Firmware 3.0 ya tiene sus dias contados, ya mucha gente (incluyéndome) espera el Firmware 3.1 o 3.0.1 que corrija estas fallas, ya que muchas funciones del iphone y del ipod touch esta estrechamente vinculadas con el Internet.

Actualizando 12.08.09:

Ya ha salido la versión 3.0.1 y el problema del Wifi sigue igual, a duras penas es posible que aparezca el Wifi de redes inalámbricas libres (sin clave) pero en muchas otras ocasiones ni siqueira salen y en caso de lograrse conectar al poco instante se desconecta de ella. Router que tenga clave, no aparecen nunca. No queda otra que esperar otra versión que logre corregir esta enorme falla puesto que gran parte de las funciones y utilidades del iPhone están ligadas a Internet, asi que no me queda otra que navegar con mi servicio pago de Internet de mi proveedor telefonico y esperar a que Apple solvente sus fallos.

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Written by jocdz

julio 15, 2009 a 1:38 pm

Publicado en Bug, iPhone, iPod, Noticia, Software

74 comentarios

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  1. la verdad es un gran mentira todo lo que comentas, ya que yo eh instalado la version 3.0 en mi iphone y el wifi jamas me ah fallado, deberias verificar tu iphone

    carlos

    agosto 14, 2009 at 6:30 pm

    • no es el iphone que falla…. es el ipod touch

      massi

      octubre 25, 2009 at 9:56 pm

      • massi yo tengo un iphone por lo que confirmaria que el problema del firmware absorve a ambos ya que referente a iphone no soy el unico con el problema, sucede tanto para ipod como para iphone.

        saludos.

        jose

        octubre 25, 2009 at 10:04 pm

    • hola te saludo y a la vez te deseo q sigas teniendo la suerte de q tu ipod siga funcionando bien con la actualizacion en mi caso desaparecio el wifi simplemente no hay y antes de esta actualizacion entraba a internet a cada momento y tenia una buena señal, asi q xq a ti no te paso no significa q hay mentirosos por ahi diciendo cosas q no son, ya q ofendes a los q realmente tienen problemas.

      omar martinez

      octubre 3, 2010 at 5:40 pm

  2. Carlos, normalmente no indico nada en mi blog si no ha sido probado más de una vez, podemos decir que la falla no la tendra todos, tu eres prueba irrefutable de ello, más no podria indicarse o dar la razón de que sea mentira, ya que mi caso es prueba de que esto aun falla (indiquemos: para algunos y para otros no) como es tu caso.

    Mi iPhone es original, no es ningun iPhone jailbreakdo, el mismo iTunes es quien me avisa de nuevas versiones del Firmware y me actualiza.

    Otra cosa es que debes tomar en cuenta es que no soy la única persona que ha reportado este fallo en el Firmware 3.0, si buscas un poco en internet verás que son varios los casos, donde se comenta que incluso Apple confirma del fallo y explica unos pasos para sovlentarlo lo cual comento en el post, esto en vista de muchas personas que han reportado esto a la misma Apple, esto está en internet.

    Lo único que puede quedar asentado es que el Firmware 3.0 y el 3.0.1 presenta algunos problemas para algunos iPhone (en mi caso) y para otros no (el tuyo).

    jose

    agosto 14, 2009 at 10:22 pm

  3. el mio esta actualizado al 3.0 con jailbroke y funciona totalmente bien jajaja tu ipod que esta dañado tan wey comprate otro 😉 jaja

    eduardo

    agosto 20, 2009 at 5:07 pm

  4. El q a vosotros no os falle no significa q el iphone o el ipod de muchísima gente esté dañado. Este problema lo ha admitido Apple, ayer mismo me dijeron los técnicos vía telefónica q el problema lo están intentando solucionar con una nueva actualización. Si mirais un poquito por Internet hay muchísima gente q tiene problemas con el wifi al poner la nueva versión 3.0. En mi caso no se abren los vídeos de youtube (tarda una eternidad) y la vista previa en el itunes va fatal, cuando con la anterior versión íba todo perfecto.
    De haberlo sabido no lo actualizo ni pago los 8 euros.
    Saludos

    Mai

    agosto 22, 2009 at 12:02 pm

  5. Totalmente deacuerdo con Mai, efectivamente el firmware 3 no fue desarrollado como debio de haber sido en comparacion con sus predecesores.

    No entiendo lo del pago de los 8 euros, seria triste que ademas de tener un iphone original, se tenga que pagar por un soporte de una empresa que fue quien con su actualizacion danase el producto legal y originalmente comprado. Como si un ServicePack de Windows danara el sistema y haya que pagar a soporte por arreglar lo que ellos han causado.

    Triste mas, que les funcione a quienes posean un iphone jailbreakdo y quienes posean uno original sean los que tengan este problema.

    jose

    septiembre 8, 2009 at 1:03 pm

  6. Hola a todos,
    Yo tengo un ipod touch 1G, y después de intentarlo continuamente, hablar con Apple, no hay manera y sigo sin poder disponer de wifi. Justamente Apple me confirma por teléfono que si han tenido quejas del mal funcionamiento de este firmware, de hecho esta misma semana ha salido la versión 3.1.2 y sigue igual. Apple me envió un e-mail donde según ellos me iban a pasar el link para descargar el firmware anterior 2.2.1, pero nada de nada no aparece por ningún lado. Si no otra que nos queda es esperar a que salga una versión que solucione este problema. Me parece muy triste que una compañía como Apple permita estos problemas y además en su día nos cobran los 7,99 €!!

    Saludos.
    David.

    David

    octubre 12, 2009 at 11:25 am

  7. Tienes toda la razon.
    Algo triste que seamos personas con iPhone originales que podamos contar con el soporte de Apple y que ellos no puedan (y hasta me atrevo a comentar, ni sepan ellos mismos) lo que lo causa ya que sino ya habrian solventado la falla. Que personas que tienen iPhone jailbreakdos (algunos o todos) si puedan disfrutar del wifi (si el problema es el firmware que asi es, deberia ser igual para iphone originales y jailbreados). Pero volviendo al caso de originales, es el colmo que Apple cobre por el soporte, al menos no deberia para iphone aun dentro de la garantia.

    Que hasta ellos mismos esten sugiriendo una version vieja del firmware, tomandose en cuenta que muchas funciones del iPhone dependen de internet y que tengan este problema ahora. Tambien el que los usuarios esten esperando nuevas versiones del Firmware que lo solventen y cuando una nueva version aparece, el problema siga.

    En lo que a mi concierne, procedere a buscar un Firmware viejo y hasta incluso, hacerle jailbreak, tendre mas libertades y aplicaciones, y sobre preocuparme de perder la garantia, cual garantia? si no me ha servido de nada en caso de tener algun soporte el cual brilla por su ausencia, dan soluciones que no son soluciones y hasta son capaces de cobrar por un soporte que deberia brindarse gratuitamente para quienes aun tengan el iphone dentro del tiempo de garantia. Apple nos estropea su propio producto y ya luego de varios meses aun no saben como solucionarlo.

    No se a que esten jugando los tecnicos de Apple, pero si el problema sigue, el iPhone perdera terreno y sera un producto no muy deseado por la actual deficiencia del producto, gracias a Apple.

    jose

    octubre 12, 2009 at 2:14 pm

  8. la actualizacion 3.1.2 si no m equivoco tambien da problemas cn youtube va mu lento x no decir q no ves el video ademas de que las descargas desde app store tmpoco avanzan , vamos es una mierda, muxas mejoras y tonterias pero para lo q se le da mas uso 0

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  42. charge and induces a voltage. The result is that the receiver sees a higher voltage than what was transmitted. 117 Chapter 7 A TERMINATION GREATER THAN CHARACTERISTIC IMPEDANCE VOLTAGE ________ ACROSS TERMINATION IT 3T 5T 7T B TERMINATION LESS THAN CHARACTER I STIC IMPEDANCE VOLTAGE ACROSS TERMINATION ——– IT 3T 5T 7T C TERMINATION EQUALS CHARACTERISTIC IMPEDANCE VOLT AGE ACROSS TERMINATION IT 3T 5T 7T T =1-WAY CABLE DELAY Figure 7-5: The initial voltages across a parallel termination vary depending on the difference between the termination and the line’s characteristic impedance. If the line has a termination but the value is greater than the characteristic impedance, the effect is similar but less extreme. Some of the initial current flows in the termination and the rest reflects. The reflected current eventually returns to the driver. The driver absorbs part of the reflection and bounces the rest back, resulting in a reduced voltage at the receiver. The reflections may continue to bounce back and forth for a few 118 Designing RS-485 Links and Networks rounds with each round of lower amplitude than previous rounds. Eventually the current settles to a final value determined mainly by the termination, the driver’s output resistance, and other series resistances. If the source’s impedance and termination are less than the characteristic impedance, the voltage on the line gradually rises to its final value. The extreme case of a termination less than the characteristic impedance is when the wires are shorted together at the far end. When the current reaches the end, there is no load, so there is no voltage drop at all. The entire transmitted voltage has to reflect back to the driver. The electric field collapses and the magnetic field increases, inducing a current. If the line has a termination but its value is less than the characteristic impedance, the effect is similar but less extreme. Some of the initial voltage drops across the termination and the res prom-electric.ru

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  43. ing to trilinear or other color arrays. Typical parts include the 3-channel Fairchild TMC1103, the Burr-Brown VST 2000 and VST 3000 series, and the Maxim MAX1101. Magnetic Sensors Hall Effect Sensors Probably the simplest magnetic sensor to use in an embedded application is a Hall effect sensor. The Hall effect was discovered by Dr. Edwin Hall in 1879. He discovered that if a magnetic field was placed perpendicular to one face of a thin gold sheet in which a current was flowing, a voltage would appear across the sheet (Figure 3.24). This voltage is proportional to the current flowing in the sheet and the magnetic flux density. A Hall effect sensor is made from silicon, and the Hall voltage produced in silicon is only a few microvolts per volt per gauss. Consequently, a high-gain amplifier is required to bring the signal from the Hall element to a useable range. Hall effect sensors integrate the amplifier into the same package as the sensor element. Hall effect sensors are available as sensors that produce an output proportional to the magnetic field, or as switches that change state when the magnetic field exceeds a certain level. Analog Hall effect sensors are suited to applications where you need to know how close a magnet is to the sensor such as sensing whether an oscillating arm is really moving. Hall effect switches are best for applications where you just need to know if a magnet is near the sensor, such as sensing whether a safety hood is closed or open. The output of an analog Hall effect sensor can be connected to a comparator or ADC like any other voltage-output sensor. One caution: some analog output sensors provide an output that is proportional to the supply voltage. For an accurate noise-free output, you must power the sensor from a noise-free, well-regulated supply. A typical analog Hall effect sensor will produce an output that is halfway between the supply voltage and ground when no magnetic field is present. When a north pole is near the sensor, the voltage moves toward ground, and when a south pole is near the sensor the voltage moves toward the positive supply. Hall effect switches produce a digital output to indicate the presence of a magnetic field. They drive the output active when a certain magnetic strength (the operate point) is sensed, then drive the output inactive when the magnetic field drops below a certain level (the release point). There is some hysteresis in the range, where the release point is less than the operate value. 82 Analog Interfacing to Embedded Microprocessors Figure 3.24 The Hall effect. Hall effect switches come in two varieties: unipolar and bipolar, which are sometimes called nonlatching and latching. Bipolar switches have a positive (south pole) operate point and a negative (north pole) release point. Unipolar switches have a positive (south pole) operate point and a less-positive release point. The operate and release points vary with temperature. Both bipolar and unipolar switches typically have an open-collector output that has to be pulled up with an external resistor. Hall effect sensors are commonly available in three-lead packages similar to the TO-92 transistor package. The three leads are power, ground, and output. Typical supply voltages are 5 to 10v, although some sensors operate up to 30v or more. When using a Hall effect sensor, remember to account for stray magnetic fields. If using a magnet on, say, a rotating shaft, be sure that Sensors 83 Figure 3.25 Geartooth Hall effect sensor. the magnet doesn t excessively magnetize the shaft itself, or this will affect the output of the sensor. Remember that the magnetic field falls off with the approximate square of the distance. Approximate because the size and shape of the magnet, as well as surrounding magnetizable objects, affect the result. In any event, the output of an analog Hall effect sensor may be linear with respect to the strength of the magnetic field, but it will not be linear with respect to distance. . For more information visit site http://www.prom-electric.ru

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  44. ance. Any common-mode noise is ignored by the current measurement circuit. One drawback to this method is the need for a pair of wires and sensing circuitry for every sensor in the system. Figure 10.1 4-20ma current loop. Standard Interfaces 231 Appendix A Opamp Basics The opamp is a very high-gain amplifier with two inputs and an output. One input is called the inverting input (V_), and the other input is called the noninverting input (V+). The formula for the output (Vo) is given by: Vo = Av(V+ _ V_) where Av is the gain of the opamp (usually very high over 100,000) and V_ and V+ are the voltages at the inverting and noninverting input pins. Four Opamp Configurations Figure A.1 shows four opamp configurations: a buffer, inverting amplifier, noninverting amplifier, and differential amplifier. We can analyze these as follows. Buffer For the buffer configuration, the output (Vo) is connected to the inverting input (V_), and the input signal is applied to the noninverting input (V+). We can write the basic opamp equations like this: Vo = Av(V+ _ V_) Where Av is the open loop gain of the opamp. Since Vo is connected to V_ Vo = V_ The input voltage, V1, is applied to the noninverting input, V+. So we can rewrite the basic opamp equation like this: Vo = Av( V+ _ V_); Vo = Av( V1 _ Vo) Solving for Vo we get: Vo = Av V1 1 + Av 233 BUFFER INVERTING AMPLIFIER NONINVERTING AMPLIFIER DIFFERENTIAL AMPLIFIER Figure A.1 Opamp configurations. 234 Appendix A Dividing by Av, we get: Vo = V1 +1 Av If Av is very large, the _term approaches zero, leaving Vo = V1. Av Inverting Amplifier Starting with the basic opamp equation: Vo = Av(V+ – V-) In this case, the noninverting pin is grounded, so V+ is zero. V- is at one point of a voltage divider made up of R1 and R2. So we can write an equation for V- as: V =(Vo-V1)R1 + V1 R2 + R1 Substituting this into the basic opamp equation: Vo = Av R2 + R1 Solving for Vo, we get: – Av x V1 x R2 R2 + R1 + AvR2 Dividing the right side by Av/Av: Vo = – V1 x R2 Av Av R2 R1 лг If Av is very large, the and terms go to zero, leaving Vo is the gain of the inverting configuration. Noninverting Configuration The formula for the noninverting configuration is: Vo = Av(V+ – V-) The V+ pin is connected to the input, V1. The V- pin is a voltage divider with Vo and ground: – V1 x R2 R1 – R2/R1 w = Vo x R1 – = R2 + R1 Appendix A 235 Substituting these values into the basic opamp equation: Vo = Av V1 – Vo x R1 ^ R1+R2 ) Expanding: Vo(R1 + R2) = (Av x V1 x R1) + (Av x V1 x R2) – (Av x Vo x R1) Solving for Vo: Vo = Av x V1 x R1 R1 + R2 + (Av x R1) Av x V1 x R2 R1 + R2 + (Av x R1) Dividing both terms on the right by Av/Av: Vo = V1 x R1 R1 Av -R2 + R1 Av + V1 x R2 R1 Av -R2 + R1 Av If Av is very large, we are left with: Vo= V1 + V1 x R2 R1 or v =v1(i+ ) Differential Amplifier The differential amplifier is a combination of the inverting and noninverting configurations. V+ and V- are both voltage dividers, so we can write the equations for them like this: (Vo – V2)R3 Rf+ R3 + V2 V1 x R2 V+ = + R1 + R2 Substituting into the basic opamp equation: Vo = Av V1 x R2 R1 + R2 Rf + R3 Expanding and solving for Vo: V1RfR2 + V1R2R3 – (V2RfRL + V2RfR2) Vo = Av R1Rf + R2R3 + R2Rf + R2R3 + AvR3R1 + AvR2R3 Dividing the right fraction by Av/Av and allowing Av to be very large: V1R2Rf + V1R2R3 – (V2RfRL + V2RfR2) Vo = R1R3 + R2R3 Vi^2 Г1 + RL VV2-R1 R1 + R21 R3) R3 If R2 = RF and R1 = R3, then we get: 236 Appendix A Vo= (V1 – V2) R3 So the differential amplifier multiplies the difference between the inputs by the gain, Rf/R3. If a voltage divider was not used on the noninverting input, and V+ was connected to the V1 input, the output would be: Vo= V1 1 +-Rf- V V2 R3) R3 Without the voltage divider on the noninverting input, the gain for V1 is greater than the gain for V2. With the voltage divider, if V1 = V2, Vo will be 0. Without the divider, this is not the case. General Opamp Design Equations In general, an opamp that is operating in the l. For more information visit site http://www.prom-electric.ru

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  45. enerate the MSB of the result. The output of this flash converter then drives a 4-bit DAC to generate the voltage represented by the 4-bit result. The output of the DAC is subtracted from the input signal, leaving a remainder that is converted by another 4-bit flash to produce the LS 4 bits of the result. Figure 2.7 Half-flash converter. Digital-to-Analog Converters 25 If the converter shown in Figure 2.7 were a 0-5v converter, converting a 3.1v input, then the conversion would look like this: Upper flash converter output = 9 DAC output = 2.8125v(9 x 16 x 19.53mv) Subtracter output = 3.1v – 2.8125v = .2875v Lower flash converter output = E(hex) Final result = 9E (hex), 158 (decimal) Half-flash converters can also use three stages instead of two; a 12-bit converter might have three stages of 4 bits each. The result of the MS 4 bits would be subtracted from the input voltage and applied to the middle 4-bit state. The result of the middle stage would be subtracted from its input and applied to the least significant 4-bit stage. A half-flash converter is slower than an equivalent flash converter, but uses fewer comparators, so it draws less current. ADC Comparison Figure 2.8 shows the range of resolutions available for integrating, sigma-delta, successive approximation, and flash converters. The maximum conversion speed for each type is shown as well. As you can see, the speed of available sigma-delta ADCs reaches into the range of the SAR ADCs, but is not as fast as even the slowest flash ADCs. What these charts do not show is tradeoffs between speed and accuracy. For instance, while you can get SAR ADCs that range from 8 to 16 bits, you won t find the 16-bit version to be the fastest in a given family of parts. The fastest flash ADC won t be the 12-bit part, it will be a 6- or 8-bit part. These charts are a snapshot of the current state of the technology. As CMOS processes have improved, SAR conversion times have moved from tens of microseconds to microseconds. Not all technology improvements affect all types of converters; CMOS process improvements speed up all families of converters, but the ability to put increasingly sophisticated DSP functionality on the ADC chip doesn t improve SAR converters. It does improve sigma-delta types. Sample and Hold ADC operation is straightforward when a DC signal is being converted. What happens when the signal is changing? Figure 2.9 shows a successive- 26 Analog Interfacing to Embedded Microprocessors Figure 2.8 ADC comparison. Figure 2.9 ADC inaccuracy caused by a changing input. approximation ADC attempting to convert a changing input. When the ADC starts the conversion, the input voltage is 2.3v. This should result in an output code of 117 (decimal) or 75 (hex). The SAR register sets the MSB, making the internal DAC voltage 2.5v. Since the signal is below 2.5v, the SAR resets bit 7 and sets bit 6 on the next clock. The ADC chases the input signal, Digital-to-Analog Converters 27 ending up with a final result of 12710 (7F16). The actual voltage at the end of the conversion is 2.8v, corresponding to a code of 14310 (8F16). The final code out of the ADC (127d) corresponds to a voltage of 2.48V. This is neither the starting voltage (2.3v) nor the ending voltage (2.8v). This example used a relatively fast input to show the effect; a slowly changing input has the same effect, but the error will be smaller. One way to reduce these errors is to place a low-pass filter ahead of the ADC. The filter parameters are selected to insure that the ADC input does not change appreciably within a conversion cycle. Another way to handle changing inputs is to add a sample-and-hold (S/H) circuit ahead of the ADC. Figure 2.10 shows how a sample-and-hold circuit works. The S/H circuit has an analog (solid state) switch with a control input. When the switch is closed, the input signal is connected to the hold capacitor and the output of the buffer follows the input. When the switch is open, the input is disconne. For more information visit site http://www.prom-electric.ru

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  46. difference are summed with the amplified error to produce the output signal. I don t want to write a book about calculus, nor do you want to read one. This book is about practical embedded control, so I want to focus on practi- 110 Analog Interfacing to Embedded Microprocessors Figure 5.5 PID control system. cal applications. However, we need to take a look at the general formula for calculating the output of a PID controller, which is: Output = G ^ e +11 edt + D ^ j where: G is the gain e is the error (difference between setpoint and actual value) I is the amount of integral to apply D is the amount of derivative to apply If I and D are 0, then the output is: G x e which is the formula for a proportional controller. If I and D are 0 and G is very large so that the output always saturates in one direction or the other, then this describes an on-off controller. The things that set the PID controller apart from the proportional controller are the integral and derivative terms. These are time-based terms: the integral is an integral over some time period, and the derivative is the derivative between two time periods. Let s see what this means in practical terms. Almost any system has some kind of inertia. When you turn on a heater, it gets hotter than whatever it is trying to heat (the load). It has to, or it will absorb heat from the load instead of transferring heat into it. When you turn the heater off, it doesn t cool off immediately. Instead, its temperature ramps down slowly. Until the heater cools down to the same temperature as the load, it will continue to raise the temperature of the load. Figure 5.6 illustrates this. The amount of difference between the heater temperature and the load temperature, and how fast each one heats up and cools off, is dependent on the mass, the amount of energy applied to the heater, and so on. Output Control Methods 111 Figure 5.6 Inertia in a control system. Figure 5.6 also shows the effect of a light load versus a heavy load on the same heater. If the heater is heating metal blocks, the heavy load might be a bigger block of metal or one that is more massive (steel versus aluminum, for example). As you can see, the heavy load heats up and cools down more slowly because it has more mass more inertia. If we were looking at the speed of a 112 Analog Interfacing to Embedded Microprocessors car instead of a heater, the heavy load might be an uphill acceleration and the light load might be a downhill acceleration. Let s say that the heater is controlled by a proportional system. Since the amount of energy put into the heater is determined only by the difference between the desired temperature and the actual temperature, the control signal will be the same for the light load as for the heavy load. This means that the light load will overshoot the desired temperature by a greater amount. Once the right temperature is reached, there will be more oscillation (bigger temperature swings) around the setpoint. Accuracy of temperature is less precise than for a heavy load. Derivative Adding a derivative term to the control equation allows better control. The derivative is a measure of how fast the temperature of the load is changing how many degrees per minute, for example. This gives the control system some indication of the size of the load. Mathematically, the derivative of a curve is the slope of a curve in this case, the slope of the error. If the error term is decreasing, it has a negative slope and the derivative will be negative. If the error term is increasing, the derivative will be positive. If the error term doesn t change at all, then the slope and derivative are both 0. Note that any error, even a very large one, will have a derivative of 0 if the error doesn t change. The original heater/load graph is shown in Figure 5.7, along with the resulting error term and the derivative term. If we make the gain smaller and then add the derivative to the Gain X error term, our proportional control s. For more information visit site http://www.prom-electric.ru

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  47. the memory? Will data be lost? Does choosing this approach mean you have to go from a 100 MHz processor to a 500 MHz processor just to keep the garbage collection interval short? Avoiding Excess Speed Choosing a bus architecture and a processor that is fast enough to do the job is important, but it can also be important to avoid too much speed. It may not seem obvious that you wouldn t always want the fastest bus and the fastest microprocessor, but there are applications where that is exactly the case. There are two basic reasons for this: cost and EMC (electromagnetic compatibility). Cost The PC/104 standard defines mechanical and electrical characteristics of PC boards, optimized for embedded applications. PC/104 CPU boards come with the original PC/104 bus, which has electrical and timing characteristics System Design 7 similar to the ISA bus used in personal computers and is capable of data transfers in the 5 Mbytes/sec range. Many CPU boards also have the PC/104 Plus bus, which has characteristics similar to the much faster (133 Mbytes/sec) PCI bus. Although it might seem that the faster bus is always preferred, it is often less expensive to design a peripheral board for the PC/104 bus than for the PC/104 Plus. PC/104, due to the slower clock rates, allows longer traces and simpler logic. If you have a relatively large analog I/O board plugged into a PC/104 CPU board, the relaxed timing constraints of PC/104 may make layout easier. Many low-volume products simply do not sell enough units to justify the higher development costs associated with PC/104 Plus. Of course, this assumes that the PC/104 bus will support the necessary data rates. Similar considerations apply to other buses, such as PCI and Compact PCI. EMC Almost every microprocessor-based design will have to undergo EMC (electromagnetic compatibility) testing before it can be sold in the United States or Europe. EMC regulations limit the amount of energy the product can emit, to prevent interference with other equipment such as televisions and radios. Generally, the higher the clock rates are, the more emissions the equipment generates. Current EMC standards test radiated emissions in the frequency range between 30 MHz and 1 GHz. A processor running with a 6 MHz clock will not have any fundamental emissions in this range; the only frequencies in the test range will be those from the fifth and higher harmonics of the processor clock. The higher harmonics typically have less energy. On the other hand, a 33 MHz processor will produce energy in the test band from its fundamental frequency and higher. In addition, a faster processor clock rate means faster logic with faster edges and correspondingly higher energy in the harmonics. Although using a 6 MHz example in an era of 1000 MHz Pentiums may seem archaic, it does illustrate the point. EMC concerns are a valid reason to limit bus and processor speeds only to what is actually needed for the application. The caution here is not to limit the design too much. If the processor can just barely keep up with the application, there is no margin left to fix problems or add enhancements. Other System Considerations Peripheral Hardware An imaging system was having problems with lost data. This particular system buffered considerable image data on a hard disk drive. The problem was 8 Analog Interfacing to Embedded Microprocessors traced to the disk drive, where the drive would just stop accepting data for a while and the image buffers would overflow. It turned out that this particular drive had a thermal compensation feature that required the on-drive CPU to go away for a few tens of milliseconds every so often. The application required continuous access to the drive. Be sure the peripheral hardware is compatible with your application and does not introduce problems. Shared Interfaces What is the impact of shared interfaces? For example, if you are continuously buffering data from two different image cameras on two disk drives. For more information visit site http://www.prom-electric.ru

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  65. Цитата:
    Букмекерскую вилку надо искать автоматическим сервисом – это очевидно. Современные сканеры в своем большинстве не просто находят нужные ситуации, но и рассчитывают актуальную сумму ставки. Это удобно и гораздо снижает «трудозатраты» вилочника. Только, в сети полным-полно жалоб игроков, что «хваленые» сканеры плохо ищут букмекерские вилки. Чего не так? Разве сервисы врут? Так категорично рассматривать нельзя. Ведь, игрок должен не только сервис выбрать надежный, однако и обдумать с какими букмекерами стоит иметь дело. Только минимум, перед вступлением в противостояние нужно ознакомиться с правилами БК.

    Что скажете об сервисах вилок? Кто-нибудь пробовал?

    Tanyanow

    julio 17, 2017 at 10:35 pm

  66. Где можно посмотреть вживую уличные тренажеры и спортивные площадки от завода-производителя с сайтом http://kiddishop.com.ua

    Victorcek

    julio 23, 2017 at 9:22 am


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