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                        JWST499-Cetinkunt
            JWST499-c07
                                                            ELECTROHYDRAULIC MOTION CONTROL SYSTEMS  513
                                  The command signal (either in speed or force control mode) is obtained either
                             from a real-time program or joystick. The real-time program can command motion to any
                             accuracy desired, for example 1 part in 10 000 depending on the data type used in the control
                             algorithm. If two-byte signed integers are used and scaled to cover the whole range, the
                             commanded signal resolution can be 1 part in (2 15  − 1). If four-byte signed long integers
                             are used in the servo control algorithm and numbers are scaled to cover the whole motion
                             range, the commanded signal resolution can be 1 part in (2 31  − 1).
                                  If the command signal is obtained from the joystick sensor, it is important to note that
                             a human operator cannot command motion without shaking his/her hand better than 1% of
                             the maximum displacement range of the joystick. Therefore, the resolution requirements
                             of the joystick command are much smaller. Even if we provided a high resolution sensor
                             and ADC, the human operator cannot actually change the command in finer resolutions
                             than about 1 part in 100. Therefore, an 8-bit ADC converter with an analog speed sensor
                             that can provide 10 VDC with an accuracy of 1% is sufficient. If an incremental encoder is
                             used for the sensor, an encoder with 512 lines∕rev also would have the sufficient range and
                             resolution for the command signal.
                                  The ECU should have resources in the following areas:

                                1. Speed of the CPU (microprocessor or DSP chip) in order to implement the control
                                  logic fast enough. Generally, this is not a problem with the current state of art.
                                2. Memory resources to store control code and sensory data for control purposes as well
                                  as for off-line analysis purposes (ROM or battery-backed RAM for program storage,
                                  RAM for real-time data storage and program execution).
                                3. I/O interface circuit: analog to digital converter (ADC), digital to analog converter
                                  (DAC), interface for discrete I/O (DIO), encoder interface, PWM signal input, and
                                  output interface. Notice that the ECU does not have to have all of these types of
                                  I/O interface, only the types required by the the selected sensor inputs and amplifier
                                  outputs.
                                4. Interrupt lines and interrupt handling software. Interrupts are key to the operation of
                                  most real-time systems, even though this application may be solved without use of
                                  interrupts.

                             In general, the higher the resolution (number of bits) of the ADC and DAC converter circuit,
                             the better it is. If the programming is to be done for the servo loop by the designer, then it is
                             preferable to have the ECU based on a microprocessor or DSP which has a C-compiler so
                             that we can use a high level programming language to implement the servo control loops
                             and use integer, long integer, or floating point data types. If we have to program the ECU
                             in assembly language, then we have to manage the data and its size explicitly. In general,
                             the speed, memory, and signal interface resources of the ECU are not limiting factors in
                             the design.
                                  Let us assume that the valve rating is specified for ΔP = 1000 psi (6.8948 MPa =
                                                                              v
                                                            2
                                                      6
                                       6
                             6.8948 × 10 Pa = 6.8948 × 10 Nt∕m pressure drop), and that the input shaft speed at the
                             pump is w = 1000 rpm. Following the component sizing algorithm, let us pick a cylinder
                                      in
                             diameter size, d = 0.05 m, which determines the cross-sectional area, A . Using Equations
                                          c                                           c
                             7.259–7.260, calculate the Q , Q , and ΔP ,
                                                    nl  r      L
                                                                        2
                                                        2
                                                  A =   d ∕4 = 0.0019635 m ≈ 0.002 m 2         (7.282)
                                                   c
                                                        c
                                                 Q = V ⋅ A = 240 l∕min                         (7.283)
                                                           c
                                                       nl
                                                  nl
                                                  Q = V ⋅ A = 180 l∕min                        (7.284)
                                                   r
                                                       r
                                                           c
                                                                   6
                                                                        2
                                                ΔP = F ∕A = 5 × 10 Nt∕m = 5 MPa                (7.285)
                                                   L
                                                          c
                                                       r
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