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The relief valves play a role of unloading when overload and the check ones solve the flow unbalance between the rod-side and the piston-side. When the piston moves downwards, the pump delivers the hydraulic oil towards piston head with less volume of oil returned from the rod side. The supplement of oil is insufficient in the inlet of pump and the pressure in rod side decreases. The check valve opens reversely and the shortage of oil is suctioned from oil tank. However, as the piston moves upwards, the quantity of outflow oil in piston side is more than outflow oil in rod side. The pilot check valve opens reversely to return the surplus to the reservoir through the pipe shown in dot line. The flow volume balance is realized for the volume control of asymmetric cylinder. 3 Mathematic model of volume control electro-hydraulic servo system 3.1 Mathematic model of SRM module The SRM and its driving system is a kind of typical complicate nonlinear system, thus it is hard to establish the exact mathematic model. In this work, according to the property of the motor and the limits of the highest rotation speed of piston pump, this module is simplified and shown in Fig.2. The model can be established inpidually from three zones. The first zone (−u1≤uc≤u1) is dead zone, in which the SRM does not turn because the lowest permissible rotating speed of SRM is 0.5 r/s. The second zone (uc≤−u2, uc≥u2) is saturation zone, in which the rotating speed of SRM is limited to the highest permissible speed of bump 25 r/s. The third zone is the Fig.2 Property curve of SRM module normal working zone of SRM. Due to the frequency band of this module is wider than that of direct-drive electro-hydraulic position servo system enormously, thus, the property of SRM in this zone can be simplified to first-order inertial element. The transfer function is presented as: p vD()()() 1Ns KGsUs Ts==+ (1) where Np is the rotation rate, r/s; Kv is the speed gain, r/(s•V); TD is the time constant, s; and U is the control voltage, V. 3.2 Mathematic model of volume control module 3.2.1 Equation of flow continuity If the leakage of pump and the pressure at low pressure terminal are ignored, the flow equation of piston side when piston moves downwards is shown as follows: 111pp 1 1e()d dddVx p xQND p Attλβ==+ + (2) where Np is the rotation rate of pump, r/s; Dp is the pump delivery, m3/r; λ is the leakage coefficient of piston cylinder, m3/(s•Pa); p1 is the oil pressure at piston side, Pa; V1(x) is the volume of piston-side (including volume of pipe and pump), m3; βe is the effective volume elastic modulus of medium, N/m2; A1 is the section area of piston-side, m2; and x is the displacement of piston, m. 3.2.2 Force equilibrium equation of piston If some nonlinear factors, such as static fiction are ignored, based on Newton motion law, the force equilibrium equation when piston moves downwards can be obtained as follows: 211 p L 2ddd dxxF Ap m B Kx Ft t== + ++ (3) where F represents the theoretical output force of hydraulic cylinder, N; m is the total mass of load and piston, kg; Bp is viscous damping coefficient, N•s/m; K is elastic stiffness, N/m; and FL is the external load, N. 3.2.3 Transfer function of volume control module If viscous damping and elasticity of load are ignored (Bp=0, K=0), the transfer function of piston displacement can be represented as follow based on Eqs.(2) and (3) when the rotation rate Np(s) and external load FL(s) act on the piston simultaneously: pp 1L 21e 12h2h h()1()()21DN s Vs FsA AXssssλλβξω ω⎛⎞−+ ⎜⎟⎝⎠=⎛⎞++ ⎜⎟ ⎜⎟⎝⎠ (4) where ωh is the system natural frequency, ωh= 2e11,()AVxmβrad/s; ξh is the damping ratio, ξh= e11 2()mA Vxβ λ. 3.3 Mathematical model of system When the piston moves downwards (upwards is similar), the open-loop transfer function is 2Ch24322 hh hhh hDDD()2 12KGss sssTTTωξω ω ξω ω=⎛⎞⎛ ⎞++++ + ⎜⎟⎜ ⎟⎝⎠⎝ ⎠ (5) where KC is open-loop gain calculated by the equation of KC=KvDp/(TDA1). According to Eq.(5), the state equation and output equation of electro-hydraulic servo system driven by SRM directly are 12340100 00010 00001 00xxxx abc d⎡⎤ ⎡ ⎤⎡⎤⎢⎥ ⎢ ⎥⎢⎥⎢⎥ ⎢ ⎥⎢⎥ =+= +⎢⎥ ⎢ ⎥⎢⎥⎢⎥ ⎢ ⎥⎢⎥⎣ ⎦⎣⎦ ⎣⎦xAxBu u & (6) []12341000xxxx⎡ ⎤⎢ ⎥⎢ ⎥ ==⎢ ⎥⎢ ⎥⎣ ⎦yCx (7) In Eqs.(6) and (7) D2hTaω − = , Dh h 2h2Tbω ξω − − = Dh h12Tc − − = ω ξ , .2h Cω K d = where A, B and C are the state matrix, input matrix and output matrix, respectively; u is the input of system; x is the state vector, x=[x1 x2 x3 x4]T=[x x & x && x &&& ]T. 4 Fuzzy ILC controller design 4.1 Principle of ILC For the controlled object with the characteristics of repetitive motion, unknown construction and parameters, it requires that output y(t) tracks expected output yd(t) when t∈[0, T]. Assuming that the expected control ud(t) exists under the initial condition x(0), then the aim of ILC is to make u(t)⇒ ud(t), y(t)⇒ yd(t) with certain learning rule through repetitive motion. ILC is classified into open-loop and closed-loop learning. Through k operations, the output error is d () () () kket yt yt =−