# Electrical Machine & Automatic Control (EEE-409) (ME-II Yr) UNIT-3 Content: Signals u(t) = 1 when t 0 = 0 when t <0

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1 Electrical Machine & Automatic Control (EEE-409) (ME-II Yr) UNIT-3 Content: Modeling of Mechanical : linear mechanical elements, force-voltage and force current analogy, and electrical analog of simple mechanical systems; concept of transfer function & its determination for simple systems. Control : Open loop & closed loop controls, servo mechanisms, concept of various types of system. Signals: Unit step, unit ramp, unit impulse and periodic signals with their mathematical representation and characteristics. Signals Name Description Mathematical representation Characteristics Laplace Value Impulse An ideal impulse function is a function that is zero everywhere but at the origin, where it is infinitely high. However, the area of the impulse is finite δ (t) = 1 when t=0 = 0 when t 0 L [ δ (t)]= 1 Unit Step It is a discontinuous function whose value is zero for negative argument and one for positive argument. u(t) = 1 when t 0 = 0 when t <0 L[u(t)] = 1/s Ramp A unit ramp is defined in terms of the unit step function, as such: [Unit Ramp ] r(t) = t u(t) r (t) = t when t 0 = 0 when t < 0 L[u(t)] = 1/s 2 Parabolic It has the quadratic expression w.r.to time. r(t) = t 2 u(t) r (t) = t 2 when t 0 = 0 when t < 0 L[u(t)] = 2/s 3 By: ALOK TRIPATHI Page 1

2 Control OPEN LOOP CONTROL SYSTEM These are called as without feedback system. Because they don t have any feedback path. Block Diagram: CLOSED LOOP CONTROL SYSTEM They have feedback path. Output is compared with the reference input and error signal is produced. Block Diagram: Advantages: Simple Economical Less maintenance is required Proper calibration is not a problem Disadvantages: Inaccurate Not reliable Slow Optimization is not possible Applications: Immersion Rod Microwave Semi automatic washing machine Advantages: More reliable Faster No of variables can be handled simultaneously Optimization is possible Disadvantages: Expensive Maintenance is difficult Complicated Installation Applications: Air conditioners Manual Traffic control Types of SISO Single Input and Single Output As its name classifies that this type of system has Only one input and one output Example: Fan Input: Voltage Supply Output: Rotation MIMO Multiple Input and Multiple Output As its name classifies that this type of system has multiple input as well as multiple output. Example: Automobile driving system Inputs: Steering and Accelerator Outputs: Direction and Speed control By: ALOK TRIPATHI Page 2

3 Types Of feedback Control s Linear/ Non Linear Control The system which follows Superposition and homogeneity is called as Linear Otherwise nonlinear. Time Variant/ Invariant A system is said to be Time Invariant if its input output characteristics do not change with time. Otherwise it is said to be Time Variant system. Continuous The system in which we use the continuous time signal for processing is called as Continuous system. These are of two types: (i)ac un-modulated (ii)dc modulated Discontinuous The system in which we use the discontinuous time signal for processing is called as Discontinuous system. These are of two types: (i)sampled Data (ii)digital Form Servomechanism A servomechanism, sometimes shortened to servo, is an automatic device that uses error-sensing negative feedback to correct the performance of a mechanism and is defined by its function. In general, a low power device that adjusts or controls a more powerful device in response to the changes detected in one or more variables. The servo motor is actually an assembly of four things: a normal DC motor, a gear reduction unit, a positionsensing device (usually a potentiometer a volume control knob), and a control circuit. Block Diagram and Transfer function of DC motor: [1] Armature Controlled DC motor: (I a is varying, I f = constant) Where: R a = Armature Current L a = Armature self Inductance I a = Armature Current If = Field Current E = Induced EMF in Armature V = Applied Voltage T = Developed Torque θ = Angular displacement of the motor shaft J = moment of inertia of motor B= friction coefficient Applying KVL in armature circuit we get V = R a I a + E + L a di a /dt (1) By: ALOK TRIPATHI Page 3

4 Flux depends upon the field current and now the field current is constant so the produced flux is also constant. Induced EMF will be E α Φw E = K b w = K b dθ/dt (where w = angular velocity, K b = back EMF constant ) (2) Torque delivered by the motor will be T α ΦI a T = KI a (where K = motor torque constant) (3) And T = J d 2 θ/dt 2 + B dθ/dt (4) Now taking Laplace of equations1, 2, 3 & 4 We get V(s) E(s) = I a (s)[r a + L a s] E(s) = K b s θ(s) T(s) = ki a (s) T(s) = [Js 2 + Bs] θ(s) Or T(s) = [Js + B] sθ(s) Now the block diagram representation of all above equations will be In this the input is V(s) and the output is θ(s). So the transfer function will be the ratio of these two values. θ(s)/v(s) = K/[{R a.s.b.(1+sl a /R a )(1+sJ/B)} +K.K b s] Field Controlled DC Motor: (I a =Constant, I f is varying) Flux depends upon the field current and field current is varying so the flux will also be variable. Applying KVL in field circuit V f = Rf a If + LfdI f /dt (1) Φ = K f I f Torque delivered by the motor will be T α ΦI a T = KK f I f I a = KK f I f (where I a = constant) (2) And T = J d 2 θ/dt 2 + B dθ/dt (3) Taking Laplace of equations 1, 2 and 3 By equation (5) and (6) [Js 2 + Bs] θ(s) = K K f I f (s) Now put the value of I f (s) from equation (4) [Js 2 + Bs] θ(s) = K K f V f (s) / [R f + L f s] V f (s) = I f (s)[r f + L f s] (4) T(s) = K K f I f (s) (5) T(s) = [Js 2 + Bs] θ(s) (6) By: ALOK TRIPATHI Page 4

5 The transfer function will be θ(s) / V f (s) = K K f / [R f.b.s(1+sl f /R f )(1+sJ/B)] MECHANICAL SYSTEM TRANSLATIONAL SYSTEM It has linear movement. ROTATIONAL SYSTEM It has angular movement. Translational : In this the motion takes place along a straight line. There are three types of forces that resist motion. Inertia Force Damping Force Spring force Consider a body of mass M and acceleration a, then according to Newton s second law of motion the inertia force will be equal to the product of Mass M and acceleration a. F m (t) = M.a = M.dv(t)/dt = M. d 2 x(t)/dt 2 For viscous friction we assume that the damping force is proportional to the velocity. F D (t) = B.v(t) = B. dx(t)/dt Where B= damping coefficient A spring stores the potential energy. The restoring of a spring is proportional to the displacement. F m (t) = K x(t) Where K= spring constant or stiffness D ALEMBERT s PRINCIPLE For any body, the algebraic sum of externally applied forces resisting motion in any given direction is zero. Let External Force = F(t) Resisting forces: (i) Inertia Force, (ii) Damping Force, (iii) Spring Force So according to D Alembert s principle External force + Inertia Force + Damping Force + Spring Force = 0 F(t) + F m (t) + F D (t) +F K (t) = 0 F(t) - M. d 2 x(t)/dt 2 - B. dx(t)/dt - K x(t) =0 By: ALOK TRIPATHI Page 5

6 ANALOGY [1] Force Voltage (F-V) Analogy Translational (Linear ) Electrical system By D Alembert principle F(t) - M. d 2 x(t)/dt 2 - B. dx(t)/dt - K x(t) =0 Force (F) Mass (M) Damping Coefficient (B) Stiffness (K) Displacement (x) By KVL V Ld 2 q/dt 2 - Rdq/dt q/c =0 Voltage (V) Inductance (L) Resistance (R) Reciprocal of capacitance (1/C) Charge (q) [2] Force Current (F-I) Analogy Translational (Linear Syatem) Electrical system By D Alembert principle F(t) - M. d 2 x(t)/dt 2 - B. dx(t)/dt - K x(t) =0 Force (F) Mass (M) Damping Coefficient (B) Stiffness (K) Displacement (x) By KCL I Cd 2 ϕ/dt 2 1/[Rdϕ/dt] ϕ/l =0 Current (I) Capacitance (C) Reciprocal of Resistance (1/R) Reciprocal of Inductance (1/L) Flux (ϕ) By: ALOK TRIPATHI Page 6

7 TRANSFER FUNCTION It is the ratio of output (in s domain) to input (in s domain) with all initial conditions zero. Transfer function = Y(S) / X(S) Where Y(S) = output in s domain X(S) = input in s domain Procedure to find Transfer function: Step 1: Write the expression for input and output in time domain. Step 2: Take the Laplace of both input and input equations. Step 3: Now manipulate the equations to find the ratio of output to input. (in s domain obtained by step 2) ** For numerical practice refer the books of control systems. By: ALOK TRIPATHI Page 7

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