Control Systems : Basics of Control Systems

Control Engineering

Basically, Control engineering is applicable to aeronautical, chemical, mechanical, environmental, civil, and electrical engineering which is based on the foundations of feedback theory and linear system analysis, and it generates the concepts of network theory and communication system theory.

Hence according to the theory of control engineering, it is not limited to any engineering discipline but applicable to the different areas which require the control process for their functioning & Stable Operations.

Open Loop Control System

• An open-loop control system consists of a control actuator or controller to receive the desired response.
• It uses a switching device to control the process directly without using any device.
• An illustration of an open-loop control system is an electric toaster.
• In Open Loop Controlling action, there is no feedback system present to sense the error in the desired output.

Closed-Loop Control System

• In a closed-loop control system, it consists of an additional measure of the actual output to compare the actual output with the desired output response.
• This additional measure of the output is called the feedback signal.
• A feedback control system that set out to maintain the relationship of one system variable to another by comparing the functions of these variables and using the difference as a means of control.
• Since the system becomes more complex, the interrelationship of many controlled variables may be considered in the control scheme.
• An example of a closed-loop control system is a person steering (or driving) an automobile by looking at the auto’s location on the road and making the appropriate adjustments.

TEMPERATURE CONTROL SYSTEMS

• In the electric furnace, the temperature is measured by a thermometer, which is an analog device.
• The analog temperature converted to digital temperature using an A/D converter. The digital temperature is then fed to a controller through an interface.
• The digital temperature is then compared with the programmed input temperature, and if there is any error, the controller then sends out a signal to the heater, through an amplifier, interface, and relay to bring the furnace temperature to the desired value.

Comparison between Open Loop & Closed Loop Control System 

Block Diagram Reduction technique 

Need for Block Diagram Reduction:- Some of the block diagrams are complex, such that the evaluation of their performance required simplification (or reduction) of the block diagrams which is done by the block diagram rearrangements.

Advantages of Block Diagram reduction

• Its very simple to construct the block diagram for complicated systems.
• Single, as well as the overall performance of the system, can be studied by using transfer functions shown in the block diagram.
• The overall closed-loop transfer function can be calculated easily using block diagram laws.
• The function of the individual element can be visualized with the help of block diagram.

Components of Linear Time-Invariant Systems 

Rules of Block Diagram reduction Technique

• Cascade series Connection 

• Parallel Connection 

• Block Diagram Algebra for Summing Junctions

• Block Diagram Algebra for Branch Point

Basic Rule For Block Diagram Transformation

Meson's Gain Formula

The objectives for this technical note are:

• Define the key terms describing signal-flow graphs.
• Draw signal flow graphs for a block diagram.
• Draw a block diagram from a signal-flow graph.
• List down the steps in the process to solve a system using Mason’s gain formula.
• Apply Mason’s gain formula to systems in block diagram or signal-flow-graph form.

Basic Definition Related to Meson's Gain Formula

• Forward paths: Forward paths are continuous paths through the graph from the input to the output. No node is passed more than once.
• Feedback loops: Feedback loops are continuous paths through the graph that starts and end at the same node.
• Path gain: Path gain is the product of the signal gains encountered on the path.
• Loop gain: Loop gain is the product of signal gains encountered in a feedback loop.
• Source node: Source nodes are nodes with only outgoing branches.
• Sink nodes: Sink nodes are nodes with only incoming branches.

    Mesons gain Formula Statement

• P_K represents the path gain for the kth forward path.
• Δ =1-(∑ Sum of all individual loop gains) +(∑ sum of products of all pairs of loop gains,(non-touching loops) )-  (∑ sum of products of all triples of loop gains,(Non-touching loops))+ ...
• Δ_k = Δ-(∑ loop gains in Δ that touch forward path k)

• Linear System:  Essentially a Linear system is one that follows the principle of Superposition & Homogeneity in their response to the system.

        Consider a system with the input f(t) and output x(t)

Now if the input is changed to g(t), the output is y(t)

If the system is linear, then an input of h(t)=g(t)+f(t) yields an output z(t)=x(t)+y(t)

• Example of a nonlinear system:  Now consider the same situation when the system is nonlinear for example a squaring function.

Now we can not find the output to the complex function h(t) by adding the responses of the systems to the simpler function.

• Time-Invariant: In the time-invariant system, the physical parameters of the system do not change with time.  The classic example of a non-time invariant (or time-variant) system is a rocket whose mass changes with time (a time-invariant rocket would have constant mass).
• Continuous-Time:  The Continuous-time systems are time is a continuous, or real-valued, variable.  On the other hand, discrete-time systems have time that moves in discrete steps. 

        Examples of discrete-time systems include weekly closing stock prices (updated weekly), the sound on a standard audio CD (updated 44,100 times per second).

Analogous System

 An analogous electrical and mechanical system has differential equations of the same kind. There are two analogies that are used to go between the electrical and mechanical systems.

To understand the analogy more clearly. The parameters for the mechanical analogous are formed by substituting the analogous parameters into the equations for the electrical elements. For example, by Ohm's law for electrical circuit e=iR. For the Mechanical analog  I, e is replaced by v, I by f and R by 1/B, which yields v=f/B.

Force-Current 'F-I' Analogy (Electrical to Mechanical)

Kirchoff's Current Law and D'Alemberts Law (with inertial forces included) are helpful for converting an electrical circuit to a mechanical system.

Procedure for Conversion from Electrical to Mechanical.

• Start with an electrical circuit. Label all node voltages.
• Write a node equation for each node voltage.
• Rewrite the equations using analogous making substitutions from the table, with each electrical node being replaced by a position.
• Draw the mechanical system that interconnects with the equations.

Example: Draw the mechanical equivalent circuit of the given system.

Solution: By following the steps in the given Procedure

Alternative method: Another way from electrical to mechanical simply redrawing the electrical circuit using mechanical components.

• Draw over the circuit, replacing electrical elements with their analogous; voltage sources by input velocities, current sources replaced by force generators, resistors with friction elements, inductors with springs, and capacitors (which must be grounded) by masses. Each node becomes a position (or velocity)
• Label positions, currents, and the mechanical elements as they were in original electrical circuits.

Force-Current 'F-I' to Electrical to Mechanical

The procedure for Mechanical to Electrical analogy is simply the reverse of Electrical to Mechanical analogy.  Either a mathematical method can be used as in the previous example, Electrical to Mechanical conversion can be understood by reading the table from bottom to top, or by the method where force generators are replaced by current sources, friction elements by resistors, springs by inductors, and masses by capacitors (which are grounded).  Each position becomes a node in the circuit.

Procedure for 'F-I' analogy for Electrical to Mechanical Conversion

• Start with the mechanical system. Label all positions.
• Draw the circuit by replacing mechanical elements with their analogous; force generators by current sources, input velocities by voltage sources,  friction elements by resistors, springs by inductors, and masses by capacitors (which are grounded). Each position becomes a node.
• Label the nodes and electrical elements as they were in the original mechanical system.

Force-Voltage 'F-V' Analogy (From Electrical to Mechanical)

The important relationship when converting from a circuit to the Mechanical analog is that between Kirchoff's Voltage Law and D'Alemberts Law (with inertial forces included).

Procedure for Conversion from Electrical to Mechanical

• Start with an electrical circuit. Label all currents. Choose the currents so that only one current flows through the inductors.
• Write loop equations for each loop.
• Rewrite the equations using analogs, making substitutions from the table, with each electrical loop being replaced by a position.
• Draw the mechanical system that interconnects with the equations.

Example: Draw the Mechanical equivalent system of the Electrical Circuit.

Solution: By following the procedure given for force voltage analogy.

Rotating Mechanical Systems

Gear System

The gear system performs many functions, here we look at the gears that increase or decrease angular velocity (while simultaneously decreasing or increasing torque, such that energy is conserved).  

• If we consider two gears in equilibrium and in contact with each other, we can obtain very useful relationships.

• First, we note the geometric relationship that concludes from the path that the arc lengths along their circumference must be equal to the gear's turn.

 Since the arc lengths (shown with a heavy blue line) must be equal ⇒ arc length⇒ r₁θ₁ = r₂θ₂  arc length

• Now we can derive the second relationship from a torque balance. Here we must define a force between the gears termed a "contact force."This force must be equal and opposite across the interface between the two gears, but its direction is arbitrary.

Since the contact force is tangent to both gears and so produces a torque equal to the radius times of the force.

We can have a torque balance on each of the two gears

 For Gear 1: torque  T₁ = f_cr_{1  or} f_c = T₁/r₁ & For Gear 2: torque  T₂ = f_cr_{2  or} f_c = T₂/r₂

From the above two equation we concluded that  f_c = T₁/r₁ = T₂/r₂

• In the system below, a torque (T_a), is applied to gear 1 (with a moment of inertia J₁).  It, in turn, is connected to gear 2 (with a moment of inertia J₂) and a rotational friction B_r.  The angle θ₁ is defined as positive clockwise, θ₂ is also defined as positive clockwise. The torque acts in the direction of θ₁.

• We start by drawing free body diagrams, including a contact force that we will arbitrarily choose to be down on J₁ and up on J₂.  The directions of the reactive forces due to inertia and friction are chosen, and as always, opposite to the defined positive direction.

This yields the two equations of motion

_Ta + f_cr₁ - J₁θ₁ =0 

f_cr₂-J₂θ₂ +B_rθ₂ = 0

• We can easily solve for f_c and eliminate it from the above equations, but we also need to eliminate θ₂. To do this, we use the relationship between θ₁ and θ₂ (from equal arc lengths).

r₁θ₁ = -r₂θ₂

• Note that we have a negative sign here because of the way θ₁ and θ₂ were defined (if θ₁ moves in the positive direction, then θ₂ is negative).  When you use the arc length expression, you must be careful of the signs.

        f_cr₂-J₂θ₂ - B_rθ₂ = 0

        and   θ₂ = - r₁/r₂(θ₁)

we can we write it as  f_cr₂-J₂{r₁/r₂(θ₁)} =0 

         fc = -J₂{r₁/(r₂)²}θ₁ - Br {r₁/(r₂)²}θ₁

We can put this into the equation for J1 and solve (in standard form with the output (θ₁) on the left, and the input (T_a) on the right.

         T_a + f_cr₁ - J₁θ₁ =0 

        or     T_a +[-J₂{r₁/(r₂)²}θ₁ - Br {r₁/(r₂)²}θ₁r₁] - J₁θ₁ =0

        or      {J₁+ J₂(r₁/r₂)²}θ₁ + Br(r1/r2)²}θ1 = Ta

        So we get  θ₁ = r₂/r₁(θ₂)  &  ω₁ = r₂/r₁(ω₂)

        & Since from the relation T₁= r₁/r₂(T₂)

we concluded that 

             T₁ω₁ = {r₁/r₂(τ₂)}{r₂/r₁(ω₂)} = T₂ω₂ 

           T₁ω₁ =T₂ω₂