Category: control Systems

Control Systems – Compensators There are three types of compensators — lag, lead and lag-lead compensators. These are most commonly used. Lag Compensator The Lag Compensator is an electrical network which produces a sinusoidal output having the phase lag when a sinusoidal input is applied. The lag compensator circuit in the ‘s’ domain is shown in the following figure. Here, the capacitor is in series with the resistor $R_2$ and the output is measured across this combination. The transfer function of this lag compensator is – $$frac{V_o(s)}{V_i(s)}=frac{1}{alpha} left( frac{s+frac{1}{tau}}{s+frac{1}{alphatau}} right )$$ Where, $$tau=R_2C$$ $$alpha=frac{R_1+R_2}{R_2}$$ From the above equation, $alpha$ is always greater than one. From the transfer function, we can conclude that the lag compensator has one pole at $s = − frac{1}{alpha tau}$ and one zero at $s = −frac{1}{tau}$ . This means, the pole will be nearer to origin in the pole-zero configuration of the lag compensator. Substitute, $s = jomega$ in the transfer function. $$frac{V_o(jomega)}{V_i(jomega)}=frac{1}{alpha}left( frac{jomega+frac{1}{tau}}{jomega+frac{1}{alphatau}}right )$$ Phase angle $phi = tan^{−1} omegatau − tan^{−1} alphaomegatau$ We know that, the phase of the output sinusoidal signal is equal to the sum of the phase angles of input sinusoidal signal and the transfer function. So, in order to produce the phase lag at the output of this compensator, the phase angle of the transfer function should be negative. This will happen when $alpha > 1$. Lead Compensator The lead compensator is an electrical network which produces a sinusoidal output having phase lead when a sinusoidal input is applied. The lead compensator circuit in the ‘s’ domain is shown in the following figure. Here, the capacitor is parallel to the resistor $R_1$ and the output is measured across resistor $R_2. The transfer function of this lead compensator is – $$frac{V_o(s)}{V_i(s)}=beta left( frac{stau+1}{beta stau+1} right )$$ Where, $$tau=R_1C$$ $$beta=frac{R_2}{R_1+R_2}$$ From the transfer function, we can conclude that the lead compensator has pole at $s = −frac{1}{beta}$ and zero at $s = − frac{1}{betatau}$. Substitute, $s = jomega$ in the transfer function. $$frac{V_o(jomega)}{V_i(jomega)}=beta left( frac{jomegatau+1}{beta j omegatau+1} right )$$ Phase angle $phi = tan^{−1}omegatau − tan^{−1}betaomegatau$ We know that, the phase of the output sinusoidal signal is equal to the sum of the phase angles of input sinusoidal signal and the transfer function. So, in order to produce the phase lead at the output of this compensator, the phase angle of the transfer function should be positive. This will happen when $0 < beta < 1$. Therefore, zero will be nearer to origin in pole-zero configuration of the lead compensator. Lag-Lead Compensator Lag-Lead compensator is an electrical network which produces phase lag at one frequency region and phase lead at other frequency region. It is a combination of both the lag and the lead compensators. The lag-lead compensator circuit in the ‘s’ domain is shown in the following figure. This circuit looks like both the compensators are cascaded. So, the transfer function of this circuit will be the product of transfer functions of the lead and the lag compensators. $$frac{V_o(s)}{V_i(s)}=beta left( frac{stau_1+1}{beta s tau_1+1} right )frac{1}{alpha} left ( frac{s+frac{1}{tau_2}}{s+frac{1}{alphatau_2}} right )$$ We know $alphabeta=1$. $$Rightarrow frac{V_o(s)}{V_i(s)}=left ( frac{s+frac{1}{tau_1}}{s+frac{1}{betatau_1}} right )left ( frac{s+frac{1}{tau_2}}{s+frac{1}{alphatau_2}} right )$$ Where, $$tau_1=R_1C_1$$ $$tau_2=R_2C_2$$ Learning working make money

Control Systems – Quick Guide Control Systems – Introduction A control system is a system, which provides the desired response by controlling the output. The following figure shows the simple block diagram of a control system. Here, the control system is represented by a single block. Since, the output is controlled by varying input, the control system got this name. We will vary this input with some mechanism. In the next section on open loop and closed loop control systems, we will study in detail about the blocks inside the control system and how to vary this input in order to get the desired response. Examples − Traffic lights control system, washing machine Traffic lights control system is an example of control system. Here, a sequence of input signal is applied to this control system and the output is one of the three lights that will be on for some duration of time. During this time, the other two lights will be off. Based on the traffic study at a particular junction, the on and off times of the lights can be determined. Accordingly, the input signal controls the output. So, the traffic lights control system operates on time basis. Classification of Control Systems Based on some parameters, we can classify the control systems into the following ways. Continuous time and Discrete-time Control Systems Control Systems can be classified as continuous time control systems and discrete time control systems based on the type of the signal used. In continuous time control systems, all the signals are continuous in time. But, in discrete time control systems, there exists one or more discrete time signals. SISO and MIMO Control Systems Control Systems can be classified as SISO control systems and MIMO control systems based on the number of inputs and outputs present. SISO (Single Input and Single Output) control systems have one input and one output. Whereas, MIMO (Multiple Inputs and Multiple Outputs) control systems have more than one input and more than one output. Open Loop and Closed Loop Control Systems Control Systems can be classified as open loop control systems and closed loop control systems based on the feedback path. In open loop control systems, output is not fed-back to the input. So, the control action is independent of the desired output. The following figure shows the block diagram of the open loop control system. Here, an input is applied to a controller and it produces an actuating signal or controlling signal. This signal is given as an input to a plant or process which is to be controlled. So, the plant produces an output, which is controlled. The traffic lights control system which we discussed earlier is an example of an open loop control system. In closed loop control systems, output is fed back to the input. So, the control action is dependent on the desired output. The following figure shows the block diagram of negative feedback closed loop control system. The error detector produces an error signal, which is the difference between the input and the feedback signal. This feedback signal is obtained from the block (feedback elements) by considering the output of the overall system as an input to this block. Instead of the direct input, the error signal is applied as an input to a controller. So, the controller produces an actuating signal which controls the plant. In this combination, the output of the control system is adjusted automatically till we get the desired response. Hence, the closed loop control systems are also called the automatic control systems. Traffic lights control system having sensor at the input is an example of a closed loop control system. The differences between the open loop and the closed loop control systems are mentioned in the following table. Open Loop Control Systems Closed Loop Control Systems Control action is independent of the desired output. Control action is dependent of the desired output. Feedback path is not present. Feedback path is present. These are also called as non-feedback control systems. These are also called as feedback control systems. Easy to design. Difficult to design. These are economical. These are costlier. Inaccurate. Accurate. Control Systems – Feedback If either the output or some part of the output is returned to the input side and utilized as part of the system input, then it is known as feedback. Feedback plays an important role in order to improve the performance of the control systems. In this chapter, let us discuss the types of feedback & effects of feedback. Types of Feedback There are two types of feedback − Positive feedback Negative feedback Positive Feedback The positive feedback adds the reference input, $R(s)$ and feedback output. The following figure shows the block diagram of positive feedback control system. The concept of transfer function will be discussed in later chapters. For the time being, consider the transfer function of positive feedback control system is, $T=frac{G}{1-GH}$ (Equation 1) Where, T is the transfer function or overall gain of positive feedback control system. G is the open loop gain, which is function of frequency. H is the gain of feedback path, which is function of frequency. Negative Feedback Negative feedback reduces the error between the reference input, $R(s)$ and system output. The following figure shows the block diagram of the negative feedback control system. Transfer function of negative feedback control system is, $T=frac{G}{1+GH}$ (Equation 2) Where, T is the transfer function or overall gain of negative feedback control system. G is the open loop gain, which is function of frequency. H is the gain of feedback path, which is function of frequency. The derivation of the above transfer function is present in later chapters. Effects of Feedback Let us now understand the effects of feedback. Effect of Feedback on Overall Gain From Equation 2, we can say that the overall gain of negative feedback closed loop control system is the ratio of ”G” and (1+GH). So, the overall gain may increase or decrease depending on the value

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Construction of Root Locus The root locus is a graphical representation in s-domain and it is symmetrical about the real axis. Because the open loop poles and zeros exist in the s-domain having the values either as real or as complex conjugate pairs. In this chapter, let us discuss how to construct (draw) the root locus. Rules for Construction of Root Locus Follow these rules for constructing a root locus. Rule 1 − Locate the open loop poles and zeros in the ‘s’ plane. Rule 2 − Find the number of root locus branches. We know that the root locus branches start at the open loop poles and end at open loop zeros. So, the number of root locus branches N is equal to the number of finite open loop poles P or the number of finite open loop zeros Z, whichever is greater. Mathematically, we can write the number of root locus branches N as $N=P$ if $Pgeq Z$ $N=Z$ if $P<Z$ Rule 3 − Identify and draw the real axis root locus branches. If the angle of the open loop transfer function at a point is an odd multiple of 1800, then that point is on the root locus. If odd number of the open loop poles and zeros exist to the left side of a point on the real axis, then that point is on the root locus branch. Therefore, the branch of points which satisfies this condition is the real axis of the root locus branch. Rule 4 − Find the centroid and the angle of asymptotes. If $P = Z$, then all the root locus branches start at finite open loop poles and end at finite open loop zeros. If $P > Z$ , then $Z$ number of root locus branches start at finite open loop poles and end at finite open loop zeros and $P − Z$ number of root locus branches start at finite open loop poles and end at infinite open loop zeros. If $P < Z$ , then P number of root locus branches start at finite open loop poles and end at finite open loop zeros and $Z − P$ number of root locus branches start at infinite open loop poles and end at finite open loop zeros. So, some of the root locus branches approach infinity, when $P neq Z$. Asymptotes give the direction of these root locus branches. The intersection point of asymptotes on the real axis is known as centroid. We can calculate the centroid α by using this formula, $alpha = frac{sum Real: part: of: finite: open: loop: poles:-sum Real: part: of: finite: open: loop: zeros}{P-Z}$ The formula for the angle of asymptotes θ is $$theta=frac{(2q+1)180^0}{P-Z}$$ Where, $$q=0,1,2,….,(P-Z)-1$$ Rule 5 − Find the intersection points of root locus branches with an imaginary axis. We can calculate the point at which the root locus branch intersects the imaginary axis and the value of K at that point by using the Routh array method and special case (ii). If all elements of any row of the Routh array are zero, then the root locus branch intersects the imaginary axis and vice-versa. Identify the row in such a way that if we make the first element as zero, then the elements of the entire row are zero. Find the value of K for this combination. Substitute this K value in the auxiliary equation. You will get the intersection point of the root locus branch with an imaginary axis. Rule 6 − Find Break-away and Break-in points. If there exists a real axis root locus branch between two open loop poles, then there will be a break-away point in between these two open loop poles. If there exists a real axis root locus branch between two open loop zeros, then there will be a break-in point in between these two open loop zeros. Note − Break-away and break-in points exist only on the real axis root locus branches. Follow these steps to find break-away and break-in points. Write $K$ in terms of $s$ from the characteristic equation $1 + G(s)H(s) = 0$. Differentiate $K$ with respect to s and make it equal to zero. Substitute these values of $s$ in the above equation. The values of $s$ for which the $K$ value is positive are the break points. Rule 7 − Find the angle of departure and the angle of arrival. The Angle of departure and the angle of arrival can be calculated at complex conjugate open loop poles and complex conjugate open loop zeros respectively. The formula for the angle of departure $phi_d$ is $$phi_d=180^0-phi$$ The formula for the angle of arrival $phi_a$ is $$phi_a=180^0+phi$$ Where, $$phi=sum phi_P-sum phi_Z$$ Example Let us now draw the root locus of the control system having open loop transfer function, $G(s)H(s)=frac{K}{s(s+1)(s+5)}$ Step 1 − The given open loop transfer function has three poles at $s = 0, s = −1$ and $s = −5$. It doesn’t have any zero. Therefore, the number of root locus branches is equal to the number of poles of the open loop transfer function. $$N=P=3$$ The three poles are located are shown in the above figure. The line segment between $s = −1$ and $s = 0$ is one branch of root locus on real axis. And the other branch of the root locus on the real axis is the line segment to the left of $s = −5$. Step 2 − We will get the values of the centroid and the angle of asymptotes by using the given formulae. Centroid $alpha = −2$ The angle of asymptotes are $theta = 60^0,180^0$ and $300^0$. The centroid and three asymptotes are shown in the following figure. Step 3 − Since two asymptotes have the angles of $60^0$ and $300^0$, two root locus branches intersect the imaginary axis. By using the Routh array method and special case(ii), the root locus branches intersects the imaginary axis at $jsqrt{5}$ and $−jsqrt{5}$. There will be one break-away point on the real axis

Control Systems – Polar Plots In the previous chapters, we discussed the Bode plots. There, we have two separate plots for both magnitude and phase as the function of frequency. Let us now discuss about polar plots. Polar plot is a plot which can be drawn between magnitude and phase. Here, the magnitudes are represented by normal values only. The polar form of $G(jomega)H(jomega)$ is $$G(jomega)H(jomega)=|G(jomega)H(jomega)| angle G(jomega)H(jomega)$$ The Polar plot is a plot, which can be drawn between the magnitude and the phase angle of $G(jomega)H(jomega)$ by varying $omega$ from zero to ∞. The polar graph sheet is shown in the following figure. This graph sheet consists of concentric circles and radial lines. The concentric circles and the radial lines represent the magnitudes and phase angles respectively. These angles are represented by positive values in anti-clock wise direction. Similarly, we can represent angles with negative values in clockwise direction. For example, the angle 2700 in anti-clock wise direction is equal to the angle −900 in clockwise direction. Rules for Drawing Polar Plots Follow these rules for plotting the polar plots. Substitute, $s = jomega$ in the open loop transfer function. Write the expressions for magnitude and the phase of $G(jomega)H(jomega)$. Find the starting magnitude and the phase of $G(jomega)H(jomega)$ by substituting $omega = 0$. So, the polar plot starts with this magnitude and the phase angle. Find the ending magnitude and the phase of $G(jomega)H(jomega)$ by substituting $omega = infty$. So, the polar plot ends with this magnitude and the phase angle. Check whether the polar plot intersects the real axis, by making the imaginary term of $G(jomega)H(jomega)$ equal to zero and find the value(s) of $omega$. Check whether the polar plot intersects the imaginary axis, by making real term of $G(jomega)H(jomega)$ equal to zero and find the value(s) of $omega$. For drawing polar plot more clearly, find the magnitude and phase of $G(jomega)H(jomega)$ by considering the other value(s) of $omega$. Example Consider the open loop transfer function of a closed loop control system. $$G(s)H(s)=frac{5}{s(s+1)(s+2)}$$ Let us draw the polar plot for this control system using the above rules. Step 1 − Substitute, $s = jomega$ in the open loop transfer function. $$G(jomega)H(jomega)=frac{5}{jomega(jomega+1)(jomega+2)}$$ The magnitude of the open loop transfer function is $$M=frac{5}{omega(sqrt{omega^2+1})(sqrt{omega^2+4})}$$ The phase angle of the open loop transfer function is $$phi=-90^0-tan^{-1}omega-tan^{-1}frac{omega}{2}$$ Step 2 − The following table shows the magnitude and the phase angle of the open loop transfer function at $omega = 0$ rad/sec and $omega = infty$ rad/sec. Frequency (rad/sec) Magnitude Phase angle(degrees) 0 ∞ -90 or 270 ∞ 0 -270 or 90 So, the polar plot starts at (∞,−900) and ends at (0,−2700). The first and the second terms within the brackets indicate the magnitude and phase angle respectively. Step 3 − Based on the starting and the ending polar co-ordinates, this polar plot will intersect the negative real axis. The phase angle corresponding to the negative real axis is −1800 or 1800. So, by equating the phase angle of the open loop transfer function to either −1800 or 1800, we will get the $omega$ value as $sqrt{2}$. By substituting $omega = sqrt{2}$ in the magnitude of the open loop transfer function, we will get $M = 0.83$. Therefore, the polar plot intersects the negative real axis when $omega = sqrt{2}$ and the polar coordinate is (0.83,−1800). So, we can draw the polar plot with the above information on the polar graph sheet. Learning working make money

Control Systems – Signal Flow Graphs Signal flow graph is a graphical representation of algebraic equations. In this chapter, let us discuss the basic concepts related signal flow graph and also learn how to draw signal flow graphs. Basic Elements of Signal Flow Graph Nodes and branches are the basic elements of signal flow graph. Node Node is a point which represents either a variable or a signal. There are three types of nodes — input node, output node and mixed node. Input Node − It is a node, which has only outgoing branches. Output Node − It is a node, which has only incoming branches. Mixed Node − It is a node, which has both incoming and outgoing branches. Example Let us consider the following signal flow graph to identify these nodes. The nodes present in this signal flow graph are y1, y2, y3 and y4. y1 and y4 are the input node and output node respectively. y2 and y3 are mixed nodes. Branch Branch is a line segment which joins two nodes. It has both gain and direction. For example, there are four branches in the above signal flow graph. These branches have gains of a, b, c and -d. Construction of Signal Flow Graph Let us construct a signal flow graph by considering the following algebraic equations − $$y_2=a_{12}y_1+a_{42}y_4$$ $$y_3=a_{23}y_2+a_{53}y_5$$ $$y_4=a_{34}y_3$$ $$y_5=a_{45}y_4+a_{35}y_3$$ $$y_6=a_{56}y_5$$ There will be six nodes (y1, y2, y3, y4, y5 and y6) and eight branches in this signal flow graph. The gains of the branches are a12, a23, a34, a45, a56, a42, a53 and a35. To get the overall signal flow graph, draw the signal flow graph for each equation, then combine all these signal flow graphs and then follow the steps given below − Step 1 − Signal flow graph for $y_2 = a_{13}y_1 + a_{42}y_4$ is shown in the following figure. Step 2 − Signal flow graph for $y_3 = a_{23}y_2 + a_{53}y_5$ is shown in the following figure. Step 3 − Signal flow graph for $y_4 = a_{34}y_3$ is shown in the following figure. Step 4 − Signal flow graph for $y_5 = a_{45}y_4 + a_{35}y_3$ is shown in the following figure. Step 5 − Signal flow graph for $y_6 = a_{56}y_5$ is shown in the following figure. Step 6 − Signal flow graph of overall system is shown in the following figure. Conversion of Block Diagrams into Signal Flow Graphs Follow these steps for converting a block diagram into its equivalent signal flow graph. Represent all the signals, variables, summing points and take-off points of block diagram as nodes in signal flow graph. Represent the blocks of block diagram as branches in signal flow graph. Represent the transfer functions inside the blocks of block diagram as gains of the branches in signal flow graph. Connect the nodes as per the block diagram. If there is connection between two nodes (but there is no block in between), then represent the gain of the branch as one. For example, between summing points, between summing point and takeoff point, between input and summing point, between take-off point and output. Example Let us convert the following block diagram into its equivalent signal flow graph. Represent the input signal $R(s)$ and output signal $C(s)$ of block diagram as input node $R(s)$ and output node $C(s)$ of signal flow graph. Just for reference, the remaining nodes (y1 to y9) are labelled in the block diagram. There are nine nodes other than input and output nodes. That is four nodes for four summing points, four nodes for four take-off points and one node for the variable between blocks $G_1$ and $G_2$. The following figure shows the equivalent signal flow graph. With the help of Mason’s gain formula (discussed in the next chapter), you can calculate the transfer function of this signal flow graph. This is the advantage of signal flow graphs. Here, we no need to simplify (reduce) the signal flow graphs for calculating the transfer function. Learning working make money

Electrical Analogies of Mechanical Systems Two systems are said to be analogous to each other if the following two conditions are satisfied. The two systems are physically different Differential equation modelling of these two systems are same Electrical systems and mechanical systems are two physically different systems. There are two types of electrical analogies of translational mechanical systems. Those are force voltage analogy and force current analogy. Force Voltage Analogy In force voltage analogy, the mathematical equations of translational mechanical system are compared with mesh equations of the electrical system. Consider the following translational mechanical system as shown in the following figure. The force balanced equation for this system is $$F=F_m+F_b+F_k$$ $Rightarrow F=Mfrac{text{d}^2x}{text{d}t^2}+Bfrac{text{d}x}{text{d}t}+Kx$ (Equation 1) Consider the following electrical system as shown in the following figure. This circuit consists of a resistor, an inductor and a capacitor. All these electrical elements are connected in a series. The input voltage applied to this circuit is $V$ volts and the current flowing through the circuit is $i$ Amps. Mesh equation for this circuit is $V=Ri+Lfrac{text{d}i}{text{d}t}+frac{1}{c}int idt$ (Equation 2) Substitute, $i=frac{text{d}q}{text{d}t}$ in Equation 2. $$V=Rfrac{text{d}q}{text{d}t}+Lfrac{text{d}^2q}{text{d}t^2}+frac{q}{C}$$ $Rightarrow V=Lfrac{text{d}^2q}{text{d}t^2}+Rfrac{text{d}q}{text{d}t}+left ( frac{1}{c} right )q$ (Equation 3) By comparing Equation 1 and Equation 3, we will get the analogous quantities of the translational mechanical system and electrical system. The following table shows these analogous quantities. Translational Mechanical System Electrical System Force(F) Voltage(V) Mass(M) Inductance(L) Frictional Coefficient(B) Resistance(R) Spring Constant(K) Reciprocal of Capacitance $(frac{1}{c})$ Displacement(x) Charge(q) Velocity(v) Current(i) Similarly, there is torque voltage analogy for rotational mechanical systems. Let us now discuss about this analogy. Torque Voltage Analogy In this analogy, the mathematical equations of rotational mechanical system are compared with mesh equations of the electrical system. Rotational mechanical system is shown in the following figure. The torque balanced equation is $$T=T_j+T_b+T_k$$ $Rightarrow T=Jfrac{text{d}^2theta}{text{d}t^2}+Bfrac{text{d}theta}{text{d}t}+ktheta$ (Equation 4) By comparing Equation 4 and Equation 3, we will get the analogous quantities of rotational mechanical system and electrical system. The following table shows these analogous quantities. Rotational Mechanical System Electrical System Torque(T) Voltage(V) Moment of Inertia(J) Inductance(L) Rotational friction coefficient(B) Resistance(R) Torsional spring constant(K) Reciprocal of Capacitance $(frac{1}{c})$ Angular Displacement(θ) Charge(q) Angular Velocity(ω) Current(i) Force Current Analogy In force current analogy, the mathematical equations of the translational mechanical system are compared with the nodal equations of the electrical system. Consider the following electrical system as shown in the following figure. This circuit consists of current source, resistor, inductor and capacitor. All these electrical elements are connected in parallel. The nodal equation is $i=frac{V}{R}+frac{1}{L}int Vdt+Cfrac{text{d}V}{text{d}t}$ (Equation 5) Substitute, $V=frac{text{d}Psi}{text{d}t}$ in Equation 5. $$i=frac{1}{R}frac{text{d}Psi}{text{d}t}+left ( frac{1}{L} right )Psi+Cfrac{text{d}^2Psi}{text{d}t^2}$$ $Rightarrow i=Cfrac{text{d}^2Psi}{text{d}t^2}+left ( frac{1}{R} right )frac{text{d}Psi}{text{d}t}+left ( frac{1}{L} right )Psi$ (Equation 6) By comparing Equation 1 and Equation 6, we will get the analogous quantities of the translational mechanical system and electrical system. The following table shows these analogous quantities. Translational Mechanical System Electrical System Force(F) Current(i) Mass(M) Capacitance(C) Frictional coefficient(B) Reciprocal of Resistance$(frac{1}{R})$ Spring constant(K) Reciprocal of Inductance$(frac{1}{L})$ Displacement(x) Magnetic Flux(ψ) Velocity(v) Voltage(V) Similarly, there is a torque current analogy for rotational mechanical systems. Let us now discuss this analogy. Torque Current Analogy In this analogy, the mathematical equations of the rotational mechanical system are compared with the nodal mesh equations of the electrical system. By comparing Equation 4 and Equation 6, we will get the analogous quantities of rotational mechanical system and electrical system. The following table shows these analogous quantities. Rotational Mechanical System Electrical System Torque(T) Current(i) Moment of inertia(J) Capacitance(C) Rotational friction coefficient(B) Reciprocal of Resistance$(frac{1}{R})$ Torsional spring constant(K) Reciprocal of Inductance$(frac{1}{L})$ Angular displacement(θ) Magnetic flux(ψ) Angular velocity(ω) Voltage(V) In this chapter, we discussed the electrical analogies of the mechanical systems. These analogies are helpful to study and analyze the non-electrical system like mechanical system from analogous electrical system. Learning working make money

Control Systems – Root Locus In the root locus diagram, we can observe the path of the closed loop poles. Hence, we can identify the nature of the control system. In this technique, we will use an open loop transfer function to know the stability of the closed loop control system. Basics of Root Locus The Root locus is the locus of the roots of the characteristic equation by varying system gain K from zero to infinity. We know that, the characteristic equation of the closed loop control system is $$1+G(s)H(s)=0$$ We can represent $G(s)H(s)$ as $$G(s)H(s)=Kfrac{N(s)}{D(s)}$$ Where, K represents the multiplying factor N(s) represents the numerator term having (factored) nth order polynomial of ‘s’. D(s) represents the denominator term having (factored) mth order polynomial of ‘s’. Substitute, $G(s)H(s)$ value in the characteristic equation. $$1+kfrac{N(s)}{D(s)}=0$$ $$Rightarrow D(s)+KN(s)=0$$ Case 1 − K = 0 If $K=0$, then $D(s)=0$. That means, the closed loop poles are equal to open loop poles when K is zero. Case 2 − K = ∞ Re-write the above characteristic equation as $$Kleft(frac{1}{K}+frac{N(s)}{D(s)} right )=0 Rightarrow frac{1}{K}+frac{N(s)}{D(s)}=0$$ Substitute, $K = infty$ in the above equation. $$frac{1}{infty}+frac{N(s)}{D(s)}=0 Rightarrow frac{N(s)}{D(s)}=0 Rightarrow N(s)=0$$ If $K=infty$, then $N(s)=0$. It means the closed loop poles are equal to the open loop zeros when K is infinity. From above two cases, we can conclude that the root locus branches start at open loop poles and end at open loop zeros. Angle Condition and Magnitude Condition The points on the root locus branches satisfy the angle condition. So, the angle condition is used to know whether the point exist on root locus branch or not. We can find the value of K for the points on the root locus branches by using magnitude condition. So, we can use the magnitude condition for the points, and this satisfies the angle condition. Characteristic equation of closed loop control system is $$1+G(s)H(s)=0$$ $$Rightarrow G(s)H(s)=-1+j0$$ The phase angle of $G(s)H(s)$ is $$angle G(s)H(s)=tan^{-1}left ( frac{0}{-1} right )=(2n+1)pi$$ The angle condition is the point at which the angle of the open loop transfer function is an odd multiple of 1800. Magnitude of $G(s)H(s)$ is – $$|G(s)H(s)|=sqrt {(-1)^2+0^2}=1$$ The magnitude condition is that the point (which satisfied the angle condition) at which the magnitude of the open loop transfer function is one. Learning working make money

Modelling of Mechanical Systems In this chapter, let us discuss the differential equation modeling of mechanical systems. There are two types of mechanical systems based on the type of motion. Translational mechanical systems Rotational mechanical systems Modeling of Translational Mechanical Systems Translational mechanical systems move along a straight line. These systems mainly consist of three basic elements. Those are mass, spring and dashpot or damper. If a force is applied to a translational mechanical system, then it is opposed by opposing forces due to mass, elasticity and friction of the system. Since the applied force and the opposing forces are in opposite directions, the algebraic sum of the forces acting on the system is zero. Let us now see the force opposed by these three elements individually. Mass Mass is the property of a body, which stores kinetic energy. If a force is applied on a body having mass M, then it is opposed by an opposing force due to mass. This opposing force is proportional to the acceleration of the body. Assume elasticity and friction are negligible. $$F_mpropto: a$$ $$Rightarrow F_m=Ma=Mfrac{text{d}^2x}{text{d}t^2}$$ $$F=F_m=Mfrac{text{d}^2x}{text{d}t^2}$$ Where, F is the applied force Fm is the opposing force due to mass M is mass a is acceleration x is displacement Spring Spring is an element, which stores potential energy. If a force is applied on spring K, then it is opposed by an opposing force due to elasticity of spring. This opposing force is proportional to the displacement of the spring. Assume mass and friction are negligible. $$Fpropto: x$$ $$Rightarrow F_k=Kx$$ $$F=F_k=Kx$$ Where, F is the applied force Fk is the opposing force due to elasticity of spring K is spring constant x is displacement Dashpot If a force is applied on dashpot B, then it is opposed by an opposing force due to friction of the dashpot. This opposing force is proportional to the velocity of the body. Assume mass and elasticity are negligible. $$F_bpropto: nu$$ $$Rightarrow F_b=Bnu=Bfrac{text{d}x}{text{d}t}$$ $$F=F_b=Bfrac{text{d}x}{text{d}t}$$ Where, Fb is the opposing force due to friction of dashpot B is the frictional coefficient v is velocity x is displacement Modeling of Rotational Mechanical Systems Rotational mechanical systems move about a fixed axis. These systems mainly consist of three basic elements. Those are moment of inertia, torsional spring and dashpot. If a torque is applied to a rotational mechanical system, then it is opposed by opposing torques due to moment of inertia, elasticity and friction of the system. Since the applied torque and the opposing torques are in opposite directions, the algebraic sum of torques acting on the system is zero. Let us now see the torque opposed by these three elements individually. Moment of Inertia In translational mechanical system, mass stores kinetic energy. Similarly, in rotational mechanical system, moment of inertia stores kinetic energy. If a torque is applied on a body having moment of inertia J, then it is opposed by an opposing torque due to the moment of inertia. This opposing torque is proportional to angular acceleration of the body. Assume elasticity and friction are negligible. $$T_jpropto: alpha$$ $$Rightarrow T_j=Jalpha=Jfrac{text{d}^2theta}{text{d}t^2}$$ $$T=T_j=Jfrac{text{d}^2theta}{text{d}t^2}$$ Where, T is the applied torque Tj is the opposing torque due to moment of inertia J is moment of inertia α is angular acceleration θ is angular displacement Torsional Spring In translational mechanical system, spring stores potential energy. Similarly, in rotational mechanical system, torsional spring stores potential energy. If a torque is applied on torsional spring K, then it is opposed by an opposing torque due to the elasticity of torsional spring. This opposing torque is proportional to the angular displacement of the torsional spring. Assume that the moment of inertia and friction are negligible. $$T_kpropto: theta$$ $$Rightarrow T_k=Ktheta$$ $$T=T_k=Ktheta$$ Where, T is the applied torque Tk is the opposing torque due to elasticity of torsional spring K is the torsional spring constant θ is angular displacement Dashpot If a torque is applied on dashpot B, then it is opposed by an opposing torque due to the rotational friction of the dashpot. This opposing torque is proportional to the angular velocity of the body. Assume the moment of inertia and elasticity are negligible. $$T_bpropto: omega$$ $$Rightarrow T_b=Bomega=Bfrac{text{d}theta}{text{d}t}$$ $$T=T_b=Bfrac{text{d}theta}{text{d}t}$$ Where, Tb is the opposing torque due to the rotational friction of the dashpot B is the rotational friction coefficient ω is the angular velocity θ is the angular displacement Learning working make money

Control Systems – Stability Analysis In this chapter, let us discuss the stability analysis in the ‘s’ domain using the RouthHurwitz stability criterion. In this criterion, we require the characteristic equation to find the stability of the closed loop control systems. Routh-Hurwitz Stability Criterion Routh-Hurwitz stability criterion is having one necessary condition and one sufficient condition for stability. If any control system doesn’t satisfy the necessary condition, then we can say that the control system is unstable. But, if the control system satisfies the necessary condition, then it may or may not be stable. So, the sufficient condition is helpful for knowing whether the control system is stable or not. Necessary Condition for Routh-Hurwitz Stability The necessary condition is that the coefficients of the characteristic polynomial should be positive. This implies that all the roots of the characteristic equation should have negative real parts. Consider the characteristic equation of the order ‘n’ is – $$a_0s^n+a_1s^{n-1}+a_2s^{n-2}+…+a_{n-1}s^1+a_ns^0=0$$ Note that, there should not be any term missing in the nth order characteristic equation. This means that the nth order characteristic equation should not have any coefficient that is of zero value. Sufficient Condition for Routh-Hurwitz Stability The sufficient condition is that all the elements of the first column of the Routh array should have the same sign. This means that all the elements of the first column of the Routh array should be either positive or negative. Routh Array Method If all the roots of the characteristic equation exist to the left half of the ‘s’ plane, then the control system is stable. If at least one root of the characteristic equation exists to the right half of the ‘s’ plane, then the control system is unstable. So, we have to find the roots of the characteristic equation to know whether the control system is stable or unstable. But, it is difficult to find the roots of the characteristic equation as order increases. So, to overcome this problem there we have the Routh array method. In this method, there is no need to calculate the roots of the characteristic equation. First formulate the Routh table and find the number of the sign changes in the first column of the Routh table. The number of sign changes in the first column of the Routh table gives the number of roots of characteristic equation that exist in the right half of the ‘s’ plane and the control system is unstable. Follow this procedure for forming the Routh table. Fill the first two rows of the Routh array with the coefficients of the characteristic polynomial as mentioned in the table below. Start with the coefficient of $s^n$ and continue up to the coefficient of $s^0$. Fill the remaining rows of the Routh array with the elements as mentioned in the table below. Continue this process till you get the first column element of row $s^0$ is $a_n$. Here, $a_n$ is the coefficient of $s^0$ in the characteristic polynomial. Note − If any row elements of the Routh table have some common factor, then you can divide the row elements with that factor for the simplification will be easy. The following table shows the Routh array of the nth order characteristic polynomial. $$a_0s^n+a_1s^{n-1}+a_2s^{n-2}+…+a_{n-1}s^1+a_ns^0$$ $s^n$ $a_0$ $a_2$ $a_4$ $a_6$ … … $s^{n-1}$ $a_1$ $a_3$ $a_5$ $a_7$ … … $s^{n-2}$ $b_1=frac{a_1a_2-a_3a_0}{a_1}$ $b_2=frac{a_1a_4-a_5a_0}{a_1}$ $b_3=frac{a_1a_6-a_7a_0}{a_1}$ … … … $s^{n-3}$ $c_1=frac{b_1a_3-b_2a_1}{b_1}$ $c_2=frac{b_1a_55-b_3a_1}{b_1}$ $vdots$ $vdots $ $vdots$ $vdots$ $vdots$ $s^1$ $vdots$ $vdots$ $s^0$ $a_n$ Example Let us find the stability of the control system having characteristic equation, $$s^4+3s^3+3s^2+2s+1=0$$ Step 1 − Verify the necessary condition for the Routh-Hurwitz stability. All the coefficients of the characteristic polynomial, $s^4+3s^3+3s^2+2s+1$ are positive. So, the control system satisfies the necessary condition. Step 2 − Form the Routh array for the given characteristic polynomial. $s^4$ $1$ $3$ $1$ $s^3$ $3$ $2$ $s^2$ $frac{(3 times 3)-(2 times 1)}{3}=frac{7}{3}$ $frac{(3 times 1)-(0 times 1)}{3}=frac{3}{3}=1$ $s^1$ $frac{left ( frac{7}{3}times 2 right )-(1 times 3)}{frac{7}{3}}=frac{5}{7}$ $s^0$ $1$ Step 3 − Verify the sufficient condition for the Routh-Hurwitz stability. All the elements of the first column of the Routh array are positive. There is no sign change in the first column of the Routh array. So, the control system is stable. Special Cases of Routh Array We may come across two types of situations, while forming the Routh table. It is difficult to complete the Routh table from these two situations. The two special cases are − The first element of any row of the Routh array is zero. All the elements of any row of the Routh array are zero. Let us now discuss how to overcome the difficulty in these two cases, one by one. First Element of any row of the Routh array is zero If any row of the Routh array contains only the first element as zero and at least one of the remaining elements have non-zero value, then replace the first element with a small positive integer, $epsilon$. And then continue the process of completing the Routh table. Now, find the number of sign changes in the first column of the Routh table by substituting $epsilon$ tends to zero. Example Let us find the stability of the control system having characteristic equation, $$s^4+2s^3+s^2+2s+1=0$$ Step 1 − Verify the necessary condition for the Routh-Hurwitz stability. All the coefficients of the characteristic polynomial, $s^4+2s^3+s^2+2s+1$ are positive. So, the control system satisfied the necessary condition. Step 2 − Form the Routh array for the given characteristic polynomial. $s^4$ $1$ $1$ $1$ $s^3$ 2 1 2 1 $s^2$ $frac{(1 times 1)-(1 times 1)}{1}=0$ $frac{(1 times 1)-(0 times 1)}{1}=1$ $s^1$ $s^0$ The row $s^3$ elements have 2 as the common factor. So, all these elements are divided by 2. Special case (i) − Only the first element of row $s^2$ is zero. So, replace it by $epsilon$ and continue the process of completing the Routh table. $s^4$ 1 1 1 $s^3$ 1 1 $s^2$ $epsilon$ 1 $s^1$ $frac{left ( epsilon times 1 right )-left ( 1 times 1 right )}{epsilon}=frac{epsilon-1}{epsilon}$ $s^0$