Category: control Systems

Control Systems – Time Response Analysis We can analyze the response of the control systems in both the time domain and the frequency domain. We will discuss frequency response analysis of control systems in later chapters. Let us now discuss about the time response analysis of control systems. What is Time Response? If the output of control system for an input varies with respect to time, then it is called the time response of the control system. The time response consists of two parts. Transient response Steady state response The response of control system in time domain is shown in the following figure. Here, both the transient and the steady states are indicated in the figure. The responses corresponding to these states are known as transient and steady state responses. Mathematically, we can write the time response c(t) as $$c(t)=c_{tr}(t)+c_{ss}(t)$$ Where, ctr(t) is the transient response css(t) is the steady state response Transient Response After applying input to the control system, output takes certain time to reach steady state. So, the output will be in transient state till it goes to a steady state. Therefore, the response of the control system during the transient state is known as transient response. The transient response will be zero for large values of ‘t’. Ideally, this value of ‘t’ is infinity and practically, it is five times constant. Mathematically, we can write it as $$lim_{trightarrow infty }c_{tr}(t)=0$$ Steady state Response The part of the time response that remains even after the transient response has zero value for large values of ‘t’ is known as steady state response. This means, the transient response will be zero even during the steady state. Example Let us find the transient and steady state terms of the time response of the control system $c(t)=10+5e^{-t}$ Here, the second term $5e^{-t}$ will be zero as t denotes infinity. So, this is the transient term. And the first term 10 remains even as t approaches infinity. So, this is the steady state term. Standard Test Signals The standard test signals are impulse, step, ramp and parabolic. These signals are used to know the performance of the control systems using time response of the output. Unit Impulse Signal A unit impulse signal, δ(t) is defined as $delta (t)=0$ for $tneq 0$ and $int_{0^-}^{0^+} delta (t)dt=1$ The following figure shows unit impulse signal. So, the unit impulse signal exists only at ‘t’ is equal to zero. The area of this signal under small interval of time around ‘t’ is equal to zero is one. The value of unit impulse signal is zero for all other values of ‘t’. Unit Step Signal A unit step signal, u(t) is defined as $$u(t)=1;tgeq 0$$ $=0; t<0$ Following figure shows unit step signal. So, the unit step signal exists for all positive values of ‘t’ including zero. And its value is one during this interval. The value of the unit step signal is zero for all negative values of ‘t’. Unit Ramp Signal A unit ramp signal, r(t) is defined as $$r(t)=t; tgeq 0$$ $=0; t<0$ We can write unit ramp signal, $r(t)$ in terms of unit step signal, $u(t)$ as $$r(t)=tu(t)$$ Following figure shows unit ramp signal. So, the unit ramp signal exists for all positive values of ‘t’ including zero. And its value increases linearly with respect to ‘t’ during this interval. The value of unit ramp signal is zero for all negative values of ‘t’. Unit Parabolic Signal A unit parabolic signal, p(t) is defined as, $$p(t)=frac{t^2}{2}; tgeq 0$$ $=0; t<0$ We can write unit parabolic signal, $p(t)$ in terms of the unit step signal, $u(t)$ as, $$p(t)=frac{t^2}{2}u(t)$$ The following figure shows the unit parabolic signal. So, the unit parabolic signal exists for all the positive values of ‘t’ including zero. And its value increases non-linearly with respect to ‘t’ during this interval. The value of the unit parabolic signal is zero for all the negative values of ‘t’. Learning working make money

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 of (1+GH). If the value of (1+GH) is less than 1, then the overall gain increases. In this case, ”GH” value is negative because the gain of the feedback path is negative. If the value of (1+GH) is greater than 1, then the overall gain decreases. In this case, ”GH” value is positive because the gain of the feedback path is positive. In general, ”G” and ”H” are functions of frequency. So, the feedback will increase the overall gain of the system in one frequency range and decrease in the other frequency range. Effect of Feedback on Sensitivity Sensitivity of the overall gain of negative feedback closed loop control system (T) to the variation in open loop gain (G) is defined as $S_{G}^{T} = frac{frac{partial T}{T}}{frac{partial G}{G}}=frac{Percentage: change : in :T}{Percentage: change : in :G}$ (Equation 3) Where, ∂T is the incremental change in T due to incremental change in G. We can rewrite Equation 3 as $S_{G}^{T}=frac{partial T}{partial G}frac{G}{T}$ (Equation 4) Do partial differentiation with respect to G on both sides of Equation 2. $frac{partial T}{partial G}=frac{partial}{partial G}left (frac{G}{1+GH} right )=frac{(1+GH).1-G(H)}{(1+GH)^2}=frac{1}{(1+GH)^2}$ (Equation 5) From Equation 2, you will get $frac{G}{T}=1+GH$ (Equation 6) Substitute Equation 5 and Equation 6 in Equation 4. $$S_{G}^{T}=frac{1}{(1+GH)^2}(1+GH)=frac{1}{1+GH}$$ So, we got the sensitivity of the overall gain of closed loop control system as the reciprocal of (1+GH). So, Sensitivity may increase or decrease depending on the value of (1+GH). If the value of (1+GH) is less than 1, then sensitivity increases. In this case, ”GH” value is negative because the gain of feedback path is negative. If the value of (1+GH) is greater than 1, then sensitivity decreases. In this case, ”GH” value is positive because the gain of feedback path is positive. In general, ”G” and ”H” are functions of frequency. So, feedback will increase the sensitivity of the system gain in one frequency range and decrease in the other frequency range. Therefore, we have to choose the values of ”GH” in such a way that the system is insensitive or less sensitive to parameter variations. Effect of Feedback on Stability A system is said to be stable, if its output is under control. Otherwise, it is said to be unstable. In Equation 2, if the denominator value is zero (i.e., GH = -1), then the output of the control system will be infinite. So, the control system becomes unstable. Therefore, we have to properly choose the feedback in order to make the control system stable. Effect of Feedback on Noise To know the effect of feedback on noise, let us compare the transfer function relations with and without feedback due to noise signal alone. Consider an open loop control system with noise signal as shown below. The open loop transfer function due to noise signal alone is $frac{C(s)}{N(s)}=G_b$ (Equation 7) It is obtained by making the other input $R(s)$ equal to zero. Consider a closed loop control system with noise signal as shown below. The closed loop transfer function due to noise signal alone is $frac{C(s)}{N(s)}=frac{G_b}{1+G_aG_bH}$ (Equation 8) It is obtained by making the other input $R(s)$ equal to zero. Compare Equation 7 and Equation 8, In the closed loop control system, the gain due to noise signal is decreased by a factor of $(1+G_a G_b H)$ provided that the term $(1+G_a G_b H)$ is greater than one. Learning working make money

Control Systems Tutorial Job Search This tutorial is meant to provide the readers the know how to analyze the control systems with the help of mathematical models. After completing this tutorial, you will be able to learn various methods and techniques in order to improve the performance of the control systems based on the requirements. Audience This tutorial is meant for all those readers who are aspiring to learn the fundamental concepts of Control Systems. Prerequisites A learner who wants to go ahead with this tutorial needs to have a basic understanding of Learning working make money

Control Systems – Nyquist Plots Nyquist plots are the continuation of polar plots for finding the stability of the closed loop control systems by varying ω from −∞ to ∞. That means, Nyquist plots are used to draw the complete frequency response of the open loop transfer function. Nyquist Stability Criterion The Nyquist stability criterion works on the principle of argument. It states that if there are P poles and Z zeros are enclosed by the ‘s’ plane closed path, then the corresponding $G(s)H(s)$ plane must encircle the origin $P − Z$ times. So, we can write the number of encirclements N as, $$N=P-Z$$ If the enclosed ‘s’ plane closed path contains only poles, then the direction of the encirclement in the $G(s)H(s)$ plane will be opposite to the direction of the enclosed closed path in the ‘s’ plane. If the enclosed ‘s’ plane closed path contains only zeros, then the direction of the encirclement in the $G(s)H(s)$ plane will be in the same direction as that of the enclosed closed path in the ‘s’ plane. Let us now apply the principle of argument to the entire right half of the ‘s’ plane by selecting it as a closed path. This selected path is called the Nyquist contour. We know that the closed loop control system is stable if all the poles of the closed loop transfer function are in the left half of the ‘s’ plane. So, the poles of the closed loop transfer function are nothing but the roots of the characteristic equation. As the order of the characteristic equation increases, it is difficult to find the roots. So, let us correlate these roots of the characteristic equation as follows. The Poles of the characteristic equation are same as that of the poles of the open loop transfer function. The zeros of the characteristic equation are same as that of the poles of the closed loop transfer function. We know that the open loop control system is stable if there is no open loop pole in the the right half of the ‘s’ plane. i.e.,$P=0 Rightarrow N=-Z$ We know that the closed loop control system is stable if there is no closed loop pole in the right half of the ‘s’ plane. i.e.,$Z=0 Rightarrow N=P$ Nyquist stability criterion states the number of encirclements about the critical point (1+j0) must be equal to the poles of characteristic equation, which is nothing but the poles of the open loop transfer function in the right half of the ‘s’ plane. The shift in origin to (1+j0) gives the characteristic equation plane. Rules for Drawing Nyquist Plots Follow these rules for plotting the Nyquist plots. Locate the poles and zeros of open loop transfer function $G(s)H(s)$ in ‘s’ plane. Draw the polar plot by varying $omega$ from zero to infinity. If pole or zero present at s = 0, then varying $omega$ from 0+ to infinity for drawing polar plot. Draw the mirror image of above polar plot for values of $omega$ ranging from −∞ to zero (0− if any pole or zero present at s=0). The number of infinite radius half circles will be equal to the number of poles or zeros at origin. The infinite radius half circle will start at the point where the mirror image of the polar plot ends. And this infinite radius half circle will end at the point where the polar plot starts. After drawing the Nyquist plot, we can find the stability of the closed loop control system using the Nyquist stability criterion. If the critical point (-1+j0) lies outside the encirclement, then the closed loop control system is absolutely stable. Stability Analysis using Nyquist Plots From the Nyquist plots, we can identify whether the control system is stable, marginally stable or unstable based on the values of these parameters. Gain cross over frequency and phase cross over frequency Gain margin and phase margin Phase Cross over Frequency The frequency at which the Nyquist plot intersects the negative real axis (phase angle is 1800) is known as the phase cross over frequency. It is denoted by $omega_{pc}$. Gain Cross over Frequency The frequency at which the Nyquist plot is having the magnitude of one is known as the gain cross over frequency. It is denoted by $omega_{gc}$. The stability of the control system based on the relation between phase cross over frequency and gain cross over frequency is listed below. If the phase cross over frequency $omega_{pc}$ is greater than the gain cross over frequency $omega_{gc}$, then the control system is stable. If the phase cross over frequency $omega_{pc}$ is equal to the gain cross over frequency $omega_{gc}$, then the control system is marginally stable. If phase cross over frequency $omega_{pc}$ is less than gain cross over frequency $omega_{gc}$, then the control system is unstable. Gain Margin The gain margin $GM$ is equal to the reciprocal of the magnitude of the Nyquist plot at the phase cross over frequency. $$GM=frac{1}{M_{pc}}$$ Where, $M_{pc}$ is the magnitude in normal scale at the phase cross over frequency. Phase Margin The phase margin $PM$ is equal to the sum of 1800 and the phase angle at the gain cross over frequency. $$PM=180^0+phi_{gc}$$ Where, $phi_{gc}$ is the phase angle at the gain cross over frequency. The stability of the control system based on the relation between the gain margin and the phase margin is listed below. If the gain margin $GM$ is greater than one and the phase margin $PM$ is positive, then the control system is stable. If the gain margin $GM$ is equal to one and the phase margin $PM$ is zero degrees, then the control system is marginally stable. If the gain margin $GM$ is less than one and / or the phase margin $PM$ is negative, then the control system is unstable. Learning working make money

Control Systems – Bode Plots The Bode plot or the Bode diagram consists of two plots − Magnitude plot Phase plot In both the plots, x-axis represents angular frequency (logarithmic scale). Whereas, yaxis represents the magnitude (linear scale) of open loop transfer function in the magnitude plot and the phase angle (linear scale) of the open loop transfer function in the phase plot. The magnitude of the open loop transfer function in dB is – $$M=20: log|G(jomega)H(jomega)|$$ The phase angle of the open loop transfer function in degrees is – $$phi=angle G(jomega)H(jomega)$$ Note − The base of logarithm is 10. Basic of Bode Plots The following table shows the slope, magnitude and the phase angle values of the terms present in the open loop transfer function. This data is useful while drawing the Bode plots. Type of term G(jω)H(jω) Slope(dB/dec) Magnitude (dB) Phase angle(degrees) Constant $K$ $0$ $20 log K$ $0$ Zero at origin $jomega$ $20$ $20 log omega$ $90$ ‘n’ zeros at origin $(jomega)^n$ $20: n$ $20: n log omega$ $90: n$ Pole at origin $frac{1}{jomega}$ $-20$ $-20 log omega$ $-90 : or : 270$ ‘n’ poles at origin $frac{1}{(jomega)^n}$ $-20: n$ $-20 : n log omega$ $-90 : n : or : 270 : n$ Simple zero $1+jomega r$ $20$ $0: for: omega < frac{1}{r}$ $20: log omega r: for : omega > frac{1}{r}$ $0 : for : omega < frac{1}{r}$ $90 : for : omega > frac{1}{r}$ Simple pole $frac{1}{1+jomega r}$ $-20$ $0: for: omega < frac{1}{r}$ $-20: log omega r: for: omega > frac{1}{r}$ $0 : for : omega < frac{1}{r}$ $-90: or : 270 : for: omega > frac{1}{r}$ Second order derivative term $omega_n^2left ( 1-frac{omega^2}{omega_n^2}+frac{2jdeltaomega}{omega_n} right )$ $40$ $40: log: omega_n: for : omega < omega_n$ $20: log:(2deltaomega_n^2): for : omega=omega_n$ $40 : log : omega:for :omega > omega_n$ $0 : for : omega < omega_n$ $90 : for : omega = omega_n$ $180 : for : omega > omega_n$ Second order integral term $frac{1}{omega_n^2left ( 1-frac{omega^2}{omega_n^2}+frac{2jdeltaomega}{omega_n} right )}$ $-40$ $-40: log: omega_n: for : omega < omega_n$ $-20: log:(2deltaomega_n^2): for : omega=omega_n$ $-40 : log : omega:for :omega > omega_n$ $-0 : for : omega < omega_n$ $-90 : for : omega = omega_n$ $-180 : for : omega > omega_n$ Consider the open loop transfer function $G(s)H(s) = K$. Magnitude $M = 20: log K$ dB Phase angle $phi = 0$ degrees If $K = 1$, then magnitude is 0 dB. If $K > 1$, then magnitude will be positive. If $K < 1$, then magnitude will be negative. The following figure shows the corresponding Bode plot. The magnitude plot is a horizontal line, which is independent of frequency. The 0 dB line itself is the magnitude plot when the value of K is one. For the positive values of K, the horizontal line will shift $20 :log K$ dB above the 0 dB line. For the negative values of K, the horizontal line will shift $20: log K$ dB below the 0 dB line. The Zero degrees line itself is the phase plot for all the positive values of K. Consider the open loop transfer function $G(s)H(s) = s$. Magnitude $M = 20 log omega$ dB Phase angle $phi = 90^0$ At $omega = 0.1$ rad/sec, the magnitude is -20 dB. At $omega = 1$ rad/sec, the magnitude is 0 dB. At $omega = 10$ rad/sec, the magnitude is 20 dB. The following figure shows the corresponding Bode plot. The magnitude plot is a line, which is having a slope of 20 dB/dec. This line started at $omega = 0.1$ rad/sec having a magnitude of -20 dB and it continues on the same slope. It is touching 0 dB line at $omega = 1$ rad/sec. In this case, the phase plot is 900 line. Consider the open loop transfer function $G(s)H(s) = 1 + stau$. Magnitude $M = 20: log sqrt{1 + omega^2tau^2}$ dB Phase angle $phi = tan^{-1}omegatau$ degrees For $ω < frac{1}{tau}$ , the magnitude is 0 dB and phase angle is 0 degrees. For $omega > frac{1}{tau}$ , the magnitude is $20: log omegatau$ dB and phase angle is 900. The following figure shows the corresponding Bode plot. The magnitude plot is having magnitude of 0 dB upto $omega=frac{1}{tau}$ rad/sec. From $omega = frac{1}{tau}$ rad/sec, it is having a slope of 20 dB/dec. In this case, the phase plot is having phase angle of 0 degrees up to $omega = frac{1}{tau}$ rad/sec and from here, it is having phase angle of 900. This Bode plot is called the asymptotic Bode plot. As the magnitude and the phase plots are represented with straight lines, the Exact Bode plots resemble the asymptotic Bode plots. The only difference is that the Exact Bode plots will have simple curves instead of straight lines. Similarly, you can draw the Bode plots for other terms of the open loop transfer function which are given in the table. Learning working make money

Control Systems – Block Diagram Reduction The concepts discussed in the previous chapter are helpful for reducing (simplifying) the block diagrams. Block Diagram Reduction Rules Follow these rules for simplifying (reducing) the block diagram, which is having many blocks, summing points and take-off points. Rule 1 − Check for the blocks connected in series and simplify. Rule 2 − Check for the blocks connected in parallel and simplify. Rule 3 − Check for the blocks connected in feedback loop and simplify. Rule 4 − If there is difficulty with take-off point while simplifying, shift it towards right. Rule 5 − If there is difficulty with summing point while simplifying, shift it towards left. Rule 6 − Repeat the above steps till you get the simplified form, i.e., single block. Note − The transfer function present in this single block is the transfer function of the overall block diagram. Example Consider the block diagram shown in the following figure. Let us simplify (reduce) this block diagram using the block diagram reduction rules. Step 1 − Use Rule 1 for blocks $G_1$ and $G_2$. Use Rule 2 for blocks $G_3$ and $G_4$. The modified block diagram is shown in the following figure. Step 2 − Use Rule 3 for blocks $G_1G_2$ and $H_1$. Use Rule 4 for shifting take-off point after the block $G_5$. The modified block diagram is shown in the following figure. Step 3 − Use Rule 1 for blocks $(G_3 + G_4)$ and $G_5$. The modified block diagram is shown in the following figure. Step 4 − Use Rule 3 for blocks $(G_3 + G_4)G_5$ and $H_3$. The modified block diagram is shown in the following figure. Step 5 − Use Rule 1 for blocks connected in series. The modified block diagram is shown in the following figure. Step 6 − Use Rule 3 for blocks connected in feedback loop. The modified block diagram is shown in the following figure. This is the simplified block diagram. Therefore, the transfer function of the system is $$frac{Y(s)}{R(s)}=frac{G_1G_2G_5^2(G_3+G_4)}{(1+G_1G_2H_1)lbrace 1+(G_3+G_4)G_5H_3rbrace G_5-G_1G_2G_5(G_3+G_4)H_2}$$ Note − Follow these steps in order to calculate the transfer function of the block diagram having multiple inputs. Step 1 − Find the transfer function of block diagram by considering one input at a time and make the remaining inputs as zero. Step 2 − Repeat step 1 for remaining inputs. Step 3 − Get the overall transfer function by adding all those transfer functions. The block diagram reduction process takes more time for complicated systems. Because, we have to draw the (partially simplified) block diagram after each step. So, to overcome this drawback, use signal flow graphs (representation). In the next two chapters, we will discuss about the concepts related to signal flow graphs, i.e., how to represent signal flow graph from a given block diagram and calculation of transfer function just by using a gain formula without doing any reduction process. Learning working make money

Control Systems – State Space Model The state space model of Linear Time-Invariant (LTI) system can be represented as, $$dot{X}=AX+BU$$ $$Y=CX+DU$$ The first and the second equations are known as state equation and output equation respectively. Where, X and $dot{X}$ are the state vector and the differential state vector respectively. U and Y are input vector and output vector respectively. A is the system matrix. B and C are the input and the output matrices. D is the feed-forward matrix. Basic Concepts of State Space Model The following basic terminology involved in this chapter. State It is a group of variables, which summarizes the history of the system in order to predict the future values (outputs). State Variable The number of the state variables required is equal to the number of the storage elements present in the system. Examples − current flowing through inductor, voltage across capacitor State Vector It is a vector, which contains the state variables as elements. In the earlier chapters, we have discussed two mathematical models of the control systems. Those are the differential equation model and the transfer function model. The state space model can be obtained from any one of these two mathematical models. Let us now discuss these two methods one by one. State Space Model from Differential Equation Consider the following series of the RLC circuit. It is having an input voltage, $v_i(t)$ and the current flowing through the circuit is $i(t)$. There are two storage elements (inductor and capacitor) in this circuit. So, the number of the state variables is equal to two and these state variables are the current flowing through the inductor, $i(t)$ and the voltage across capacitor, $v_c(t)$. From the circuit, the output voltage, $v_0(t)$ is equal to the voltage across capacitor, $v_c(t)$. $$v_0(t)=v_c(t)$$ Apply KVL around the loop. $$v_i(t)=Ri(t)+Lfrac{text{d}i(t)}{text{d}t}+v_c(t)$$ $$Rightarrow frac{text{d}i(t)}{text{d}t}=-frac{Ri(t)}{L}-frac{v_c(t)}{L}+frac{v_i(t)}{L}$$ The voltage across the capacitor is – $$v_c(t)=frac{1}{C} int i(t) dt$$ Differentiate the above equation with respect to time. $$frac{text{d}v_c(t)}{text{d}t}=frac{i(t)}{C}$$ State vector, $X=begin{bmatrix}i(t) \v_c(t) end{bmatrix}$ Differential state vector, $dot{X}=begin{bmatrix}frac{text{d}i(t)}{text{d}t} \frac{text{d}v_c(t)}{text{d}t} end{bmatrix}$ We can arrange the differential equations and output equation into the standard form of state space model as, $$dot{X}=begin{bmatrix}frac{text{d}i(t)}{text{d}t} \frac{text{d}v_c(t)}{text{d}t} end{bmatrix}=begin{bmatrix}-frac{R}{L} & -frac{1}{L} \frac{1}{C} & 0 end{bmatrix}begin{bmatrix}i(t) \v_c(t) end{bmatrix}+begin{bmatrix}frac{1}{L} \0 end{bmatrix}begin{bmatrix}v_i(t) end{bmatrix}$$ $$Y=begin{bmatrix}0 & 1 end{bmatrix}begin{bmatrix}i(t) \v_c(t) end{bmatrix}$$ Where, $$A=begin{bmatrix}-frac{R}{L} & -frac{1}{L} \frac{1}{C} & 0 end{bmatrix}, : B=begin{bmatrix}frac{1}{L} \0 end{bmatrix}, : C=begin{bmatrix}0 & 1 end{bmatrix} : and : D=begin{bmatrix}0 end{bmatrix}$$ State Space Model from Transfer Function Consider the two types of transfer functions based on the type of terms present in the numerator. Transfer function having constant term in Numerator. Transfer function having polynomial function of ‘s’ in Numerator. Transfer function having constant term in Numerator Consider the following transfer function of a system $$frac{Y(s)}{U(s)}=frac{b_0}{s^n+a_{n-1}s^{n-1}+…+a_1s+a_0}$$ Rearrange, the above equation as $$(s^n+a_{n-1}s^{n-1}+…+a_0)Y(s)=b_0 U(s)$$ Apply inverse Laplace transform on both sides. $$frac{text{d}^ny(t)}{text{d}t^n}+a_{n-1}frac{text{d}^{n-1}y(t)}{text{d}t^{n-1}}+…+a_1frac{text{d}y(t)}{text{d}t}+a_0y(t)=b_0 u(t)$$ Let $$y(t)=x_1$$ $$frac{text{d}y(t)}{text{d}t}=x_2=dot{x}_1$$ $$frac{text{d}^2y(t)}{text{d}t^2}=x_3=dot{x}_2$$ $$.$$ $$.$$ $$.$$ $$frac{text{d}^{n-1}y(t)}{text{d}t^{n-1}}=x_n=dot{x}_{n-1}$$ $$frac{text{d}^ny(t)}{text{d}t^n}=dot{x}_n$$ and $u(t)=u$ Then, $$dot{x}_n+a_{n-1}x_n+…+a_1x_2+a_0x_1=b_0 u$$ From the above equation, we can write the following state equation. $$dot{x}_n=-a_0x_1-a_1x_2-…-a_{n-1}x_n+b_0 u$$ The output equation is – $$y(t)=y=x_1$$ The state space model is – $dot{X}=begin{bmatrix}dot{x}_1 \dot{x}_2 \vdots \dot{x}_{n-1} \dot{x}_n end{bmatrix}$ $$=begin{bmatrix}0 & 1 & 0 & dotso & 0 & 0 \0 & 0 & 1 & dotso & 0 & 0 \vdots & vdots & vdots & dotso & vdots & vdots \ 0 & 0 & 0 & dotso & 0 & 1 \-a_0 & -a_1 & -a_2 & dotso & -a_{n-2} & -a_{n-1} end{bmatrix} begin{bmatrix}x_1 \x_2 \vdots \x_{n-1} \x_n end{bmatrix}+begin{bmatrix}0 \0 \vdots \0 \b_0 end{bmatrix}begin{bmatrix}u end{bmatrix}$$ $$Y=begin{bmatrix}1 & 0 & dotso & 0 & 0 end{bmatrix}begin{bmatrix}x_1 \x_2 \vdots \x_{n-1} \x_n end{bmatrix}$$ Here, $D=left [ 0 right ].$ Example Find the state space model for the system having transfer function. $$frac{Y(s)}{U(s)}=frac{1}{s^2+s+1}$$ Rearrange, the above equation as, $$(s^2+s+1)Y(s)=U(s)$$ Apply inverse Laplace transform on both the sides. $$frac{text{d}^2y(t)}{text{d}t^2}+frac{text{d}y(t)}{text{d}t}+y(t)=u(t)$$ Let $$y(t)=x_1$$ $$frac{text{d}y(t)}{text{d}t}=x_2=dot{x}_1$$ and $u(t)=u$ Then, the state equation is $$dot{x}_2=-x_1-x_2+u$$ The output equation is $$y(t)=y=x_1$$ The state space model is $$dot{X}=begin{bmatrix}dot{x}_1 \dot{x}_2 end{bmatrix}=begin{bmatrix}0 & 1 \-1 & -1 end{bmatrix}begin{bmatrix}x_1 \x_2 end{bmatrix}+begin{bmatrix}0 \1 end{bmatrix}left [u right ]$$ $$Y=begin{bmatrix}1 & 0 end{bmatrix}begin{bmatrix}x_1 \x_2 end{bmatrix}$$ Transfer function having polynomial function of ‘s’ in Numerator Consider the following transfer function of a system $$frac{Y(s)}{U(s)}=frac{b_n s^n+b_{n-1}s^{n-1}+…+b_1s+b_0}{s^n+a_{n-1}s^{n-1}+…+a_1 s+a_0}$$ $$Rightarrow frac{Y(s)}{U(s)}=left( frac{1}{s^n+a_{n-1}s^{n-1}+…+a_1 s+a_0} right )(b_n s^n+b_{n-1}s^{n-1}+…+b_1s+b_0)$$ The above equation is in the form of product of transfer functions of two blocks, which are cascaded. $$frac{Y(s)}{U(s)}=left(frac{V(s)}{U(s)} right ) left(frac{Y(s)}{V(s)} right )$$ Here, $$frac{V(s)}{U(s)}=frac{1}{s^n+a_{n-1}s^{n-1}+…+a_1 s+a_0}$$ Rearrange, the above equation as $$(s^n+a_{n-1}s^{n-1}+…+a_0)V(s)=U(s)$$ Apply inverse Laplace transform on both the sides. $$frac{text{d}^nv(t)}{text{d}t^n}+a_{n-1}frac{text{d}^{n-1}v(t)}{text{d}t^{n-1}}+…+a_1 frac{text{d}v(t)}{text{d}t}+a_0v(t)=u(t)$$ Let $$v(t)=x_1$$ $$frac{text{d}v((t)}{text{d}t}=x_2=dot{x}_1$$ $$frac{text{d}^2v(t)}{text{d}t^2}=x_3=dot{x}_2$$ $$.$$ $$.$$ $$.$$ $$frac{text{d}^{n-1}v(t)}{text{d}t^{n-1}}=x_n=dot{x}_{n-1}$$ $$frac{text{d}^nv(t)}{text{d}t^n}=dot{x}_n$$ and $u(t)=u$ Then, the state equation is $$dot{x}_n=-a_0x_1-a_1x_2-…-a_{n-1}x_n+u$$ Consider, $$frac{Y(s)}{V(s)}=b_ns^n+b_{n-1}s^{n-1}+…+b_1s+b_0$$ Rearrange, the above equation as $$Y(s)=(b_ns^n+b_{n-1}s^{n-1}+…+b_1s+b_0)V(s)$$ Apply inverse Laplace transform on both the sides. $$y(t)=b_nfrac{text{d}^nv(t)}{text{d}t^n}+b_{n-1}frac{text{d}^{n-1}v(t)}{text{d}t^{n-1}}+…+b_1frac{text{d}v(t)}{text{d}t}+b_0v(t)$$ By substituting the state variables and $y(t)=y$ in the above equation, will get the output equation as, $$y=b_ndot{x}_n+b_{n-1}x_n+…+b_1x_2+b_0x_1$$ Substitute, $dot{x}_n$ value in the above equation. $$y=b_n(-a_0x_1-a_1x_2-…-a_{n-1}x_n+u)+b_{n-1}x_n+…+b_1x_2+b_0x_1$$ $$y=(b_0-b_na_0)x_1+(b_1-b_na_1)x_2+…+(b_{n-1}-b_na_{n-1})x_n+b_n u$$ The state space model is $dot{X}=begin{bmatrix}dot{x}_1 \dot{x}_2 \vdots \dot{x}_{n-1} \dot{x}_n end{bmatrix}$ $$=begin{bmatrix}0 & 1 & 0 & dotso & 0 & 0 \0 & 0 & 1 & dotso & 0 & 0 \vdots & vdots & vdots & dotso & vdots & vdots \ 0 & 0 & 0 & dotso & 0 & 1 \-a_0 & -a_1 & -a_2 & dotso & -a_{n-2} & -a_{n-1} end{bmatrix} begin{bmatrix}x_1 \x_2 \vdots \x_{n-1} \x_n end{bmatrix}+begin{bmatrix}0 \0 \vdots \0 \b_0 end{bmatrix}begin{bmatrix}u end{bmatrix}$$ $$Y=[b_0-b_na_0 quad b_1-b_na_1 quad … quad b_{n-2}-b_na_{n-2} quad b_{n-1}-b_na_{n-1}]begin{bmatrix}x_1 \x_2 \vdots \x_{n-1} \x_n end{bmatrix}$$ If $b_n = 0$, then, $$Y=[b_0 quad b_1 quad …quad b_{n-2} quad b_{n-1}]begin{bmatrix}x_1 \x_2 \vdots \x_{n-1} \x_n end{bmatrix}$$ Learning working make money

Control Systems – State Space Analysis In the previous chapter, we learnt how to obtain the state space model from differential equation and transfer function. In this chapter, let us discuss how to obtain transfer function from the state space model. Transfer Function from State Space Model We know the state space model of a Linear Time-Invariant (LTI) system is – $$dot{X}=AX+BU$$ $$Y=CX+DU$$ Apply Laplace Transform on both sides of the state equation. $$sX(s)=AX(s)+BU(s)$$ $$Rightarrow (sI-A)X(s)=BU(s)$$ $$Rightarrow X(s)=(sI-A)^{-1}BU(s)$$ Apply Laplace Transform on both sides of the output equation. $$Y(s)=CX(s)+DU(s)$$ Substitute, X(s) value in the above equation. $$Rightarrow Y(s)=C(sI-A)^{-1}BU(s)+DU(s)$$ $$Rightarrow Y(s)=[C(sI-A)^{-1}B+D]U(s)$$ $$Rightarrow frac{Y(s)}{U(s)}=C(sI-A)^{-1}B+D$$ The above equation represents the transfer function of the system. So, we can calculate the transfer function of the system by using this formula for the system represented in the state space model. Note − When $D = [0]$, the transfer function will be $$frac{Y(s)}{U(s)}=C(sI-A)^{-1}B$$ Example Let us calculate the transfer function of the system represented in the state space model as, $$dot{X}=begin{bmatrix}dot{x}_1 \dot{x}_2 end{bmatrix}=begin{bmatrix}-1 & -1 \1 & 0 end{bmatrix}begin{bmatrix}x_1 \x_2 end{bmatrix}+begin{bmatrix}1 \0 end{bmatrix}[u]$$ $$Y=begin{bmatrix}0 & 1 end{bmatrix}begin{bmatrix}x_1 \x_2 end{bmatrix}$$ Here, $$A=begin{bmatrix}-1 & -1 \1 & 0 end{bmatrix}, quad B=begin{bmatrix}1 \0 end{bmatrix}, quad C=begin{bmatrix}0 & 1 end{bmatrix} quad and quad D=[0]$$ The formula for the transfer function when $D = [0]$ is – $$frac{Y(s)}{U(s)}=C(sI-A)^{-1}B$$ Substitute, A, B & C matrices in the above equation. $$frac{Y(s)}{U(s)}=begin{bmatrix}0 & 1 end{bmatrix}begin{bmatrix}s+1 & 1 \-1 & s end{bmatrix}^{-1}begin{bmatrix}1 \0 end{bmatrix}$$ $$Rightarrow frac{Y(s)}{U(s)}=begin{bmatrix}0 & 1 end{bmatrix} frac{begin{bmatrix}s & -1 \1 & s+1 end{bmatrix}}{(s+1)s-1(-1)}begin{bmatrix}1 \0 end{bmatrix}$$ $$Rightarrow frac{Y(s)}{U(s)}=frac{begin{bmatrix}0 & 1 end{bmatrix}begin{bmatrix}s \1 end{bmatrix}}{s^2+s+1}=frac{1}{s^2+s+1}$$ Therefore, the transfer function of the system for the given state space model is $$frac{Y(s)}{U(s)}=frac{1}{s^2+s+1}$$ State Transition Matrix and its Properties If the system is having initial conditions, then it will produce an output. Since, this output is present even in the absence of input, it is called zero input response $x_{ZIR}(t)$. Mathematically, we can write it as, $$x_{ZIR}(t)=e^{At}X(0)=L^{-1}left { left [ sI-A right ]^{-1}X(0) right }$$ From the above relation, we can write the state transition matrix $phi(t)$ as $$phi(t)=e^{At}=L^{-1}[sI-A]^{-1}$$ So, the zero input response can be obtained by multiplying the state transition matrix $phi(t)$ with the initial conditions matrix. Following are the properties of the state transition matrix. If $t = 0$, then state transition matrix will be equal to an Identity matrix. $$phi(0) = I$$ Inverse of state transition matrix will be same as that of state transition matrix just by replcing ‘t’ by ‘-t’. $$phi^{-1}(t) = phi(−t)$$ If $t = t_1 + t_2$ , then the corresponding state transition matrix is equal to the multiplication of the two state transition matrices at $t = t_1$ and $t = t_2$. $$phi(t_1 + t_2) = phi(t_1) phi(t_2)$$ Controllability and Observability Let us now discuss controllability and observability of control system one by one. Controllability A control system is said to be controllable if the initial states of the control system are transferred (changed) to some other desired states by a controlled input in finite duration of time. We can check the controllability of a control system by using Kalman’s test. Write the matrix $Q_c$ in the following form. $$Q_c=left [ B quad AB quad A^2B quad …quad A^{n-1}B right ]$$ Find the determinant of matrix $Q_c$ and if it is not equal to zero, then the control system is controllable. Observability A control system is said to be observable if it is able to determine the initial states of the control system by observing the outputs in finite duration of time. We can check the observability of a control system by using Kalman’s test. Write the matrix $Q_o$ in following form. $$Q_o=left [ C^T quad A^TC^T quad (A^T)^2C^T quad …quad (A^T)^{n-1}C^T right ]$$ Find the determinant of matrix $Q_o$ and if it is not equal to zero, then the control system is observable. Example Let us verify the controllability and observability of a control system which is represented in the state space model as, $$dot{x}=begin{bmatrix}dot{x}_1 \dot{x}_2 end{bmatrix}=begin{bmatrix}-1 & -1 \1 & 0 end{bmatrix}begin{bmatrix}x_1 \x_2 end{bmatrix}+begin{bmatrix}1 \0 end{bmatrix} [u]$$ $$Y=begin{bmatrix}0 & 1 end{bmatrix}begin{bmatrix}x_1 \x_2 end{bmatrix}$$ Here, $$A=begin{bmatrix}-1 & -1 \1 & 0 end{bmatrix}, quad B=begin{bmatrix}1 \0 end{bmatrix}, quad begin{bmatrix}0 & 1 end{bmatrix}, D=[0]quad and quad n=2$$ For $n = 2$, the matrix $Q_c$ will be $$Q_c=left [B quad AB right ]$$ We will get the product of matrices A and B as, $$AB=begin{bmatrix}-1 \1 end{bmatrix}$$ $$Rightarrow Q_c =begin{bmatrix}1 & -1 \0 & 1 end{bmatrix}$$ $$|Q_c|=1 neq 0$$ Since the determinant of matrix $Q_c$ is not equal to zero, the given control system is controllable. For $n = 2$, the matrix $Q_o$ will be – $$Q_o=left [C^T quad A^TC^T right ]$$ Here, $$A^T=begin{bmatrix}-1 & 1 \-1 & 0 end{bmatrix} quad and quad C^T=begin{bmatrix}0 \1 end{bmatrix}$$ We will get the product of matrices $A^T$ and $C^T$ as $$A^TC^T=begin{bmatrix}1 \0 end{bmatrix}$$ $$Rightarrow Q_o=begin{bmatrix}0 & 1 \1 & 0 end{bmatrix}$$ $$Rightarrow |Q_o|=-1 quad neq 0$$ Since, the determinant of matrix $Q_o$ is not equal to zero, the given control system is observable. Therefore, the given control system is both controllable and observable. Learning working make money

Control Systems – Controllers The various types of controllers are used to improve the performance of control systems. In this chapter, we will discuss the basic controllers such as the proportional, the derivative and the integral controllers. Proportional Controller The proportional controller produces an output, which is proportional to error signal. $$u(t) propto e(t) $$ $$Rightarrow u(t)=K_P e(t)$$ Apply Laplace transform on both the sides – $$U(s)=K_P E(s)$$ $$frac{U(s)}{E(s)}=K_P$$ Therefore, the transfer function of the proportional controller is $K_P$. Where, U(s) is the Laplace transform of the actuating signal u(t) E(s) is the Laplace transform of the error signal e(t) KP is the proportionality constant The block diagram of the unity negative feedback closed loop control system along with the proportional controller is shown in the following figure. The proportional controller is used to change the transient response as per the requirement. Derivative Controller The derivative controller produces an output, which is derivative of the error signal. $$u(t)=K_D frac{text{d}e(t)}{text{d}t}$$ Apply Laplace transform on both sides. $$U(s)=K_D sE(s)$$ $$frac{U(s)}{E(s)}=K_D s$$ Therefore, the transfer function of the derivative controller is $K_D s$. Where, $K_D$ is the derivative constant. The block diagram of the unity negative feedback closed loop control system along with the derivative controller is shown in the following figure. The derivative controller is used to make the unstable control system into a stable one. Integral Controller The integral controller produces an output, which is integral of the error signal. $$u(t)=K_I int e(t) dt$$ Apply Laplace transform on both the sides – $$U(s)=frac{K_I E(s)}{s}$$ $$frac{U(s)}{E(s)}=frac{K_I}{s}$$ Therefore, the transfer function of the integral controller is $frac{K_I}{s}$. Where, $K_I$ is the integral constant. The block diagram of the unity negative feedback closed loop control system along with the integral controller is shown in the following figure. The integral controller is used to decrease the steady state error. Let us now discuss about the combination of basic controllers. Proportional Derivative (PD) Controller The proportional derivative controller produces an output, which is the combination of the outputs of proportional and derivative controllers. $$u(t)=K_P e(t)+K_D frac{text{d}e(t)}{text{d}t}$$ Apply Laplace transform on both sides – $$U(s)=(K_P+K_D s)E(s)$$ $$frac{U(s)}{E(s)}=K_P+K_D s$$ Therefore, the transfer function of the proportional derivative controller is $K_P + K_D s$. The block diagram of the unity negative feedback closed loop control system along with the proportional derivative controller is shown in the following figure. The proportional derivative controller is used to improve the stability of control system without affecting the steady state error. Proportional Integral (PI) Controller The proportional integral controller produces an output, which is the combination of outputs of the proportional and integral controllers. $$u(t)=K_P e(t)+K_I int e(t) dt$$ Apply Laplace transform on both sides – $$U(s)=left(K_P+frac{K_I}{s} right )E(s)$$ $$frac{U(s)}{E(s)}=K_P+frac{K_I}{s}$$ Therefore, the transfer function of proportional integral controller is $K_P + frac{K_I} {s}$. The block diagram of the unity negative feedback closed loop control system along with the proportional integral controller is shown in the following figure. The proportional integral controller is used to decrease the steady state error without affecting the stability of the control system. Proportional Integral Derivative (PID) Controller The proportional integral derivative controller produces an output, which is the combination of the outputs of proportional, integral and derivative controllers. $$u(t)=K_P e(t)+K_I int e(t) dt+K_D frac{text{d}e(t)}{text{d}t}$$ Apply Laplace transform on both sides – $$U(s)=left(K_P+frac{K_I}{s}+K_D s right )E(s)$$ $$frac{U(s)}{E(s)}=K_P+frac{K_I}{s}+K_D s$$ Therefore, the transfer function of the proportional integral derivative controller is $K_P + frac{K_I} {s} + K_D s$. The block diagram of the unity negative feedback closed loop control system along with the proportional integral derivative controller is shown in the following figure. The proportional integral derivative controller is used to improve the stability of the control system and to decrease steady state error. Learning working make money

Control Systems – Construction of Bode Plots In this chapter, let us understand in detail how to construct (draw) Bode plots. Rules for Construction of Bode Plots Follow these rules while constructing a Bode plot. Represent the open loop transfer function in the standard time constant form. Substitute, $s=jomega$ in the above equation. Find the corner frequencies and arrange them in ascending order. Consider the starting frequency of the Bode plot as 1/10th of the minimum corner frequency or 0.1 rad/sec whichever is smaller value and draw the Bode plot upto 10 times maximum corner frequency. Draw the magnitude plots for each term and combine these plots properly. Draw the phase plots for each term and combine these plots properly. Note − The corner frequency is the frequency at which there is a change in the slope of the magnitude plot. Example Consider the open loop transfer function of a closed loop control system $$G(s)H(s)=frac{10s}{(s+2)(s+5)}$$ Let us convert this open loop transfer function into standard time constant form. $$G(s)H(s)=frac{10s}{2left( frac{s}{2}+1 right )5 left( frac{s}{5}+1 right )}$$ $$Rightarrow G(s)H(s)=frac{s}{left( 1+frac{s}{2} right )left( 1+frac{s}{5} right )}$$ So, we can draw the Bode plot in semi log sheet using the rules mentioned earlier. Stability Analysis using Bode Plots From the Bode plots, we can say whether the control system is stable, marginally stable or unstable based on the values of these parameters. Gain cross over frequency and phase cross over frequency Gain margin and phase margin Phase Cross over Frequency The frequency at which the phase plot is having the phase of -1800 is known as phase cross over frequency. It is denoted by $omega_{pc}$. The unit of phase cross over frequency is rad/sec. Gain Cross over Frequency The frequency at which the magnitude plot is having the magnitude of zero dB is known as gain cross over frequency. It is denoted by $omega_{gc}$. The unit of gain cross over frequency is rad/sec. The stability of the control system based on the relation between the phase cross over frequency and the gain cross over frequency is listed below. If the phase cross over frequency $omega_{pc}$ is greater than the gain cross over frequency $omega_{gc}$, then the control system is stable. If the phase cross over frequency $omega_{pc}$ is equal to the gain cross over frequency $omega_{gc}$, then the control system is marginally stable. If the phase cross over frequency $omega_{pc}$ is less than the gain cross over frequency $omega_{gc}$, then the control system is unstable. Gain Margin Gain margin $GM$ is equal to negative of the magnitude in dB at phase cross over frequency. $$GM=20logleft( frac{1}{M_{pc}}right )=20logM_{pc}$$ Where, $M_{pc}$ is the magnitude at phase cross over frequency. The unit of gain margin (GM) is dB. Phase Margin The formula for phase margin $PM$ is $$PM=180^0+phi_{gc}$$ Where, $phi_{gc}$ is the phase angle at gain cross over frequency. The unit of phase margin is degrees. The stability of the control system based on the relation between gain margin and phase margin is listed below. If both the gain margin $GM$ and the phase margin $PM$ are positive, then the control system is stable. If both the gain margin $GM$ and the phase margin $PM$ are equal to zero, then the control system is marginally stable. If the gain margin $GM$ and / or the phase margin $PM$ are/is negative, then the control system is unstable. Learning working make money