Discuss Control Systems 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. Learning working make money
Mason”s Gain Formula Let us now discuss the Mason’s Gain Formula. Suppose there are ‘N’ forward paths in a signal flow graph. The gain between the input and the output nodes of a signal flow graph is nothing but the transfer function of the system. It can be calculated by using Mason’s gain formula. Mason’s gain formula is $$T=frac{C(s)}{R(s)}=frac{Sigma ^N _{i=1}P_iDelta _i}{Delta}$$ Where, C(s) is the output node R(s) is the input node T is the transfer function or gain between $R(s)$ and $C(s)$ Pi is the ith forward path gain $Delta =1-(sum : of : all : individual : loop : gains)$ $+(sum : of : gain : products : of : all : possible : two :nontouching : loops)$ $$-(sum : of : gain : products : of : all : possible : three : nontouching : loops)+…$$ Δi is obtained from Δ by removing the loops which are touching the ith forward path. Consider the following signal flow graph in order to understand the basic terminology involved here. Path It is a traversal of branches from one node to any other node in the direction of branch arrows. It should not traverse any node more than once. Examples − $y_2 rightarrow y_3 rightarrow y_4 rightarrow y_5$ and $y_5 rightarrow y_3 rightarrow y_2$ Forward Path The path that exists from the input node to the output node is known as forward path. Examples − $y_1 rightarrow y_2 rightarrow y_3 rightarrow y_4 rightarrow y_5 rightarrow y_6$ and $y_1 rightarrow y_2 rightarrow y_3 rightarrow y_5 rightarrow y_6$. Forward Path Gain It is obtained by calculating the product of all branch gains of the forward path. Examples − $abcde$ is the forward path gain of $y_1 rightarrow y_2 rightarrow y_3 rightarrow y_4 rightarrow y_5 rightarrow y_6$ and abge is the forward path gain of $y_1 rightarrow y_2 rightarrow y_3 rightarrow y_5 rightarrow y_6$. Loop The path that starts from one node and ends at the same node is known as loop. Hence, it is a closed path. Examples − $y_2 rightarrow y_3 rightarrow y_2$ and $y_3 rightarrow y_5 rightarrow y_3$. Loop Gain It is obtained by calculating the product of all branch gains of a loop. Examples − $b_j$ is the loop gain of $y_2 rightarrow y_3 rightarrow y_2$ and $g_h$ is the loop gain of $y_3 rightarrow y_5 rightarrow y_3$. Non-touching Loops These are the loops, which should not have any common node. Examples − The loops, $y_2 rightarrow y_3 rightarrow y_2$ and $y_4 rightarrow y_5 rightarrow y_4$ are non-touching. Calculation of Transfer Function using Mason’s Gain Formula Let us consider the same signal flow graph for finding transfer function. Number of forward paths, N = 2. First forward path is – $y_1 rightarrow y_2 rightarrow y_3 rightarrow y_4 rightarrow y_5 rightarrow y_6$. First forward path gain, $p_1 = abcde$. Second forward path is – $y_1 rightarrow y_2 rightarrow y_3 rightarrow y_5 rightarrow y_6$. Second forward path gain, $p_2 = abge$. Number of individual loops, L = 5. Loops are – $y_2 rightarrow y_3 rightarrow y_2$, $y_3 rightarrow y_5 rightarrow y_3$, $y_3 rightarrow y_4 rightarrow y_5 rightarrow y_3$, $y_4 rightarrow y_5 rightarrow y_4$ and $y_5 rightarrow y_5$. Loop gains are – $l_1 = bj$, $l_2 = gh$, $l_3 = cdh$, $l_4 = di$ and $l_5 = f$. Number of two non-touching loops = 2. First non-touching loops pair is – $y_2 rightarrow y_3 rightarrow y_2$, $y_4 rightarrow y_5 rightarrow y_4$. Gain product of first non-touching loops pair, $l_1l_4 = bjdi$ Second non-touching loops pair is – $y_2 rightarrow y_3 rightarrow y_2$, $y_5 rightarrow y_5$. Gain product of second non-touching loops pair is – $l_1l_5 = bjf$ Higher number of (more than two) non-touching loops are not present in this signal flow graph. We know, $Delta =1-(sum : of : all : individual : loop : gains)$ $+(sum : of : gain : products : of : all : possible : two :nontouching : loops)$ $$-(sum : of : gain : products : of : all : possible : three : nontouching : loops)+…$$ Substitute the values in the above equation, $Delta =1-(bj+gh+cdh+di+f)+(bjdi+bjf)-(0)$ $Rightarrow Delta=1-(bj+gh+cdh+di+f)+bjdi+bjf$ There is no loop which is non-touching to the first forward path. So, $Delta_1=1$. Similarly, $Delta_2=1$. Since, no loop which is non-touching to the second forward path. Substitute, N = 2 in Mason’s gain formula $$T=frac{C(s)}{R(s)}=frac{Sigma ^2 _{i=1}P_iDelta _i}{Delta}$$ $$T=frac{C(s)}{R(s)}=frac{P_1Delta_1+P_2Delta_2}{Delta}$$ Substitute all the necessary values in the above equation. $$T=frac{C(s)}{R(s)}=frac{(abcde)1+(abge)1}{1-(bj+gh+cdh+di+f)+bjdi+bjf}$$ $$Rightarrow T=frac{C(s)}{R(s)}=frac{(abcde)+(abge)}{1-(bj+gh+cdh+di+f)+bjdi+bjf}$$ Therefore, the transfer function is – $$T=frac{C(s)}{R(s)}=frac{(abcde)+(abge)}{1-(bj+gh+cdh+di+f)+bjdi+bjf}$$ Learning working make money
Response of the First Order System In this chapter, let us discuss the time response of the first order system. Consider the following block diagram of the closed loop control system. Here, an open loop transfer function, $frac{1}{sT}$ is connected with a unity negative feedback. We know that the transfer function of the closed loop control system has unity negative feedback as, $$frac{C(s)}{R(s)}=frac{G(s)}{1+G(s)}$$ Substitute, $G(s)=frac{1}{sT}$ in the above equation. $$frac{C(s)}{R(s)}=frac{frac{1}{sT}}{1+frac{1}{sT}}=frac{1}{sT+1}$$ The power of s is one in the denominator term. Hence, the above transfer function is of the first order and the system is said to be the first order system. We can re-write the above equation as $$C(s)=left ( frac{1}{sT+1} right )R(s)$$ Where, C(s) is the Laplace transform of the output signal c(t), R(s) is the Laplace transform of the input signal r(t), and T is the time constant. Follow these steps to get the response (output) of the first order system in the time domain. Take the Laplace transform of the input signal $r(t)$. Consider the equation, $C(s)=left ( frac{1}{sT+1} right )R(s)$ Substitute $R(s)$ value in the above equation. Do partial fractions of $C(s)$ if required. Apply inverse Laplace transform to $C(s)$. In the previous chapter, we have seen the standard test signals like impulse, step, ramp and parabolic. Let us now find out the responses of the first order system for each input, one by one. The name of the response is given as per the name of the input signal. For example, the response of the system for an impulse input is called as impulse response. Impulse Response of First Order System Consider the unit impulse signal as an input to the first order system. So, $r(t)=delta (t)$ Apply Laplace transform on both the sides. $R(s)=1$ Consider the equation, $C(s)=left ( frac{1}{sT+1} right )R(s)$ Substitute, $R(s) = 1$ in the above equation. $$C(s)=left ( frac{1}{sT+1} right )(1)=frac{1}{sT+1}$$ Rearrange the above equation in one of the standard forms of Laplace transforms. $$C(s)=frac{1}{Tleft ( s+frac{1}{T} right )} Rightarrow C(s)=frac{1}{T}left ( frac{1}{s+frac{1}{T}} right )$$ Apply inverse Laplace transform on both sides. $$c(t)=frac{1}{T}e^left ( {-frac{t}{T}} right )u(t)$$ The unit impulse response is shown in the following figure. The unit impulse response, c(t) is an exponential decaying signal for positive values of ‘t’ and it is zero for negative values of ‘t’. Step Response of First Order System Consider the unit step signal as an input to first order system. So, $r(t)=u(t)$ Apply Laplace transform on both the sides. $$R(s)=frac{1}{s}$$ Consider the equation, $C(s)=left ( frac{1}{sT+1} right )R(s)$ Substitute, $R(s)=frac{1}{s}$ in the above equation. $$C(s)=left ( frac{1}{sT+1} right )left ( frac{1}{s} right )=frac{1}{sleft ( sT+1 right )}$$ Do partial fractions of C(s). $$C(s)=frac{1}{sleft ( sT+1 right )}=frac{A}{s}+frac{B}{sT+1}$$ $$Rightarrow frac{1}{sleft ( sT+1 right )}=frac{Aleft ( sT+1 right )+Bs}{sleft ( sT+1 right )}$$ On both the sides, the denominator term is the same. So, they will get cancelled by each other. Hence, equate the numerator terms. $$1=Aleft ( sT+1 right )+Bs$$ By equating the constant terms on both the sides, you will get A = 1. Substitute, A = 1 and equate the coefficient of the s terms on both the sides. $$0=T+B Rightarrow B=-T$$ Substitute, A = 1 and B = −T in partial fraction expansion of $C(s)$. $$C(s)=frac{1}{s}-frac{T}{sT+1}=frac{1}{s}-frac{T}{Tleft ( s+frac{1}{T} right )}$$ $$Rightarrow C(s)=frac{1}{s}-frac{1}{s+frac{1}{T}}$$ Apply inverse Laplace transform on both the sides. $$c(t)=left ( 1-e^{-left ( frac{t}{T} right )} right )u(t)$$ The unit step response, c(t) has both the transient and the steady state terms. The transient term in the unit step response is – $$c_{tr}(t)=-e^{-left ( frac{t}{T} right )}u(t)$$ The steady state term in the unit step response is – $$c_{ss}(t)=u(t)$$ The following figure shows the unit step response. The value of the unit step response, c(t) is zero at t = 0 and for all negative values of t. It is gradually increasing from zero value and finally reaches to one in steady state. So, the steady state value depends on the magnitude of the input. Ramp Response of First Order System Consider the unit ramp signal as an input to the first order system. $So, r(t)=tu(t)$ Apply Laplace transform on both the sides. $$R(s)=frac{1}{s^2}$$ Consider the equation, $C(s)=left ( frac{1}{sT+1} right )R(s)$ Substitute, $R(s)=frac{1}{s^2}$ in the above equation. $$C(s)=left ( frac{1}{sT+1} right )left ( frac{1}{s^2} right )=frac{1}{s^2(sT+1)}$$ Do partial fractions of $C(s)$. $$C(s)=frac{1}{s^2(sT+1)}=frac{A}{s^2}+frac{B}{s}+frac{C}{sT+1}$$ $$Rightarrow frac{1}{s^2(sT+1)}=frac{A(sT+1)+Bs(sT+1)+Cs^2}{s^2(sT+1)}$$ On both the sides, the denominator term is the same. So, they will get cancelled by each other. Hence, equate the numerator terms. $$1=A(sT+1)+Bs(sT+1)+Cs^2$$ By equating the constant terms on both the sides, you will get A = 1. Substitute, A = 1 and equate the coefficient of the s terms on both the sides. $$0=T+B Rightarrow B=-T$$ Similarly, substitute B = −T and equate the coefficient of $s^2$ terms on both the sides. You will get $C=T^2$. Substitute A = 1, B = −T and $C = T^2$ in the partial fraction expansion of $C(s)$. $$C(s)=frac{1}{s^2}-frac{T}{s}+frac{T^2}{sT+1}=frac{1}{s^2}-frac{T}{s}+frac{T^2}{Tleft ( s+frac{1}{T} right )}$$ $$Rightarrow C(s)=frac{1}{s^2}-frac{T}{s}+frac{T}{s+frac{1}{T}}$$ Apply inverse Laplace transform on both the sides. $$c(t)=left ( t-T+Te^{-left ( frac{t}{T} right )} right )u(t)$$ The unit ramp response, c(t) has both the transient and the steady state terms. The transient term in the unit ramp response is – $$c_{tr}(t)=Te^{-left ( frac{t}{T} right )}u(t)$$ The steady state term in the unit ramp response is – $$c_{ss}(t)=(t-T)u(t)$$ The following figure shows the unit ramp response. The unit ramp response, c(t) follows the unit ramp input signal for all positive values of t. But, there is a deviation of T units from the input signal. Parabolic Response of First Order System Consider the unit parabolic signal as an input to the first order system. So, $r(t)=frac{t^2}{2}u(t)$ Apply Laplace transform on both the sides. $$R(s)=frac{1}{s^3}$$ Consider the equation, $C(s)=left ( frac{1}{sT+1} right )R(s)$ Substitute $R(s)=frac{1}{s^3}$ in the above equation. $$C(s)=left ( frac{1}{sT+1} right )left( frac{1}{s^3} right )=frac{1}{s^3(sT+1)}$$ Do partial fractions of $C(s)$. $$C(s)=frac{1}{s^3(sT+1)}=frac{A}{s^3}+frac{B}{s^2}+frac{C}{s}+frac{D}{sT+1}$$ After simplifying, you will get the values of A, B, C and D as 1, $-T, : T^2: and : −T^3$ respectively. Substitute these values in the
Control Systems – Stability Stability is an important concept. In this chapter, let us discuss the stability of system and types of systems based on stability. What is Stability? A system is said to be stable, if its output is under control. Otherwise, it is said to be unstable. A stable system produces a bounded output for a given bounded input. The following figure shows the response of a stable system. This is the response of first order control system for unit step input. This response has the values between 0 and 1. So, it is bounded output. We know that the unit step signal has the value of one for all positive values of t including zero. So, it is bounded input. Therefore, the first order control system is stable since both the input and the output are bounded. Types of Systems based on Stability We can classify the systems based on stability as follows. Absolutely stable system Conditionally stable system Marginally stable system Absolutely Stable System If the system is stable for all the range of system component values, then it is known as the absolutely stable system. The open loop control system is absolutely stable if all the poles of the open loop transfer function present in left half of ‘s’ plane. Similarly, the closed loop control system is absolutely stable if all the poles of the closed loop transfer function present in the left half of the ‘s’ plane. Conditionally Stable System If the system is stable for a certain range of system component values, then it is known as conditionally stable system. Marginally Stable System If the system is stable by producing an output signal with constant amplitude and constant frequency of oscillations for bounded input, then it is known as marginally stable system. The open loop control system is marginally stable if any two poles of the open loop transfer function is present on the imaginary axis. Similarly, the closed loop control system is marginally stable if any two poles of the closed loop transfer function is present on the imaginary axis. Learning working make money
Frequency Response Analysis We have already discussed time response analysis of the control systems and the time domain specifications of the second order control systems. In this chapter, let us discuss the frequency response analysis of the control systems and the frequency domain specifications of the second order control systems. What is Frequency Response? The response of a system can be partitioned into both the transient response and the steady state response. We can find the transient response by using Fourier integrals. The steady state response of a system for an input sinusoidal signal is known as the frequency response. In this chapter, we will focus only on the steady state response. If a sinusoidal signal is applied as an input to a Linear Time-Invariant (LTI) system, then it produces the steady state output, which is also a sinusoidal signal. The input and output sinusoidal signals have the same frequency, but different amplitudes and phase angles. Let the input signal be − $$r(t)=Asin(omega_0t)$$ The open loop transfer function will be − $$G(s)=G(jomega)$$ We can represent $G(jomega)$ in terms of magnitude and phase as shown below. $$G(jomega)=|G(jomega)| angle G(jomega)$$ Substitute, $omega = omega_0$ in the above equation. $$G(jomega_0)=|G(jomega_0)| angle G(jomega_0)$$ The output signal is $$c(t)=A|G(jomega_0)|sin(omega_0t + angle G(jomega_0))$$ The amplitude of the output sinusoidal signal is obtained by multiplying the amplitude of the input sinusoidal signal and the magnitude of $G(jomega)$ at $omega = omega_0$. The phase of the output sinusoidal signal is obtained by adding the phase of the input sinusoidal signal and the phase of $G(jomega)$ at $omega = omega_0$. Where, A is the amplitude of the input sinusoidal signal. ω0 is angular frequency of the input sinusoidal signal. We can write, angular frequency $omega_0$ as shown below. $$omega_0=2pi f_0$$ Here, $f_0$ is the frequency of the input sinusoidal signal. Similarly, you can follow the same procedure for closed loop control system. Frequency Domain Specifications The frequency domain specifications are resonant peak, resonant frequency and bandwidth. Consider the transfer function of the second order closed loop control system as, $$T(s)=frac{C(s)}{R(s)}=frac{omega_n^2}{s^2+2deltaomega_ns+omega_n^2}$$ Substitute, $s = jomega$ in the above equation. $$T(jomega)=frac{omega_n^2}{(jomega)^2+2deltaomega_n(jomega)+omega_n^2}$$ $$Rightarrow T(jomega)=frac{omega_n^2}{-omega^2+2jdeltaomegaomega_n+omega_n^2}=frac{omega_n^2}{omega_n^2left ( 1-frac{omega^2}{omega_n^2}+frac{2jdeltaomega}{omega_n} right )}$$ $$Rightarrow T(jomega)=frac{1}{left ( 1-frac{omega^2}{omega_n^2} right )+jleft ( frac{2deltaomega}{omega_n} right )}$$ Let, $frac{omega}{omega_n}=u$ Substitute this value in the above equation. $$T(jomega)=frac{1}{(1-u^2)+j(2delta u)}$$ Magnitude of $T(jomega)$ is – $$M=|T(jomega)|=frac{1}{sqrt {(1-u^2)^2+(2delta u)^2}}$$ Phase of $T(jomega)$ is – $$angle T(jomega)=-tan^{-1}left( frac{2delta u}{1-u^2} right )$$ Resonant Frequency It is the frequency at which the magnitude of the frequency response has peak value for the first time. It is denoted by $omega_r$. At $omega = omega_r$, the first derivate of the magnitude of $T(jomega)$ is zero. Differentiate $M$ with respect to $u$. $$frac{text{d}M}{text{d}u}=-frac{1}{2}left [ (1-u^2)^2+(2delta u)^2 right ]^{frac{-3}{2}} left [2(1-u^2)(-2u)+2(2delta u)(2delta) right ]$$ $$Rightarrow frac{text{d}M}{text{d}u}=-frac{1}{2}left [ (1-u^2)^2+(2delta u)^2 right ]^{frac{-3}{2}} left [4u(u^2-1 +2delta^2) right ]$$ Substitute, $u=u_r$ and $frac{text{d}M}{text{d}u}==0$ in the above equation. $$0=-frac{1}{2}left [ (1-u_r^2)^2+(2delta u_r)^2 right ]^{-frac{3}{2}}left [ 4u_r(u_r^2-1 +2delta^2) right ]$$ $$Rightarrow 4u_r(u_r^2-1 +2delta^2)=0$$ $$Rightarrow u_r^2-1+2delta^2=0$$ $$Rightarrow u_r^2=1-2delta^2$$ $$Rightarrow u_r=sqrt{1-2delta^2}$$ Substitute, $u_r=frac{omega_r}{omega_n}$ in the above equation. $$frac{omega_r}{omega_n}=sqrt{1-2delta^2}$$ $$Rightarrow omega_r=omega_n sqrt{1-2delta^2}$$ Resonant Peak It is the peak (maximum) value of the magnitude of $T(jomega)$. It is denoted by $M_r$. At $u = u_r$, the Magnitude of $T(jomega)$ is – $$M_r=frac{1}{sqrt{(1-u_r^2)^2+(2delta u_r)^2}}$$ Substitute, $u_r = sqrt{1 − 2delta^2}$ and $1 − u_r^2 = 2delta^2$ in the above equation. $$M_r=frac{1}{sqrt{(2delta^2)^2+(2delta sqrt{1-2delta^2})^2}}$$ $$Rightarrow M_r=frac{1}{2delta sqrt {1-delta^2}}$$ Resonant peak in frequency response corresponds to the peak overshoot in the time domain transient response for certain values of damping ratio $delta$. So, the resonant peak and peak overshoot are correlated to each other. Bandwidth It is the range of frequencies over which, the magnitude of $T(jomega)$ drops to 70.7% from its zero frequency value. At $omega = 0$, the value of $u$ will be zero. Substitute, $u = 0$ in M. $$M=frac{1}{sqrt {(1-0^2)^2+(2delta(0))^2}}=1$$ Therefore, the magnitude of $T(jomega)$ is one at $omega = 0$. At 3-dB frequency, the magnitude of $T(jomega)$ will be 70.7% of magnitude of $T(jomega)$ at $omega = 0$. i.e., at $omega = omega_B, M = 0.707(1) = frac{1}{sqrt{2}}$ $$Rightarrow M=frac{1}{sqrt{2}}=frac{1}{sqrt{(1-u_b^2)^2+(2delta u_b)^2}}$$ $$Rightarrow 2=(1-u_b^2)^2+(2delta)^2 u_b^2$$ Let, $u_b^2=x$ $$Rightarrow 2=(1-x)^2+(2delta)^2 x$$ $$Rightarrow x^2+(4delta^2-2)x-1=0$$ $$Rightarrow x=frac{-(4delta^2 -2)pm sqrt{(4delta^2-2)^2+4}}{2}$$ Consider only the positive value of x. $$x=1-2delta^2+sqrt {(2delta^2-1)^2+1}$$ $$Rightarrow x=1-2delta^2+sqrt {(2-4delta^2+4delta^4)}$$ Substitute, $x=u_b^2=frac{omega_b^2}{omega_n^2}$ $$frac{omega_b^2}{omega_n^2}=1-2delta^2+sqrt {(2-4delta^2+4delta^4)}$$ $$Rightarrow omega_b=omega_n sqrt {1-2delta^2+sqrt {(2-4delta^2+4delta^4)}}$$ Bandwidth $omega_b$ in the frequency response is inversely proportional to the rise time $t_r$ in the time domain transient response. Learning working make money
Control Systems – Steady State Errors The deviation of the output of control system from desired response during steady state is known as steady state error. It is represented as $e_{ss}$. We can find steady state error using the final value theorem as follows. $$e_{ss}=lim_{t to infty}e(t)=lim_{s to 0}sE(s)$$ Where, E(s) is the Laplace transform of the error signal, $e(t)$ Let us discuss how to find steady state errors for unity feedback and non-unity feedback control systems one by one. Steady State Errors for Unity Feedback Systems Consider the following block diagram of closed loop control system, which is having unity negative feedback. Where, R(s) is the Laplace transform of the reference Input signal $r(t)$ C(s) is the Laplace transform of the output signal $c(t)$ We know the transfer function of the unity negative feedback closed loop control system as $$frac{C(s)}{R(s)}=frac{G(s)}{1+G(s)}$$ $$Rightarrow C(s)=frac{R(s)G(s)}{1+G(s)}$$ The output of the summing point is – $$E(s)=R(s)-C(s)$$ Substitute $C(s)$ value in the above equation. $$E(s)=R(s)-frac{R(s)G(s)}{1+G(s)}$$ $$Rightarrow E(s)=frac{R(s)+R(s)G(s)-R(s)G(s)}{1+G(s)}$$ $$Rightarrow E(s)=frac{R(s)}{1+G(s)}$$ Substitute $E(s)$ value in the steady state error formula $$e_{ss}=lim_{s to 0} frac{sR(s)}{1+G(s)}$$ The following table shows the steady state errors and the error constants for standard input signals like unit step, unit ramp & unit parabolic signals. Input signal Steady state error $e_{ss}$ Error constant unit step signal $frac{1}{1+k_p}$ $K_p=lim_{s to 0}G(s)$ unit ramp signal $frac{1}{K_v}$ $K_v=lim_{s to 0}sG(s)$ unit parabolic signal $frac{1}{K_a}$ $K_a=lim_{s to 0}s^2G(s)$ Where, $K_p$, $K_v$ and $K_a$ are position error constant, velocity error constant and acceleration error constant respectively. Note − If any of the above input signals has the amplitude other than unity, then multiply corresponding steady state error with that amplitude. Note − We can’t define the steady state error for the unit impulse signal because, it exists only at origin. So, we can’t compare the impulse response with the unit impulse input as t denotes infinity. Example Let us find the steady state error for an input signal $r(t)=left( 5+2t+frac{t^2}{2} right )u(t)$ of unity negative feedback control system with $G(s)=frac{5(s+4)}{s^2(s+1)(s+20)}$ The given input signal is a combination of three signals step, ramp and parabolic. The following table shows the error constants and steady state error values for these three signals. Input signal Error constant Steady state error $r_1(t)=5u(t)$ $K_p=lim_{s to 0}G(s)=infty$ $e_{ss1}=frac{5}{1+k_p}=0$ $r_2(t)=2tu(t)$ $K_v=lim_{s to 0}sG(s)=infty$ $e_{ss2}=frac{2}{K_v}=0$ $r_3(t)=frac{t^2}{2}u(t)$ $K_a=lim_{s to 0}s^2G(s)=1$ $e_{ss3}=frac{1}{k_a}=1$ We will get the overall steady state error, by adding the above three steady state errors. $$e_{ss}=e_{ss1}+e_{ss2}+e_{ss3}$$ $$Rightarrow e_{ss}=0+0+1=1$$ Therefore, we got the steady state error $e_{ss}$ as 1 for this example. Steady State Errors for Non-Unity Feedback Systems Consider the following block diagram of closed loop control system, which is having nonunity negative feedback. We can find the steady state errors only for the unity feedback systems. So, we have to convert the non-unity feedback system into unity feedback system. For this, include one unity positive feedback path and one unity negative feedback path in the above block diagram. The new block diagram looks like as shown below. Simplify the above block diagram by keeping the unity negative feedback as it is. The following is the simplified block diagram. This block diagram resembles the block diagram of the unity negative feedback closed loop control system. Here, the single block is having the transfer function $frac{G(s)}{1+G(s)H(s)-G(s)}$ instead of $G(s)$. You can now calculate the steady state errors by using steady state error formula given for the unity negative feedback systems. Note − It is meaningless to find the steady state errors for unstable closed loop systems. So, we have to calculate the steady state errors only for closed loop stable systems. This means we need to check whether the control system is stable or not before finding the steady state errors. In the next chapter, we will discuss the concepts-related stability. Learning working make money
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. Learning working make money
Time Domain Specifications In this chapter, let us discuss the time domain specifications of the second order system. The step response of the second order system for the underdamped case is shown in the following figure. All the time domain specifications are represented in this figure. The response up to the settling time is known as transient response and the response after the settling time is known as steady state response. Delay Time It is the time required for the response to reach half of its final value from the zero instant. It is denoted by $t_d$. Consider the step response of the second order system for t ≥ 0, when ‘δ’ lies between zero and one. $$c(t)=1-left ( frac{e^{-delta omega_nt}}{sqrt{1-delta^2}} right )sin(omega_dt+theta)$$ The final value of the step response is one. Therefore, at $t=t_d$, the value of the step response will be 0.5. Substitute, these values in the above equation. $$c(t_d)=0.5=1-left ( frac{e^{-deltaomega_nt_d}}{sqrt{1-delta^2}} right )sin(omega_dt_d+theta)$$ $$Rightarrow left ( frac{e^{-deltaomega_nt_d}}{sqrt{1-delta^2}} right )sin(omega_dt_d+theta)=0.5$$ By using linear approximation, you will get the delay time td as $$t_d=frac{1+0.7delta}{omega_n}$$ Rise Time It is the time required for the response to rise from 0% to 100% of its final value. This is applicable for the under-damped systems. For the over-damped systems, consider the duration from 10% to 90% of the final value. Rise time is denoted by tr. At t = t1 = 0, c(t) = 0. We know that the final value of the step response is one. Therefore, at $t = t_2$, the value of step response is one. Substitute, these values in the following equation. $$c(t)=1-left ( frac{e^{-delta omega_nt}}{sqrt{1-delta^2}} right )sin(omega_dt+theta)$$ $$c(t_2)=1=1-left ( frac{e^{-deltaomega_nt_2}}{sqrt{1-delta^2}} right )sin(omega_dt_2+theta)$$ $$Rightarrow left ( frac{e^{-deltaomega_nt_2}}{sqrt{1-delta^2}} right )sin(omega_dt_2+theta)=0$$ $$Rightarrow sin(omega_dt_2+theta)=0$$ $$Rightarrow omega_dt_2+theta=pi$$ $$Rightarrow t_2=frac{pi-theta}{omega_d}$$ Substitute t1 and t2 values in the following equation of rise time, $$t_r=t_2-t_1$$ $$therefore : t_r=frac{pi-theta}{omega_d}$$ From above equation, we can conclude that the rise time $t_r$ and the damped frequency $omega_d$ are inversely proportional to each other. Peak Time It is the time required for the response to reach the peak value for the first time. It is denoted by $t_p$. At $t = t_p$, the first derivate of the response is zero. We know the step response of second order system for under-damped case is $$c(t)=1-left ( frac{e^{-delta omega_nt}}{sqrt{1-delta^2}} right )sin(omega_dt+theta)$$ Differentiate $c(t)$ with respect to ‘t’. $$frac{text{d}c(t)}{text{d}t}=-left ( frac{e^{-deltaomega_nt}}{sqrt{1-delta^2}} right )omega_dcos(omega_dt+theta)-left ( frac{-deltaomega_ne^{-deltaomega_nt}}{sqrt{1-delta^2}} right )sin(omega_dt+theta)$$ Substitute, $t=t_p$ and $frac{text{d}c(t)}{text{d}t}=0$ in the above equation. $$0=-left ( frac{e^{-deltaomega_nt_p}}{sqrt{1-delta^2}} right )left [ omega_dcos(omega_dt_p+theta)-deltaomega_nsin(omega_dt_p+theta) right ]$$ $$Rightarrow omega_nsqrt{1-delta^2}cos(omega_dt_p+theta)-deltaomega_nsin(omega_dt_p+theta)=0$$ $$Rightarrow sqrt{1-delta^2}cos(omega_dt_p+theta)-deltasin(omega_dt_p+theta)=0$$ $$Rightarrow sin(theta)cos(omega_dt_p+theta)-cos(theta)sin(omega_dt_p+theta)=0$$ $$Rightarrow sin(theta-omega_dt_p-theta)=0$$ $$Rightarrow sin(-omega_dt_p)=0Rightarrow -sin(omega_dt_p)=0Rightarrow sin(omega_dt_p)=0$$ $$Rightarrow omega_dt_p=pi$$ $$Rightarrow t_p=frac{pi}{omega_d}$$ From the above equation, we can conclude that the peak time $t_p$ and the damped frequency $omega_d$ are inversely proportional to each other. Peak Overshoot Peak overshoot Mp is defined as the deviation of the response at peak time from the final value of response. It is also called the maximum overshoot. Mathematically, we can write it as $$M_p=c(t_p)-c(infty)$$ Where, c(tp) is the peak value of the response. c(∞) is the final (steady state) value of the response. At $t = t_p$, the response c(t) is – $$c(t_p)=1-left ( frac{e^{-deltaomega_nt_p}}{sqrt{1-delta^2}} right )sin(omega_dt_p+theta)$$ Substitute, $t_p=frac{pi}{omega_d}$ in the right hand side of the above equation. $$c(t_P)=1-left ( frac{e^{-deltaomega_nleft ( frac{pi}{omega_d} right )}}{sqrt{1-delta^2}} right )sinleft ( omega_dleft ( frac{pi}{omega_d} right ) +thetaright )$$ $$Rightarrow c(t_p)=1-left ( frac{e^{-left ( frac{deltapi}{sqrt{1-delta^2}} right )}}{sqrt{1-delta^2}} right )(-sin(theta))$$ We know that $$sin(theta)=sqrt{1-delta^2}$$ So, we will get $c(t_p)$ as $$c(t_p)=1+e^{-left ( frac{deltapi}{sqrt{1-delta^2}} right )}$$ Substitute the values of $c(t_p)$ and $c(infty)$ in the peak overshoot equation. $$M_p=1+e^{-left ( frac{deltapi}{sqrt{1-delta^2}} right )}-1$$ $$Rightarrow M_p=e^{-left ( frac{deltapi}{sqrt{1-delta^2}} right )}$$ Percentage of peak overshoot % $M_p$ can be calculated by using this formula. $$%M_p=frac{M_p}{c(infty )}times 100%$$ By substituting the values of $M_p$ and $c(infty)$ in above formula, we will get the Percentage of the peak overshoot $%M_p$ as $$%M_p=left ( e^ {-left ( frac{deltapi}{sqrt{1-delta^2}} right )} right )times 100%$$ From the above equation, we can conclude that the percentage of peak overshoot $% M_p$ will decrease if the damping ratio $delta$ increases. Settling time It is the time required for the response to reach the steady state and stay within the specified tolerance bands around the final value. In general, the tolerance bands are 2% and 5%. The settling time is denoted by $t_s$. The settling time for 5% tolerance band is – $$t_s=frac{3}{deltaomega_n}=3tau$$ The settling time for 2% tolerance band is – $$t_s=frac{4}{deltaomega_n}=4tau$$ Where, $tau$ is the time constant and is equal to $frac{1}{deltaomega_n}$. Both the settling time $t_s$ and the time constant $tau$ are inversely proportional to the damping ratio $delta$. Both the settling time $t_s$ and the time constant $tau$ are independent of the system gain. That means even the system gain changes, the settling time $t_s$ and time constant $tau$ will never change. Example Let us now find the time domain specifications of a control system having the closed loop transfer function $frac{4}{s^2+2s+4}$ when the unit step signal is applied as an input to this control system. We know that the standard form of the transfer function of the second order closed loop control system as $$frac{omega_n^2}{s^2+2deltaomega_ns+omega_n^2}$$ By equating these two transfer functions, we will get the un-damped natural frequency $omega_n$ as 2 rad/sec and the damping ratio $delta$ as 0.5. We know the formula for damped frequency $omega_d$ as $$omega_d=omega_nsqrt{1-delta^2}$$ Substitute, $omega_n$ and $delta$ values in the above formula. $$Rightarrow omega_d=2sqrt{1-(0.5)^2}$$ $$Rightarrow omega_d=1.732 : rad/sec$$ Substitute, $delta$ value in following relation $$theta=cos^{-1}delta$$ $$Rightarrow theta=cos^{-1}(0.5)=frac{pi}{3}:rad$$ Substitute the above necessary values in the formula of each time domain specification and simplify in order to get the values of time domain specifications for given transfer function. The following table shows the formulae of time domain specifications, substitution of necessary values and the final values. Time domain specification Formula Substitution of values in Formula Final value Delay time $t_d=frac{1+0.7delta}{omega_n}$ $t_d=frac{1+0.7(0.5)}{2}$ $t_d$=0.675 sec Rise time $t_r=frac{pi-theta}{omega_d}$ $t_r=frac{pi-(frac{pi}{3})}{1.732}$ $t_r$=1.207 sec Peak time $t_p=frac{pi}{omega_d}$ $t_p=frac{pi}{1.732}$ $t_p$=1.813 sec % Peak overshoot $%M_p=left( e^{-left (frac{deltapi}{sqrt{1-delta^2}} right ) }right )times 100%$ $%M_p=left( e^{-left (frac{0.5pi}{sqrt{1-(0.5)^2}} right ) }right )times 100%$
Learning working make money
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