Learning Response of Second Order System work project make money

Response of Second Order System In this chapter, let us discuss the time response of second order system. Consider the following block diagram of closed loop control system. Here, an open loop transfer function, $frac{omega ^2_n}{s(s+2delta omega_n)}$ is connected with a unity negative feedback. We know that the transfer function of the closed loop control system having unity negative feedback as $$frac{C(s)}{R(s)}=frac{G(s)}{1+G(s)}$$ Substitute, $G(s)=frac{omega ^2_n}{s(s+2delta omega_n)}$ in the above equation. $$frac{C(s)}{R(s)}=frac{left (frac{omega ^2_n}{s(s+2delta omega_n)} right )}{1+ left ( frac{omega ^2_n}{s(s+2delta omega_n)} right )}=frac{omega _n^2}{s^2+2delta omega _ns+omega _n^2}$$ The power of ‘s’ is two in the denominator term. Hence, the above transfer function is of the second order and the system is said to be the second order system. The characteristic equation is – $$s^2+2delta omega _ns+omega _n^2=0$$ The roots of characteristic equation are – $$s=frac{-2omega delta _npm sqrt{(2deltaomega _n)^2-4omega _n^2}}{2}=frac{-2(deltaomega _npm omega _nsqrt{delta ^2-1})}{2}$$ $$Rightarrow s=-delta omega_n pm omega _nsqrt{delta ^2-1}$$ The two roots are imaginary when δ = 0. The two roots are real and equal when δ = 1. The two roots are real but not equal when δ > 1. The two roots are complex conjugate when 0 < δ < 1. We can write $C(s)$ equation as, $$C(s)=left ( frac{omega _n^2}{s^2+2deltaomega_ns+omega_n^2} right )R(s)$$ 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) ωn is the natural frequency δ is the damping ratio. Follow these steps to get the response (output) of the second order system in the time domain. Take Laplace transform of the input signal, $r(t)$. Consider the equation, $C(s)=left ( frac{omega _n^2}{s^2+2deltaomega_ns+omega_n^2} right )R(s)$ Substitute $R(s)$ value in the above equation. Do partial fractions of $C(s)$ if required. Apply inverse Laplace transform to $C(s)$. Step Response of Second Order System Consider the unit step signal as an input to the second order system. Laplace transform of the unit step signal is, $$R(s)=frac{1}{s}$$ We know the transfer function of the second order closed loop control system is, $$frac{C(s)}{R(s)}=frac{omega _n^2}{s^2+2deltaomega_ns+omega_n^2}$$ Case 1: δ = 0 Substitute, $delta = 0$ in the transfer function. $$frac{C(s)}{R(s)}=frac{omega_n^2}{s^2+omega_n^2}$$ $$Rightarrow C(s)=left( frac{omega_n^2}{s^2+omega_n^2} right )R(s)$$ Substitute, $R(s) = frac{1}{s}$ in the above equation. $$C(s)=left( frac{omega_n^2}{s^2+omega_n^2} right )left( frac{1}{s} right )=frac{omega_n^2}{s(s^2+omega_n^2)}$$ Apply inverse Laplace transform on both the sides. $$c(t)=left ( 1-cos(omega_n t) right )u(t)$$ So, the unit step response of the second order system when $/delta = 0$ will be a continuous time signal with constant amplitude and frequency. Case 2: δ = 1 Substitute, $/delta = 1$ in the transfer function. $$frac{C(s)}{R(s)}=frac{omega_n^2}{s^2+2omega_ns+omega_n^2}$$ $$Rightarrow C(s)=left( frac{omega_n^2}{(s+omega_n)^2} right)R(s)$$ Substitute, $R(s) = frac{1}{s}$ in the above equation. $$C(s)=left( frac{omega_n^2}{(s+omega_n)^2} right)left ( frac{1}{s} right)=frac{omega_n^2}{s(s+omega_n)^2}$$ Do partial fractions of $C(s)$. $$C(s)=frac{omega_n^2}{s(s+omega_n)^2}=frac{A}{s}+frac{B}{s+omega_n}+frac{C}{(s+omega_n)^2}$$ After simplifying, you will get the values of A, B and C as $1,: -1: and : −omega _n$ respectively. Substitute these values in the above partial fraction expansion of $C(s)$. $$C(s)=frac{1}{s}-frac{1}{s+omega_n}-frac{omega_n}{(s+omega_n)^2}$$ Apply inverse Laplace transform on both the sides. $$c(t)=(1-e^{-omega_nt}-omega _nte^{-omega_nt})u(t)$$ So, the unit step response of the second order system will try to reach the step input in steady state. Case 3: 0 < δ < 1 We can modify the denominator term of the transfer function as follows − $$s^2+2deltaomega_ns+omega_n^2=left { s^2+2(s)(delta omega_n)+(delta omega_n)^2 right }+omega_n^2-(deltaomega_n)^2$$ $$=(s+deltaomega_n)^2+omega_n^2(1-delta^2)$$ The transfer function becomes, $$frac{C(s)}{R(s)}=frac{omega_n^2}{(s+deltaomega_n)^2+omega_n^2(1-delta^2)}$$ $$Rightarrow C(s)=left( frac{omega_n^2}{(s+deltaomega_n)^2+omega_n^2(1-delta^2)} right )R(s)$$ Substitute, $R(s) = frac{1}{s}$ in the above equation. $$C(s)=left( frac{omega_n^2}{(s+deltaomega_n)^2+omega_n^2(1-delta^2)} right )left( frac{1}{s} right )=frac{omega_n^2}{sleft ((s+deltaomega_n)^2+omega_n^2(1-delta^2) right)}$$ Do partial fractions of $C(s)$. $$C(s)=frac{omega_n^2}{sleft ((s+deltaomega_n)^2+omega_n^2(1-delta^2) right)}=frac{A}{s}+frac{Bs+C}{(s+deltaomega_n)^2+omega_n^2(1-delta^2)}$$ After simplifying, you will get the values of A, B and C as $1,: -1 : and : −2delta omega _n$ respectively. Substitute these values in the above partial fraction expansion of C(s). $$C(s)=frac{1}{s}-frac{s+2deltaomega_n}{(s+deltaomega_n)^2+omega_n^2(1-delta^2)}$$ $$C(s)=frac{1}{s}-frac{s+deltaomega_n}{(s+deltaomega_n)^2+omega_n^2(1-delta^2)}-frac{deltaomega_n}{(s+deltaomega_n)^2+omega_n^2(1-delta^2)}$$ $C(s)=frac{1}{s}-frac{(s+deltaomega_n)}{(s+deltaomega_n)^2+(omega_nsqrt{1-delta^2})^2}-frac{delta}{sqrt{1-delta^2}}left ( frac{omega_nsqrt{1-delta^2}}{(s+deltaomega_n)^2+(omega_nsqrt{1-delta^2})^2} right )$ Substitute, $omega_nsqrt{1-delta^2}$ as $omega_d$ in the above equation. $$C(s)=frac{1}{s}-frac{(s+deltaomega_n)}{(s+deltaomega_n)^2+omega_d^2}-frac{delta}{sqrt{1-delta^2}}left ( frac{omega_d}{(s+deltaomega_n)^2+omega_d^2} right )$$ Apply inverse Laplace transform on both the sides. $$c(t)=left ( 1-e^{-delta omega_nt}cos(omega_dt)-frac{delta}{sqrt{1-delta^2}}e^{-deltaomega_nt}sin(omega_dt) right )u(t)$$ $$c(t)=left ( 1-frac{e^{-deltaomega_nt}}{sqrt{1-delta^2}}left ( (sqrt{1-delta^2})cos(omega_dt)+delta sin(omega_dt) right ) right )u(t)$$ If $sqrt{1-delta^2}=sin(theta)$, then ‘δ’ will be cos(θ). Substitute these values in the above equation. $$c(t)=left ( 1-frac{e^{-deltaomega_nt}}{sqrt{1-delta^2}}(sin(theta)cos(omega_dt)+cos(theta)sin(omega_dt)) right )u(t)$$ $$Rightarrow c(t)=left ( 1-left ( frac{e^{-deltaomega_nt}}{sqrt{1-delta^2}} right )sin(omega_dt+theta) right )u(t)$$ So, the unit step response of the second order system is having damped oscillations (decreasing amplitude) when ‘δ’ lies between zero and one. Case 4: δ > 1 We can modify the denominator term of the transfer function as follows − $$s^2+2deltaomega_ns+omega_n^2=left { s^2+2(s)(deltaomega_n)+(deltaomega_n)^2 right }+omega_n^2-(deltaomega_n)^2$$ $$=left ( s+deltaomega_n right )^2-omega_n^2left ( delta^2-1 right )$$ The transfer function becomes, $$frac{C(s)}{R(s)}=frac{omega_n^2}{(s+deltaomega_n)^2-omega_n^2(delta^2-1)}$$ $$Rightarrow C(s)=left ( frac{omega_n^2}{(s+deltaomega_n)^2-omega_n^2(delta^2-1)} right )R(s)$$ Substitute, $R(s) = frac{1}{s}$ in the above equation. $C(s)=left ( frac{omega_n^2}{(s+deltaomega_n)^2-(omega_nsqrt{delta^2-1})^2} right )left ( frac{1}{s} right )=frac{omega_n^2}{s(s+deltaomega_n+omega_nsqrt{delta^2-1})(s+deltaomega_n-omega_nsqrt{delta^2-1})}$ Do partial fractions of $C(s)$. $$C(s)=frac{omega_n^2}{s(s+deltaomega_n+omega_nsqrt{delta^2-1})(s+deltaomega_n-omega_nsqrt{delta^2-1})}$$ $$=frac{A}{s}+frac{B}{s+deltaomega_n+omega_nsqrt{delta^2-1}}+frac{C}{s+deltaomega_n-omega_nsqrt{delta^2-1}}$$ After simplifying, you will get the values of A, B and C as 1, $frac{1}{2(delta+sqrt{delta^2-1})(sqrt{delta^2-1})}$ and $frac{-1}{2(delta-sqrt{delta^2-1})(sqrt{delta^2-1})}$ respectively. Substitute these values in above partial fraction expansion of $C(s)$. $$C(s)=frac{1}{s}+frac{1}{2(delta+sqrt{delta^2-1})(sqrt{delta^2-1})}left ( frac{1}{s+deltaomega_n+omega_nsqrt{delta^2-1}} right )-left ( frac{1}{2(delta-sqrt{delta^2-1})(sqrt{delta^2-1})} right )left ( frac{1}{s+deltaomega_n-omega_nsqrt{delta^2-1}} right )$$ Apply inverse Laplace transform on both the sides. $c(t)=left ( 1+left ( frac{1}{2(delta+sqrt{delta^2-1})(sqrt{delta^2-1})} right )e^{-(deltaomega_n+omega_nsqrt{delta^2-1})t}-left ( frac{1}{2(delta-sqrt{delta^2-1})(sqrt{delta^2-1})} right )e^{-(deltaomega_n-omega_nsqrt{delta^2-1})t} right )u(t)$ Since it is over damped, the unit step response of the second order system when δ > 1 will never reach step input in the steady state. Impulse Response of Second Order System The impulse response of the second order system can be obtained by using any one of these two methods. Follow the procedure involved while deriving step response by considering the value of $R(s)$ as 1 instead of $frac{1}{s}$. Do the differentiation of the step response. The following table shows the impulse response of the second order system for 4 cases of the damping ratio. Condition of Damping ratio Impulse response for t ≥ 0 δ = 0 $omega_nsin(omega_nt)$ δ = 1 $omega_n^2te^{-omega_nt}$ 0 < δ < 1 $left ( frac{omega_ne^{-deltaomega_nt}}{sqrt{1-delta^2}} right )sin(omega_dt)$ δ > 1 $left ( frac{omega_n}{2sqrt{delta^2-1}} right )left ( e^{-(deltaomega_n-omega_nsqrt{delta^2-1})t}-e^{-(deltaomega_n+omega_nsqrt{delta^2-1})t} right )$ Learning working make money

Learning Construction of Root Locus work project make money

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