Category: control Systems

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

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