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Root Locus Method In Control System – The Ultimate Guide !!? All Answers

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Root locus Example 1, #RootLocus, #RootLocusProblem, #ControlSystem, #ControlEngineeing

Root locus Example 1, #RootLocus, #RootLocusProblem, #ControlSystem, #ControlEngineeing
Root locus Example 1, #RootLocus, #RootLocusProblem, #ControlSystem, #ControlEngineeing

Images related to the topicRoot locus Example 1, #RootLocus, #RootLocusProblem, #ControlSystem, #ControlEngineeing

Root Locus Example 1, #Rootlocus, #Rootlocusproblem, #Controlsystem, #Controlengineeing
Root Locus Example 1, #Rootlocus, #Rootlocusproblem, #Controlsystem, #Controlengineeing

Spoiler: This post is going to be long but fun! ! In my previous post, we looked at the Routh-Hurwitz standard in detail. It gives us information about the number of poles in the right half of the s-plane of the unstable system and the location of the poles on the imaginary axis in the case of an edge-stable system. So the biggest flaw of the Rolls criterion is that it doesn’t give us the remaining pole position. The root locus Evans introduced by W.R. is a graphical representation of how the “closed loop poles” move as the gain of the system changes. This technique is often used in control engineering because it reflects the entire dynamics of the system. The most important advantage of the root locus is to check the system behavior by adjusting the gain value “K”. Therefore, adjusting to a value of “K” will track the origin of the root on the R-H plane.

Angle and Magnitude Condition

These are the two important conditions that we would need to know before we can proceed. We know the characteristic equation is given by: 1 + G(s)H(s) = 0 G(s)H(s) = -1 , As the s plane is complex , we can write G(s)H(s) = -1 + 0j This equation has magnitude  as well as phase, | G(s)H(s) | =  |-1 + 0j | =  1 , ∠ G(s)H(s)  =   180 ,  but since -1 + 0j point can be traced with an angle of   180  , 540 ,  900  ……   we thus have a generalized angle condition as : ∠ G(s)H(s)  =  ( 2q + 1 ) 180    q = 0 , 1 , 2 …………. Construction of Root Locus(General method for drawing) We construct root locus by following a certain set of rules. Implementing each of the below-mentioned rules gives us the final root locus. Rule no. Statement Comments Rule 1 The root locus is symmetric about the real axis Root locus that we construct will always be symmetric about the real axis (irrespective of system) Rule 2 Total number of loci The total number of loci will be equal to max(p,z) where p is no: of the open loop poles and z is number of open loop zeros Rule 3 Real axis loci A point that lies on real axis basically lies on the root locus only if the total number of real open loop poles and open loop zeros present in the RHS of this point is odd Rule 4 Angle of asymptotes Total number of branches of the root locus tending towards infinity is equal to  p – z. The angle of asymptotes gives us the direction along which these p-z branches approach infinity. Rule 5 centro Centro is a point on real axis , through which the asymptotes pass. Rule 6 Breakaway point A break away point on the root locus is a point where the two poles will meet. Once they meet,they dive (split) i.e break away from the real axis. Rule 7 Angle of departure / arrival This gives us angles along which complex poles will depart( and complex zeros arrive) from their original position Rule 8 Intersection with the imaginary axis This gives us points on imaginary axis which the given root locus cut through while moving to right half of the s plane We shall take a simple example and explain each rule , Let  G(s)H(s) =     ,    H(s) = 1 Rule 1 :  The root locus is symmetric about the real axis Irrespective of the system, the root locus that we construct will always be symmetric about the real axis. Rule 2 : Total number of loci The total number of loci will be equal to max(p,z) where p is no: of the open loop poles and z is number of open loop zeros. Total number of branches of the root locus tending towards infinity is equal to p – z In this case we have, G(s) =    ,   H(s) = 1           p = 3  ,  z = 1 Total number of loci = max ( p , z )  =  max ( 3 , 1 ) =  3 Total number of loci tending to infinity = p – z = 3 – 1 = 2 All branches of the root locus start from open-loop poles . In the given example , of the 3 loci, 2 will go to infinity , while one of them will go to the  zero.  Rule 3 : Real axis loci Any point on the real axis will lie on the root locus only if the total number of real open loop zeros and the open loop poles to the right-hand se of the given point is odd We plot the open loop poles and zeros in the s-plane first: In this conser any point ‘a’  between 0 and -2. Total number of  real poles and zeros on right of this point is odd.(only 1 pole ). Now similarly conser another point ‘b’ in beween -2 and -5 . Total number of real poles and zeros to the right of this point is even(two poles) . Hence this point doesn’t lie on root locus. Applying the same principles we can state that points lying between -5 and -6 will be a part of root locus. Rule 4 :   Angle of asymptotes Total number of branches of the root locus tending towards infinity is equal to  p – z. The angle of asymptotes gives us the direction along which these p-z branches approach infinity. Angle of asymptotes is given by the formula         where x = 0 , 1 , 2 , 3 ……p-z-1 In the given example   G(s) =     ,   H(s) = 1             Here  p = 3 , z = 1 Total number of branches approaching infinity are= p – z = 3 – 1 = 2 Hence    =  90    ,    =  270  , Rule 5 : Centro is a point on real axis , through which the asymptotes pass. Where ‘OLTF’  is  Open loop transfer function The centro is always real and it may or may not be part of root locus.In this example, G(s) =     ,   H(s) = 1             Here  p = 3 , z = 1 , = 0 – 2 – 5 = – 7 = -6 Hence ,        Once centro is obtained , we draw the asymptotes passing through this centro. Note: Asymptotes only gives us direction . It is not final root locus and hence should be drawn with light pencil (or dotted line) Rule 6 : The Break away point A break away point on the root locus is a point where the two poles colle (meet) .Once they meet ( or colle),they dive (split) that is break away from real axis. We have to follow following steps for finding out the break away point, Compute the characteristic equ. Write it in terms of k. Compute and equate to 0. Now let’s take our example, G(s) =     ,   H(s) = 1             Here  p = 3 , z = 1 Step 1: Characteristic equation :     1 +  G(s)H(s)  =  0 , We get ,      s(s + 2) (s + 5)   +   k(s + 6)  =  0 Step 2: We write this equation in terms of  k k  =    =  Step 3: Compute   =  0    =  0 We use   rule  ,    d( )  =         (division rule of derivatives)     =  =   0 ,   On solving we get, 2  +  25  +  84s +  60  =  0 ,   The roots of this equation are -0.9753, -4.1988 , -7.32 Breakaway points are points where two poles come and meet. Hence breakaway point have to lie on the root locus . Out of three root, only -0.9753 lies on real axis of the root locus(see rule number 3) Rule 8 :  Intersection with the imaginary axis(we will see rule 8 first and then rule 7 will be explained) This proves us the points that lies in the imaginary that the root locus cut through while moving towards the right half of s plane. For the intersection with the imaginary axis, we follow the following steps: Conser characteristic equation 1 + G(s)H(s) = 0,. Construct Routh array. Determine ( row of zeros). Then obtain the Auxiliary equation. Finally find the roots of the Auxiliary equation. These will give the intersection points with the  imaginary axis . Let’s take our example, G(s) =     ,   H(s) = 1 , Characteristic equation = 1 + G(s) H(s) =  + 7  + (10+ k)s + 6k  = 0 We now construct Routh array, 1 10 + k 7 6k 0 6k Now for poles to lie on imaginary axis, the system has to be marginally stable. For marginal stability one row of the routh array must be zero.Thus we take one row zero as: =  0 , On solving,   =  -70 , Now we know ‘k’ is the gain of the system, so it can’t be negative. Thus ,  k = -70  cannot be true, hence there is no  ( i.e there is no intersection with the imaginary axis and the two poles will colle at breakaway point and do not move towards the right half plane). Rule 7 : Angle of departure/arrival This gives us angles along which complex poles will depart( and complex zeros arrive) from their original position. This rule is used only if the system has complex pole or zero. Since our example had only real poles and zero, we should take new example to compute the angle of departure/arrival, conser a system having G(s) =    ,   H(s) = 1 Given system has complex pole and thus we need to use rule 7. The root locus arrives at a complex zero while a root locus departs from a complex pole i.e Angle of departure is associated with complex poles while  angle of arrival is associated with complex zeros. We encounter complex poles more often and hence angle of departure is what we need to use more often. Angle of departure  () is given by the formula: =  180 –  , Where  =  ,     Similarly Angle of arrival is given by  ,    =   180 +  We use graphical method to compute  . We join all poles and zeros to the complex pole whose angle of departure needs to be  calculated. We then  compute the angle made by poles and angles made by zeros. Let’s take our example, G(s) =     ,   H(s) = 1 , We plot the open loop poles  and zeros,  we join all remaining poles and zeros to this complex pole. = 135   ,  =  90   ,     =    = 18.43  , =    ,     Hence  ,  = + –     =   225 –  18.43  =  206.57 , =  180 –   = -26.57 ,   Hence the Angle of departure is -26.57  ,    What is the significance of this angle? This means that the complex pole at s = -1 + j will depart from its original position at an angle of  -26.57  and then intersect the imaginary axis at a value obtained in rule 8. Angle of departure of other complex pole at s = -1 – j will be 26.57  as shown below, .

Determining the value of ‘k’ from the Damping Ratio (\delta):

We saw the concept of damping in the previous post on timing analysis. The damping ratio is controlled by the dominant root. Now to determine the value of “k”, we can use the following steps: Draw the root locus. Given, calculate, . Draws a line with an angle from the origin measured from the negative real axis. Determine the point where this line intersects the root locus, the point s = -a + bj . Substitute this value of s into the quantity condition , |G(s)H(s)| = 1 and get the value of “k”. So, in fact, it is clear that the root locus method involves evaluating the change of the poles of the CLTF (closed loop transfer function) with respect to the system control gain (k). If the characteristic equation of the closed-loop system can be solved, this change can be directly represented, and the root locus method is not needed. However, for larger order characteristic equations, it is often difficult to factor polynomials. This is where the root locus method comes into play. In my next article, I’ll prove the various numbers you can get from this topic. So, go back and learn about all the rules and concepts in this post. We will meet soon in my next article. report this ad


What is root locus in control system?

In control theory and stability theory, root locus analysis is a graphical method for examining how the roots of a system change with variation of a certain system parameter, commonly a gain within a feedback system.

Why do we need the root locus method in controlling the system?

Root locus is helping us to map graphically as graph all possible locations of the poles within the system on the s-plane. The different locations of the poles are obtained under the effect of gain changes (proportional gain).

How do you draw a root locus in a control system?

Follow these rules for constructing a root locus.
  1. Rule 1 − Locate the open loop poles and zeros in the ‘s’ plane.
  2. Rule 2 − Find the number of root locus branches.
  3. Rule 3 − Identify and draw the real axis root locus branches.
  4. Rule 4 − Find the centroid and the angle of asymptotes.

What are the advantages of root locus method?

Advantages of Root Locus Technique
  • Root locus technique in control system is easy to implement as compared to other methods.
  • With the help of root locus we can easily predict the performance of the whole system.
  • Root locus provides the better way to indicate the parameters.

What are the two conditions of a root locus?

Angle and Magnitude Condition of Root Locus

Thus, the above-given equation must be satisfied for each individual value of s in order to be present on the root locus. Further, the two conditions of root locus are: Angle condition. Magnitude condition.

What are the applications for root locus?

The Root Locus Plot technique can be applied to determine the dynamic response of the system. This method associates itself with the transient response of the system and is particularly useful in the investigation of stability characteristics of the system.

How can a root locus be used to determine stability?

The root locus procedure should produce a graph of where the poles of the system are for all values of gain K. When any or all of the roots of D (denominator) are in the unstable region, the system is unstable. When any of the roots are in the marginally stable region, the system is marginally stable (oscillatory).

How do you draw root locus examples?

Example of root locus
  1. Find the number of poles, zeroes, number of branches, etc., from the given transfer functions.
  2. Draw the plot that shows the poles and zeroes marked on it.
  3. Calculate the angle of asymptotes and draw a separate sketch.
  4. Find the centroid and draw a separate sketch.
  5. Find the breakaway points.

How do you read a root locus?

Interpreting a Root Locus Diagram

If the real components of all poles are negative, then the system is said to be stable for that value of Kc. If the real component of the pole is positive, the system is unstable for that value of Kc, meaning the output signal will diverge from the set point.

What are the limitations of root locus?

The limitations of root locus method for tuning PID controllers are: Not perform well on a nonlinear system. Loses significance at high frequencies or high degrees of damping. The designs are susceptible to noise .

What is phase criteria for root locus?

If K=∞, then N(s)=0. It means the closed loop poles are equal to the open loop zeros when K is infinity. From above two cases, we can conclude that the root locus branches start at open loop poles and end at open loop zeros.

What is the difference between root locus and Bode plot?

Bode plots tells us how many resonant peaks are there if I am not wrong (magnitude response). Phase response is absolutely important. Root locus tells about response when gain changes (stability is clear from response).


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