How to draw root locus graph with simple steps

We know that the stability of the given closed loop system depends upon the location of the roots of the characteristics equation. That is the location of the closed loop poles. If we change some parameter of a system, then the location of closed loop pole changes in ‘s’ plane. The study of this locus (of moving pole in ‘s’ plane) because of variation of any parameter of the system is very important while designing any closed loop system.

This movement of poles in ‘s’ plane is called as ‘Root Locus’.

This method was invented by W.R. Evans in 1948.

Root Locus is a simple graphical method for determining the roots of the characteristic equation. It can be drawn by varying the parameter (generally gain of the system but there are also other parameters that can be varied) from zero to infinity.

Root Locus Method with step by step solution

General steps for drawing the Root Locus of the given system:

  1. Determine the open loop poles, zeros and a number of branches from given G(s)H(s).
  2. Draw the pole-zero plot and determine the region of real axis for which the root locus exists. Also, determine the number of breakaway points (This will be explained while solving the problems).
  3. Calculate the angle of asymptotes.
  4. Determine the centroid.
  5. Calculate the breakaway points (if any).
  6. Calculate the intersection point of root locus with the imaginary axis.
  7. Calculate the angle of departure or angle of arrivals if any.
  8. From above steps draw the overall sketch of the root locus.
  9. Predict the stability and performance of the given system by the root locus.

Solved Problem on Root Locus:

Question: For a unity feedback system, G(s) = K/[s(s+4)(s+2)]. Sketch the nature of root locus showing all details on it. Comment on the stability of the system.


Given system is unity feedback system. Therefore H(s) = 1.

Therefore G(s) H(s) = K/[s(s+4)(s+2)].

Step 1:

Poles = 0, -4, -2. Therefore P=3.

Zeros = there are no zeros. Z=0.

So all (P-Z=3) branches terminate at infinity.

Step 2: Pole-zero plot and sections of the real axis.

The pole-zero plot of the system is shown in the figure below. Here RL denotes Root Locus existence region and NRL denotes the non-existence region of root locus.

These sections of real axis identified as a part of the root locus as to the right sum of poles and zeros is odd for those sections.

Step 3: Angle of asymptotes

‘A line to which root locus touches at infinity is called asymptotes.’

Number of asymptotes = P-Z = 3. Therefore 3 asymptotes are approaching to infinity.

Step 4: Centroid or Centre of asymptotes.

Asymptote touches real axis at a point called centroid.

Branches will approach infinity along these lines which are asymptotes.

centroid 2

Step 5: To find breakaway point, we have characteristic equation as,

As there is no root locus between -2 to -4, -3.15 can not be a breakaway point. it also can be confirmed by calculating ‘K’ for s = -3.15. It will be negative that confirms s = -3.15 is not a breakaway point.

For s = -3.15, K = -3.079 (Substituting in equation for K). But as there has to be breakaway point between’0’and ‘-2’, s = – 0.845 is a valid breakaway point.

For s = -0.845 K = +3.079. As K is positive s = – 0.845 is valid breakaway point.

Step 6: Intersection point with the imaginary axis.

Characteristic equation

s^3+6s^2 +8s+K = 0

Routh’s array:

Intersection of root locus with imaginary axis is at ±j2.828 and corresponding value of K(marginal) = 48.

Step 7 : As there are no complex conjugate poles or zeros, no angles of departures or arrivals are required to be calculated.

Step 8: The complete root locus is as shown in the figure below.

Step 9: Prediction about stability:

For 0 < K < 48, all the roots are in left half of s-plane hence system is absolutely stable.

For K(marginal) = +48, a pair of dominant roots on imaginary axis with remaining root in left half. So the system is marginally stable oscillating at 2.82 rad/sec. For 48 < K < ∞, dominant roots are located in right half of s-plane hence system is unstable.

Stability is predicted by locations of dominant roots. Dominant roots are those which are located closest to the imaginary axis.

MATLAB Program for Root Locus Method:

%AIM: To plot the root locus of given transfer function
%Created by :
clear all;
close all;

Root locus method Matlab program
Root locus method Matlab program

Root locus method matlab program 1
Root locus method Matlab program

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How to draw bode plot in control system

How to draw a Bode plot on semi-log graph paper

Question: A unity control feedback system has G(s) =80/s(s+2)(s+20). Draw the bode plot. Determine the gain margin and phase margin. Also determine gain cross-over frequency and phase cross-over frequency. Comment on the system stability using this Bode plot.


Step 1:

Arrange G(s)H(s)in time constant form as follows:

bode plot 1

Step 2:

Identify factors of given H(s)G(s):

1)      K=2

2)      One pole is at origin. (Because there is ‘s’ in the denominator).

3)      Simple pole 1/(1+s/2) with T1 = ½. Hence wc1 = 1/T1 = 2.

4)      Similarly simple pole 1/(1+s/20) with T2 = 1/20. wc2 = 1/T2 = 20.

Step 3: Magnitude plot analysis:

1)      For K=2, draw a line of 20logK = 20log2 = 6 dB.

2)      For one pole at origin. Straight line of slope -20 dB/decade passing through intersection point of w=1 and 0 dB. ( Trick to remember: if there is no pole at origin draw a straight line. For simple pole draw -20dB/decade line, for second order pole i.e. ‘s^2 term in denominator’ draw -40dB/decade line, for third order pole (i.e. S^3)draw -60dB/decade line and so on. And always take intersection point of w=1 and 0 dB. )

3)      Shift intersection point of w=1 and 0 dB on 20logK line and draw parallel to -20 dB/decade line drawn. This will continue as a resultant of K and 1/s till first corner frequency occurs. i.e. wc2=2.

4)      At wc1 = 2, as there is simple pole it will contribute the rate of -20 dB/decade hence resultant slope after wc1 = 2 becomes -20-20=-40 dB/decade. This is addition of K, 1/s and 1/(1+s/2). This will continue till it intersects next corner frequency line i.e. wc2=20.

5)      At wc2 = 20, there is simple pole contributing -20dB/ decade and hence resultant slope after wc2=20 becomes -40-20=-60 dB/decade. This is resultant of overall G(s)H(s). i.e. G(jw)H(jw) the final slope is -60dB/decade, as there is no other factor present.

Step 4: Phase angle plot: Convert G(s)H(s) to G(jw)H(jw)

To draw straight line of -40 dB/decade and -60 dB/decade from wc1 = 2 and wc2 = 20, draw faint lines of slope -20, -40, -60 dB/decade from intersection point of w=1 and 0 dB line and just draw parallel to them from respective points .

bode plot 2

bode plot 3

Step 5: Bode plot and solution:


bode plot
bode plot

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Mason’s gain formula for determining overall gain of system (with example)

Mason’s gain formula for determining overall gain of system


In the last article we have seen block diagram algebra. For studying the complete behavior of the system we should know its transfer function. Using block diagram algebra we can reduce the block diagram of a control system into a single block representing the transfer function of that system. But in case of complicated block diagrams it will become difficult to reduce the block diagram into its equivalent single block having same input and output as that of given control system.

So to avoid this complexity of reducing the block diagram we use one simple method. In this method we first construct a signal flow graph of the given system using its ‘describing equations’. And using this signal flow graph and one formula (known as Mason’s gain formula) we can easily find out the overall gain of the given control system.

Construction of signal flow graph:

As discussed earlier, for the construction of signal flow graph we need its set of describing equations. For the clear understanding of construction of signal flow graph, we are using following set of equations:


Where x1 is the input variable and x5 is the output variable.

Steps to draw signal flow graph:

1)      In the given set of instruction there are five nodes (x1, x2… x5), therefore mark five nodes as shown in diagram a) as below.

construction of signal flow graph

2)      Now consider the first equation, in this equation x2 is equal to the sum of four signals having different coefficients.  When we connect x1 with x2 the direction should be indicated as (1→2) because of a12, similarly for x2 and x3 the arrow should be in the direction of (3→2) because of a32. The signal flow graph for the first equation is shown in the diagram b).

3)      Similarly, we have to follow the same procedure for all remaining equations, which are shown in figure c), d) and e).

construction of signal flow graph 2

4)      Finally figure f) shows the complete representation of the signal flow graph of the given control system.

Mason’s gain formula:

As we know the relationship between the input and output of the given control system is given by its net gain. This net gain is also known as overall gain of the system. For determining the overall gain of the control system Mason’s derived one formula, this formula is used for calculating the overall gain of the system, which is given by,



  • Pk represents path gain of kth forward path.
  • ∆ represents determinant of the graph which is given by:   

∆ = 1- (sum of loop gains of all individual loops) + (sum of gain products of all possible combinations of two non-touching loops) – (sum of gain products of all possible combinations of three non-touching loops) + …


  • ‘∆k’ represents the value of ∆ for that part of the graph which does not touch the kth forward path.
  • ‘T’ is the overall gain of the given control system.


In the given signal flow graph x1 is the input variable, whereas x5 is the output variable.

By using Mason’s gain formula, overall gain of the system is given by,

1)      In the first step, we have to determine the number of forward path. Forward path means the path from input node to the output node. So in this case we have 2 forward path (i.e. k=2) with path gains:

 there are two forward path

2)      There are 5 individual loops with loop gains:

 there are five

3)      Now we have to find the all possible combinations of two non-touching loops (i.e. those loops which do not contain common node) and their gain products. So such non-touching loops with there product gain is given by:

 there are two possible

4)      Similarly, we have to find all possible combination of three non-touching loops. But in this case there are no such combination exist.


 there are no combination

forward path equation


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