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UNIT 5
TRANSFER FUNCTION AND STATE
VARIABLE ANALYSIS
5.1 Frequency response of second order system
Second order systems are the systems or networks which contain two or more storage elements and have
describing equations that are second order differential equations.
In this project great emphasis will be given on second order filters. These filters are very important for two main
reasons:
1.
2.
They provide an inexpensive approximation to ideal filters
They are used as building blocks for more expensive filters that provide more accurate
approximations to ideal filters.
The frequency response of second order filters is characterised by three filter parameters: the gain k, the corner
frequency
and the quality factor Q.
A second order filter is a circuit that has a transfer function of the form:
For better understanding of the above a worked out example is explained below.
Example of a second order system
Consider the circuit below,
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To deduce the transfer function of the system, as for first order systems, we use the voltage divider rule,
We also know for an inductor
We also know that
and for a capacitor
and
and therefore
. Applying these to the above equation, and get,
………..(1)
The denominator of the transfer function if made equal to zero in order to calculate the poles of the system
become a second order equation with a discriminant D.
For two different and real roots to exist
We can set the quality factor
, thus,
and clearly for this type of roots we must have
.
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In the case that Q is equal to ½ then the two roots are real and equal.
In the last case where Q is smaller than ½ then the two roots are complex and conjugate. In this case the system
has no resonant frequency thus on peak in the response. This is a significant deviation from the straight line
approximation.
When plotting the frequency response graph of a second order system we should have in mind that:
·
Each pole derived from the transfer function will result, as in first order, a +6db/octave
declination on the asymptotic response line, at a frequency calculated for each pole.
·
Each zero derived from the transfer function will result, as in first order, a -6db/octave
inclination on the asymptotic response line, at a frequency calculated for each zero.
·
For the case where we have two identical poles or zeroes then the declination or inclination
on the response line would be
-/+12db/octave.
Gain response plot for various values of Q
Phase response plots for different values of Q:
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5.2 Bode Plots
Bode plots were first introduced by H.W. Bode, when he was working at Bell labs in the United States. Now
before I describe what are this plots it is very essential here to discuss a few advantages over other stability
criteria. Some of the advantages of this plot are written below:
Advantages of Bode Plot
It is based on the asymptotic approximation, which provides a simple method to plot the logarithmic
magnitude curve.
2. The multiplication of various magnitude appears in the transfer function can be treated as an addition, while
division can be treated as subtraction as we are using a logarithmic scale.
3. With the help of this plot only we can directly comment on the stability of the system without doing any
calculations.
4. Bode plots provides relative stability in terms of gain margin and phase margin.
5. It also covers from low frequency to high frequency range.
1.
Now there are various terms related to this plot that we will use frequently in this article.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Gain Margin: Greater will the gain margin greater will be the stability of the system. It refers to the amount
of gain, which can be increased or decreased without making the system unstable. It is usually expressed in
dB.
Phase Margin: Greater will the phase margin greater will be the stability of the system. It refers to the phase
which can be increased or decreased without making the system unstable. It is usually expressed in phase.
Gain Cross Over Frequency: It refers to the frequency at which magnitude curve cuts the zero dB axis in the
bode plot.
Phase Cross Over Frequency: It refers to the frequency at which phase curve cuts the negative times the 180
degree axis in this plot.
Corner Frequency: The frequency at which the two asymptotes cuts or meet each other is known as break
frequency or corner frequency.
Resonant Frequency: The value of frequency at which the modulus of G (jω) has a peak value is known as
resonant frequency.
Factors: Every loop transfer function (i.e. G(s) × H(s)) product of various factors like constant term K,
Integral factors (jω), first order factors ( 1 + jωT) (± n) where n is an integer, second order or quadratic factors.
Slope: There is a slope corresponding to each factor and slope for each factor is expressed in the dB per
decade.
Angle: There is an angle corresponding to each factor and angle for each factor is expressed in the degrees.
Bode Plot
These are also known as logarithmic plot (because we draw these plots on semi-log papers) and are used for
determining the relative stabilities of the given system. Now in order to determine the stability of the system
using bode plot we draw two curves, one is for magnitude called magnitude curve another for phase called Bode
phase plot.
Now there are some results that one should remember in order to plot the Bode curve. These results are written
below:
Constant term K: This factor has a slope of zero dB per decade. There is no corner frequency corresponding
to this constant term. The phase angle associated with this constant term is also zero.
Integral factor 1/(jω)n: This factor has a slope of -20 × n (where n is any integer)dB per decade. There is no
corner frequency corresponding to this integral factor. The phase angle associated with this integral factor is -90
× n here n is also an integer.
First order factor 1/ (1+jωT): This factor has a slope of -20 dB per decade. The corner frequency
corresponding to this factor is 1/T radian per second. The phase angle associated with this first factor is -tan1
(ωT).
First order factor (1+jωT): This factor has a slope of 20 dB per decade. The corner frequency
corresponding to this factor is 1/T radian per second. The phase angle associated with this first factor is tan 1
(ωT) .
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Second order or quadratic factor : [{1/(1+(2ζ/ω)} × (jω) + {(1/ω2)} × (jω)2)]: This factor has a slope of 40 dB per decade. The corner frequency corresponding to this factor is ω n radian per second. The phase angle
associated with this first factor is - tan-1{ (2ζω / ωn) / (1-(ω / ωn)2)} .
Keeping all these points in mind we are able to draw the plot for any kind of system. Now let us discuss the
procedure of making a bode plot:
Substitute the s = jω in the open loop transfer function G(s) × H(s).
Find the corresponding corner frequencies and tabulate them.
Now we are required one semi-log graph chooses a frequency range such that the plot should start with the
frequency which is lower than the lowest corner frequency. Mark angular frequencies on the x-axis, mark
slopes on the left hand side of the y-axis by marking a zero slope in the middle and on the right hand side
mark phase angle by taking -180 degrees in the middle.
4. Calculate the gain factor and the type or order of the system.
5. Now calculate slope corresponding to each factor.
For drawing the Magnitude curve :
 Mark the corner frequency on the semi log graph paper.
 Tabulate these factors moving from top to bottom in the given sequence.
1. Constant term K.
2. Integral factor 1/(jω)n.
3. First order factor 1/ (1+jωT).
4. First order factor (1+jωT).
5. Second order or quadratic factor : [{1/(1+(2ζ/ω)} × (jω) + {(1/ω 2)} × (jω)2)]
 Now sketch the line with the help of corresponding slope of the given factor. Change the slope at every corner
frequency by adding the slope of the next factor. You will get magnitude plot.
 Calculate the gain margin. For drawing the Bode phase plot :
1. Calculate the phase function adding all the phases of factors.
2. Substitute various values to above function in order to find out the phase at different points and plot a curve.
You will get a phase curve.
3. Calculate the phase margin.
1.
2.
3.
Stability Conditions of Bode Plots
Stability conditions are given below :
1. For Stable System: Both the margins should be positive. Or phase margin should be greater than the gain
margin.
2. For Marginal Stable System : Both the margins should be zero. Or phase margin should be equal to the gain
margin.
3. For Unstable System : If any of them is negative. Or phase margin should be less than the gain margin.
5.3 Nyquist Stability criterion
The stability analysis of a feedback control system is based on identifying the location of the roots of the
characteristic equation on s-plane. The system is stable if the roots lie on left hand side of s-plane. Relative
stability of a system can be determined by using frequency response methods like Nyquist plot and Bode plot.
Nyquist criterion is used to identify the presence of roots of a characteristic equation in a specified region of splane. To understand Nyquist plot we need to know about some of the terminologies. Contour : Closed path in
a complex plane is called contour.
Nyquist path or Nyquist contour
The Nyquist contour is a closed contour in the s-plane which completely encloses the entire right hand half of
s-plane. In order to enclose the complete RHS of s-plane a large semicircle path is drawn with diameter along jω
axis and centre at origin. The radius of the semicircle is treated as infinity.
Nyquist Encirclement
A point is said to be encircled by a contour if it is found inside the contour.
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Nyquist Mapping
The process by which a point in s-plane transformed into a point in F(s) plane is called mapping and F(s) is
called mapping function.








Steps of drawing the Nyquist path
Step 1 - Check for the poles of G(s) H(s) of jω axis including that at origin.
Step 2 - Select the proper Nyquist contour – a) Include the entire right half of s-plane by drawing a semicircle
of radius R with R tends to infinity.
Step 3 - Identify the various segments on the contour with reference to Nyquist path
Step 4 - Perform the mapping segment by segment substituting the equation for respective segment in the
mapping function. Basically we have to sketch the polar plots of the respective segment.
Step 5 - Mapping of the segments are usually mirror images of mapping of respective path of +ve imaginary
axis.
Step 6 - The semicircular path which covers the right half of s plane generally maps into a point in G(s) H(s)
plane.
Step 7- Interconnect all the mapping of different segments to yield the required Nyquist diagram.
Step 8 - Note the number of clockwise encirclement about (-1, 0) and decide stability by N = Z – P
is the Open loop transfer function (O.L.T.F)
is the Closed loop transfer function (C.L.T.F). N(s) = 0 is the open loop zero and D(s) is the open loop pole
From stability point of view no closed loop poles should lie in the RH side of s-plane . Characteristics equation
1 + G(s) H(s) = 0 denotes closed loop poles .
Now as 1+ G(s) H(s) = 0 hence q(s) should also be zero.
Therefore , from the stability point of view zeroes of q(s) should not lie in RHP of s-plane. To define the
stability entire RHP (Right Hand Plane) is considered. We assume a semicircle which encloses all points in the
RHP by considering the radius of the semicircle R tends to infinity. [R → ∞]. The first step to understand the
application of Nyquist criterion in relation to determination of stability of control systems is mapping from splane to G(s) H(s) - plane. s is considered as independent complex variable and corresponding value of G(s)
H(s) being the dependent variable plotted in another complex plane called G(s) H(s) - plane. Thus for every
point in s-plane there exists a corresponding point in G(s) H(s) - plane. During the process of mapping the
independent variable s is varied along a specified path in s - plane and the corresponding points in G(s)H(s)
plane are joined. This completes the process of mapping from s-plane to G(s)H(s) - plane. Nyquist stability
criterion says that N = Z - P. Where, N is the total no. of encirclement about the origin, P is the total no. of
poles and Z is the total no. of zeroes.
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Case 1:- N = 0 (no encirclement), so Z = P = 0 & Z = P If N = 0, P must be zero therefore system is stable.
Case 2:- N > 0 (clockwise encirclement), so P = 0 , Z ≠0 & Z > P For both cases system is unstable.
Case 3 :- N < 0 (counterclockwise encirclement), so Z = 0, P ≠0 & P > Z System is stable.
5.4 Concept of state variable
Before I introduce you about the concept of state space analysis of control system, it is very important to
discuss here the differences between the conventional theory of control system and modern theory of control
system.
1.
The conventional control theory is completely based on the frequency domain approach while the
modern control system approach is based on time domain approach.
2.
In the conventional theory of control system we have linear and time invariant single input single
output (SISO) systems only but with the help of theory of modern control system we can easily do the analysis
of even non linear and time variant multiple inputs multiple outputs (MIMO) systems also.
3. In the modern theory of control system the stability analysis and time response analysis can be done by
both graphical and analytically method very easily.
Now state space analysis of control system is based on the modern theory which is applicable to all types of
systems like single input single output systems, multiple inputs and multiple outputs systems, linear and non
linear systems, time varying and time invariant systems. Let us consider few basic terms related to state space
analysis of modern theory of control systems.
1.
2.
3.
4.
State in State Space Analysis : It refers to smallest set of variables whose knowledge at t=t0 together with the
knowledge of input for t ≥ t0 gives the complete knowledge of the behavior of the system at any time t ≥ t 0.
State Variables in State Space analysis : It refers to the smallest set of variables which help us to determine
the state of the dynamic system. State variables are defined by x1(t), x2(t)........Xn(t).
State Vector : Suppose there is a requirement of n state variables in order to describe the complete behavior
of the given system, then these n state variables are considered to be n components of a vector x(t). Such a
vector is known as state vector.
State Space : It refers to the n dimensional space which has x1 axis, x2 axis .........xn axis.
State Space Equations
Let us derive state space equations for the system which is linear and time invariant. Let us consider multiple
inputs and multiple outputs system which has r inputs and m outputs. Where r=u 1, u2, u3 ........... ur. And m = y1,
y2 ........... ym. Now we are taking n state variables to describe the given system hence n = x1, x2, ........... xn. Also
we define input and output vectors as, Transpose of input vectors,
Where, T is transpose of the matrix. Transpose of output vectors,
Where, T is transpose of the matrix. Transpose of state vectors,
Where, T is transpose of the matrix.
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These variables are related by a set of equations which are written below and are known as state space equations
Representation of State Model using Transfer Function
Decomposition : It is defined as the process of obtaining the state model from the given transfer function. Now
we can decompose the transfer function using three different ways:
1.
2.
3.
Direct decomposition,
Cascade or series decomposition,
Parallel decomposition.
In all the above decomposition methods we first convert the given transfer function into the differential
equations also called the dynamic equations. After converting into differential equations we will take inverse
Laplace transform of the above equation then corresponding to the type of decomposition we can create model.
We can represent any type of transfer function in state model. We have various types of model like electrical
model, mechanical model etc. Expression of Transfer Matrix in terms of A, B, C and D. We define transfer
matrix as the Laplace transform of output to the Laplace transform of input. On writing the state equations again
and taking the Laplace transform of both the state equation (assuming initial conditions equal to zero) we have
We can write the equation as
Where, I is an identity matrix.
Now substituting the value of X(s) in the equation Y(s) and putting D = 0 (means is a null matrix) we have
Inverse of matrix can substitute by adj of matrix divided by the determinant of the matrix, now on rewriting the
expression we have of
|sI-A| is also known as characteristic equation when equated to zero.
Concept of Eigen Values and Eigen Vectors
The roots of characteristic equation that we have described above are known as eigen values or eigen values of
matrix A. Now there are some properties related to eigen values and these properties are written below1.
2.
3.
4.
Any square matrix A and its transpose At have the same eigen values.
Sum of eigen values of any matrix A is equal to the trace of the matrix A.
Product of the eigen values of any matrix A is equal to the determinant of the matrix A.
If we multiply a scalar quantity to matrix A then the eigen values are also get multiplied by the same value of
scalar.
5. If we inverse the given matrix A then its eigen values are also get inverses.
6. If all the elements of the matrix are real then the eigen values corresponding to that matrix are either real or
exists in complex conjugate pair.
Now there exists one eigen vector corresponding to one Eigen value, if it satisfy the following condition ( ek × I
- A )Pk = 0. Where k = 1, 2, 3, ........n.
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State Transition Matrix and Zero State Response
We are here interested in deriving the expressions for the state transition matrix and zero state response. Again
taking the state equations that we have derived above and taking their Laplace transformation we have,
Now on rewriting the above equation we have
Let [sI-A]-1 = θ(s) and taking the inverse Laplace of the above equation we have
The expression θ(t) is known as state transition matrix. L-1.θ(t)BU(s) = zero state response. Now let us discuss
some of the properties of the state transition matrix.
1.
2.
3.
If we substitute t = 0 in the above equation then we will get 1. Mathematically we can write θ(0) =1.
If we substitute t = -t in the θ(t) then we will get inverse of θ(t). Mathematically we can write θ(-t) = [θ(t)]-1.
We also another important property [θ(t)]n = θ(nt).
A Simple Example
Consider an 4th order system represented by a single 4th order differential equation with input x and
output y.
We can define 4 new variables, q1 through q4.
but
We can now rewrite the 4th order differential equation as 4 first order equations
This is compactly written in state space format as
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with
For this problem a state space representation was easy to find. In many cases (e.g., if there are
derivatives on the right side of the differential equation) this problem can be much more difficult. Such
cases are explained in the discussion of transformations between system representations.
Numerical 1: Derive a state space model for the system shown. The input is fa and the output is y.
We can write free body equations for the system at x and at y.
Freebody Diagram
Equation
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There are three energy storage elements, so we expect three state equations. The energy storage elements
are the spring, k2, the mass, m, and the spring, k1. Therefore we choose as our state variables x (the energy in
spring k2 is ½k2x²), the velocity at x (the energy in the mass m is ½mv², where v is the first derivative of x),
and y (the energy in spring k1 is ½k1(y-x)² , so we could pick y-x as a state variable, but we'll just use y (since
x is already a state variable; recall that the choice of state variables is not unique). Our state
variablesbecome:
Now we want equations for their derivatives. The equations of motion from the free body diagrams yield
or
with the input u=fa.
Numerical 2: Derive a state space model for the system shown. The input is ia and the output is e2.
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There are three energy storage elements, so we expect three state equations. Try choosing i1, i2 and e1 as
state variables. Now we want equations for their derivatives. The voltage across the inductor L2 is e1 (which
is one of our state variables)
so our first state variable equation is
If we sum currents into the node labeled n1 we get
This equation has our input (ia) and two state variable (iL2 and iL1) and the current through the
capacitor. So from this we can get our second state equation
Our third, and final, state equation we get by writing an equation for the voltage across L 1 (which is e2) in
terms of our other state variables
We also need an output equation:
So our state space representation becomes