Bsa ppt 48

BASIC SYSTEM
ANALYSIS
(NEE-403)

BASIC SYSTEM ANALYSIS
UNIT -1
(Introduction to Continuous Time Signals and Systems)

SIGNAL
• A signal is a physical quantity that varies with
time,space or any other independent
variable.by which information can be
conveyed“
• "A signal is a source of information generally a
physical quantity which varies with respect to
time, space, temperature like any
independent variable"

TYPE OF SIGNAL
Continuous Time Signal
If the independent variable (t) iscontinuous,
then the corresponding signal is continuous
time signal.

SIGNAL(cont’d)
Discrete Time Signal
If the independent variable (t) takes on
onlydiscrete values, for
example t = ±1, ±2, ±3, ...

SIGNAL(cont’d)
Periodic Signal
If the transformed signal is same as x(t+nT),
then the signal is periodic. where T is
fundamental period
(the smallest period) of signal x(t). i.e.
x(t)= x(t+nT)

SIGNAL(cont’d)
• Non-Periodic Signal
• If the transformed signal is not same as
x(t+nT), then the signal is periodic. where T
Is fundamental period (the smallest period)
of signal x(t). i.e.
x(t) is not equal to x(t+nT)

Periodic And Non-Periodic
Signal

SIGNAL(cont’d)
Even and Odd Signal
• One of characteristics of signal is symmetry
that may be useful for signal analysis.
• Even signals are symmetric around vertical
axis, and Odd signals are symmetric about
origin.

SIGNAL(cont’d)
Even Signal:
A signal is referred to as an even if it is identical
to its time-reversed
counterparts; x(t) = x(-t).
Odd Signal:
A signal is odd if x(t) = -x(-t). An odd signal
must be 0 at t=0, in other words odd signal
passes the origin.

Deterministic signal And Random
Signal
Deterministic signal
• A deterministic signal about which there is no
uncertainity with respect to its value at any
time.
• Therefore ,a deterministic signal can be
defined as completely specified function of
time

Deterministic signal And Random
Signal
Random Signal
• Random Signal is defined as a signal above
which there is no uncertainity before its actual
occurrence.

Energy And Power Signal
• In an electrical signal, the instantaneous
power for the voltage across the resistance R
and its energy are defined;

Energy And Power Signal(Cont’d)

• where |x(t)| denotes the magnitude of x(t).
• It is necessary to get a scalar quantity for
complex signal, because magnitude of
complex number is defined as:

• And, it is also squared because of common
convention to use similar terminology for any
signal (look at the definitions of kinetic and
electrical signal energy).
• Therefore, the energy of a signal is defined as
a sum of square of magnitude.
• The average power of signal is defined by;

Energy And Power Signal(Cont’d
• A signal can be categorized into energy signal
or power signal:
• An energy signal has a finite energy, 0 < E <
∞. In other words, energy signals have values
only in the limited time duration.

Energy And Power Signal(Cont’d
• For example, a signal having only one square
pulse is energy signal.
• A signal that decays exponentially has finite
energy, so, it is also an energy signal. The
power of an energy signal is 0, because of
dividing finite energy by infinite time (or
length).

• On the contrary, the power signal is not
limited in time.
• It always exists from beginning to end and it
never ends.
• For example, sine wave in infinite length is
power signal. Since the energy of a power
signal is infinite, it has no meaning to us.

• Thus, we use power (energy per given time)
for power signal, because the power of power
signal is finite, 0 < P < ∞.

Analog And Digital Signal
• Digital signals must have a finite set of
possible values. ... That's the big difference
between analog digital waves.
Analog waves are smooth and
continuous, digital waves are stepping,
square, and discrete.

UNIT STEP SIGNAL
• Unit step signal means the signal has unit
amplitude for positive axis and has zero
amplitude for negative axis.
• There are two types of unit step signal as
follows:
• 1) Discrete time unit step signal
• 2) Continuous time unit step signal
• Let us discuss these two types one by one:

UNIT STEP SIGNAL(Cont’d)
• 1) Discrete time unit step signal
A discrete time unit step signal is denoted by
u(n). its value is unity (1) for all positive values
of n. that means its value is one for n = 0.
While for other values of n, its value is zero.
• u(n)= 1 for n = 0
• u(n)= 0 for n < 0
• In the form of sequence it can written as,
• u(n) = {1,1,1,1,….}
• Graphically it can be represented as follows:

unit step u(n)

• 2) Continuous time unit step signal
A continuous type unit step signal is denoted
by u(t). mathematically it can be expressed as,
• u(t)= 1 for t = 0
• u(t) = 0 for t < 0
• it is shown in figure below:

unit step u(t)

Unit ramp signal
• A discrete time unit ramp signal is denoted by
ur(n). its value increases linearly with sample
number n. mathematically it is defined as,
• Ur(n)= n for n = 0
• Ur(n) = 0 for n < 0
• From above equation, it is clear that the value
of signal at a particular interval is equal to the
number of interval at that instant.

Unit ramp signal
• for example; for first interval signal has
amplitude 1, for second it has amplitude 2, for
third it is 3, and so on.
• Graphically it is represented in figure below:

Unit ramp signal
unit ramp signal

Unit ramp signal
• A continuous time ramp type signal is denoted
by r(t). mathematically it is expressed as,
• r(t) = 1 for t = 0
• r(t) = 0 for t < 0
• From above equation, it is clear that the value
of signal at a particular time is equal to the
time at that instant.

Unit ramp signal
• for example; for one second signal has
amplitude 1, for two second it has amplitude
2, for third it is 3, and so on.
• It is shown in figure below:

Delta or unit impulse
function:
• Delta or unit impulse function:
A discrete time unit impulse function is
denoted by d(n). Its amplitude is 1 at n=0 and
for all other values of n; its amplitude is zero.
• d(n)= 1 for n=0

function:
Unit impulse function

function:
• A continuous time delta function is denoted
by d(t). mathematically it is expressed as
follows:
• d(t)=1 for t=0
• d(t)=0 for t?0
• The graphical representation of delta function
for C.T. signal is shown in figure below:

Linear And Non-Linear
System
• A Linear system is said to be linear if
superposition theorem applies to that system.
• Consider the two system as defined as
followsw:
• Y1(t)= f{x1(t)}
• Y2(t)= f{x2(t)}

Linear And Non-Linear System
• Then for the linear system:
f{a1 x1(t)+a2 x2(t)} =a1y1(t) + a2y2(t)
Here a1 and a2 are constants.

Causal And Non-Causal
System
• A system is said to be causal if its output at
any time depends upon present and past
inputs only i:e.
y(to)=f{x(t);t<=to}
• A system is said to be non-Causal if its
output depends upon future inputs also.
such system are not physically realizable.

Time –Invariant and Time –Variant
system
• A system is called time invariant if the time
shifts in the i/p signal results in corresponding
time shift in the output.
• Let Y(t) = f{x(t)} : y(t) response for x(t)
If x(t) is delayed by time t1 then o/p
y(t) will also delayed by the same time i:e

system
• f{x(t-t1)}=y(t-t1)
• The time invariant system do not satisfy the
above relation.The time invariant is also
called fixed system.

Stable and Unstable System
• the stability of control systems is an important
property.
• Considering any bounded input signal of a
system, and if the output signal of the system
to such a signal is also bounded, then the
system is calledbounded-input-bounded-
output stable.
• If the output signal does not show this
property, the system is unstable

Stable and Unstable System
he stability of control systems is an important property. Considering any bounded input signal of a system, and if the output signal of the system to such a signal is also bounded, then the system ishe stability of control systems is an important property. Considering any bounded input signal of a system, and if the output signal of the system to such a signal is also bounded, then the system isa) Stable and (b) unstable system response to a bounded input signal

system
• f{x(t-t1)}=y(t-t1)
The time invariant system do not satisfy
the above relation.The time invariant is
also called fixed system.

Static And Dynamic System
• A system is said to be static if its o/p depends upon present input
only.For example
v(t)=i(t)R ∞
Also v(t)=1/c∫ t i(t) dt ,Thus the voltage
-∞
across the capacitor depends upon present as well as
past values.Such systems are called Dynamic system.

Inversion ,Shifting and Scaling
of System
• Time Shifting is simply shifting the signal in
time. When we add a constant to the time,
we obtain the advanced signal, & when we
decrease the time, we get the delayed signal.

Time Scaling
• Time Scaling is compressing or dilating the signal.
• Time Inversion is simply flipping the signal about
the y-axis.

Analogous System(Linear
(Mechanical Element)
Description Trans Mech
Damper
(a.k.a. Dashpot or Linear Friction)
f= = ±B(V1 ± V2)
Power dissipation in Damper P = fv =f^2.1/B=v^2B
Spring f =±K(x1± x2)
Energy stored in spring E =1/2 K( ∆x)^2 or E=1/2 (1/k) f^2
Mass f = Mdv/dt
or dv/dt= f /M
= , where f is the sum of all forces,
each taken with the appropriate sign.
Energy stored in mass E = ½Mv^2

Force –Voltage Analogy
Mechanical System Electrical System
Force; f Voltage ; V
Velocity ;v Current ; i
Displacement ;x Charge ;q
Mass ;M Inductance ;L
Damping Coefficient ;D Resistance ;R
Compliance; 1/K Capacitance ;C

Force –Current Analogy
Mechanical System Electrical System
Force; f Current; I
Velocity ;v Voltage ;V
Displacement ;x Flux linkage ;ф
Mass ;M Capacitance; C
Damping Coefficient ;D Conductance ; 1/R
Compliance; 1/K Inductance ; L

Modeling of mechanical and electro-
mechanical systems
• There are two types of mechanical systems based on the
type of motion.
• Translational mechanical systems
• Rotational mechanical systems
• Modeling of Translational Mechanical Systems
• Translational mechanical systems move along a straight
line. These systems mainly consist of three basic
elements. Those are mass, spring and dashpot or
damper.

Modeling of mechanical systems
• If a force is applied to a translational mechanical system, then
it is opposed by opposing forces due to mass, elasticity and
friction of the system.
• Since the applied force and the opposing forces are in
opposite directions, the algebraic sum of the forces acting on
the system is zero.
• Let us now see the force opposed by these three elements
individually.

• Mass
Mass is the property of a body, which stores kinetic energy.
If a force is applied on a body having mass M, then it is
opposed by an opposing force due to mass.
This opposing force is proportional to the acceleration of the
body. Assume elasticity and friction are negligible.

• F α a
Fm=Ma=Md2x/dt2
⇒Fm=Ma=Md2x/dt2
F=Fm=Md2x/dt2 Where,
• F is the applied force
• Fm is the opposing force due to mass
• M is mass
• a is acceleration
• x is displacement

• Spring:
Spring is an element, which stores potential energy.
If a force is applied on spring K, then it is opposed by
an opposing force due to elasticity of spring.
This opposing force is proportional to the
displacement of the spring. Assume mass and
friction are negligible.

F∝x
⇒Fk=Kx⇒Fk=Kx
F=Fk=Kx
• Where,
• F is the applied force
• Fk is the opposing force due to elasticity of spring
• K is spring constant

• Dashpot:
If a force is applied on dashpot B, then it is opposed
by an opposing force due to friction of the dashpot.
This opposing force is proportional to the velocity of
the body. Assume mass and elasticity are negligible.

Fb∝ν
⇒Fb=Bν=Bdx/dt
F=Fb=Bdx/dt
• Where,
• Fb is the opposing force due to friction of dashpot
• B is the frictional coefficient
• v is velocity

Modeling of Rotational Mechanical
Systems
• Modeling of Rotational Mechanical Systems
Rotational mechanical systems move about a fixed axis.
These systems mainly consist of three basic elements.
Those are moment of inertia, torsional
spring and dashpot.
• If a torque is applied to a rotational mechanical system, then
it is opposed by opposing torques due to moment of inertia,
elasticity and friction of the system. Since the applied torque
and the opposing torques are in opposite directions, the
algebraic sum of torques acting on the system is zero. Let us
now see the torque opposed by these three elements
individually.

Systems
• Moment of Inertia
In translational mechanical system, mass stores kinetic energy.
Similarly, in rotational mechanical system, moment of inertia
stores kinetic energy.
• If a torque is applied on a body having moment of
inertia J, then it is opposed by an opposing torque due to
the moment of inertia. This opposing torque is
proportional to angular acceleration of the body
• Assume elasticity and friction are negligible.

Systems

Systems
Tk∝θ
⇒Tk=Kθ
T=Tk=Kθ
• Where,
• T is the applied torque
• Tk is the opposing torque due to elasticity of
torsional spring
• K is the torsional spring constant
• θ is angular displacement

Systems
• Dashpot
If a torque is applied on dashpot B, then it isopposed by an
opposing torque due to the rotational friction of the
dashpot.
This opposing torque is proportional to the angular velocity
of the body.
Assume the moment of inertia and elasticity are negligible.

Systems
Tb∝ω
⇒Tb=Bω=Bdθdt
T=Tb=Bdθ/dt
Where,
• Tb is the opposing torque due to the rotational friction of the dashpot.
• B is the rotational friction coefficient.
• ω is the angular velocity.
• θ is the angular displacement

Modeling Of Electromechanical
System
• The purpose of Electro-Mechanical Modeling is to model
and simulate an electro-mechanical system, such that its
physical parameters can be examined before the actual
system is built.
• Parameter estimation and physical realization of the overall
system is the major design objective of Electro-Mechanical
modeling

Modeling Of Electro-mechanical
System
• Theory driven mathematical model can be used or
applied to other system to judge the performance of
the joint system as a whole.
• This is a well known & proven technique for designing
large control system for industrial as well as academic
multi-disciplinary complex system.[2]

Modeling Of Electro-mechanical
System

UNIT-2
(FOURIER Transform Analysis)

Exponential Form Of Fourier
Series
• A continuous time signal x(t) is said to be
periodic if there is a positive non-zero value of
T for which
x(t+T)=X(t) for all t
As we know any periodic signal can be
classified into harmonically related sinusoids
or complex exponential, provided it satisfies
the Dirichlet’s Conditions

Exponential Form Of Fourier Series
• This decomposed representation is called
FOURIER SERIES.
• Two type of Fourier Series representation are
there. Both are equivalent to each other.
• Exponential Fourier Series
• Trigonometric Fourier Series

• Exponential Fourier Series
• A periodic signal is analyzed in terms
of Exponential Fourier Series in the following
three stages:Representation of Periodic
Signal.
• Amplitude and Phase Spectra of a Periodic
Signal.
• Power Content of a Periodic Signal.

∞
x(t)=∑Cke^jkw0t ω0=2 π /T
k=-∞
Where, C is known as the Complex Fourier
Coefficient and is given by,
T
CK=1/T0∫ x(t)e^-jkw0t dt
0

T0 T0 ∞
• ∫ x(t)e^-jkω0t dt =∫ ∑Cke^jkω0t .e^-jkω0t
0 ∞ 0 K=-∞ T0
= ∑Ck[e^j(k-1)ω0t/j(k-1)w0t]
∞ T0 k=-∞ 0
=∑ Ck ∫e^j(k-1)ω0t
K=-∞ 0

∞ T0
=∑ Ck[e^j(k-1)ω0t/j(k-1)w0t]
k=-∞ 0
T0 TO
= ∫ dt=tІ =TO
0 0

T0
=∫ x(t).e^-jkω0t =C1T0
0 T0
Ck=1/T0∫ x(t).e^-jkwot
0
when k=0
TO
Ck=1/T0∫ x(t)
0

• which indicates average value of x(t) over a
period.
When x (t) is real
• Ck=C-K*
• Where, * indicates conjugate.

• Representation of Periodic Signal
A periodic signal in Fourier Series may be
represented in two different time
domains:Continuous Time Domain.
• Discrete Time Domain.
• Continuous Time Domain
• The complex Exponential Fourier
Series representation of a periodic signal x(t) with
fundamental period To is given by

• Discrete Time Domain
• Fourier representation in discrete is very much
similar to Fourier representation of periodic
signal of continuous time domain.
•
The discrete Fourier series representation of a
periodic sequence x[n] with fundamental
period No is given by

N-1
• X[n]= ∑ Cke^jkΩ0n Ω0=2 π /N0
k=0
N-1
Ck=1/N0∑x[n]e^-jkΩ0n
C k=ΙCkΙe^jФk
C-k=Ck*

• ΙC-kІ= ΙCkІ =Фk= Ф-k
• Hence, the amplitude spectrum is an even
function of ω, and the phase spectrum is an
odd function of 0 for a real periodic signal.
• Power Content of a Periodic Signal
• Average Power Content of a Periodic Signal is
given by

• TO
• P=1/TO∫Іx(t)І^2dt
∞ 0
• P=∑ ІCkІ^2
k=-∞
This equation is known as Parseval’s identity
or Parseval’s Theorem.

Trigonometric Form Of Series
∞ ∞
• X(t)=a0+∑a(k)cos k ω0t +∑bk sink ω0t
• k=1 k=1
Where a0=1/T∫x(t)dt
<T>
a(k)=2/T ∫x(t)cosk ω0 tdt
<T>
b(k)=2/T ∫x(t)sink ω0 tdt ω0 =2 π/T
<T>

Compact Trigonometric Form Of
Series
∞
• X(t)=D(0)+ ∑ D(K)cos(kω0t +ф(k)
• k=1
• D(0)=a0=1/T∫x(t)dt
<T>
D(k)=[a(k)^2+b(k)^2]^2
and
Ф(k)=tan^-1{(b(k)}/(a(K))

Fourier Symmetry
• Even and odd functions
When discussing the Fourier series, we have
distinguished between even and odd
functions because odd functions require only
the sine terms and even functions require only
the cosine terms in the series approximation.
• There is no need to calculate all the
Fourier coefficients and find out that half
of them are zero if we know that anyway.

Fourier Symmetry
Therefore, exploiting symmetries can save some
effort (or computing power).
An even function is symmetric with respect to
the y axis, i.e. if you fold the plot over along
they axis, the function maps onto itself. The value
of the function at any negative value is the same
as that at the corresponding positive value: f(-x) =
f(x). Even powers are even functions (hence the
name): x2, x4, x6... and so is cos(x).

Fourier Symmetry
• An odd function is one where the symmetry is
one of inversion at the origin. Any point in the
top right quadrant maps onto one at the
bottom left etc., i.e. f(-x) = -f(x). Odd powers
such as x, x3, x5... and sin(x) are odd functions

Fourier Symmetry
• while, in general, functions are neither even
nor odd, any function can be represented as a
sum of an even and an odd part:
• f(x)=e(x)+o(x), where e(-x)=e(x) and o(-x)=-
o(x).

Fourier Symmetry
• Multiplying even and odd functions is not like
multiplying even and odd numbers
• Because of its symmetry, the integral of the
positive and the negative halves of an even
function are the same (left), therefore .in fig.
• (1)
On the other hand, the integrals of the
positive and negative sides of an odd function
(right) cancel out; i.e. . In fig.(2)

Fourier Symmetry
• Fourier transformation of even and odd
functions
• A general function is a sum of an even and an
odd one:
• f(x)=e(x)+o(x)
• the Fourier transform of f(x) is

Fourier Symmetry
∞
• F(q)=∫f(x)e^-iqx
∞ -∞ ∞ ∞
∫e(x)cos(qx)dx-i∫e(x)sin(qx)dx +∫o(x)cos(qx)
- ∞ ∞
∞ -∞ -∞
-i∫ o(x)sin(x)q
∞

Fourier Symmetry
• ∞ ∞ ∞
∞
• =2∫(e(x)cos(qx)dx-2
0
• ∞
i∫o(x)sin(qx)dx
0

Fourier Symmetry
• Real and imaginary functions
• In general, both the input and the output functions
of the Fourier transformation are complex functions.
If either the imaginary or the real part of the input
function is zero, this will result in a symmetric Fourier
transform just as the even/odd symmetry does.
• Do not confuse the terms imaginary and complex: A
complex function has both real and imaginary parts,
but the real part of an imaginary function is zero:

Fourier Symmetry
• Real function: f(x) = re(x)
Imaginary function: f(x) = i im(x)
Complex function: f(x) = re(x) + i im(x)
• If f(x) is a complex function,f(x) = re(x)
+ i im(x),

Fourier Symmetry
• Do not confuse the
terms imaginary and complex: A complex
function has both real and imaginary parts, but
the real part of an imaginary function is zero:
• Real function: f(x) = re(x)
Imaginary function: f(x) = i im(x)
Complex function: f(x) = re(x) + i im(x)
• If f(x) is a complex function,f(x) = re(x) + i im(x),

Fourier Transform:PROPERTIES
• Linearty:
• Z(t)=Ax(t)+BY(t)
• ↔Z(jw)=AX(jw)+BY(jw)
where F[x(t)]=X(jw) and F[Y(t)]=F(jw)=Z(jw)
Time Shifting:
x(t)↔X(jw)

• Then x(t-t0)↔e^-jwt0X(jw)
• Conjugation and conjugate symmetry
If x(t)↔X(jw)
then x*(t)↔ X*(-jw)
Even {x(t)↔Re{X(jw)}
odd{x(t)}↔jIm{X(jw)}

• Differentiation and Integration:
∞
x(t)=1/2π∫X(jw)e^jwt dw
- ∞

Application To Netwoork
Analysis
• There are mainly two applications of fourier
representation:
• 1)Analysis of the interaction between signals and
• system
• 2)Numerical evaluation of signal properties and
• system behaviour.
• Frequency Response of LTI Systems
• The convolution is given as

Application To Netwoork Analysis
∞
• Y(t)=∫ h(τ)x(t-τ)dτ
- ∞
∞
y(t)=∫h(τ) e^jω(t-τ)dτ
- ∞
∞
=e^jωt∑ h(τ)e^-jωtdτ
-∞
Y(t)=e^jωt H(ω)

Application To Netwoork Analysis
• Again consider the convolution:
• Y(t)=x(t)*h(t)
• Y(ω)=x(ω)H(ω)
• H(ω)=Y(ω)/X(ω)

UNIT-3
(Laplace Transform)

Initial And Final Value Theorem
A right sided signal's initial value
x(0)=Δ limt→0 x(t) and final value x(0)=Δ limt→∞x(t) (if
finite) can be fou∞nd from its Laplace transform by the
following theorems:
Initial value theorem:
x(0)=lims→∞ SX(S)
Final value theorem:
x(∞)=lims→0 SX(S)

Proof For Initial Value Theorem

Application of Laplace Transform and
Analysis Of Networks.
• For circuit below, calculate the initial charging
current of capacitor using Laplace Transform
technique.
• Solve the electric circuit by using Laplace
transformation for final steady-state current

The above figure can be redrawn in Laplace
form,

Now, initial charging current,

The above circuit can be analyzed by
using Kirchhoff Voltage Law and then we get

Final value of steady-state current is

Waveform synthesis and Laplace
Transform of complex waveforms

Solution:
To determine the Laplace transform of a
function given its graph (see the attached
photo) - a square wave - using the theorem
that
p
F(S)=1/(1-e^-ps)∫e ^-st f(t)dt
0

• where f(t)f(t) is a periodic function with
period pp. From the graph and the
information in the theorem, I deduce that the
Laplace transform of the function can be
calculated as follows:
2a
F(S)=1/(1-e^-2as)∫f(t) e^-stdt
o

a
• F(S)=1/(1-e^-2as)∫f(t) e^-stdt
0
because f(t)= 1 f(t)=1 for 0≤t≤a0≤t≤a,
and f(t)=0 f(t)=0 for a≤t≤2aa≤t≤2a.
however;
F(S)=1/s(1+e^-as)

• Obtain the laplace of triangular waveform
shown in fig:
Solution : f1(t)=tu(t)
f2(a)=-(t-1)u(t-1)
f3(t)=-(t-1)u(t-1)
f4(t)=(t-2)u(t-2)
F(t)=tu(t)-(t-1)u(t-1)+(t-2)u(t-2)

UNIT -4
(State Variable Analysis)

State Variable Analysis
Introduction
Before Introducing 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.

• The conventional control theory is completely
based on the frequency domain approach while
the modern control system approach is based on
time domain approach.
• 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.

• 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.
• Let us consider few basic terms related to
state space analysis of modern theory of
control systems.

• 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 ≥ t0.
• 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 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=u1, 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.

• Transpose of output vectors, Where, T is transpose
of the matrix. Transpose of state vectors,
X(t)=[x1(t)x2(t)………….xn(t)]^T
• Where, T is transpose of the matrix.
• These variables are related by a set of equations
which are written below and are known as state
space equations

• X(t)=Ax(t)+Bu(t)+Ew(t)
Y(t)=Cx(t) +Du(t) +Fw(t)
• Representation of State Model using Transfer
Function:
x.=Ax(t)+Bu(t)……………………………………(1)
y.=Cx(t)+Du(t)……………………………………(2)
Taking Laplace Transform of eqn 1 and 2
SX(S)-X(0)=AX(S)+BU(S)

Representation of State Model using
Transfer Function:
• We can write the equation as:
X(S)=[SI-A]^-1BU(S)
Putting this value in eqn 2
y(s)/x(s)=C[SI-A]^-1B
G(S)= C[SI-A]^-1B
Where I is the Identity matrix.

Representation of State Model using
Transfer Function
G(S) = C( Adj[SI-A])/ΙSI-AΙ
Where |sI-A| is also known as characteristic equation
when equated to zero.
Concept of Eigen Values and EigenVectors
The roots of characteristic equation that we have
described above are known as eigen values or eigen
values of matrix A.

Transfer Function Using State
Model
• The General State model of multivariable
system is :
x(t)=Ax(t)+Bu(t)
And y(t)=Cx
Taking Laplace Transform
SX(S)-X(0)=AX(S) +BU(S)
X(S)=[SI-A]^-1

Transfer Function Using State Model
• Putting value in y(s)
T(S)=y(s)/U(S)=C[SI-A]^- 1B

State Transition Matrix
• The State Transition matrix is defined as the
matrix that satisfy the linear homogeneous
state equation:
.
x(t)=Ax(t)
Let ф(t) be the nxn matrix that represents the
STM; then it must satisfy the equation

.
• Ф(t)= Aф(t)
furthermore ,let x(0) denote the initial state at t=0
then ф(t) is also defined by the matrix
equation:
x(t)=ф(t)x(0)…………………………………………(1)
which is the solution of the homogeneous state
equation for t>=0

• Taking laplace transform both side
sX(S)-x(0)=AX(S)
Solving for X(S) we get,
X(S)=[SI-A]^-1x(0)
where it is assumed that matrix [SI-A] is
a non-singular.
Taking the inverse laplace transform

• Taking the inverse laplace transform
x(t)=L^-1{[SI-A]^-1}x(0): t>=0…………(2)
comparing eqn (1) and (2)
ф(t)=L^-1{[SI-A]^-1}
• ф(t) completely defines the transition of
the states from the initial time t=0 to any
time when input are zero.

Solution of Non-Homogeneous state
equation
• Let us now determine the solution of the Non-
homogeneous state equation
x(t)=Ax(t) +Bu(t) ; x(t)Ιt=0
Rewrite this eqn in the form
.
x(t)-Ax(t)=Bu(t)
Multiplying both side by e^-At,we have
.
e^-at[x(t)-Ax(t)]=d/dt[e^-At x(t)]=e^-AtBu(t) ]
Integrating both sides with respect to t between the limits 0 and t, we get
t t
e^-AtΙ =∫e^-AτBu(τ) dτ
0 0
e

equation
• Or t
e^-At x(t)-x(0)=∫e^-AτBu(τ)dτ
0
now premultiplying by both by eÂt ,we
have t
x(t)=eÂtx(0) + ∫eÂ(t-τ)Bu(τ)dτ
0

equation
• If the initial state is known at t=t0 rather than
t=0 then eqn becomes
t
x(t)=e^A(t-t0)x(t0)+∫e^A(t-τ)Bu(τ)dτ
to
In terms of ф(t), t
x(t)=ф(t-t0)x(t0) +∫ Ф(t-τ)Bu(τ)dτ ; x(t)І t=t0
t0 =x(t0)

Solution of Homogeneous Equation
Consider the system given by :
.
x = Ax
with x^n belongs to R
The solution of equation (1) can be written
as :
x(t) = eA^(t-t0)x(t0)
The initial time is given by t0.

Application Of State Variable
Technique to the Analysis of
Linear Sytem

Application Of State Variable
Technique to the Analysis of Linear
Sytem

Basic system analysis
UNIT-5
(Z-Transform)

Concept Of Z-TRANSFORM
• Z-transform. ... In mathematics and signal
processing, theZ-transform converts a discrete-time
signal, which is a sequence of real or complex
numbers, into a complex frequency domain
representation.
• It can be considered as a discrete-time equivalent of
the Laplace transform.

Z-Transform of common functions
• The z-transform of the δ(n) is defined to be :
• δ(n)= 1 for n=0
• 0 for n is not equal to zero.
• ∞
X(z)=∑ δ(n)z^-n=1.z^0=1
-∞

• The z-transform of the u(n) is defined to be
• U(n)= 1 for n>=0
• 0 for n<0
• ∞
X(z)=∑ x(n)z^-n
-∞
∞ ∞
=∑ 1.z^-n =∑(z^-1)^n
-∞ n=0
=1+z^-1+(z^-1)^2+(z^-1)^3+…………………………..

• Then
• X(z)=1/(1-z^-1) ,Ιz^-1Ι
• Find the Z-transform of right hand sided
x(n)=a^nu(n)
∞
∑ a^nu(n)z^-n
-∞
∞
=∑ a^nz^-n= 1+(az^-1)+(az^-1)^2+………….
n=0
X(z)=1/(1-az^-1) ,Ιaz^-1Ι<1

Initial Value Theorem
• Initial Value Theorem:
If the limit exists then
x[0]=lim n→0x[n]=lim z→∞ X(z)
(For a casual signal i.e. x[n]=0 for n<0
Proof:For a causal signal x[n],

Initial Value Theorem
∞
• X(z)=∑x[n].z^-n= x[0]+x[1]/z+x[2]/z^2………..
n=0
As z→∞,z^-n→0 for n>0 where as n=0,z^-n=1
Thus limZ→∞X(z)=X[0]

Final Value Theorem
• X[∞]=limn→∞ x[n]=lim z→1 (1-z^-1)x(z)
For a causal signal and the function X(z) has
all its pole strictly inside the unit circle except
possibly for the first order pole at z=1)
Proof:
Using linearity and Time shifting
property,we have

Final Value Theorem
• X[n]-x[n-1]↔(1-Z^-1)X(z)
If we now let z→1,then
lim n→∞x[n]=lim z→1(1-z^-1

Application to solution of difference
Equation
• Solve the difference equation using Z-
transform Method.
x(n-2)-9x(n-1)+18x(n)=0
Initial conditions are x(-1)=1 ,x(-2)=9
Consider the difference equation,
x(n-2)-9x(n-1)+18x(n)=0
Taking z –transform of above equation

Equation
• Z^-2X(z)+x(-1)z^-1+x(-2)]-9[z^-1X(z)+x(-1)]+
• 18X(z)=0
• [ Z^-2X(z)+z^-1+9]-9[z^-1X(z)+1]18X(z)=0
X(z)= z^-1/(z^-2-9z^-1+18)
=z/1-9z+18z^2
X(z)/z=-1/18(z^2-1/2z+1/18)

Equation
• =1/18(z-1/3)(z-1/6)
• =[(1/3)/z-1/6]-[(1/3)/(z-1/3)]
• X(z)=[(1/3)/1-1/6z^-1]-[(1/3)/(1-1/3z^-1)]
• X(n)=1/3(1/6)^n u(n)-1/3(1/3)^nu(n)
• =[2(1/6)n+1-(1/3)^n+1]u(n)

Pulse Transfer Function
• Pulse transfer function :
Pulse transfer function relates z-transform of the output at
the sampling instants to the Z- transform of the sampled
input. When the same system is subject to a sampled data or
digital signal r∗(t).
• the corresponding block diagram is given in Figure 1

Bsa ppt 48

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