Download Passive Network Synthesis Hurwitz polynomial

Initial Conditions
&
Passive Network Synthesis
Sarvajanik College of
Engineering & Technology

Made by:

Dhruvita Shah
130420117051

Khushbu Shah
130420117052

Salman Mister
140423117003
Raj Patel
140423117010
INITIAL CONDITIONS IN
ELEMENTS:
RESISTOR:
In an ideal resistor , current and voltage
are related by Ohm’s law, v=iRR. On
application of a step voltage across the
resistor , the resulting current iR has the
same shape as the applied voltage but
altered by the scale factor 1/R.
At t=0-
At t=0+

Thus, as the current through a resistor
will change instantaneously if the voltage
changes instantaneously. Similarly, voltage
will change instantaneously if the current
through it changes instantaneously.
Inductor:
The current cannot change instantaneously in a
system of constant inductance. Consequently, closing a
switch to connect an inductor to a source of energy
will not cause current to flow at the initial instant. Thus
inductor will act as if it were an open circuit
independent of the voltage at the terminals of the
inductor.
If a current of value I0 flows in the inductor at the
instant(t=0-) switching takes place, current ( i0 )Will
continue to flow just after switching (t=0+). Thus for
the initial instant, the inductor can be thought of as a
current source of amp. i.e. iL(0-)=il(0+)=I0
Capacitor:
The voltage cannot change instantaneously in
a system of fixed capacitance. If an uncharged
capacitor is connected to an energy source, a
current will flow instantaneously, the capacitor
being equivalent to a short circuit . This follows
because voltage and charge are proportional in a
capacitive system ,Vc= Q/C , so that zero charge
corresponds to zero voltage across the terminal
of the capacitor.
Uncharged capacitor is equivalent to a
short circuit i.e.Vc(0-)=0as q(0)=0.
PROCEDURE FOR EVALUATING
INITIAL CONDITION :

There is no unique procedure that must be
followed for evaluating the initial conditions.
The procedure generally followed consists of
the following two steps in sequence:
1) Solving
for the initial values of the variables
namely voltages and currents t= 0+ and
2) Solving
t= 0+.
for the derivatives of the variables at

(1)Solving for initial values of the variables at t= 0+ :

Initial values of current or voltage may be found directly from a
study of the equivalent network schematic at t= 0+. For each
element in the network, we must determine just what will happen
when the switching action takes place. From this analysis, anew
schematic of an equivalent network for t= 0+ may be constructed
according to the following rules :
i)
Resistors are left in the network without any change.
ii)
Replace all inductors with open circuits or with current generators
having the value of current flowing at t= 0+.
iii)
Replace all capacitors with short circuit or with a voltage source of
the value v(0+)= q(0+)/C if there is an initial charge, q(0+)

(2) Solving for the derivatives at t = 0+ :
The details and order of multiplication will
be different for each different network. A
successful approach will not be obvious at all,
a fact that adds interest and offers a challenge
in the solution of initial-value problem. It can
be best understood from the method
followed in the different networks studied
hereafter.
Example
In the network of the fig. the switch k is closed at t=0 with the capacitor
uncharged and with zero current in the inductor. find the values of I, di/dt and
d²i/dt² at t=0+. If v=100v, L=1H, R=10 ohm, C=10micro farad.
At t=0+
Since there is no initial voltage on the capacitor,
it may be replaced by a sort circuit. Similarly
inductor may be replaced by an open circuit ,
being no initial current. The resulting
equivalent network at t=0+ is shown in fig.
In this particular case , there is no need to
write equation for the network. By inspection.
i(0+)=0 because of open circuit due to L
Ri+L di/dt+1/C
idt  V
………1
Voltage across capacitor is zero at t=0+
So, Ri(0+)+L di(0+)/dt+0=V………..2
Substituting the values,
di/dt (0+) = 100amp/sec……….3
Differentiating eqn 1,
R di/dt + L d²i/dt² + i/c = 0……..4
Writing above eqn at t = 0+
R di(0+)/dt + L d²i(0+)/dt² + i(0+)/c = 0
Substituting values,
d²i/dt² (0+) = -1000 amp/sec²
Passive Network
Synthesis
Hurwitz polynomial

In mathematics, a Hurwitz polynomial,
named after Adolf Hurwitz, is a polynomial
whose coefficients are positive real numbers
and whose roots (zeros) are located in the left
half-plane of the complex plane or on the jω
axis, that is, the real part of every root is zero
or negative. The term is sometimes restricted
to polynomials whose roots have real parts
that are strictly negative, excluding the axis
(i.e., a Hurwitz stable polynomial)

A polynomial function P(s) of a complex variables is said
to be Hurwitz if the following conditions are satisfied:
1. P(s) is real when s is real.
2. The roots of P(s) have real parts which are zero or
negative.
Hurwitz polynomials are important in control
systems theory because they represent the characteristic
equations of stable linear systems. Whether a polynomial
is Hurwitz can be determined by solving the equation to
find the roots, or from the coefficients without solving
the equation by the Routh-Hurwitz stability criterion.
Properties of Hurwitz Polynomial

Hurwitz polynomial P(s) is given as
n
n

1
a
s

a
s

....

a
s

a
s
P(s) = n
n

1
1
0
1. All the coefficients of the polynomial must be
real and positive.
2. The roots of odd and even parts of the
polynomial P(s) lie on the imaginary axis.
3. If P(s) is either even or odd, all its roots are
on the imaginary axis.
4. The continued fraction expansion of the ratio
of odd to even parts or even to add parts of
P(s) yields all positive quotients.
5. If the polynomial P(s) is completely even or
completely odd, then differentiate P(s) is with
respect to s and obtain the polynomial P’(s) =
(d P(s)/ds). The continued fraction expansion
of P(s)/P’(s) gives all positive quotients for a
polynomial P(s) to be Hurwitz.
Example:
A simple example of a Hurwitz
polynomial is the following:
x^2 + 2x + 1.
The only real solution is −1, as it factors
to
(x+1)^2.
All the coefficients of the polynomial
are real & positive.
As the real part of the root is negative,
it is a hurwitz polynomial.
Positive Real Function

Definition

The term positive-real function was originally
defined by Otto Brune to describe any
function Z(s) which is rational (the quotient of
two polynomials), is real when s is real has
positive real part when s has a positive real
part

Many authors strictly adhere to this definition by
explicitly requiring rationality, or by restricting
attention to rational functions, at least in the first
instance. However, a similar more general condition,
not restricted to rational functions had earlier been
considered by Cauer, and some authors ascribe the
term positive-real to this type of condition, while
other consider it to be a generalization of the basic
definition.
Properties of Positive Real
Functions

The sum of two PR functions is PR.

The composition of two PR functions is PR. In
particular, if Z(s) is PR, then so are 1/Z(s) and Z(1/s).

All the poles and zeros of a PR function are in the
left half plane or on its boundary the imaginary axis.

Any poles and zeroes on the imaginary axis are
simple (have a multiplicity of one).

Any poles on the imaginary axis have real strictly
positive residues, and similarly at any zeroes on the
imaginary axis, the function has a real strictly positive

Over the right half plane, the minimum value of the
real part of a PR function occurs on the imaginary
axis (because the real part of an analytic function
constitutes a harmonic function over the plane, and
therefore satisfies the maximum principle).

For a rational PR function, the number of poles and
number of zeroes differ by at most one.