circuit and system analysis ehb 232e · the convolution theorem the convolution operation: let f...

Circuit and System AnalysisEHB 232E

Prof. Dr. Mustak E. Yalcın

Istanbul Technical UniversityFaculty of Electrical and Electronic Engineering

mustak.yalcin@itu.edu.tr

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 1 / 30

Outline I

1 Laplace Transform in Circuit AnalysisLaplace TransformInverse Laplace Transform

Analysis of state space equationCharacteristic polynomialStability and the Routh-Hurwitz CriterionRouth-Hurwitz Criterion

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 2 / 30

Why Laplace Transform

Laplace transform (F (s)) of f (t) function is given by

F (s) = L{f (t)} =

∫ ∞0−

e−st f (t)dt.

The Laplace transform converts linear differential equations into algebraicequations. These are linear equations with polynomial coefficients. Thesolution of these linear equations therefore leads to rational functionexpressions for the variables involved.

Initial Conditions, Generalized Functions, and the Laplace Transform, by Kent H.

Lundberg; Haynes R. Miller ; David L. Trumper, IEEE Control Systems Link

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 3 / 30

Laplace Transform

Signal Laplace Transform

δ(t) 1u(t) 1

stu(t) 1

s2

tn

n!u(t) 1sn+1

e−atu(t) 1s+a

cos(at) ss2+a

sin(at) as2+a

2|k |e−αt cos(βt + θ) k(s+α−jβ) + k

(s+α+jβ)

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 4 / 30

Properties of the Laplace Transform

L{k1f1(t) + k2f2(t)} = k1F1(s) + k2F2(s)

L{df (t)

dt

}= sF (s)− f (0−)

and

L{dnf (t)

dtn

}= snF (s)− sn−1f (0)− sn−2f (0)− ...− f (0)

L{∫ t

0f (τ)dτ

}=

F (s)

s

L{f (t − t0)u(t − t0)} = F (s)e−t0s

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 5 / 30

L{

df (t)dt

}= sF (s)− f (0−)

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 6 / 30

Properties of the Laplace Transform

L{e−at f (t)

}= F (s + a)

L{tnf (t)} = (−1)ndnF (s)

dsn

Inverse Laplace Transform

f (t) = L−1 {F (s)} =

∫ c+j∞

c−j∞estF (s)ds

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 7 / 30

Partial-Fraction Expansion Method

s-transform of a linear time invariant system is often of the form (n > m)

F (s) =P(s)

Q(s)=

amsm + am−1s

m−1 + ...+ a1s + a0

bnsn + bn−1sn−1 + ...+ b1s + b0

which is ratio of two polynomials. The value(s) for s where P(s) = 0 arecalled zeros. The value(s) for s where Q(s) = 0 are called poles.

If spi 6= spj , poles distinct.

if lims→∞ F (s)(s − spi ) =∞ and lims→∞ F (s)(s − spi )k is constant then

s = spi is a k-multiple pole.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 8 / 30

Lets assume that poles distinct

L−1 {F (s)} = L−1

{P(s)

Q(s)

}= L−1

{k1

(s − sp1)+

k2

(s − sp2). . .+

kn(s − spn)

}ki is the residue located at the corresponding pole spi which is

ki = F (s) (s − spi )|s=spi

L−1 {F (s)} = k1esp1tu(t) + k2e

sp2tu(t) + . . .+ knespntu(t)

L−1 {k0 + F (s)} = k0δ(t) + k1esp1tu(t) + k2e

sp2tu(t) + . . .+ knespntu(t)

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 9 / 30

Y (s) =−3s2 + 23s − 38

(s − 1)(s − 2)(s − 3)=−9

s − 1+

2

s − 2+

2

s − 3

y(t) = −9et + 4e2t + 2e3t for t > 0

Y (s) =s2 + 2s + 3

(s2 + 2s + 2)(s2 + 2s + 5)=

1/6j

s + 1− j− 1/6j

s + 1 + j+

1/6j

s + 1− 2j− 1/6j

s + 1 + 2j

y(t) =1

3e−t sin(t) +

1

3e−t sin(2t) for t > 0

x = 2x − 3y , x(0) = 8

y = −2x + y , y(0) = 3

Using Laplace transform

sX − 8 = 2X − 3Y → (s − 2)X + 3Y = 8

sY − 3 = −2x + y → 2X + (s − 1)Y = 3

X =8s − 17

s2 − 3s − 4=

5

s + 1+

3

s − 4−→ x(t) = 5e−t + 3e4t

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 10 / 30

Q(s) has a multiple pole.

L−1

{P(s)

Q(s)

}= L−1

{ki1

(s − spi )+

ki2(s − spi )2

. . .+kik

(s − spi )k+

P1(s)

Q1(s)

}where

kik = F (s) (s − spi )k∣∣∣s=spi

and

kil =1

(k − l)!

dk−lF (s)(s − spi )k

dsk−l

∣∣∣∣s=spi

for l = 1, 2, . . . k − 1.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 11 / 30

Y (s) =2s2 − 25s − 33

(s + 1)2(s − 5)=

k1

s + 1+

k2

(s + 1)2+

k

s − 5

k =2s2 − 25s − 33

(s + 1)2|s=5 = −3

k1 =1

1!

d

ds

2s2 − 25s − 33

(s − 5)|s=−1 = 5

k2 =2s2 − 25s − 33

(s − 5)|s=−1 = 1

y(t) = 5e−t + te−t − 3e5t

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 12 / 30

s-plane

F (s) =(s + 3)(s2 + 2s + 2)

s2(s − 4)(s + 2)2

zeros ”o” and poles ”x” on s-plane:

Re{s}

Img{s}

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 13 / 30

The Convolution Theorem

The convolution operation: Let f1(t) and f2(t) be functions defined on[0,∞) , and let us take them to be equal to zero for t < 0: Theconvolution of the time functions f1 and f2 is a new time function denotedby (f1 ? f2)(t) and defined for all t by

f (t) = f1(t) ? f2(t) =

∫ t

0f1(τ)f2(t − τ)dτ =

∫ t

0f2(t − τ)f2(τ)dτ

The Convolution Theorem

Let f1(t) and f2(t) have F1(s) and F2(s) as Laplace transforms. Weassume that for i = 1, 2, f (t) = 0 for t < 0. Laplace transform of theconvolution of f1 and f2 is given by

L{f1(t) ? f2(t)} = F1(s)F2(s)

Thus, the operation of convolution in the time domain is equivalent tomultiplication in the frequency domain.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 14 / 30

Example

Using The Convolution Theorem, lets find inverse Laplace transform ofF (s) = 1

s2(s+2)

1

s2(s + 2)=

1

s2

1

(s + 2)= L

{tu(t) ? e−2tu(t)

}tu(t) ? e−2tu(t) =

∫ t

0τe−2(t−τ)dτ

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 15 / 30

Analysis of state space equation

Linear time invariant system

x = Ax + Bey = Cx + De

where x state variable and y is output, u is input vectors. Using Laplacetransform

sX (s)− x(0) = AX (s) + BE (s)Y (s) = CX (s) + DE (s)

we have Laplace transform state variable

X (s) = (sI − A)−1x(0) + (sI − A)−1BE (s)

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 16 / 30

Output

Y (s) = C (sI − A)−1x(0) + (C (sI − A)−1B + D)E (s)

X (s) = (sI − A)−1x(0) + (sI − A)−1BE (s) = Q(s)x(0) + Q(s)BE (s)

where Q(s) = (sI − A)−1 and q(t) = L−1{(sI − A)−1}.Using The Convolution Theorem, we have

Q(s)x(0) + Q(s)BE (s) = L{q(t)x(0) +

∫ t

0q(t − τ)Be(τ)dτ

}the state variable in time domain

x(t) = q(t)x(0) +

∫ t

0q(t − τ)Be(τ)dτ

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 17 / 30

we know that

x(t) = Φ(t)x(0)︸ ︷︷ ︸zero-input response

+

∫ t

0Φ(t − τ)Be(τ)dτ︸ ︷︷ ︸zero-state response

In this caseΦ(s) = (sI − A)−1

andΦ(t) = L{(sI − A)−1}

X (s) = Φ(s)x(0) + Φ(s)BE (s)

Y (s) = CΦ(s)x(0) + (CΦ(s)B + D)E (s)

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 18 / 30

Network Functions

Network Function =L{zero − state response}

L{input}Y (s)

E (s)= (CΦ(s)B + D) = H(s)

Transfer (Network) functions : voltage transfer functions, transferadmittances, current transfer function, transfer impedance.

y(t) = L−1{H(s)E (s)} = h(t) ? e(t) =

∫ t

0h(t − τ)e(τ)dτ

if we chose e(t) = δ(t)

y(t) = L−1{H(s) 1} =

∫ t

0h(t − τ)δ(τ)dτ = h(t)

NETWORK FUNCTION = L{IMPULS RESPONSE}Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 19 / 30

H(s)

h(t)

E(s)

e(t)

H(s)E(s)

h(t)*e(t)

impuls response = h(t) = L−1{H(s)}step response = h(t)u(t) = L−1{H(s) 1

s }! zero-input response = 0 !

what happen if zero-input response is not zero (or system is not stable)?

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 20 / 30

Characteristic polynomial

The characteristic polynomial of A is defined by

p(s) = det{sI − A} = 0

The solutions of the characteristic equation are precisely the eigenvalues ofthe matrix A. The roots of the characteristic equation is spi = σi + jwi

if σi > 0 unstable

if σi < 0 stable

σi = 0 check the repeated eigenvalue !

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 21 / 30

Example : The transfer function for a linear time-invariant circuit isH(s) = 1

s+1 . If E (t) = 3 cos t what is the steady-state expression of theoutput?

Y (s) =3s

s2 + 1

1

s + 1and in time domain

y(t) = L−1

{3

2

s + 1

s2 + 1− 3

2

1

s + 1

}=

3

2(cos t + sin t − e−t)

Using convolution

h(t) = L{ 1

s + 1} = e−t

y(t) =

∫ t

0e−t+τ3 cos(τ)dτ

= 3e−t[eτ

2(cos(τ) + sin(τ))

∣∣∣∣t0

]=

3

2(cos t + sin t − e−t)

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 22 / 30

We would like to determine the stability of a transfer function such as

e y

The stability of

H(s) =P(s)

Q(s)

is determined by the roots of Q(s) = 0.

to find the roots of the nth-order polynomials !

There is an effective test for determining stability that does not require anexplicit solution of the algebraic equation.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 23 / 30

Routh-Hurwitz Criterion

LetQ(s) = ans

n + an−1sn−1 + an−2s

n−2 + ...a1s + a0

be a polynomial with real coefficients, with an 6= 0. Starting with theleading coefficient an, fill two rows of a table, as follows:

sn an an−2 an−4 ...sn−1 an−1 an−3 an−5 ...sn−2 b1 b3 b5 ...sn−3 c1 c3 c5 ...

. . . . ...

. . . . ...s1 . . . ...s0 . . . ...

Thus, the first row contains all coefficients with even subscript and thesecond those with odd subscripts.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 24 / 30

Now a Routh-Hurwitz Table (RHT) with (n + 1) rows is formed by analgorithmic process, described below:

b1 =an−1an−2 − anan−3

an−1

b3 =an−1an−4 − anan−5

an−1

c1 =b1an−3 − b3an−1

b1

c3 =b1an−5 − b5an−1

b1

The polynomial Q(s) is (Hurwitz) stable if and only if all the elements ofthe first column of its Routh-Hurwitz Table are non-zero and of the samesign.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 25 / 30

Example

M(s) = s5 + 4s4 + 2s3 + 5s2 + 3s + 6

s5 1 2 3s4 4 5 6s3 0.75 1.5s2 −3 6s1 3s0 6

we have −3 in first column. This polynomial is not Hurwitz! (The numberof unstable roots of p(s) is equal to the number of changes of sign)What is the stability condition of the Quadratic Polynomial ? (is stable ifand only if all its coefficients are of the same sign)What is the stability condition of the cubic polynomial ?

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 26 / 30

Example

Singular Cases : m(s) = s3 − 3s + 2

s3 1 −3s2 0(ε) 2s1 (−3− 2)/εs0 2

We have a zero in the first column. We have two sign changes, confirmingthat the polynomial has two unstable roots. (m(s) = (s − 1)2(s + 2)) .

Replace the zero by a small ε (of arbitrary sign) and continue filling in theentries of the RHT.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 27 / 30

Example

Pure imaginary pair of roots: m(s) = s3 + 2s2 + s + 2

s3 1 1s2 2 2s1 0(ε)s0 2

we have a zero in the first column. Replace the zero by a small ε (ofarbitrary sign) and continue filling in the entries of the RHT.(m(s) = (s2 + 1)(s + 2)).

if Q(s) has a simple pair of imaginary roots at and no other imaginaryroots, then this can be detected in the RHT, in that it will have the wholeof the s1 row equal to zero and only two non-zero elements in the s2 row,of the same sign.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 28 / 30

Example

m(s) = s5 + 2s4 + 24s3 + 48s2 − 25s − 50

s5 1 24 −25s4 2 48 −50s3 0(8) 0(96) d(2s4 + 48s2 − 50)/ds = 8s3 + 96s2 24 −50s1 112.7 0s0 −50

the polynomial has unstable root.(s = 1)

All the elements of a particular row are zero. The zero row is replaced bytaking the coefficients of the derivative of the the auxiliary polynomialwhich is obtained from the values in the row above the zero row.

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 29 / 30

m(s) = s5 + 2s3 + s

s5 1 2 1s4 5 6 1 d(s5 + 2s3 + s)/ds = 5s4 + 6s2 + 1s3 4/5 4/5s2 1 1s1 0(2) d(s2 + 1)/ds = 2ss0 1

Two times all the elements of a particular row are zero. Unstable!(m(s) = s(s2 + 1)2).

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 30 / 30