Lecture Notes Week11

EEE2102
Circuit Theory II
Week 11
Laplace Transform
Asst. Prof. Fatih Küçüktezcan
Applying the Laplace Transform
Let’s use the Laplace transform to solve the ordinary integrodifferential equations
that describe the behavior of lumped parameter circuits.
equation that involves both
integrals and derivatives
Assumption: No initial energy is stored in the

circuit at the instant when switch is opened.
Let’s apply KCL to the circuit to find the time-domain expression for v(t)
The initial voltage on the capacitor v(0-)=0
The next step is to find the inverse

transform of the s-domain expression
Inverse Transforms
• The expression for V(s) is a rational function of s.

• It can be expressed as ratio of two polynomials in s.
• If we can inverse-transform rational functions of s, we can solve for the time-domain expressions
for the voltages and currents.
In general, we need to find the inverse transform of a function that has the form
• The coefficients a and b are real constants.

• The exponents m and n are positive integers.
• If m>n, the ratio is called proper rational function.

• If m<n, the ratio is called improper rational function.
• Only a proper rational function can be expanded as a sum of partial fractions.

Inverse Transforms
Partial Fraction Expansion: Proper Rational Functions
Target: Writing a term or a series of terms for each root of D(s)
D(s) must be in factored form before we can make a partial fraction expansion.
For the given example, D(s) has four roots.
• Two of these roots are distinct: s=0 and s=-3 and
• A multiple root of multiplicity 2 occurs at s=-1 and
After finding K coefficients, V(t) will be expressed as
Let’s find the coefficients

(K1, K2, K3 and K4).
Inverse Transforms
The roots of D(s)
(1) Real and distinct (2)Complex and distinct (3)Real and repeated (3)Complex and repeated
Distinct Real Roots of D(s)

Multiply both sides of the identity by a factor equal to the denominator beneath the desired K.
When we evaluate both sides of the identity at the root corresponding to the multiplying factor,
• The right-hand side is always the desired K
• The left-hand side is always its numerical value.
Example:
To find K1, multiply both sides by s and then evaluate both sides at s=0
Inverse Transforms
Example:
To find K2, multiply both sides by s+8 and then evaluate both sides at s=-8
To find K3, multiply both sides by s+6 and then evaluate both sides at s=-6
Then,
Let’s check if the result is correct

Inverse Transforms
For s=-5 and s=-12, the left side of the equation is zero.
Let’s repeat this procedure for the right side of the equation.
Thus, the result is correct. Finally,

Inverse Transforms
Inverse Transforms
Complex and Distinct of D(s)
The procedure applied to find K coefficient in the case of Distinct Real Roots is same.
The only difference is dealing with the complex numbers.
Example:
Let’s find the roots of the quadratic term

To find K1, K2 and K3, we use the same process as before complex roots always appear in conjugate pairs
= (-3+j4) and (-3-j4)
Then,
Inverse Transforms
Let’s check if the result is correct
For s=-3, the left side of the equation is zero, and the right side of the equation is
Thus, the result is correct.

Inverse Transforms
Let’s simplify this result.
Finally,
Generally, when D(s) contains
appears in the partial fraction expansion.

Inverse Transforms
Inverse Transforms
Repeated Real Roots of D(s)
To find the coefficients associated with the terms generated by a multiple root of multiplicity r,
we multiply both sides of the identity by the multiple root raised to its rth power.
We find the K appearing over the factor raised to the rth power by evaluating both sides of
the identity at the multiple root.
To find the remaining (r-1) coefficients, we differentiate both sides of the identity (r-1) times.
Finally, we evaluate both sides of the identity at the multiple root. The right-hand side is always
the desired K, and the left-hand side is always its numerical value.
Example:
K1 is found as previously described:
To find K2 we multiply both sides by and then evaluate both sides at s=-5
Inverse Transforms
To find K3,
• multiply both sides by
• differentiate both sides once with respect to s
• then evaluate both sides at s=-5
To find K4,
• multiply both sides by
• differentiate both sides twice with respect to s
• then evaluate both sides at s=-5
Inverse Transforms
After finding the coefficients, the partial expansion can be represented as,
When the result is checked at s=-25, both sides of the equation are zero.
Finally,
Inverse Transforms
Repeated Complex Roots of D(s)
We handle repeated complex roots in the same way that we did repeated real roots; the only
difference is that the algebra involves complex numbers.
Example:
Then,
Then,
Inverse Transforms
Repeated Complex Roots of D(s)
Finally, inverse transform of F(s) is

Inverse Transforms
Inverse Transforms
Improper Rational Functions
An improper rational function can be expanded into a polynomial plus a proper rational function.
Polynomial is then inverse-transformed into impulse functions and derivatives of impulse functions.
The proper rational function is inverse-transformed by the mentioned techniques.
N
Example:
D
Divide the denominator into the numerator until the remainder is a proper rational function
Finally,
Poles and Zeros of F(s)
This rational function may be expressed as the ratio of two factored polynomials.
Thus,
Roots of numerator = Zeros of F(s)
Roots of denominator = Poles of F(s)
Example:
The zeros are -5 and -10.

The poles are -1, -2, -3 and -4.
Poles and Zeros of F(s)
Example:
The zeros are -5, -3+j4, and -3-j4.

The poles are 0, -10, -6+j8, and -6-j8.

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EEE2102

Assumption: No initial energy is stored in the

The initial voltage on the capacitor v(0-)=0

The next step is to find the inverse

• The expression for V(s) is a rational function of s.

• The coefficients a and b are real constants.

• If m>n, the ratio is called proper rational function.

• Only a proper rational function can be expanded as a sum of partial fractions.

Target: Writing a term or a series of terms for each root of D(s)

For the given example, D(s) has four roots.

• Two of these roots are distinct: s=0 and s=-3 and

• A multiple root of multiplicity 2 occurs at s=-1 and

After finding K coefficients, V(t) will be expressed as

Let’s find the coefficients

Distinct Real Roots of D(s)

Let’s check if the result is correct

Thus, the result is correct. Finally,

Let’s find the roots of the quadratic term

= (-3+j4) and (-3-j4)

Let’s check if the result is correct

Thus, the result is correct.

Let’s simplify this result.

Generally, when D(s) contains

appears in the partial fraction expansion.

K1 is found as previously described:

Finally, inverse transform of F(s) is

The proper rational function is inverse-transformed by the mentioned techniques.

Roots of denominator = Poles of F(s)

The zeros are -5 and -10.

The zeros are -5, -3+j4, and -3-j4.

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