INTRODUCTION TO LAPLACE TRANSFORM Advanced Circuit Analysis Technique

TOPIC COVERAGE WEEKTOPICSLABS Week 6Introduction to the Laplace Transform: Waveform Analysis Lab 3: Intro to Laplace Transform Week 7Laplace Transform in Circuit Analysis: Transfer Function. Lab 4: Laplace Transform Analysis Week 8Frequency Response: Bode Plot Week 9Frequency Selective Circuits: Passive Filters Lab 5: RL and RC Filter Lab 6: Bandpass and Bandreject Filter

TOPIC 1: LAPLACE TRANSFORM

AREA OF COVERAGE WEEKTOPICSDURATION Week 6 Intro LT: Definition, Excitation function, Functional transform, Operational transform. 2 hrs Intro LT: Operational transform, Partial fraction. 1 hr Week 7 Circuit Analysis: S-domain circuit component, Transfer function. 3 hrs

LEARNING OUTCOMES Be able to calculate the Laplace transform of a function using the definition and/or Laplace transform table. Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace table. Be able to perform circuit analysis in the s-domain. Know how to use a circuit’s transfer function to calculate the circuit’s impulse, unit, and step response.

OVERVIEW Analyze a “linear circuit” problem, in the frequency domain instead of in the time domain. Convert a set of differential equations into a corresponding set of algebraic equations, which are much easier to solve. Analyze the bandwidth, phase, and transfer characteristics important for circuit analysis and design.

OVERVIEW Analyze both the steady-state and “transient” responses of a linear circuit. Analyze the response under any types of excitation (e.g. switching on and off at any given time(s), sinusoidal, impulse, square wave excitations, etc.)

DEFINITION OF LAPLACE TRANSFORM s: complex frequency Called “The One-sided or unilateral Laplace Transform”. In the two-sided or bilateral LT, the lower limit is - . We do not use this.

DISCONTUNITY FUNCTION When f(t) has a finite discontinuity at the origin, the Laplace transform formula is rewritten as:

STEP FUNCTION The symbol for the step function is Ku(t). Mathematical definition of the step function:

f(t) = K u(t)

STEP FUNCTION A discontinuity of the step function may occur at some time other than t=0. A step that occurs at t=a is expressed as:

f(t) = K u(t-a)

Ex:

Three linear functions at t=0, t=1, t=3, and t=4

Expression of step functions Linear function +2t: on at t=0, off at t=1 Linear function -2t+4: on at t=1, off at t=3 Linear function +2t-8: on at t=3, off at t=4 Step function can be used to turn on and turn off these functions

Step functions

IMPULSE FUNCTION The symbol for the impulse function is (t). Mathematical definition of the impulse function:

f(t) = K (t)

IMPULSE FUNCTION The area under the impulse function is constant and represents the strength of the impulse. The impulse is zero everywhere except at t=0. An impulse that occurs at t=a is denoted K  (t-a)

FUNCTIONAL TRANSFORM

TYPEf(t) (t>0 - )F(s) Impulse Step Ramp Exponential Sine Cosine

TYPEf(t) (t>0 - )F(s) Damped ramp Damped sine Damped cosine

OPERATIONAL TRANSFORM

OPERATIONAL TRANSFORMS Indicate how mathematical operations performed on either f(t) or F(s) are converted into the opposite domain. The operations of primary interest are:  Multiplying by a constant  Addition/subtraction  Differentiation  Integration  Translation in the time domain  Translation in the frequency domain  Scale changing

OPERATIONf(t)F(s) Multiplication by a constant Addition/Subtr action First derivative (time) Second derivative (time)

OPERATIONf(t)F(s) n th derivative (time) Time integral Translation in time Translation in frequency

OPERATIONf(t)F(s) Scale changing First derivative (s) n th derivative s integral

Translation in time domain If we start with any function: we can represent the same function translated in time by the constant a, as: In frequency domain:

Ex:

Translation in frequency domain Translation in the frequency domain is defined as:

Ex:

Scale changing The relationship between f(t) and F(s) when the time variable is multiplied by a constant:

Ex:

APPLICATION

Problem Assumed no initial energy is stored in the circuit at the instant when the switch is opened. Find the time domain expression for v(t) when t≥0.

Integrodifferential Equation A single node voltage equation:

s-domain transformation =0

INVERSE LAPLACE TRANSFORM (L -1 ) PARTIAL FRACTION EXPANSION (PFE)

Rational function of s: expressed in a form of a ratio of two polynomials. Called proper rational function if m≥n Inverse transform rational functions of F(s), can solve for v(t) or i(t). RATIONAL FUNCTIONS

PARTIAL FRACTION EXPANSION 1) Distinct Real Roots of D(s) s 1 = 0, s 2 = -8 s 3 = -6

1) Distinct Real Roots To find K 1 : multiply both sides by s and evaluates both sides at s=0 To find K 2 : multiply both sides by s+8 and evaluates both sides at s=-8 To find K 3 : multiply both sides by s+6 and evaluates both sides at s=-6

Find K 1

Find K 2

Find K 3

Inverse Laplace of F(s)

2) Distinct Complex Roots S 1 = -6 S 2 = -3+j4S 3 = -3-j4

Partial Fraction Expansion Complex roots appears in conjugate pairs. Complex roots appears in conjugate pairs.

Find K 1

Find K 2 and K 2 * Coefficients associated with conjugate pairs are themselves conjugates.

Inverse Laplace of F(s)

Undesirable imaginary components in the time domain

Inverse Laplace of F(s)

3) Repeated Real Roots

Partial Fraction Expansion

Find K 3 First, multiply both sides by (s+5) 3. Next, differentiate both sides once with respect to s and evaluate at s=-5.

K3K3K3K3

K3K3K3K3

Find K 4 First, multiply both sides by (s+5) 3. Next, differentiate both sides twice with respect to s and evaluate at s=-5.

K4K4K4K4 Simplifying the first derivative, the second derivative becomes:

F(s) and f(t)

4) REPEATED COMPLEX ROOTS

Partial Fraction Expansion

Find K 1 and K 2

F(s) and f(t)

USEFUL TRANSFORM PAIRS

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