APPLICATION OF THE LAPLACE TRANSFORM

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Presentation transcript:

APPLICATION OF THE LAPLACE TRANSFORM TO CIRCUIT ANALYSIS LEARNING GOALS Laplace circuit solutions Showing the usefulness of the Laplace transform Circuit Element Models Transforming circuits into the Laplace domain Analysis Techniques All standard analysis techniques, KVL, KCL, node, loop analysis, Thevenin’s theorem are applicable Transfer Function The concept is revisited and given a formal meaning Pole-Zero Plots/Bode Plots Establishing the connection between them Steady State Response AC analysis revisited

LAPLACE CIRCUIT SOLUTIONS We compare a conventional approach to solve differential equations with a technique using the Laplace transform Comple mentary “Take Laplace transform” of the equation Initial conditions are automatically included P a r t i c u l Only algebra is needed No need to search for particular or comple- mentary solutions

LEARNING BY DOING In the Laplace domain the differential equation is now an algebraic equation Use partial fractions to determine inverse Initial condition given in implicit form

CIRCUIT ELEMENT MODELS The method used so far follows the steps: 1. Write the differential equation model 2. Use Laplace transform to convert the model to an algebraic form For a more efficient approach: 1. Develop s-domain models for circuit elements 2. Draw the “Laplace equivalent circuit” keeping the interconnections and replacing the elements by their s-domain models 3. Analyze the Laplace equivalent circuit. All usual circuit tools are applicable and all equations are algebraic. Resistor

Capacitor: Model 1 Source transformation Impedance in series with voltage source Capacitor: Model 2 Impedance in parallel with current source

Inductor Models

LEARNING BY DOING Determine the model in the s-domain and the expression for the voltage across the inductor Steady state for t<0 Inductor with initial current Equivalent circuit in s-domain

ANALYSIS TECHNIQUES All the analysis techniques are applicable in the s-domain LEARNING EXAMPLE Draw the s-domain equivalent and find the voltage in both s-domain and time domain One needs to determine the initial voltage across the capacitor

LEARNING EXAMPLE Assume all initial conditions are zero Could have used voltage divider here Node Analysis

Loop Analysis

Applying current source Source Superposition Current divider Applying voltage source Voltage divider

Combine the sources and use current divider Source Transformation The resistance is redundant

Using Thevenin’s Theorem Reduce this part Voltage divider Only independent sources

Using Norton’s Theorem Reduce this part Current division

LEARNING EXAMPLE Selecting the analysis technique: . Three loops, three non-reference nodes . One voltage source between non-reference nodes - supernode . One current source. One loop current known or supermesh . If v_2 is known, v_o can be obtained with a voltage divider Transforming the circuit to s-domain

Continued ... -keep dependent source and controlling variable in the same sub-circuit -Make sub-circuit to be reduced as simple as possible -Try to leave a simple voltage divider after reduction to Thevenin equivalent

Continued … Computing the inverse Laplace transform Analysis in the s-domain has established that the Laplace transform of the output voltage is One can also use quadratic factors...

LEARNING EXTENSION Assume zero initial conditions Implicit circuit transformation to s-domain supernode KCL at supernode

LEARNING EXTENSION supermesh Determine inverse transform

TRANSIENT CIRCUIT ANALYSIS USING LAPLACE TRANSFORM For the study of transients, especially transients due to switching, it is important to determine initial conditions. For this determination, one relies on the properties: 1. Voltage across capacitors cannot change discontinuously 2. Current through inductors cannot change discontinuously LEARNING EXAMPLE Assume steady state for t<0 and determine voltage across capacitors and currents through inductors Circuit for t>0

Circuit for t>0 Now determine the inverse transform Laplace Use mesh analysis

LEARNING EXTENSION Initial current through inductor Current divider

LEARNING EXTENSION Determine initial current through inductor Use source superposition

TRANSFER FUNCTION H(s) can also be interpreted as the Laplace System with all initial conditions set to zero H(s) can also be interpreted as the Laplace transform of the output when the input is an impulse and all initial conditions are zero The inverse transform of H(s) is also called the impulse response of the system If the impulse response is known then one can determine the response of the system to ANY other input

LEARNING EXAMPLE In the Laplace domain, Y(s)=H(s)X(s)

Impulse response of first and second order systems First order system Normalized second order system

LEARNING EXAMPLE Transform the circuit to the Laplace domain. All initial conditions set to zero Mesh analysis

LEARNING EXTENSION Determine the pole-zero plot, the type of damping and the unit step response x O

Second order networks: variation of poles with damping ratio LEARNING EXAMPLE Normalized second order system

POLE-ZERO PLOT/BODE PLOT CONNECTION Bode plots display magnitude and phase information of They show a cross section of G(s) If the poles get closer to imaginary axis the peaks and valleys are more pronounced Cross section shown by Bode

Due to symmetry show only positive frequencies Cross section Front view Amplitude Bode plot Uses log scales

STEADY STATE RESPONSE Response when all initial conditions are zero Laplace uses positive time functions. Even for sinusoids the response contains transitory terms EXAMPLE If interested in the steady state response only, then don’t determine residues associated with transient terms Steady state response transient For the general case

LEARNING EXAMPLE Determine the steady state response Transform the circuit to the Laplace domain. Assume all initial conditions are zero

LEARNING EXTENSION Transform circuit to Laplace domain. Assume all initial conditions are zero Thevenin APPLICATION LAPLACE