LaPlace Transform In Circuit Analysis

LaPlace Transform in Circuit AnalysisObjectives: Calculate the Laplace transform of common functionsusing the definition and the Laplace transform tables Laplace-transform a circuit, including components withnon-zero initial conditions. Analyze a circuit in the s-domain Check your s-domain answers using the initial valuetheorem (IVT) and final value theorem (FVT) Inverse Laplace-transform the result to get the timedomain solutions; be able to identify the forced andnatural response components of the time-domain solution.(Note – this material is covered in Chapter 12 and Sections13.1 – 13.3)

LaPlace Transform in Circuit AnalysisWhat types of circuits can we analyze? Circuits with any number and type of DC sources andany number of resistors. First-order (RL and RC) circuits with no source and with aDC source. Second-order (series and parallel RLC) circuits with nosource and with a DC source. Circuits with sinusoidal sources and any number ofresistors, inductors, capacitors (and a transformer or opamp), but can generate only the steady-state response.

LaPlace Transform in Circuit AnalysisWhat types of circuits will Laplace methods allow us toanalyze? Circuits with any type of source (so long as the functiondescribing the source has a Laplace transform), resistors,inductors, capacitors, transformers, and/or op amps; theLaplace methods produce the complete response!

LaPlace Transform in Circuit AnalysisDefinition of the Laplace transform: L { f (t )} F ( s) f (t )e st dt0Note that there are limitations on the types of functions forwhich a Laplace transform exists, but those functions are“pathological”, and not generally of interest to engineers!

LaPlace Transform in Circuit AnalysisAside – formally define the “step function”, which is oftenmodeled in a circuit by a voltage source in series with aswitch.Kf(t) Ku(t)f (t ) 0, K,t 0t 0tWhen K 1, f(t) u(t), which we call the unit step function

LaPlace Transform in Circuit AnalysisMore step functions:The step function shifted in timeKf(t) Ku(t-a)aThe “window” functionKtf(t) Ku(t-a1) - Ku(t-a2)a1a2t

Which of these expressions describesthe function plotted here?5A.B.C.D.u(t – 5)5u(t 15)5u(t – 15)15u(t – 5)15t

Which of these expressions describes thefunction plotted here?8A.B.C.8u(t 4)4u(t – 8)8u(t – 4)-4t

Which of these expressions describes thefunction plotted here?2A.B.C.2u(t 5) 2u(t – 10)2u(t – 5) 2u(t 10)2u(t 5) – 2u(t – 10)-510t

LaPlace Transform in Circuit AnalysisUse “window” functions toexpress this piecewise linearfunction as a single functionvalid for all time.0,t 02t ,0 t 1s[u(t ) u(t 1)]f (t ) 2t 4, 0 t 1 s [u(t 1) u(t 3)]2t 8, 0 t 1 s [u(t 3) u(t 4)]0,t 4sf (t ) 2t[u(t ) u(t 1)] ( 2t 4)[u(t 1) u(t 3)] (2t 8)[u(t 3) u(t 4)] 2tu(t ) 4(t 1)u(t 1) 4(t 3)u(t 3) 2(t 4)u(t 4)

LaPlace Transform in Circuit AnalysisThe impulse function, created so that the step function’s derivative isdefined for all time:The step function1The first derivative of the step functiondf(t)/dtf(t) u(t)1ttThe value of thederivative at theorigin is undefined!

LaPlace Transform in Circuit AnalysisUse a limiting function to define the step function and its firstderivative!The step functionThe first derivative of the step functiong(t)dg(t)/dt1- 1/2 tg(t) f(t)as 0- t[dg/dt](0) (t)as 0

LaPlace Transform in Circuit AnalysisThe unit impulse function is represented symbolically as (t).Definition: (t ) 0fort 0and (t )dt 1(Note that the area under the g (t ) function is1( ), which approaches 1 as 0)2 Note also that any limiting function with the following characteristicscan be used to generate the unit impulse function: Height as 0 Width 0 as 0 Area is constant for all values of

LaPlace Transform in Circuit AnalysisAnother definition: (t ) du(t )dt (t)1K (t)K (t-a)KtKtatThe sifting property is an important property of the impulse function: f (t ) (t a)dt f (a)

Evaluate the following integral, using thesifting property of the impulse function.10 10A.B.C.24273(6t 2 3) (t 2)dt6(2) 3 272

LaPlace Transform in Circuit AnalysisUse the definition of Laplace transform to calculate the Laplacetransforms of some functions of interest: L{ (t )} (t )e dt (t 0)e st dt e s ( 0 ) 1 st00 00 00L{u(t )} u(t )e st dt 1 st1 11e st dt e 0 s ss0 L{e at } e at e st dt e ( s a ) t dt L{sin t} 0 e j t e j t st e dt 2j 111e ( s a ) t 0 (s a) (s a) (s a)0 1j2 0 [e ( s j ) t e ( s j ) t ]dt 1 e ( s j ) t 1 e ( s j ) t 1 11 j 2 ( s j ) 0 j 2 ( s j ) 0j 2 ( s j ) ( s j ) s 2 2

Look at the Functional Transforms table. Basedon the pattern that exists relating the step andramp transforms, and the exponential anddamped-ramp transforms, what do you predictthe Laplace transform of t2 is?A.B.C.1/(s a)s1/s3

LaPlace Transform in Circuit AnalysisUsing the definition of the Laplace transform, determine the effect ofvarious operations on time-domain functions when the result isLaplace-transformed. These are collected in the OperationalTransform table. L{K1 f1 (t ) K 2 f 2 (t ) K 3 f 3 (t )} [ K1 f1 (t )e st K 2 f 2 (t )e st K 3 f 3 (t )e st ]dt0 K1 f1 (t )e dt K 2 f 2 (t )e dt K 3 f 3 (t )e st dt st0 st0 0 K1 f1 (t )e dt K 2 f 2 (t )e dt K 3 f 3 (t )e st dt st0 st00 K1F1 ( s ) K 2 F2 ( s ) K 2 F2 ( s ) df (t ) st stL e f (t ) 0 0 f (t )[ se ]dt dt f (0) s f (t )e st dt sF ( s ) f (0)0(integrati on by parts!)

Now lets use the operational transformtable to find the correct value of theLaplace transform of t2, given that1L{t} 2sA.B.C.1/s32/s3-2/s3

LaPlace Transform in Circuit AnalysisExample – Find the Laplace transform of t2e at.ndF ( s)Use the operational transform: L t f (t ) ( 1)ds n1 atUse the functional transform: L e (s a)nnd 2 1 d 1 2L t e ( 1) ds 2 s a ds ( s a )2 ( s a )32 at2Alternatively, atUse the operational transform: L e f (t ) F ( s a)Use the functional transform:L t 2e at 2( s a )3L t 2 2s3

LaPlace Transform in Circuit AnalysisHow can we use the Laplace transform to solve circuitproblems? Option 1: Write the set of differential equations in the timedomain that describe the relationship between voltageand current for the circuit. Use KVL, KCL, and the laws governing voltage andcurrent for resistors, inductors (and coupled coils) andcapacitors. Laplace transform the equations to eliminate theintegrals and derivatives, and solve these equations forV(s) and I(s). Inverse-Laplace transform to get v(t) and i(t).

LaPlace Transform in Circuit AnalysisHow can we use the Laplace transform to solve circuitproblems? Option 2: Laplace transform the circuit (following the process weused in the phasor transform) and use DC circuitanalysis to find V(s) and I(s). Inverse-Laplace transform to get v(t) and i(t).

LaPlace Transform in Circuit AnalysisTime-domainv(t ) Ri(t )Laplace transform – resistors:s-domain (Laplace)L V ( s) RI ( s)

LaPlace Transform in Circuit AnalysisTime-domaindi (t )v (t ) Ldti ( 0) I 0Laplace transform – inductors:s-domain (Laplace)L V ( s ) sLI ( s ) LI 0V ( s) I 0I ( s) sLs

LaPlace Transform in Circuit AnalysisTime-domaindv(t )i (t ) Cdtv(0) V0Laplace transform – resistors:s-domain (Laplace)L I ( s) sCV ( s) CV0

Find the value of the complex impedance andthe series-connected voltage source, representingthe Laplace transform of a capacitor.A.B.C.sC, V0/s1/sC, V0/s1/sC, V0/sI ( s) sCV ( s) CV0

LaPlace Transform in Circuit AnalysisRecipe for Laplace transform circuit analysis:1. Redraw the circuit (nothing about the Laplace transformchanges the types of elements or their interconnections).2. Any voltages or currents with values given are Laplacetransformed using the functional and operational tables.3. Any voltages or currents represented symbolically, usingi(t) and v(t), are replaced with the symbols I(s) and V(s).4. All component values are replaced with the correspondingcomplex impedance, Z(s).5. Use DC circuit analysis techniques to write the s-domainequations and solve them.6. Inverse-Laplace transform s-domain solutions to get timedomain solutions.

LaPlace Transform in Circuit AnalysisExample:There is no initial energystored in this circuit. Findi1(t) and i2(t) for t 0. 336 ( 42 8.4 s ) I1 42 I 2 0s10s 90I242336 ( 42 8.4 s )(10s 90) Substituting, 42 I 2 0s42 336(42)168 I 2 ( s) s[( 42 8.4 s )(10s 90) 422 s 3 14 s 2 24 s(10s 90) I 2 42 I1 0I1 ( s ) I1 10s 90 16840s 360 42 s 3 14 s 2 24 s s 3 14 s 2 24 s

LaPlace Transform in Circuit AnalysisRecipe for Laplace transform circuit analysis:1. Redraw the circuit (nothing about the Laplace transformchanges the types of elements or their interconnections).2. Any voltages or currents with values given are Laplacetransformed using the functional and operational tables.3. Any voltages or currents represented symbolically, usingi(t) and v(t), are replaced with the symbols I(s) and V(s).4. All component values are replaced with the correspondingcomplex impedance, Z(s).5. Use DC circuit analysis techniques to write the s-domainequations and solve them.6. Inverse-Laplace transform s-domain solutions to get timedomain solutions.

LaPlace Transform in Circuit AnalysisFinding the inverse Laplace transform:1f (t ) j 2 c j c j F ( s)e st dst 0This is a contour integral in the complex plane, where the complexnumber c must be chosen such that the path of integration is in theconvergence area along a line parallel to the imaginary axis atdistance c from it, where c must be larger than the real parts ofall singular values of F(s)!There must be a better way

LaPlace Transform in Circuit AnalysisInverse Laplace transform using partial fraction expansion: Every s-domain quantity, V(s) and I(s), will be in the formN ( s)D( s )where N(s) is the numerator polynomial in s, and has realcoefficients, and D(s) is the denominator polynomial in s, andalso has real coefficients, andO{N ( s)} O{D( s)} Since D(s) has real coefficients, it can always be factored,where the factors can be in the following forms: Real and distinct Real and repeated Complex conjugates and distinct Complex conjugates and repeated

LaPlace Transform in Circuit AnalysisInverse Laplace transform using partial fraction expansion: The roots of D(s) (the values of s that make D(s) 0) arecalled poles. The roots of N(s) (the values of s that make N(s) 0) arecalled zeros.Back to the example:40s 36040( s 9)I1 ( s ) 3 s 14s 2 24s s( s 2)( s 12)168168I 2 ( s) 3 2s 14s 24s s( s 2)( s 12)

Find the zeros of I1(s).40( s 9)I1 ( s ) s( s 2)( s 12)A.B.C.s 9 rad/ss 9 rad/sThere aren’t any zeros

Find the poles of I1(s).I1 ( s ) A.B.C.40( s 9)s( s 2)( s 12)s 2 rad/s, s 12 rad/ss 2 rad/s, s 12 rad/ss 0 rad/s, s 2 rad/s, s 12 rad/s

LaPlace Transform in Circuit AnalysisExample:There is no initial energystored in this circuit. Findi1(t) and i2(t) for t 0.40s 360I1 ( s ) s( s 2)( s 12)KKK3 1 2 s s 2 s 12K1 40s 360 15;( s 2)( s 12) s 0I1 ( s ) 15 14 1 s s 2 s 12K2 40s 360 14;s( s 12) s 2K3 40s 360 1s( s 2) s 12

LaPlace Transform in Circuit AnalysisExample:There is no initial energystored in this circuit. Findi1(t) and i2(t) for t 0. 1 15 14i1(t) L-1 s s 2 s 12 [15 14e 2 t e 12 t ]u(t ) AThe forced response is 15u(t ) A;The natural response is [ 14e 2 t e 12 t ]u(t ) A.

LaPlace Transform in Circuit AnalysisExample:There is no initial energystored in this circuit. Findi1(t) and i2(t) for t 0.168I 2 ( s) s( s 2)( s 12)KKK3 1 2 s s 2 s 12K1 168 7;( s 2)( s 12) s 0I 2 ( s) 7 8.41.4 s s 2 s 12K2 168 8.4;s( s 12) s 2K3 168 1.4s( s 2) s 12

LaPlace Transform in Circuit AnalysisExample:There is no initial energystored in this circuit. Findi1(t) and i2(t) for t 0.1.4 7 8.4i2(t) L-1 s s 2 s 12 [7 8.4e 2 t 1.4e 12 t ]u(t ) AThe forced response is 7u(t ) A;The natural response is [ 8.4e 2 t 1.4e 12 t ]u(t ) A.

LaPlace Transform in Circuit AnalysisExample:There is no initial energystored in this circuit. Findi1(t) and i2(t) for t 0.i1(t) (15 14e 2 t e 12 t )u(t ) Ai2(t) (7 8.4e 2 t 1.4e 12 t )u(t ) ACheck the answers at t 0 and t to make sure the circuitand the equations match!