Laplace Transform Part 1: Introduction (I&N Chap 13)

Transcription

1 Laplace Transform Part 1: Introduction (I&N Chap 13) Definition of the L.T. L.T. of Singularity Functions L.T. Pairs Properties of the L.T. Inverse L.T. Convolution IVT(initial value theorem) & FVT (final value theorem) Slide 1.1

2 Slide 1.2 Lessons from Phasor Analysis What good are phasors? dt t di L t Ri t v t v t v L R ) ( ) ( ) ( ) ( ) ( + + I I V V V L j R L R ω + + ( ) R L M t L R V t i ω ω ω tan cos ) ( ) ( + ( ) R L M L R V L j R ω ω ω tan 0 ) ( + + I V I Time domain How might we solve i(t)? Phasor domain We d much rather solve algebraic equations in the phasor domain than DEs in the time domain.

3 Motivation for Laplace Transforms Laplace transforms: generalized tool for circuit analysis(transient response, steady state, ). Circuit Input : f ( t) Time domain (t-domain) Laplace/complex frequency domain (s-domain) Input : F( Output : g( t) Output : G( Laplace Xforms: Change linear DEs into algebraic equations (easier to solve) Handle a wider variety of inputs than just sinuosoids Incorporate ICs in the solution automatically Provide the total response (natural+forced) in one operation Slide 1.3

4 Laplace Transform Slide 1.4

5 Definition of Laplace Transform Given f(t), its (one-sided) Laplace transform (if it exist is given by: L[ f ( t )] F( s ) f ( t )e Examples: Recall singularity functions are discontinuous or have discontinuous derivatives (useful for modeling switching in signal. 0 if t E.g.1: Unit step(a.k.a."heaviside Function") : u( t) 1 if t 0 st dt < if t < t E.g. 2 :Shifted Unit step: u t (t ) u( t t0 ) 0 1 if t t 0 0 E.g.3: Unit impulse (a.k.a."dirac Delta Function") : δ ( t ) 0 for t 0 but δ ( t) dt 1 δ ( t t ) f ( t) dt f ( 0 t 0 ) Slide 1.5

6 More Examples: ramp, exp & sine E.g. 4:Unit ramp: r(t ) u( t ) t E.g.5: Decaying exponential : e at u(t ) 2πt E.g. 6:Sinusoid : u(t ) sin( ωt ) u(t ) sin( T ) Slide 1.6

7 Comments about the LT Not all functions have a LT. For most circuits (including any you re required to analyse for EECE 253), the LT will exist in some region of convergence. (f 1 (t)f 2 (t)) (F 1 (F 2 (). Is the converse true? f(t): r r. What about F(? Slide 1.7

8 Laplace Transform Pairs *Defined for t 0; f(t)0 for t<0. Slide 1.8

9 Properties of the L.T. 11. Initial Value Theorem (IVT) :lim f ( t) lim sf( t Final Value Theorem (FVT):lim f ( t) lim sf( t s s 0 Slide 1.9

10 Comments about Using the LT in 253 You won t be asked to compute the LT from the definition. Instead, use the look-up tables of the preceding two slides (on an exam, these will be provided so no need to memorise them now). The LT is interesting mathematically but also takes much time to understand. For now, focus more on using this tool rather than understanding why/how it works. The power of this tool largely depends on the properties of time differentiation/integration. Observe what happens in the s-domain. We ll increasingly see that the poles of a LT (i.e., roots of the denominator) are quite important. You ve already used this fact earlier this term where? Slide 1.10

11 Examples E.g.: f ( t) δ ( t) + 2u( t) 3e 2t E.g.: 2 f ( t) t sin ( 2t)u(t) E.g.: g( t) 10 0 for 2 t 3 otherwise Slide 1.11

12 Inverse Laplace Transform If the region of convergence for F( is Re(>σ c, then the inverse Laplace transform is given by: -1 L [ F( ] f ( t) 1 2πj σ + j 1 σ j 1 F( e st ds Fortunately, in 253, this computation isn t required but you ll need to generate a partial fraction expansion (PFE) and use look-up tables. Algorithm to find inverse LT: 1. Find all poles of F(. ID them as simple vs. repeated vs. complex. 2. Find partial fraction expansion (PFE) in basic terms. 3. Look up inverse of each basic term in tables. Consider F(N(/D( where N( & D( are polynomials in s with degree (N()<degree(D()n. Poles of F( are the roots p i of D(0 so we can write: D((s-p 1 ) (s-p 2 ) (s-p n ) Slide 1.12

13 Examples 4 5s Given F( 1+, find f ( t). 2 s + 3 s + 16 Given F( 6( s + 2), find ( s + 1)( s + 3)( s + 4) f ( t). Slide 1.13

14 Poles of F( There are 3 relatively distinct types of poles that F( may have: Simple: p i is real and negative (p i <0), occurs with degree 1. Repeated: p i <0, occurs with degree m 2. Complex-Conjugate Pair: p i σ+jω with σ<0 p i+1 σ-jω p i * Slide 1.14

15 F( Partial Fraction Expansion Given F(N(/D( and the poles of D(, you often need to find the coefficients in the PFE. The text demonstrates the Residue Method for all coefficients but I recommend using this only for a pole s highest degree (i.e., if simple, if repeated). For the others (complex poles and lower degrees of a pole), I recommend a form of the Algebraic Method (examples on next two slide. Note subtle differences in my choice of notation compared to I&N (and other textbook. Consider what reasons I might have for these differences. textbook uses (s+p i ) as a factor of D( whereas I prefer (s-p i ). textbook uses {(s+α) 2 +β 2 } as a factor but I prefer {(s-σ) 2 +ω 2 }. I specified the poles must be in the LHP. Why? Slide 1.15

16 3 s + 2s + 6 Given G( 2 s ( s + 1) ( s + 3) Example, find g( t). Slide 1.16

17 ( s + 1)( s + 4s + 13) Example 10 Given G(, find g( t). 2 Slide 1.17

18 Convolution Integral Consider a linear time-invariant (LTI) system having impulse response h(t). If the system excitation (or input) is x(t), the response (or output) y(t) can be computed from the convolution integral: t y( t) x( λ) h( t λ) dλ 0 x( t) h( t) This formula is explained at great length in the text. However, I mainly require that you know the following: L { y ( t) } L{ x( t) h( t) } X( H( Y( Slide 1.18

19 Convolution Integral Notes The convolution integral applies to systems which are causal, linear and time-invariant. Suppose you have a system with zero ICs and you know its impulse (Dirac Delta) response is h(t). Causal h(t)0 for t<0 (i.e., there is no response before the input stimulu. Linear Superposition applies (i.e., if inputs r 1 (t) and r 2 (t) yield forced responses of y 1 (t) and y 2 (t), respectively, then an input of r(t) αr 1 (t)+βr 2 (t) yields the forced response y(t) αy 1 (t)+βy 2 (t)). Time-invariant The response to δ(t-t 0 ) is h(t-t 0 ). That is, if the input stimulus is shifted by time t 0 then so is the response. Unless otherwise specified, all systems that you see in this course assume these properties (as is the case for most systems you see as an undergraduate). However, do not take them for granted. Nonlinearities and time variance sometimes need to be considered. Slide 1.19

20 Linear Integrodifferential Equations Linear integrodes can be Xformed by the LT into s-domain, solved algebraically (include any IC and Xformed back into t-domain. t Example : Solve v + 4v + 4v e where v(0) v (0) 1 Example: Solve y + 3y( t) + 2 t 0 y( τ ) dτ 2e 3t where y(0) 0 Slide 1.20

21 If the network is in steady state prior to t0, find i(t) for t>0. (NB: Already solved this type of question before but now can solve using L.T.) Example Slide 1.21

22 INITIAL AND FINAL VALUE THEOREMS These results relate behavior of a function in the time domain with the behavior of the Laplace transform in the s-domain INITIAL VALUE THEOREM Assume that both f ( t), df, have Laplace dt transform. Then lim 0 f ( t) lims sf( t df L[ ] sf( f (0) dt And if the derivative is lim s df L[ ] 0 dt transformable then lim FINAL VALUE THEOREM Assume that both f ( t), df transform and that t NOTE: lim f ( t) lim t s lim t 0 sf(, have Laplace dt f ( t) exists. Then f ( t) will exist if F ( has poles withnegative real part and at most a single pole at s 0 0 st Taking limits as s 0 0 df ( t) e dt df ( t) dt dt dt sf( f (0) lim s 0 sf( f (0)