lecture notes, me 340 fall 2013 14Lesson 3 – Linearity, phasors, and Laplace transformsA component of a dynamic system is said to be linear if it satisfiesthe superposition principle:The output corresponding to a linear combination of individual inputsis the same linear combination of the outputs corresponding to theindividual inputs.Example 15. For a differentiatorddt!α f (t)+g(t)"= αdfdt(t)+dgdt(t) (53)and, similarly, for an integrator with zero initial value What happens if the initial value isdifferent from zero?#t0(α f (τ)+g(τ))dτ = (54)The amplifier and delay line components introduced previously are alsolinear.Now recall Euler’s formula Can you derive the rather appealingEuler identity ejπ+ 1 = 0 from Euler’sformula?ejx= cos x + j sin x (55)where j is the imaginary unit for which j2= −1. Here, the left-handside is the polar form of a complex number with magnitude 1 andargument x and the right-hand side is the rectangular form of thesame complex number with real part cos x and imaginary part sin x.The exponential ejxsatisfies the same rules12of calculus as eαxfor12For example, dejx/dx = jejxandejx· ejy= ej(x+y).real α.It follows directly from Euler’s formula thate−jx= (56)and, therefore thatcos x = (57)sin x = (58)A harmonic signal with amplitude A, angular frequency ω, andWhat are the corresponding formula forcosh x and sinh x? What conclusion doyou draw about the meaning of cos jxand sin jx?!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 15phase shift θ, can then be written as the linear combinationA cos(ωt − θ)=A2ej(ωt−θ)+A2e−j(ωt−θ)(59)of the unit phasor13ej(ωt−θ)and its complex conjugate e−j(ωt−θ).13A unit phasor is a complex-valuedfunction of time that correspondsto a vector from the origin of a two-dimensional coordinate system to apoint on the unit circle that rotatesabout the origin with the angularfrequency ω and makes a clockwiseangle θ with the horizontal axis at t = 0.Example 16. If the input i(t) to a differentiator is the phasor ej(ωt−θ), thenits output o(t) equalsddtej(ωt−θ)= jωej(ωt−θ)(60)i.e., jω times the input. But, in polar form,jω = ωej·(61)i.e.,o(t)=ωej(ωt−(θ−))(62)corresponding to an amplitude gain of ω and a reduction in the phase shift What is the output if the input equalsthe conjugate e−j(ωt−θ)?by .Example 17. If the input i( t) to an integrator with zero initial value is thephasor ej(ωt−θ), then its output o(t) equals#t0ej(ωt−θ)=1jω$ej(ωt−θ)− e−jθ%(63)i.e., 1/jω times the input (ignoring the constant term). But, in polar form,1jω= ej·(64)i.e.,o(t)=&ej(ωt−(θ−))− e−j(θ−)'(65)corresponding to an amplitude gain of and a reduction in the phaseshift by .Inspired by the action of differentiators and integrators (with zeroinitial value) on phasors, we represent these graphically by blockswith the corresponding complex impedances.!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 16Figure 20: An impedance-based blockrepresentation of differentiators andintegrators.Example 18. If the input i(t) to a delay line with delay a is the phasorej(ωt−θ), then its output o(t) equalsej(ω(t−a)−θ)= ej(ωt−θ)· e−jωa(66)An impedance-based representation of a delay line is shown in the figure.Figure 21: An impedance-based blockrepresentation of a delay line.Example 19. If the input i(t) to an integrator with feedback α and zeroinitial value is the phasor ej(ωt−θ), then its output o(t) equals#t0ej(ωτ−θ)· e−α(t−τ)dτ = (67)which converges onto the steady-state output signalWhat is the amplitude and phase shiftof this phasor?oss(t)=ej(ωt−θ)jω + α(68)An impedance-based representation of the steady-state behavior of anintegrator with feedback and zero initial value is shown in Fig. 22. From thefigure, we find that the steady-state output oss(t) satisfies the equationoss(t)=1jω$ej(ωt−θ)− αoss(t)%(69)in agreement with (68).!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 17Figure 22: An impedance-based blockrepresentation of an integrator withfeedback.In the impedance representation, the action on phasors of differ-entiators, integrators, delay lines, amplifiers, and integrators withfeedback corresponds to multiplication with an appropriate functionof the purely imaginary quantity jω.We seek to generalize this to the corresponding action on arbitrarytime signals. To this end, we seek to transform a time-dependent sig-nal f (t) to an equivalent14representation F(s) in terms of a “complex14By equivalent we mean that there is aunique F(s) for each f (t) and vice versa.frequency” s, such that• amplification of f (t) by a gain K corresponds to multiplication ofF(s) by K;• addition of two signals f1(t) and f2(t) correspond to addition ofF1(s) and F2(s).• time delay of f (t) by a delay α corresponds to multiplication ofF(s) by e−αs;• differentiation of f (t) corresponds to multiplication of F(s) by s;• integration of f (t) corresponds to multiplication of F(s) by 1/s;and• the response to f (t) of an integrator with zero initial value andfeedback corresponds to multiplication of F(s) by 1/(s + α).Except for a slight modification to the expressions for a delay lineand a differentiator, we are able to do this in terms of an integraltransfor m of the formF(s)= (70)for some interval Ω and some kernel K(s, t). Since integrals satisfythe superposition principle, the first two items in the list above areautomatically satisfied.!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 18The integral transform over the interval Ω =[0, ∞) and withkernel K(s, t)=e−stis called the Laplace transform15and denoted15Can you give examples of othertransforms?byL(f (t))(s) :=#∞0f (t)e−stdt (71)For piecewise continuous functions f (t) such that | f (t)|≤Meatforall t ∈ Ω and some a > 0, the Laplace transform L(f (t))exists for alls with real part greater than a. In this case, it follows thatDraw the graphs of f (t) and f (t)u(t)and explain the difference!L(f (t))(s)=L(f (t)u(t))(s) (72)Example 20. The constant function f (t)=1 satisfies the inequality| f (t)|≤Meatfor all t ∈ Ω for M = 1 and a = 0. It follows that itsLaplace transformL(1)(s)=#∞0e−stdt = (73)provided that the real part of s is positive.Example 21. The exponential function f (t)=Meatsatisfies the inequality| f (t)|≤Meattrivially for all t ∈ Ω. In this case, the Laplace transformL!Meat"(s)=#∞0Me(a−s)tdt = (74)provided that the