Lecture notes, ME 340 Fall 2013Harry DankowiczThis document contains partial lecture notes for ME 340, Dynamics ofMechanical Systems. For further detail, please attend class and referto the couse textbook Modeling and Analysis of Dynamic Systems,byEsfandiari and Lu1.1Ramin S. Esfandiari and Bei Lu. Mod-eling and Analysis of Dynamic Systems.Taylor and Francis Group, LLC, 2010.ISBN 978-1-4398-0845-0Lesson 1 – Dynamic systems, signals, and componentsInthisclass,adynamic(al) system2is a composition of compo-2Bold faced characters are used forterminology that you should memorize.nents that process, modify, and/or generate signals.Each component may have one or several inputs i1(t), i2(t), . . . andone or several outputs o1(t), o2(t), . . . Here, t denotes time and thenotation (t) identifies time-dependent functions.Figure 1: A block representation of acomponent of a dynamical system withmultiple inputs and multiple outputs.Example 1. An amplifier is a single-input-single-output3component that3We refer to single-input-single-outputcomponents or systems using theacronym SISO. We use the acronymMIMO for multi-input-multi-outputcomponents or systems.increases the strength of a signal by a given positive factor, known as the(amplifier) gain.Figure 2: A block representation of anamplifier.If the input signal i(t) received by an amplifier equals e−tsin t and thegain equals 5, then the output signal o(t) returned by the component equalsWhat is the difference between the casewhen the gain exceeds 1 and the casewhen it is smaller than 1?5i(t)=5e−tsin t.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 2Example 2. A delay line is a SISO4component that delays a signal in4See previous sidenote.time by a given amount, known as the (time) delay.Figure 3: A block representation of adelay line.If the input signal i(t) equals e−tsin t and the delay is 1, then the output What is the meaning of a negativedelay?signal o(t) equals i(t − 1)=e−(t−1)sin(t − 1).A signal of the formA cos(ωt − θ)(1)for A, ω, θ ∈ R is said to be harmonic with amplitude A, angularfrequency ω, and phase shift θ. A harmonic signal with angularfrequency5ω is periodic with period T = 2π/ω.5The angular frequency tells you thenumber of revolutions per unit time of asignal. The frequency f = 1/ T tells youthe number of periods per unit time.Figure 4: The graph of a harmonicsignal.Example 3. The functionf (t) := a cos ωt + b sin ωt (2)is a sum of two harmonic signals with the same angular frequency (andtherefore the same period) and phase shifts 0 and π/2, respectively, sinceCan you derive these equations fromthe definition of the trigonometricfunctions?sin ωt = cos!π2− ωt"= cos!ωt −π2"(3)Every harmonic signal A cos(ωt − θ)can be written in the form (2), sinceWhat is name of this identity?cos(ωt − θ)= cos ωt cos θ + sin ωt sin θ (4)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 3Figure 5: The relationship between Aand θ, on the one hand, and a and b onthe other hand.It follows that a signal of the form (2) can be written as a harmonic signalwith period 2π/ω and with amplitude and phase shift given in terms of aand b.A SISO component whose output is approximately proportional tothe derivative of the input is a differentiator.Figure 6: A block representation of adifferentiator.Example 4. The derivative of the input signali(t)=a cos ωt + b sin ωt (5)equalsdidt(t)=−ωa sin ωt + ωb cos ωt (6)But,− sin ωt = sin(−ωt)=cos!π2+ ωt"= cos ω#t −#−T4$$(7)andCan you prove that π/2 = ωT/4?cos ωt = (8)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 4from which we conclude thatdidt(t)=ω%a cos ω#t −#−T4$$+ b sin ω#t −#−T4$$&= (9)This equals the original signal amplified by ω and delayed6by −T/4 .We6A signal that is delayed by a negativequantity is by theabsolute value of the quantity.say that the derivative of the input signal is ahead of the input by a quarterof a period (or by π/2 = 90◦in phase).A SISO component whose output is approximately proportional tothe integral of the input from some reference time plus some initialvalue is an integrator.Figure 7: A block representation of anintegrator.Example 5. The definite integral over the interval [0, t] with zero initialvalue of the input signali(t)=a cos ωt + b sin ωt (10)equals't0i(τ) dτ =aωsin ωτ −bωcos ωτ((((t0= (11)But,sin ωt = cos()(12)and− cos ωt = sin()(13)from which we conclude that't0i(τ) dτ = (14)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 5This equals the original signal amplified by ω−1, delayed by T/4, and addedto the constant signal b/ω. We say that the integral of the input signal lagsbehind the input by a quarter of a period.Consider the electrical circuit shown in the figure consisting of aRepeat this analysis for the case of alinear spring and dashpot attachedin parallel at one end to a stationarysupport and acted upon at the otherend by a force F.capacitor and a resistor in series.Figure 8: An electrical circuitWith the given sense of the current i , the voltage drop Vabacross We sometimes omit the (t), whenpossible time-dependence can beinferred from the context.the capacitor equals Q/C and the voltage drop Vbcacross the resistorequals Ri. Using Kirchhoff’s laws7, we conclude that7Kirchhoff’s 1st law: The voltagechange between two nodes in a circuitis the same independently of the pathbetween the points. Kirchhoff’s 2ndlaw: Charge cannot accumulate atnodes in a circuit.Q(t)=Q(0)+'t0i(τ) dτ (15)equals the charge accumulated on the capacitor andVac= Vab+ Vbc(16)equals the total voltage drop. It follows from these results and thedefinitions of Vaband VbcthatdVabdt(t)= (17)Q(t)=Q(0)+'T0dτ (18)Example 6. If we design the circuit in Fig. (8), such that RC $ 1, then Why do we use the word design here?Vac= Vab+ RCdVabdt≈ Vab(19)and, consequently,Vbc≈ RCdVacdt(20)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 6It follows thatExample 7. If we design the circuit in Fig. (8), such that RC & 1 and suchthat Q(0)=0, thenVac=1RC't0Vbc(τ) dτ + Vbc≈ Vbc(21)and, consequently,Vab≈ (22)i.e.,