lecture notes, me 340 fall 2013 7Lesson 2 – Block diagrams, system representations, and impulses“A block diagram representation of a system is an interconnectionof blocks, each of which represents some mathematical operationperformed by a certain component, in such a way that the diagram,in its entirety, agrees with the system’s mathematical model.”88Hung V. Vu and Ramin S. Esfandiari.Dynamic Systems: Modeling and Analysis.The McGraw-Hill Companies, Inc.,1997. ISBN 0-07-021673-8Amplifiers, line delays, differentiators, and integrators are ex-amples of system components with block representations shown inLecture 1. In addition, we consider multi-input-single-output sum-ming junctions and single-input-multiple-output splitting junctions.Figure 9: Block representations ofsumming and splitting junctions.By suitable wiring, a block diagram may be made to represent acomposite relationship between inputs and outputs that could not becaptured by the simple components consider thus far.Example 8. Consider the following block diagram.Figure 10: Differentiation followed byintegration.Here,x(t)=didt(t), o(t)=c +!t0x(τ) dτ (23)and thuso(t)= (24)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 8such that o( t)=i(t) if and only if c = i(0).Example 9. Consider the following block diagram.Figure 11: Integration followed bydifferentation.Here,x(t)= , o(t)= (25)and thuso(t)=i(t) (26)as expected.The input-output relationship for an integratoro(t)=c +!t0i(τ) dτ (27)is equivalent to the initial-value problemo(0)=c,dodt(t)=i(t) (28)More complicated initial-value problems may be obtained by addingconnections and additional components to the basic integrator.Example 10. Consider the following block diagram.Figure 12: Integration with feedback.Here,o(t)=c +!t0(i(τ) − αo(τ))dτ (29)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 9which is equivalent to the initial-value problemo(0)= ,dodt(t)= (30)i.e., the general form of a first-order, scalar, constant-coefficient, linear,nonhomogeneous differential equation.Example 11. Consider the following block diagramFigure 13: Two integrators and feed-back.Here,x(t)= (31)o(t)=d +!t0x(τ) dτ (32)which implies that x(0)=c and o(0)=d anddxdt= ,dodt= (33)or, equivalently,d2odt2+ αdodt+ βo = i, o(0)=d,dodt(0)=c (34)i.e., the general form of a second-order, scalar, constant-coefficient, linear,nonhomogeneous differential equation.The block diagram in Example 10 corresponds to the first-order,scalar, constant-coefficient, linear, nonhomogeneous initial-valueproblem99We sometimes use the notation˙x and¨x to denote first and second derivativeswith respect to t.˙y + αy = f , y(0)=c (35)where we have written y for the output and f for the input. Thesolution to this initial-value problem is given byy(t)=ce−αt+!t0f (τ) e−α(t−τ)dτ (36)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 10ory(t)=ce−αt+ f (t) ∗ e−αt(37)wheref (t) ∗ e−αt:=!t0f (τ) e−α(t−τ)dτ (38)denotes the convolution of f with the exponential function e−αt.To verify that this is the correct solution, we substitute back into You may want to refer back to yourcalculus book to review differentiationof a function defined by an integral.the initial-value problem:Now suppose that α > 0.Example 12. If f = 0, i.e., in the absence of any input, the solution in (36)becomesy(t)=ce−αt(39)The output decays to a fraction e−1of its original value after the characteris-tic time α−1, known as the time constant of the system.Figure 14: The response in the absenceof input.Example 13. If f = 1, the solution10in (36) becomes10The solution of the initial-valueproblem to a constant input is a stepresponse.!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 11y(t)=ce−αt+=== (40)The output converges to a steady-state value yss= α−1in such a way thaty(t) − yssdecays as a first-order system with time constant α−1.A unit step is a signal of the formu(t)="1 t > 00 t < 0(41)where its value at t = 0 is left undefined (and is of no importance).Figure 15: The graph of a unit stepsignal.Figure 16: The graph of a shifted unitstep signal.The signal u(t) − u(t − a) for some a > 0 equals 0 for t < 0 andt > a and 1 for 0 < t < a. This is a pulse with amplitude 1 and pulsewidth a.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 12Figure 17: The graph of a pulse signal.For a > 0, the integral of the signalu(t) − u(t − a)a(42)over the interval [0, t] for t > 0 equals!t0u(τ) − u(τ − a)adτ = min#ta,1$(43)In particular, for t > 0,lima↓0!t0u(τ) − u(τ − a)adτ = (44)We use the notation δ(t) inside an integral of the form!t0f (τ) δ(τ) dτ (45)to represent the limitlima↓0!t0f (τ)u(τ) − u(τ − a)adτ (46)which equalslima↓01a!dτ = (47)In particular,!t0δ(τ) dτ = limt↓0u(t)= (48)and we refer to δ(t) as a unit impulse11.11The “function” δ(t) is known asthe Dirac delta function. It’s notreally a function, but an example of adistribution. Here, we just use it as ashorthand.Example 14. The output of an integrator with initial value x0and inputi(t) equalso(t)=x0+!t0i(τ) dτ (49)But, since!t0x0δ(τ) dτ = x0(50)!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 13Figure 18: The graphical representationof a unit impulse.it follows thato(t)=!t0%&dτ (51)i.e., the output of an integrator with initial value and inputi(t)+ (52)As shown in the figure, we can modify the block representation of an inte-grator accordingly.Figure 19: A modified block representa-tion of an integrator.By the analysis in Example 14 it follows that the solution of thefirst-order system (35) in the absence of input equals the solutionwith zero initial conditions to an impulse with amplitude c. Thesolution obtained in Example 13 with c = 1 is thus called the