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Unformatted text preview: Second Order Systems Do not distribute these notes 41 Prof. R.T. M’Closkey, UCLA Second Order Linear Differential Equations The second order linear time invariant ODE is given by: y + a1 y + a2 y = b1 u + b2 u, ¨ ˙ ˙ (1) where y= t= u= a1 , a2 , b1 , b2 = dependent variable independent variable “forcing" function specified on interval I constant coefficients (may be real or complex) One block diagram that corresponds to this ODE is u b2 b1 ￿ −a2 ￿ y −a1 There are other “topologies" that also implement (1) but their study will be deferred. Do not distribute these notes 42 Prof. R.T. M’Closkey, UCLA Homogeneous and particular solutions are also associated with this ODE: 1. Homogeneous solution. The homogeneous differential equation is the ODE in which the forcing function is zero: y + a1 y + a2 y = 0, ¨ ˙ A non-zero solution to this ODE, denoted yh , is called a homogeneous solution. A basis for all homogeneous solutions can be generated by computing the roots of the associated characteristic polynomial, λ2 + a1 λ + a2 = (λ − λ1 )(λ − λ2 ) = λ2 −(λ1 + λ2 ) λ + λ1 λ2 , a1 a2 where the roots have been denoted λ1 and λ2 . The roots may be real or complex. Note that if the coefficients a1 and a2 are real, then the roots, if complex, appear as conjugate pairs. There are two cases to consider: (a) Distinct roots. The roots are distinct when λ1 = λ2 . In this case any homogeneous solution can be expressed as yh (t) = α1 eλ1 t + α2 eλ2 t , where α1 and α2 are constants. (b) Repeated roots. If the roots are repeated, that is λ1 = λ2 , then any homogeneous solution can be expressed as yh (t) = α1 eλ1 t + α2 teλ1 t , where α1 and α2 are constants. 2. Particular solution. For the non-homogeneous ODE, a particular solution, denoted yp , is any function that satisfies yp (t) + a1 yp (t) + a2 yp (t) = b1 u(t) + b2 u(t), ∀t ∈ I . ¨ ˙ ˙ Particular solutions are not unique because any homogeneous solution can be added to a particular solution to yield another particular solution. Do not distribute these notes 43 Prof. R.T. M’Closkey, UCLA Initial Value Problem (IVP). Let the initial conditions associated with the ODE by denoted y0 and y0 . These ICs ˙ are specified at time t0 (in other words, y (t0 ) = y0 and y (t0 ) = y0 are specified) and the forcing function u is known ˙ ˙ for time in the interval t ∈ I (note that the beginning of the time interval coincides with the time when the initial condition is specified). Then, find y over the same time interval. Solution of the IVP. The solution of the IVP is unique and may be found by several different methods. 1. Ad hoc approach. The ad hoc approach relies on finding any particular solution for the IVP and then adding to it a homogeneous solution with the two free parameters that are used to match the initial conditions at t0 . In the case of distinct roots, y (t) = yp (t) + α1 eλ1 t + α2 eλ2 t . One IC gives the algebraic relation y0 = yp (t0 ) + α1 eλ1 t0 + α2 eλ2 t0 . Differentiating the solution and evaluating at t0 gives the second algebraic relation, y0 = ˙ d (yp (t)) dt t=t0 + α1 λ1 eλ1 t0 + α2 λ2 eλ2 t0 . Combining the algebraic relations gives two equations in the two unknown parameters α1 and α2 , eλ1 t0 eλ2 t0 λ1 eλ1 t0 λ2 eλ2 t0 α1 y − yp (t0 ) =0 . α2 y0 − yp (t0 ) ˙ ˙ The determinant of the 2 × 2 matrix multiplying the parameters is (λ2 − λ1 )(λ1 +λ2 )t0 = 0 since λ1 = λ2 (the distinctness assumption). Thus, the matrix is inverted and α1 and α2 are computed. 2. Variation of parameters. The variation of parameters approach assumes that the solution of the IVP may be expressed as (assuming distinct roots again) y (t) = c1 (t)eλ1 t + c2 (t)eλ2 t , for all t ∈ I , (2) for some suitable functions c1 and c2 . First order differential equations are computed for c1 and c2 , however, we will see that these equations can be integrated directly. Differentiating (2) yields, y (t) = c1 (t)eλ1 t + c2 (t)eλ2 t + c1 (t)λ1 eλ1 t + c2 (t)λ2 eλ2 t . ˙ ˙ ˙ Do not distribute these notes 44 Prof. R.T. M’Closkey, UCLA There is some freedom in this representation of the solution so the following expression is set to zero c1 (t)eλ1 t + c2 (t)eλ2 t = 0 for all t ∈ I , ˙ ˙ (3) y (t) = c1 (t)λ1 eλ1 t + c2 (t)λ2 eλ2 t . ˙ (4) y (t) = c1 (t)λ1 eλ1 t + c2 (t)λ2 eλ2 t + c1 (t)λ2 eλ1 t + c2 (t)λ2 eλ2 t . ¨ ˙ ˙ 1 2 (5) which gives Differentiating once more, Substituting (2), (4) and (5) into (1) yields, c1 (t)λ1 eλ1 t + c2 (t)λ2 eλ2 t + c1 (t)λ2 eλ1 t + c2 (t)λ2 eλ2 t + a1 c1 (t)λ1 eλ1 t + c2 (t)λ2 eλ2 t ˙ ˙ 1 2 ˙ + a2 c1 (t)eλ1 t + c2 (t)eλ2 t = b1 u(t) + b2 u(t), Note that t∈I λ2 + a1 λ1 + a2 c1 (t)eλ1 t = 0, 1 λ2 + a1 λ2 + a2 c2 (t)eλ2 t = 0, 2 because λ1 and λ2 are roots of the characteristic polynomial. Thus, t∈I t∈I c1 (t)λ1 eλ1 t + c2 (t)λ2 eλ2 t = b1 u(t) + b2 u(t), ˙ ˙ ˙ t ∈ I. (6) Combining (3) and (6) yields two coupled first order ODEs, eλ1 t eλ2 t λ1 eλ1 t λ2 eλ2 t c1 (t) ˙ 0 = c2 (t) ˙ b1 u(t) + b2 u(t) ˙ t ∈ I. Solving for the parameters, 1 λ2 eλ2 t −eλ2 t 0 c1 (t) ˙ = λ1 t λ1 t c2 (t) ˙ e b1 u(t) + b2 u(t) ˙ W (t) −λ1 e 1 −e−λ2 t (b1 u(t) + b2 u(t)) ˙ = W (t) eλ1 t 1 −e−λ1 t = (b1 u(t) + b2 u(t)) , ˙ λ2 − λ1 e−λ2 t Do not distribute these notes 45 (7) Prof. R.T. M’Closkey, UCLA where W (t) = (λ2 − λ1 )e(λ1 +λ2 )t is the Wronskian. Note W (t) = 0 for all t. These expressions for the parameters may be integrated but first we can calculate the initial conditions associated with the parameters based on y0 and y0 . Evaluating (2) and (4) at t0 yields, ˙ Integrating (7) yields for t ≥ t0 , y0 eλ1 t0 eλ2 t0 c1 (t0 ) = λ 1 t0 λ 2 t0 y0 ˙ λ1 e λ2 e c2 (t0 ) 1 (λ2 y0 − y0 )eλ2 t0 ˙ c1 (t0 ) = ˙ c2 (t0 ) W (t0 ) (−λ1 y0 + y0 )eλ1 t0 =⇒ 1 (λ2 y0 − y0 )e−λ1 t0 ˙ = ˙ λ2 − λ1 (−λ1 y0 + y0 )e−λ2 t0 t 1 e−λ1 τ (b1 u(τ ) + b2 u(τ )) dτ ˙ c1 (t) = c1 (t0 ) − λ2 − λ2 t0 t (λ2 y0 − y0 ) −λ1 t0 ˙ 1 = e − e−λ1 τ (b1 u(τ ) + b2 u(τ )) dτ ˙ λ2 − λ1 λ2 − λ2 t0 t 1 (−λ1 y0 + y0 ) −λ2 t0 ˙ e + e−λ2 τ (b1 u(τ ) + b2 u(τ )) dτ. ˙ c2 (t) = λ2 − λ1 λ2 − λ1 t0 The final expression for y is obtained by substituting these results into (2), y (t) = (λ2 y0 − y0 ) λ1 (t−t0 ) (−λ1 y0 + y0 ) λ2 (t−t0 ) ˙ ˙ e + e λ2 − λ1 λ2 − λ1 t 1 + −eλ1 (t−τ ) + eλ2 (t−τ ) (b1 u(τ ) + b2 u(τ )) dτ, ˙ t0 λ2 − λ1 ˜ h(t−τ ) = (8) 1 1 λ2 eλ1 (t−t0 ) − λ1 eλ2 (t−t0 ) y0 + −eλ1 (t−t0 ) + eλ2 (t−t0 ) y0 ˙ λ2 − λ1 λ2 − λ1 t + t0 Do not distribute these notes t∈I ˜ h(t − τ ) (b1 u(τ ) + b2 u(τ )) dτ, ˙ 46 Prof. R.T. M’Closkey, UCLA ˜ where h is defined to be ˜ h(t) = 1 −eλ1 t + eλ2 t , λ2 − λ1 t ≥ 0. Superposition It is clear that since the initial conditions and input (and its derivatives) appear linearly in the solution that the superposition property discussed for first order systems extends to this case. Time-invariance The solution also satisfies the time-invariance property: if the “starting time" t0 is shifted, then the solution shifts a corresponding amount. Do not distribute these notes 47 Prof. R.T. M’Closkey, UCLA Example. Consider the damped spring-mass system shown below, Equation of motion: y my + cy + ky = u ¨ ˙ u fixed base The mass, spring, and damping constants are 1, 2, and 2, respectively, and u represents the force applied to mass. The mass moves in the vertical direction and its displacement relative to its rest position is denoted y . In comparing with (1) we see a1 = 2, a2 = 2, b1 = 0 and b2 = 1. The characteristic roots are λ1 = −1 + j, λ2 = −1 − j. Since λ1 = λ2 , (8) can be applied (set t0 = 0 without loss of generality), −t t −t y (t) = e (sin(t) + cos(t)) y0 + e sin(t)y0 + ˙ 0 ˜ h(t − τ )u(τ )dτ, t≥0 where ˜ h(t) = e−t sin(t), t ≥ 0, and the initial position and velocity are denoted y0 and y0 , respectively. ˙ Do not distribute these notes 48 Prof. R.T. M’Closkey, UCLA Convolution Representation Despite the fact that the time derivative appears has a non-homogeneous term in (1), the signal u is considered the “input" to the system as indicated by the block diagram: 2nd order ODE u y Thus, we wish to find a function h so that the integral in (8) can be represented as t 0 h(t − τ )u(τ )dτ. The term that needs massaging is t t0 ˜ ˜ ˜ h(t − τ )u(τ )dτ = h(0)u(t − t0 ) − h(t − t0 )u(t0 ) + ˙ t t0 ˙ ˜ h(t − τ )u(τ )dτ, ˜ where the integration-by-parts formula has been employed. Note that h(0) = 0, however, it will be shown later that u(t0 ) = 0 is also required. Thus, the IVP solution can also be expressed as y (t) = 1 1 λ2 eλ1 (t−t0 ) − λ1 eλ2 (t−t0 ) y0 + −eλ1 (t−t0 ) + eλ2 (t−t0 ) y0 ˙ λ2 − λ1 λ2 − λ1 t ˙ ˜ ˜ + b h(t − τ ) + b h(t − τ ) u(τ )dτ 1 2 t0 This result shows that the portion of the system response that is due to the input u can be computed by convolving u with the function ˙ ˜ ˜ h(t) = b1 h(t) + b2 h(t) b1 b2 = −λ1 eλ1 t + λ2 eλ2 t + −eλ1 t + eλ2 t , λ2 − λ1 λ2 − λ1 Do not distribute these notes 49 t ≥ 0. Prof. R.T. M’Closkey, UCLA The function h will be revealed to be the impulse response of the system. As in the study of the first order ODE, we do two things: • define the input for t < 0 in order to dispense with the initial conditions • extend the domain of definition of h to include t < 0, in which h is specified to be zero there With these modifications, the dependent variable can be compactly written as the convolution y (t) = ∞ −∞ h(t − τ )u(τ )dτ, t ∈ (−∞, ∞) (9) where h(t) = 0 b1 λ2 −λ1 −λ1 eλ1 t + λ2 eλ2 t + b2 λ2 −λ1 −eλ1 t + eλ2 t t<0 t≥0 (10) When we introduce more advanced material it will become apparent how the input can be defined for t < 0 in order to manipulate y (0) and y (0) to any desired values as long as −b2 /b1 is not equal to a characteristic root, i.e. the ˙ input for t < 0 can be used to “set up" initial conditions at t = 0 assuming −b2 /b1 = λ1 , λ2 (if −b2 /b1 = λ1 , for example, then it is easy to show h(t) = b1 eλ2 t , t ≥ 0, and in this case it is not possible to independently specify y0 and y0 ). ˙ Do not distribute these notes 50 Prof. R.T. M’Closkey, UCLA Example. Consider the modified damped spring-mass system shown below, y Equation of motion: my + cy + ky = cu + ku ¨ ˙ ˙ moving base u in which the base now moves and it’s position is denoted u. The position of the mass is still measured with respect to an inertial frame. Using the same mass, spring rate and damping values as the previous example yields a1 = 2, a2 = 2, b1 = 2 and b2 = 2. The position of the mass at any time can be computed from (9) where h(t) = Do not distribute these notes 0 t<0 . 2e−t cos(t) t ≥ 0 51 Prof. R.T. M’Closkey, UCLA Transfer Function The transfer function of the second order ODE is obtained by setting u(t) = est for all t, where s ∈ C. A particular solution is sought in which yp (t) = Hest . Substituting yp into (1) yields, s2 + a1 s + a2 H = b1 s + b2 . H can be computed if s is not a root of the characteristic polynomial (if this is the case, then the assumed form of yp is not correct). Thus, the transfer function associated with (1) is H (s) = b1 s + b2 , + a1 s + a2 s2 s ∈ C, s = characteristic root. (11) assuming there is no common root between the numerator polynomial and the denominator polynomial. Common factors between numerator and denominator polynomials are always cancelled so transfer functions are always written as a ratio of coprime polynomials (coprime = no common roots). The roots of the denominator polynomial are called the characteristic roots of the system. Let’s suppose those roots are λ1 and λ2 , so s2 + a1 s + a2 = (s − λ1 )(s − λ2 ). The root of the numerator polynomial is b1 s + b2 = b1 (s + b2 /b1 ) =⇒ root = − b2 /b1 . It is possible that this root is equal to one of the characteristic roots. As an example, suppose −b2 /b1 = λ1 . In this case the transfer function is given by H (s) = Do not distribute these notes b1 s + b2 b1 (s + b2 /b1 ) b1 = = . s 2 + a1 s + a2 (s − λ1 )(s − λ2 ) s − λ2 52 Prof. R.T. M’Closkey, UCLA Such cancellations indicate that a lower order differential equation defines the relationship between dependent variable and the input. In the case of the present example this differential equation is y = λ2 y + b1 u. ˙ Note that this differential equation corresponds to the impulse response computed several slides ago when we also assumed −b2 /b1 = λ1 . Example. Consider fixed base 2 The transfer function is H (s) = 1 y y + 2y + 2y = 2u ¨ ˙ 2 , s2 + 2 s + 2 and the corresponding impulse response is 2 h(t) = moving base 0 t<0 . 2e−t sin t t ≥ 0 u Do not distribute these notes 53 Prof. R.T. M’Closkey, UCLA Example. Now consider The transfer function is fixed base H (s) = 2 1 , s+1 and the corresponding impulse response is y + 3y + 2y = u + 2u ¨ ˙ ˙ 1 y 2 1 moving base u h(t) = 0 t<0 . e−t t ≥ 0 In other words, the relationship between y and u is governed by a first order differential equation despite the fact that the homogeneous ODE for the spring-mass system is second order. Frequency Response Function The frequency response function associated with the ODE is obtained when the input is constrained to be a sinusoid and a sinusoidal particular solution is sought. The frequency response function can be computed from the transfer function by substituting s = jω , where ω ∈ R, is the frequency of the sinusoid. It is assumed for any particular ω that jω is not a characteristic root. If u(t) = cos(ωt) then yp (t) = R H (jω )ejωt = |H (jω )| cos(ωt + ∠H (jω )). The graphical plots of |H | and ∠H versus frequency are the most informative way to convey the frequency response function. Do not distribute these notes 54 Prof. R.T. M’Closkey, UCLA Example. Consider the modified damped spring-mass system with governing ODE y + 2y + 2y = 2u + 2u. ¨ ˙ ˙ The frequency response function is H (jω ) = j 2ω + 2 . −ω 2 + 2 + j 2ω A plot of the magnitude and phase of H (jω ) are shown below, 1 10 100 80 60 40 0 magnitude phase (degrees) 10 10 1 20 0 20 40 60 80 2 10 10 2 10 1 Do not distribute these notes 0 10 rad/s 1 10 2 10 55 100 10 2 10 1 0 10 rad/s 1 10 2 10 Prof. R.T. M’Closkey, UCLA Stability of Second Order ODEs The second order system y + a1 y + a2 y = b1 u + b2 u, ¨ ˙ ˙ is asymptotically stable if the characteristic roots (which may be complex) have negative real part. This ensures that all solutions of the homogeneous problem exponentially decay to zero. In other words, asymptotically stable linear systems “forget" their initial conditions with some minimum exponential decay rate. Do not distribute these notes 56 Prof. R.T. M’Closkey, UCLA Block Diagrams for Second Order ODEs Block diagrams for the second order ODE are easily derived by manipulating the ODE assuming u(t) = est and y (t) = H (s)est , s2 H (s)est +a1 s H (s)est +a2 H (s)est = b1 s est +b2 est . y (t) The relation y (t) u(t) y (t) u(t) (s2 + a1 s + a2 )y (t) = (b1 s + b2 )u(t), can be rearranged to y (t) = 1 1 −a1 y (t) + b1 u(t) + (−a2 y (t) + b2 u(t)) . s s This yields the block diagram, u b2 b1 ￿ −a2 ￿ y −a1 Other block diagrams are possible, one of which will be shown below when it is demonstrated how to decompose the second order ODE into two “parallel" first order ODEs. Do not distribute these notes 57 Prof. R.T. M’Closkey, UCLA Connection with First Order Systems: Partial Fraction Expansion The transfer function (11) can be represented as the sum of two first order systems. We will assume that the denominator polynomial has been factored such that s2 + a1 s + a2 = (s − λ1 )(s − λ2 ), where λ1 = λ2 (the case where the roots are equal can be separately addressed).The idea is to find constants β1 and β2 such that, b1 s + b2 β1 β2 = + for all s ∈ C, s = λ1 , λ2 . (s − λ1 )(s − λ2 ) s − λ1 s − λ2 Combining the right hand side into one expression with a common denominator yields, β2 β1 (s − λ2 ) + β2 (s − λ1 ) β1 + = s − λ1 s − λ2 (s − λ1 )(s − λ2 ) =⇒ β1 (s − λ2 ) + β2 (s − λ1 ) = b1 s + b2 . The last expression is equivalent to two algebraic equations in the yet-to-be-determined parameters β1 and β2 , s0 coefficients: s1 coefficients: β1 + β2 = b1 =⇒ −λ2 β1 − λ1 β2 = b2 1 1 −λ2 −λ1 β1 b = 1. β2 b2 Solving for the parameters yields, β1 = − λ1 b1 + b2 , λ2 − λ1 β2 = λ2 b1 + b2 . λ2 − λ1 The process of decomposing a transfer function into a sum of transfer functions associated with first order systems is called partial fraction expansion. Recall from the analysis of first order systems that we associated the following transfer function and impulse response: b ⇐⇒ h(t) = s−a Do not distribute these notes 58 0 t<0 . eat b t ≥ 0 Prof. R.T. M’Closkey, UCLA Applying this association to the two first order transfer functions gives us another way to compute the impulse response of the system, b1 s + b2 β1 β2 = + (s − λ1 )(s − λ2 ) s − λ1 s − λ2 ⇐⇒ h(t) = β1 eλ1 t + β2 eλ2 t λ1 b1 + b2 λ1 t λ2 b1 + b2 λ2 t =− e+ e λ2 − λ1 λ2 − λ1 b2 b1 −λ1 eλ1 t + λ2 eλ2 t + eλ1 t + eλ2 t , = λ2 − λ1 λ2 − λ1 t ≥ 0. For t < 0, h(t) = 0. Note that this expression for h is equal to (10) which was derived by manipulating the convolution expression. A block diagram representation showing the original second order system and the two first order systems derived from the partial fraction expansion is x1 − λ1 x1 = β1 u ˙ u y + a1 y + a2 y = b1 u + b2 u ¨ ˙ ˙ x1 y u y x2 − λ2 x2 = β2 u ˙ x2 These block diagrams represent the same system, although in the case of the partial fraction expansion, the “identity" of the original dependent variable y seems to be lost since we have introduced two new dependent variables, denoted x1 and x2 , that are associated with the first order ODEs. Although it is not initially clear how the initial conditions y (0) and y (0) map to the initial conditions x1 (0) and x2 (0), the relationship can be figured out with ˙ little effort. To wit, since y (t) = x1 (t) + x2 (t), then y = x1 (t) + x2 (t) = λ1 x1 + β1 u + λ2 x2 + β2 u, so these two ˙ ˙ ˙ equations yield the relationship between y and y , and x1 and x2 : ˙ y (t) 11 = y (t) ˙ λ1 λ2 Do not distribute these notes x1 (t) β + 1 u. x2 (t) β2 59 Prof. R.T. M’Closkey, UCLA Since the matrix multiplying x1 and x2 is non-singular (recall λ1 = λ2 ), we conclude that y and y are uniquely ˙ related to x1 and x2 . Block Diagram The block diagram associated with the “parallel" form of the system representation is λ1 ￿ β1 x1 λ2 u ￿ β2 y x2 If any of the coefficients are complex-valued then this representation is not so useful for realizing, say, and analog electrical circuit. Nevertheless, it does represent at least mathematically the second order differential equation. The special case we considered for second order systems is the case when there was a cancellation between numerator and denominator polynomials when computing the transfer function. The case we considered was when −b2 /b1 = λ1 . In the parallel form block diagram β1 and β2 are computed to be λ1 b1 + b2 (−b2 /b1 )b1 + b2 =− = 0, λ2 − λ1 λ2 − λ1 λ2 b1 + b2 λ2 b1 − λ1 b1 β2 = = = b1 . λ2 − λ1 λ2 − λ1 β1 = − Do not distribute these notes 60 Prof. R.T. M’Closkey, UCLA Thus, the associated block diagram for this special case is λ1 ￿ x1 λ2 u ￿ b1 y x2 It is easy to see in this representation that the relationship between the dependent variable y and the input u is governed by the first order ODE y = λ2 y + b1 u. ˙ Do not distribute these notes 61 Prof. R.T. M’Closkey, UCLA Summary for 2nd Order Systems Second-order, linear, constant coefficient differential equation y +a1 y + a2 y = b1 u + b2 u ¨ ˙ ˙ y0 , y0 defined at t0 = 0 ˙ u(t) defined for t ≥ 0 λ1 ￿= λ2 y (t) = =⇒ 0 ￿ λ1 t ￿ 1 ˜ −e + eλ2 t , h(t) = λ 2 − λ1 y (t) = ∞ −∞ Transfer Function Representation impulse response of system h(t − τ )u(τ )dτ Second-order, linear, time-invariant system u defined for t ∈ (−∞, ∞) Impulse response formula: h(t) = ￿ 0 b1 λ2 −λ1 t≥0 -often a result of first-principles modeling -useful form for numerical analysis including numerical solutions Convolution representation ￿ ￿ λ1 t ￿ ￿ λ1 t ￿ 1 1 λ2 e − λ1 eλ2 t y0 + −e + eλ2 t y0 ˙ λ 2 − λ1 λ2 − λ1 ￿t ˜ (t ≥ 0) + h(t − τ ) (b1 u(τ ) + b2 u(τ )) dτ ˙ “transfer function” s = jω “frequency response function” ￿ λ1 t −λ1 e + λ2 e λ2 t ￿ + b2 λ2 −λ1 ￿ λ1 t −e +e λ2 t ￿ t<0 t≥0 s2 b1 s + b 2 + a1 s + a2 (no num-denom cancellation) H (j ω ) = jb1 ω + b2 −ω 2 + ja1 ω + a2 -use in finding particular solutions when the input is of exponential form: u(t) = est =⇒ yp (t) = H (s)est for any s ￿= λ1 , λ2 -the impulse response function is often obtained as the result of a test on a physical system -more general interpretation possible once Laplace transforms are studied; useful for manipulating block diagrams -for causal systems, the impulse response is zero for negative time (all physical systems are causal) -frequency response functions give graphical insight into system behavior; more general interpretation once Fourier transforms are studied; frequency response testing is reliable method for identifying models of physical systems -the convolution representation may also be used to define linear systems that cannot be described by ODEs Do not distribute these notes H (s) = 62 Prof. R.T. M’Closkey, UCLA ...
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This note was uploaded on 08/08/2011 for the course MAE 107 taught by Professor Tsao during the Spring '06 term at UCLA.

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