{
  "id": "enwiki-001376758248",
  "slug": "final-value-theorem",
  "title": "Final value theorem",
  "qid": "",
  "lead": "In mathematical analysis, the final value theorem (FVT) is one of several similar theorems used to relate frequency domain expressions to the time domain behavior as time approaches infinity. Mathematically, if f(t) in continuous time has (unilateral) Laplace transform F(s) , then a final value theorem establishes conditions under which \\lim_{t\\,\\to\\,\\infty}f(t) = \\lim_{s\\,\\to\\, 0}{sF(s)}. Likewise, if f[k] in discrete time has (unilateral) Z-transform F(z) , then a final value theorem establishes conditions under which \\lim_{k\\,\\to\\,\\infty}f[k] = \\lim_{z\\,\\to\\, 1}{(z-1)F(z)}. An Abelian final value theorem makes assumptions about the time-domain behavior of f(t) \\text{ (or }f[k]) to calculate \\lim_{s\\,\\to\\, 0}{sF(s)}. Conversely, a Tauberian final value theorem makes assumptions about the frequency-domain behaviour of F(s) to calculate \\lim_{t\\to\\infty}f(t) \\text{(or }\\lim_{k\\to\\infty}f[k]) (see Abelian and Tauberian theorems for integral transforms).",
  "body": "In mathematical analysis, the final value theorem (FVT) is one of several similar theorems used to relate frequency domain expressions to the time domain behavior as time approaches infinity. Mathematically, if f(t) in continuous time has (unilateral) Laplace transform F(s) , then a final value theorem establishes conditions under which \\lim_{t\\,\\to\\,\\infty}f(t) = \\lim_{s\\,\\to\\, 0}{sF(s)}. Likewise, if f[k] in discrete time has (unilateral) Z-transform F(z) , then a final value theorem establishes conditions under which \\lim_{k\\,\\to\\,\\infty}f[k] = \\lim_{z\\,\\to\\, 1}{(z-1)F(z)}. An Abelian final value theorem makes assumptions about the time-domain behavior of f(t) \\text{ (or }f[k]) to calculate \\lim_{s\\,\\to\\, 0}{sF(s)}. Conversely, a Tauberian final value theorem makes assumptions about the frequency-domain behaviour of F(s) to calculate \\lim_{t\\to\\infty}f(t) \\text{(or }\\lim_{k\\to\\infty}f[k]) (see Abelian and Tauberian theorems for integral transforms). Suppose that every pole of F(s) is either in the open left half plane or at the origin, and that F(s) has at most a single pole at the origin. Then \\lim_{t\\to\\infty}f(t) = \\lim_{s\\to 0}sF(s). Suppose that f(t) and f'(t) both have Laplace transforms that exist for all s > 0. If \\lim_{t\\to\\infty}f(t) exists and \\lim_{s\\,\\to\\, 0}{sF(s)} exists then \\lim_{t\\to\\infty}f(t) = \\lim_{s\\,\\to\\, 0}{sF(s)}. Remark Both limits must exist for the theorem to hold. For example, if f(t) = \\sin(t) then \\lim_{t\\to\\infty}f(t) does not exist, but \\lim_{s\\,\\to\\, 0}{sF(s)} = \\lim_{s\\,\\to\\, 0}{\\frac{s}{s^2+1}} = 0. t f'(t) is also bounded on (0,\\infty) . If sF(s) \\to L \\in \\mathbb{C} as s \\to 0 then \\lim_{t\\to\\infty}f(t) = L. sF(s) \\to L \\in \\mathbb{R} as s \\downarrow 0, and \\lim_{t\\to\\infty}f(t) = L. sF(s) \\to +\\infty \\in \\mathbb{R} as s \\downarrow 0, and f(t) \\to +\\infty as t \\to \\infty. sF(s) \\to -\\infty \\in \\mathbb{R} as s \\downarrow 0, and f(t) \\to -\\infty as t \\to \\infty. In particular, if s = 0 is a multiple pole of F(s) then case 2 or 3 applies (f(t) \\to +\\infty\\text{ or }f(t) \\to -\\infty). Suppose that f(t) is Laplace transformable. Let \\lambda > -1 . If \\lim_{t\\to\\infty}\\frac{f(t)}{t^\\lambda} exists and \\lim_{s\\downarrow0}{s^{\\lambda+1}F(s)} exists then \\lim_{t\\to\\infty}\\frac{f(t)}{t^\\lambda} = \\frac{1}{\\Gamma(\\lambda+1)} \\lim_{s\\downarrow0}{s^{\\lambda+1}F(s)}, where \\Gamma(x) denotes the Gamma function. Suppose that f : (0,\\infty) \\to \\mathbb{C} is bounded and measurable and \\lim_{t\\to\\infty}f(t) = \\alpha \\in \\mathbb{C}. Then F(s) exists for all s > 0 and \\lim_{s\\,\\downarrow\\, 0}{sF(s)} = \\alpha. Elementary proof Suppose for convenience that |f(t)|\\le1 on (0,\\infty), and let \\alpha=\\lim_{t\\to\\infty}f(t) . Let \\epsilon>0, and choose A so that |f(t)-\\alpha| for all t > A. Since s\\int_0^\\infty e^{-st}\\,\\mathrm dt=1, for every s>0 we have sF(s)-\\alpha=s\\int_0^\\infty(f(t)-\\alpha)e^{-st}\\,\\mathrm dt; hence |sF(s)-\\alpha|\\le s\\int_0^A|f(t)-\\alpha|e^{-st}\\,\\mathrm dt+s\\int_A^\\infty |f(t)-\\alpha|e^{-st}\\,\\mathrm dt \\le2s\\int_0^Ae^{-st}\\,\\mathrm dt+\\epsilon s\\int_A^\\infty e^{-st}\\,\\mathrm dt \\equiv I+II. Now for every s>0 we have II On the other hand, since A is fixed it is clear that \\lim_{s\\to 0}I=0 , and so |sF(s)-\\alpha| if s>0 is small enough. \\lim_{s \\to 0^{+}} sF(s) = \\lim_{t\\to\\infty} f(t). Remark The proof uses the dominated convergence theorem. \\lim_{T\\to\\infty} \\frac{1}{T} \\int_{0}^{T} f(t) \\, dt = \\alpha \\in \\mathbb{C} Then \\lim_{s\\,\\to\\, 0, \\, s>0}{sF(s)} = \\alpha. \\lim_{s \\to 0}sF(s) = \\lim_{t \\to \\infty} \\frac{1}{t} \\int_{0}^{t} f(x) \\, \\mathrm dx. Let h : [0,\\infty) \\to \\mathbb{R} be measurable and such that the (possibly improper) integral f(x) := \\int_0^x h(t)\\,\\mathrm dt converges for x\\to\\infty. Then \\int_0^\\infty h(t)\\, \\mathrm dt := \\lim_{x\\to\\infty} f(x) = \\lim_{s\\downarrow 0}\\int_0^\\infty e^{-st}h(t)\\,\\mathrm dt. This is a version of Abel's theorem. To see this, notice that f'(t) = h(t) and apply the final value theorem to f after an integration by parts: For s > 0, s\\int_0^\\infty e^{-st}f(t)\\, \\mathrm dt = \\Big[- e^{-st}f(t)\\Big]_{t=o}^\\infty + \\int_0^\\infty e^{-st} f'(t) \\, \\mathrm dt = \\int_0^\\infty e^{-st} h(t) \\, \\mathrm dt. By the final value theorem, the left-hand side converges to \\lim_{x\\to\\infty} f(x) for s\\to 0. To establish the convergence of the improper integral \\lim_{x\\to\\infty}f(x) in practice, Dirichlet's test for improper integrals is often helpful. An example is the Dirichlet integral. E[X^n] = (-1)^n\\left.\\frac{d^n\\rho(s)}{ds^n}\\right|_{s=0}. The strategy is to write \\frac{d^n\\rho(s)}{ds^n} = \\mathcal{F}\\bigl(G_1(s), G_2(s), \\dots, G_k(s), \\dots\\bigr), where \\mathcal{F}(\\dots) is continuous and for each k, G_k(s) = sF_k(s) for a function F_k(s). For each k, put f_k(t) as the inverse Laplace transform of F_k(s), obtain \\lim_{t\\to\\infty}f_k(t), and apply a final value theorem to deduce \\lim_{s\\,\\to\\, 0}{G_k(s)} =\\lim_{s\\,\\to\\, 0}{sF_k(s)} = \\lim_{t\\to\\infty}f_k(t). Then \\left.\\frac{d^n\\rho(s)}{ds^n}\\right|_{s=0} = \\mathcal{F}\\Bigl(\\lim_{s\\,\\to\\, 0} G_1(s), \\lim_{s\\,\\to\\, 0} G_2(s), \\dots, \\lim_{s\\,\\to\\, 0} G_k(s), \\dots\\Bigr), and hence E[X^n] is obtained. H(s) = \\frac{ 6 }{s + 2}, the impulse response converges to \\lim_{t \\to \\infty} h(t) = \\lim_{s \\to 0} \\frac{6s}{s+2} = 0. That is, the system returns to zero after being disturbed by a short impulse. However, the Laplace transform of the unit step response is G(s) = \\frac{1}{s} \\frac{6}{s+2} and so the step response converges to \\lim_{t \\to \\infty} g(t) = \\lim_{s \\to 0} \\frac{s}{s} \\frac{6}{s+2} = \\frac{6}{2} = 3 So a zero-state system will follow an exponential rise to a final value of 3. H(s) = \\frac{9}{s^2 + 9}, the final value theorem appears to predict the final value of the impulse response to be 0 and the final value of the step response to be 1. However, neither time-domain limit exists, and so the final value theorem predictions are not valid. In fact, both the impulse response and step response oscillate, and (in this special case) the final value theorem describes the average values around which the responses oscillate. There are two checks performed in Control theory which confirm valid results for the Final Value Theorem: All non-zero roots of the denominator of H(s) must have negative real parts. H(s) must not have more than one pole at the origin. Rule 1 was not satisfied in this example, in that the roots of the denominator are 0+j3 and 0-j3. If \\lim_{k\\to\\infty}f[k] exists and \\lim_{z\\,\\to\\, 1}{(z-1)F(z)} exists then \\lim_{k\\to\\infty}f[k] = \\lim_{z\\,\\to\\, 1}{(z-1)F(z)}. \\dot{\\mathbf{x}}(t) = \\mathbf{A} \\mathbf{x}(t) + \\mathbf{B} \\mathbf{u}(t) \\mathbf{y}(t) = \\mathbf{C} \\mathbf{x}(t) in response to a step input \\mathbf{u}(t) with amplitude R is: \\lim_{t\\to\\infty}\\mathbf{y}(t) = -\\mathbf{CA}^{-1}\\mathbf{B}R The sampled-data system of the above continuous-time LTI system at the aperiodic sampling times t_{i}, i=1,2,... is the discrete-time system {\\mathbf{x}}(t_{i+1}) = \\mathbf{\\Phi}(h_{i}) \\mathbf{x}(t_{i}) + \\mathbf{\\Gamma}(h_{i}) \\mathbf{u}(t_{i}) \\mathbf{y}(t_{i}) = \\mathbf{C} \\mathbf{x}(t_{i}) where h_{i} = t_{i+1}-t_{i} and \\mathbf{\\Phi}(h_{i})=e^{\\mathbf{A}h_{i}} , \\mathbf{\\Gamma}(h_{i})=\\int_0^{h_{i}} e^{\\mathbf{A}s} \\,\\mathrm ds The final value of this system in response to a step input \\mathbf{u}(t) with amplitude R is the same as the final value of its original continuous-time system. ==See also== Initial value theorem Z-transform Laplace Transform Abelian and Tauberian theorems ==Notes== ==External links== https://web.archive.org/web/20101225034508/http://wikis.controltheorypro.com/index.php?title=Final_Value_Theorem http://fourier.eng.hmc.edu/e102/lectures/Laplace_Transform/node17.html : final value for Laplace https://web.archive.org/web/20110719222313/http://www.engr.iupui.edu/~skoskie/ECE595s7/handouts/fvt_proof.pdf: final value proof for Z-transforms it:Teorema del valore iniziale",
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    "title": "Final value theorem",
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    "publisher": "Anticloud FZ LLE",
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