The Definition — Question 8

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Question 8

For real ss, use F(s)=∫0∞e−stf(t)dtF(s)=\int_0^\infty e^{-st}f(t)\,dt whenever this ordinary improper integral converges.

An unknown signal belongs to the family f(t)=0f(t)=0 for 0≤t<a0\le t<a and f(t)=Af(t)=A for t≥at\ge a, where A>0A>0 and a≥0a\ge 0. Exact measurements are F(1)=2/e,F(2)=e−2.F(1)=2/e,\qquad F(2)=e^{-2}.

Tasks

  1. Derive the transform and its real convergence set directly, then recover AA and aa from the measurements.

  2. For arbitrary positive measurements M1=F(1)M_1=F(1) and M2=F(2)M_2=F(2), give a necessary and sufficient consistency condition for this family and explicit recovery formulas.

  3. For h(t)=e−t(t2−5t/3+1/3)h(t)=e^{-t}(t^2-5t/3+1/3), compute its transform at s>0s>0 directly by integration by parts and show that it vanishes at both measured parameters.

  4. Use hh to explain precisely why the two measurements identify the signal within the stated family but do not identify an arbitrary real-valued signal.

Original worksheet page 1: question and worked solution for 4-1-008
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Question 8 – Solution

Strategy. Distinguish uniqueness of a few model parameters from uniqueness among all possible functions.

Step 1: Integrate and calibrate. For s>0s>0, F(s)=A∫a∞e−stdt=Ae−ass.F(s)=A\int_a^\infty e^{-st}\,dt=\frac{Ae^{-as}}s. For s≤0s\le 0, the positive tail integral diverges. Dividing the two measured equations gives 2F(2)/F(1)=e−a=e−12F(2)/F(1)=e^{-a}=e^{-1}, so a=1,A=2\boxed{a=1,\ A=2}. These values reproduce both measurements.

Step 2: Characterize all consistent data. For positive M1,M2M_1,M_2, the family requires 2M2/M1=e−a∈(0,1]2M_2/M_1=e^{-a}\in(0,1]. Thus consistency is exactly 2M2≤M1\boxed{2M_2\le M_1}. In that case the unique parameters are a=ln⁡M12M2,A=M122M2.\boxed{a=\ln\frac{M_1}{2M_2},\qquad A=\frac{M_1^2}{2M_2}.} They satisfy a≥0a\ge 0, A>0A>0 and both equations, proving sufficiency as well as necessity. Equality corresponds to onset at zero.

Step 3: Construct an invisible perturbation. Set b=s+1>0b=s+1>0. Integration by parts gives the three integrals of 1,t,t21,t,t^2 against e−bte^{-bt} as 1/b,1/b2,2/b31/b,1/b^2,2/b^3. Therefore H(s)=2(s+1)3−53(s+1)2+13(s+1)=(s−1)(s−2)3(s+1)3.H(s)=\frac 2{(s+1)^3}-\frac 5{3(s+1)^2}+\frac 1{3(s+1)} =\boxed{\frac{(s-1)(s-2)}{3(s+1)^3}}. The exponential boundary terms vanish, justifying those integrations. In particular, H(1)=H(2)=0H(1)=H(2)=0, although hh is not the zero function.

Step 4: State the identification limit. For every real λ\lambda, the signal f+λhf+\lambda h has a convergent transform for s>0s>0. Splitting its defining integral gives F(s)+λH(s)F(s)+\lambda H(s), so both recorded measurements are unchanged. Distinct λ\lambda give distinct real-valued signals; for nonzero λ\lambda they are not members of the original two-parameter family. The recovery formulas establish uniqueness only under that family assumption, not among arbitrary signals.

Original worksheet page 2: question and worked solution for 4-1-008

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