The differential equation model for a particular chemical process has been obtained from experimental test data, τ1τ2 d^2y/dt^2 + (τ1 + τ2) dy/dt + y = Ku(t) where τ1 and τ2 are constants and u(t) is the input function.What are the functions of time (e.g., e^−t) in the solution for output y(t) for

Question

The differential equation model for a particular chemical process has been obtained from experimental test data,

[latex] au_{1} au_{2} \frac{d^{2} y}{d t^{2}}+\left( au_{1}+ au_{2} ight) \frac{d y}{d t}+y=K u(t)[/latex]

where [latex] au_{1}[/latex] and [latex] au_{2}[/latex] are constants and u(t) is the input function. What are the functions of time [latex]\left( ext { e.g. }, e^{-t} ight)[/latex] in the solution for output y(t) for the following cases? (Optional: find the solutions for y(t).)

(a) [latex]u(t)=a S(t)[/latex] step change of magnitude, A

(b) [latex]u(t)=b e^{-t / au} \quad au eq au_{1} eq au_{2}[/latex]

(c) [latex]u(t)=c e^{-t / au} \quad au= au_{1} eq au_{2}[/latex]

(d) [latex]u(t)=d \sin \omega t \quad au_{1} eq au_{2}[/latex]

Accepted Answer

Since the time function in the solution is not a function of initial conditions, take the Laplace transform with:

[latex]x(0)=\frac{d x(0)}{d t}=0[/latex]

[latex] au_{1} au_{2} s^{2} X(s)+\left( au_{1}+ au_{2} ight) s X(s)+X(s)=K U(s)[/latex]

[latex]X(s)=\frac{K}{ au_{1} au_{2} s^{2}+\left( au_{1}+ au_{2} ight) s+1} U(s)[/latex]

Factoring the denominator

[latex]X(s)=\frac{K}{\left( au_{1} s+1 ight)\left( au_{2} s+1 ight)} U(s)[/latex]

a) If [latex]u(t)=a S(t)[/latex] then [latex]U(s)=\frac{a}{s}[/latex]

[latex]X_{a}(s)=\frac{K a}{s\left( au_{1} s+1 ight)\left( au_{2} s+1 ight)} \quad au_{1} eq au_{2}[/latex]

[latex]x_{a}(t)=f_{a}\left(S(t), \quad e^{-t / au_{1}}, e^{-t / au_{2}} ight)[/latex]

b) If [latex]u(t)=b e^{-t / au}[/latex] then [latex]U(s)=\frac{b au}{ au_{s}+1}[/latex]

[latex]X_{b}(s)=\frac{K b au}{( au s+1)\left( au_{1} s+1 ight)\left( au_{2} s+1 ight)} \quad au eq au_{1} eq au_{2}[/latex]

[latex]x_{b}(t)=f_{b}\left(e^{-t / au}, e^{-t / au_{1}}, \quad e^{-t / au_{2}} ight)[/latex]

c) If [latex]u(t)=c e^{-t / au}[/latex] where [latex] au= au_{1}[/latex], then [latex]U(s)=\frac{ au c}{ au_{1} s+1}[/latex]

[latex]X_{c}(s)=\frac{K c au}{\left( au_{1} s+1 ight)^{2}\left( au_{2} s+1 ight)}[/latex]

[latex]x_{c}(t)=f_{c}\left(e^{-t / au_{1}}, t e^{-t / au_{1}}, e^{-t / au_{2}} ight)[/latex]

d) If [latex]u(t)=d \sin \omega t[/latex] then [latex]U(s)=\frac{d \omega}{s^{2}+\omega^{2}}[/latex]

[latex]X_{d}(s)=\frac{K d}{\left(s^{2}+\omega^{2} ight)\left( au_{1} s+1 ight)\left( au_{2} s+1 ight)}[/latex]

[latex]x_{d}(t)=f_{d}\left(e^{-t / au_{1}}, e^{-t / au_{2}}, \sin \omega t, \cos \omega t ight)[/latex]

Question 3.11: The differential equation model for a particular chemical pr...

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