Feedback System with Pooling Delay. Many physical and biological systems involve time delays. A pure time delay has its output the same as its input but shifted in time. A more common type of delay is pooling delay. An example of such a feedback system is shown in Figure 5.3 on page 251. Here the level of fluid in tank B determines the rate at which fluid enters tank A. Suppose this rate is given by${\mathbf{R}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{\alpha }\left[\mathbf{V}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}\left(\mathbf{t}\right)\right]$ where$\mathbf{\alpha }$ and V are positive constants and${\mathbf{V}}_{\mathbf{2}}\left(\mathbf{t}\right)$ is the volume of fluid in tank B at time t.

1. If the outflow rate from tank B is constant and the flow rate from tank A into B is${\mathbf{R}}_{\mathbf{2}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{KV}}_{\mathbf{1}}\left(\mathbf{t}\right)$ where K is a positive constant and${\mathbf{V}}_1\left(\mathbf{t}\right)$ is the volume of fluid in tank A at time t, then show that this feedback system is governed by the system

$$\begin{gathered} \frac{{\mathbf{dV}}_{\mathbf{1}}}{\mathbf{dt}}\mathbf{=}\mathbf{\alpha }\left(\mathbf{V}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}\left(\mathbf{t}\right)\right)\mathbf{-}{\mathbf{KV}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{,} \\ \frac{{\mathbf{dV}}_{\mathbf{2}}}{\mathbf{dt}}\mathbf{=}{\mathbf{KV}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{-}{\mathbf{R}}_{\mathbf{3}} \end{gathered}$$

b. Find a general solution for the system in part (a) when$\mathbf{\alpha }\mathbf{=}\mathbf{5}{\left(\mathbf{min}\right)}^{\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{V}\mathbf{=}\mathbf{20}\mathbf{L}\mathbf{,}\mathbf{K}\mathbf{=}\mathbf{2}{\left(\mathbf{min}\right)}^{\mathbf{-}\mathbf{1}}\mathbf{,}$ and ${\mathbf{R}}_{\mathbf{3}}\mathbf{=}\mathbf{10} \mathbf{L}\mathbf{/}\mathbf{min}$.

c. Using the general solution obtained in part (b), what can be said about the volume of fluid in each of the tanks as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$?

Elimination Procedure for 2 x 2 Systems:

To find a general solution for the system

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$

Where${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$ and L₄are polynomials in $D=\frac{d}{dt}$:

1. Make sure that the system is written in operator form.
2. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.
3. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.
4. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants- twice as many as needed.]
5. Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

Vieta’s formulas for finding roots:

For function y(t) to be bounded when$t\to +\infty$ we need for both roots of the

the auxiliary equation to be non-positive if they are reals and, if they are complex, then the real part has to be non-positive. In other words,

1. If $r_1,r_2\in R$, then $r_1·r_2⩾0,r_1+r_2⩽0$,
2. If $r_1,r_2=\alpha \pm \beta i$, $\beta \ne 0$, then $\alpha =\frac{r_1+r_2}{2}⩽0$.

Given that, the volume of fluid in tank A is $V_1\left(t\right)$and the volume of fluid in tank B is $V_2\left(t\right)$:

R₁is the rate at which the fluid enters tank A.

R₂is the rate at which the fluid exits tank A and enters tank B.

R₃is the rate at which the fluid exits tank B.

And$R_1\left(t\right)=\alpha \left[V-V_2\left(t\right)\right]$.

Given:$R_2\left(t\right)=K{V}_1\left(t\right)$.

Then,

$$\begin{gathered} \frac{d{V}_1}{dt}=\alpha \left(V-V_2\left(t\right)\right)-K{V}_1\left(t\right), \\ \frac{d{V}_2}{dt}=K{V}_1\left(t\right)-R_3 \end{gathered}$$

The above equations can be rewritten as,

role="math" $$\begin{gathered} \frac{d{V}_1}{dt}=R_1-R_2 \\ =\alpha \left(V-V_2\left(t\right)\right)-K{V}_1\left(t\right) \\ \frac{d{V}_2}{dt}=R_2-R_3 \\ =K{V}_1\left(t\right)-R_3 \end{gathered}$$

Given,$\alpha =5{\left(\min \right)}^{-1},V=20L,K=2{\left(\min \right)}^{-1},$ and $R_3=10 L/\min$.

Referring to part (a):

$\frac{d{V}_1}{dt}=\alpha \left(V-V_2\left(t\right)\right)-K{V}_1\left(t\right) ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

$\frac{d{V}_2}{dt}=K{V}_1\left(t\right)-R_3 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Rewrite the system in operator form:

$$\begin{gathered} D{V}_1=\alpha \left(V-V_2\left(t\right)\right)-K{V}_1\left(t\right) \\ D{V}_2=K{V}_1\left(t\right)-R_3 \end{gathered}$$

Substitute the values in equations (1) and (2).

$$\begin{gathered} D{V}_1=5\left(20-V_2\right)-2V_1 \\ =100-5V_2-2V_1 \\ \left(D+2\right)V_1+5V_2=100 \\ \left(D+2\right)V_1+5V_2=100 ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

$$\begin{gathered} D{V}_2=2V_1-10 \\ 2V_1-D{v}_2=10 \\ 2V_1-D{v}_2=10 ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Multiply D on equation (3) and multiply 5 on equation (4). Then, subtract them together.

$$\begin{gathered} D\left(D+2\right)V_1+5D{V}_2+\left(10V_1-D5v_2\right)=0+50 \\ \left(D^2+2D+10\right)\left[V_1\right]=50 \\ \left(D^2+2D+10\right)\left[V_1\right]=50 ......\mathbf{\left(}\mathbf{5}\mathbf{\right)} \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is:

$r^2+2r+10=0$.

Then,

$$\begin{gathered} r=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\times 10}}{2} \\ =\frac{-2\pm \sqrt{4-40}}{2} \\ =\frac{-2\pm \sqrt{-36}}{2} \\ =\frac{-2\pm 6i}{2} \\ =-1\pm 3i \end{gathered}$$

So, the roots are$r=-1+3i$and $r=-1-3i$.

Then, the general solution of y is${\left(V_1\right)}_h\left(t\right)=A{e}^{-t}\sin 3t+B{e}^{-t}\cos 3t ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Let us assume that, ${\left(V_1\right)}_p\left(t\right)=C ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Substitute equation (7) in equation (5).

$$\begin{gathered} \left(D^2+2D+10\right)\left[V_1\right]=50 \\ \left(D^2+2D+10\right)\left[C\right]=50 \\ 10C=50 \\ C=5 \end{gathered}$$

Substitute the value of C in equation (7).

$$\begin{gathered} V_1\left(t\right)={\left(V_1\right)}_h\left(t\right)+{\left(V_2\right)}_p\left(t\right) \\ =A{e}^{-t}\sin 3t+B{e}^{-t}\cos 3t+5 \end{gathered}$$

So,$V_1\left(t\right)=A{e}^{-t}\sin 3t+B{e}^{-t}\cos 3t+5$

Now substitute equation (8) in equation (3).

$$\begin{gathered} \left(D+2\right)V_1+5V_2=100 \\ 5V_2=100-\left(D+2\right)\left[V_1\right] \\ 5V_2=100-\left(D+2\right)\left[A{e}^{-t}\sin 3t+B{e}^{-t}\cos 3t+5\right] \\ =100-\left(-A{e}^{-t}\sin 3t+3A{e}^{-t}\cos 3t-B{e}^{-t}\cos 3t-3B{e}^{-t}\sin 3t+2A{e}^{-t}\sin 3t+2B{e}^{-t}\cos 3t\right)-10 \end{gathered}$$

$$\begin{gathered} =90-\left(A{e}^{-t}\sin 3t+B{e}^{-t}\cos 3t+3A{e}^{-t}\cos 3t-3B{e}^{-t}\sin 3t\right) \\ {V}_2=18-\frac{\left[\left(A-3B\right)e^{-t}\sin 3t+\left(3A+B\right)e^{-t}\cos 3t\right]}{5} \end{gathered}$$

Hence,$V_2=18-\frac{\left[\left(A-3B\right)e^{-t}\sin 3t+\left(3A+B\right)e^{-t}\cos 3t\right]}{5}$

To find:$\underset{t\to \infty }{lim}{V}_1$ and $\underset{t\to \infty }{lim}{V}_2$.

Referring to part (b):

$$\begin{gathered} V_1\left(t\right)=A{e}^{-t}\sin 3t+B{e}^{-t}\cos 3t+5 ......\mathbf{\left(}\mathbf{8}\mathbf{\right)} \\ {V}_2\left(t\right)=18-\frac{\left[\left(A-3B\right)e^{-t}\sin 3t+\left(3A+B\right)e^{-t}\cos 3t\right]}{5} ......\mathbf{\left(}\mathbf{9}\mathbf{\right)} \end{gathered}$$

Implement the limits on equations (8) and (9).

role="math" localid="1664084322908" $$\begin{gathered} \underset{t\to \infty }{lim}{V}_1=\underset{t\to \infty }{lim}\left[A{e}^{-t}\sin 3t+B{e}^{-t}\cos 3t+5\right] \\ =5 \end{gathered}$$

role="math" localid="1664084386450" $$\begin{gathered} \underset{t\to \infty }{lim}{V}_2=\underset{t\to \infty }{lim}\left[18-\frac{\left[\left(A-3B\right)e^{-t}\sin 3t+\left(3A+B\right)e^{-t}\cos 3t\right]}{5}\right] \\ =18 \end{gathered}$$

Therefore, the solution is found.