Let's solve the liquid level system **step by step** and find the transfer function \(\frac{Q_o(s)}{Q(s)}\). --- ### **Step 1: System Modeling** Each tank can be modeled as a capacitance (\(C\)), and the valves between tanks as resistances (\(R\)). The heights \(h_1(t)\), \(h_2(t)\), \(h_3(t)\) are analogous to voltages, and the flow rates (\(q_1(t)\), \(q_2(t)\), \(q_o(t)\)) are analogous to currents. #### **For each tank:** - **Tank 1:** \(C_1 \frac{dh_1}{dt} = q(t) - q_1(t)\) - **Tank 2:** \(C_2 \frac{dh_2}{dt} = q_1(t) - q_2(t)\) - **Tank 3:** \(C_3 \frac{dh_3}{dt} = q_2(t) - q_o(t)\) #### **Ohm’s Law for hydraulic resistance:** - \(q_1(t) = \frac{h_1(t) - h_2(t)}{R_1}\) - \(q_2(t) = \frac{h_2(t) - h_3(t)}{R_2}\) - \(q_o(t) = \frac{h_3(t)}{R_3}\) (assuming output reference at 0 height) --- ### **Step 2: Write the Equations** #### **Tank 1:** \[ C_1 \frac{dh_1}{dt} = q(t) - \frac{h_1 - h_2}{R_1} \] #### **Tank 2:** \[ C_2 \frac{dh_2}{dt} = \frac{h_1 - h_2}{R_1} - \frac{h_2 - h_3}{R_2} \] #### **Tank 3:** \[ C_3 \frac{dh_3}{dt} = \frac{h_2 - h_3}{R_2} - \frac{h_3}{R_3} \] --- ### **Step 3: Take Laplace Transform** Let \(H_1(s), H_2(s), H_3(s)\) be the Laplace transforms of \(h_1(t), h_2(t), h_3(t)\), and \(Q(s)\) is the Laplace of \(q(t)\). Assume zero initial conditions. #### **Tank 1:** \[ C_1 s H_1(s) = Q(s) - \frac{H_1(s) - H_2(s)}{R_1} \] \[ C_1 s H_1(s) + \frac{H_1(s)}{R_1} = Q(s) + \frac{H_2(s)}{R_1} \] \[ \left(C_1 s + \frac{1}{R_1}\right) H_1(s) - \frac{1}{R_1} H_2(s) = Q(s) \] #### **Tank 2:** \[ C_2 s H_2(s) = \frac{H_1(s) - H_2(s)}{R_1} - \frac{H_2(s) - H_3(s)}{R_2} \] \[ C_2 s H_2(s) + \frac{H_2(s)}{R_1} + \frac{H_2(s)}{R_2} = \frac{H_1(s)}{R_1} + \frac{H_3(s)}{R_2} \] \[ \left(C_2 s + \frac{1}{R_1} + \frac{1}{R_2}\right) H_2(s) - \frac{1}{R_1} H_1(s) - \frac{1}{R_2} H_3(s) = 0 \] #### **Tank 3:** \[ C_3 s H_3(s) = \frac{H_2(s) - H_3(s)}{R_2} - \frac{H_3(s)}{R_3} \] \[ C_3 s H_3(s) + \frac{H_3(s)}{R_2} + \frac{H_3(s)}{R_3} = \frac{H_2(s)}{R_2} \] \[ \left(C_3 s + \frac{1}{R_2} + \frac{1}{R_3}\right) H_3(s) - \frac{1}{R_2} H_2(s) = 0 \] --- ### **Step 4: Output Equation** \[ Q_o(s) = \frac{H_3(s)}{R_3} \] --- ### **Step 5: Write in Matrix Form** \[ \begin{bmatrix} C_1 s + \frac{1}{R_1} & -\frac{1}{R_1} & 0 \\ -\frac{1}{R_1} & C_2 s + \frac{1}{R_1} + \frac{1}{R_2} & -\frac{1}{R_2} \\ 0 & -\frac{1}{R_2} & C_3 s + \frac{1}{R_2} + \frac{1}{R_3} \end{bmatrix} \begin{bmatrix} H_1(s) \\ H_2(s) \\ H_3(s) \end{bmatrix} = \begin{bmatrix} Q(s) \\ 0 \\ 0 \end{bmatrix} \] --- ### **Step 6: Solve for \( H_3(s) \) in terms of \( Q(s) \)** This is a system of equations. Generally, for transfer functions, we want \( \frac{Q_o(s)}{Q(s)} \). Given \( Q_o(s) = \frac{H_3(s)}{R_3} \): \[ \frac{Q_o(s)}{Q(s)} = \frac{H_3(s)}{R_3 Q(s)} \] You can solve this system using Cramer’s Rule or block elimination. For compactness, let's write the answer: --- ### **Step 7: Final Transfer Function** \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac { 1 } { R_3 \left[ \left(C_3 s + \frac{1}{R_2} + \frac{1}{R_3}\right) - \frac{\frac{1}{R_2} \left( -\frac{1}{R_1} \right)}{C_2 s + \frac{1}{R_1} + \frac{1}{R_2} - \frac{\frac{1}{R_1} \left(-\frac{1}{R_1}\right)}{C_1 s + \frac{1}{R_1}} } \right] } } \] **Or, more generally, the transfer function is:** \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{1}{R_3} \cdot \frac{ 1 }{ C_3 s + \frac{1}{R_2} + \frac{1}{R_3} - \frac{ \frac{1}{R_2} }{ C_2 s + \frac{1}{R_1} + \frac{1}{R_2} - \frac{ \frac{1}{R_1} }{ C_1 s + \frac{1}{R_1}} } } } \] You can substitute the values of \(R_1, R_2, R_3, C_1, C_2, C_3\) for a numerical answer. --- ## **Summary of Steps** 1. Write dynamic equations for each tank. 2. Express flows using hydraulic resistances. 3. Take Laplace transforms. 4. Write equations in matrix form. 5. Solve for \(H_3(s)\) in terms of \(Q(s)\). 6. Use \(Q_o(s) = \frac{H_3(s)}{R_3}\) to get the transfer function. --- **If you need the explicit polynomial form, let me know!**

Let's carefully derive the transfer function of the given liquid level system step by step. --- ### **Given System Overview** - The system consists of **three interconnected tanks** with flows controlled via resistances \( R_1, R_2, R_3 \). - Input flow: \( q(t) \) - Output flow: \( q_o(t) \) - Heights (liquid levels): \( h_1(t), h_2(t), h_3(t) \) --- ### **Step 1: Model the System** Each tank behaves like a capacitor (volume-to-height relationship), and the flow between tanks acts like a resistor. **Assumptions:** - \( C_1, C_2, C_3 \): Capacitances related to tank cross-sectional areas. - \( R_1, R_2, R_3 \): Hydraulic resistances of the valves. - Zero initial conditions. --- ### **Step 2: Write Differential Equations** For each tank: \[ \begin{aligned} C_1 \frac{dh_1}{dt} &= q(t) - \frac{h_1 - h_2}{R_1} \quad (1) \\ C_2 \frac{dh_2}{dt} &= \frac{h_1 - h_2}{R_1} - \frac{h_2 - h_3}{R_2} \quad (2) \\ C_3 \frac{dh_3}{dt} &= \frac{h_2 - h_3}{R_2} - \frac{h_3}{R_3} \quad (3) \end{aligned} \] --- ### **Step 3: Take Laplace Transforms** Transform with zero initial conditions: \[ \begin{aligned} C_1 s H_1(s) &= Q(s) - \frac{H_1(s) - H_2(s)}{R_1} \\ C_2 s H_2(s) &= \frac{H_1(s) - H_2(s)}{R_1} - \frac{H_2(s) - H_3(s)}{R_2} \\ C_3 s H_3(s) &= \frac{H_2(s) - H_3(s)}{R_2} - \frac{H_3(s)}{R_3} \end{aligned} \] Rearranged: \[ \begin{aligned} \left( C_1 s + \frac{1}{R_1} \right) H_1(s) - \frac{1}{R_1} H_2(s) &= Q(s) \quad (4) \\ -\frac{1}{R_1} H_1(s) + \left( C_2 s + \frac{1}{R_1} + \frac{1}{R_2} \right) H_2(s) - \frac{1}{R_2} H_3(s) &= 0 \quad (5) \\ -\frac{1}{R_2} H_2(s) + \left( C_3 s + \frac{1}{R_2} + \frac{1}{R_3} \right) H_3(s) &= 0 \quad (6) \end{aligned} \] --- ### **Step 4: Express in Matrix Form** \[ \begin{bmatrix} A_{11} & A_{12} & 0 \\ A_{21} & A_{22} & A_{23} \\ 0 & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} H_1(s) \\ H_2(s) \\ H_3(s) \end{bmatrix} = \begin{bmatrix} Q(s) \\ 0 \\ 0 \end{bmatrix} \] Where: \[ \begin{aligned} A_{11} &= C_1 s + \frac{1}{R_1} \\ A_{12} &= - \frac{1}{R_1} \\ A_{21} &= - \frac{1}{R_1} \\ A_{22} &= C_2 s + \frac{1}{R_1} + \frac{1}{R_2} \\ A_{23} &= - \frac{1}{R_2} \\ A_{32} &= - \frac{1}{R_2} \\ A_{33} &= C_3 s + \frac{1}{R_2} + \frac{1}{R_3} \end{aligned} \] --- ### **Step 5: Solve for \( H_3(s) \)** We are interested in the transfer function: \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{H_3(s)}{R_3 Q(s)} } \] **Method:** - Use Cramer's rule or matrix algebra to solve for \( H_3(s) \) in terms of \( Q(s) \). The solution for \( H_3(s) \): \[ H_3(s) = \frac{\det(M_3)}{\det(M)} \] where: - \( M \): the coefficient matrix. - \( M_3 \): matrix formed by replacing the third column of \( M \) with the RHS vector. --- ### **Step 6: Final Expression** The transfer function becomes: \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{H_3(s)}{Q(s) R_3} = \frac{1}{R_3} \cdot \frac{\det(M_3)}{\det(M)} } \] **Explicitly:** \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{1}{R_3} \times \frac{ \text{determinant of matrix with last column replaced by RHS} }{ \text{determinant of the coefficient matrix} } \] --- ### **Summary:** - The transfer function \(\frac{Q_o(s)}{Q(s)}\) is a ratio of determinants involving system parameters \( R_1, R_2, R_3, C_1, C_2, C_3 \). - It encapsulates the dynamic behavior of the interconnected tanks. --- ### **Optional: Numerical Example** If specific values are provided for resistances and capacitances, substitute them into the matrices to compute the determinants explicitly, resulting in a rational transfer function. --- **This completes the step-by-step derivation of the transfer function for the liquid level system.**

Let's clarify and verify the derivation of the transfer function for the given 3-tank liquid level system. --- ### **Understanding the system:** - The system has **three interconnected tanks**. - The **input flow** is \( q(t) \) into **Tank 1** (middle tank). - The **liquid levels** are \( h_1(t), h_2(t), h_3(t) \), corresponding to each tank. - The **flows** between tanks and to the output are governed by resistances \( R_1, R_2, R_3 \). --- ### **Step 1: Write the differential equations** Using mass balance and hydraulic analogy (flow = height difference / resistance): \[ \begin{aligned} C_1 \frac{dh_1}{dt} &= q(t) - \frac{h_1 - h_2}{R_1} \quad (1) \\ C_2 \frac{dh_2}{dt} &= \frac{h_1 - h_2}{R_1} - \frac{h_2 - h_3}{R_2} \quad (2) \\ C_3 \frac{dh_3}{dt} &= \frac{h_2 - h_3}{R_2} - \frac{h_3}{R_3} \quad (3) \end{aligned} \] --- ### **Step 2: Take Laplace Transform** Assuming zero initial conditions: \[ \begin{aligned} C_1 s H_1(s) &= Q(s) - \frac{H_1(s) - H_2(s)}{R_1} \\ C_2 s H_2(s) &= \frac{H_1(s) - H_2(s)}{R_1} - \frac{H_2(s) - H_3(s)}{R_2} \\ C_3 s H_3(s) &= \frac{H_2(s) - H_3(s)}{R_2} - \frac{H_3(s)}{R_3} \end{aligned} \] --- ### **Step 3: Express in matrix form** Define: \[ \begin{bmatrix} A_{11} & A_{12} & 0 \\ A_{21} & A_{22} & A_{23} \\ 0 & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} H_1(s) \\ H_2(s) \\ H_3(s) \end{bmatrix} = \begin{bmatrix} Q(s) \\ 0 \\ 0 \end{bmatrix} \] with: \[ \begin{aligned} A_{11} &= C_1 s + \frac{1}{R_1} \\ A_{12} &= - \frac{1}{R_1} \\ A_{21} &= - \frac{1}{R_1} \\ A_{22} &= C_2 s + \frac{1}{R_1} + \frac{1}{R_2} \\ A_{23} &= - \frac{1}{R_2} \\ A_{32} &= - \frac{1}{R_2} \\ A_{33} &= C_3 s + \frac{1}{R_2} + \frac{1}{R_3} \end{aligned} \] --- ### **Step 4: Solve for \( H_3(s) \)** \[ H_3(s) = \text{(cofactor expansion)} \div \det(A) \] **Since**: \[ Q_o(s) = \frac{H_3(s)}{R_3} \] the **transfer function**: \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{H_3(s)}{Q(s)} \times \frac{1}{R_3} } \] which simplifies to: \[ \frac{Q_o(s)}{Q(s)} = \frac{1}{R_3} \times \frac{\det(\text{matrix with last column replaced by RHS vector})}{\det(A)} \] --- ### **Step 5: Final expression** This ratio of determinants yields a rational function in \(s\), representing the **overall transfer function** of the interconnected tanks. --- ### **Summary:** - The **transfer function** relates the **input flow** \(Q(s)\) to the **output flow** \(Q_o(s)\). - It is obtained by solving the system of equations in the Laplace domain, expressed in matrix form. - The explicit form involves calculating determinants, which results in a rational polynomial in \(s\). --- ### **Conclusion:** Your approach—forming the equations, transforming to Laplace domain, representing as a matrix, and solving for \(H_3(s)\)—is correct. The key is to evaluate the determinants accordingly to get the explicit transfer function. --- **In essence:** \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{1}{R_3} \times \frac{\det(\text{modified matrix})}{\det(\text{original matrix})} } \] This is the correct systematic method to derive the transfer function for the given system.

Let's clarify and verify the derivation of the transfer function for the given 3-tank liquid level system. --- ### **Understanding the system:** - The system has **three interconnected tanks**. - The **input flow** is \( q(t) \) into **Tank 1**. - Liquid flows **from Tank 1 → Tank 2 → Tank 3**. - The **liquid levels** are \( h_1(t), h_2(t), h_3(t) \). - The flows between tanks and to the output are controlled by resistances \( R_1, R_2, R_3 \). --- ### **Step 1: Write the dynamic equations** Using the hydraulic analogy (flow = height difference / resistance), and applying mass balance: \[ \begin{aligned} C_1 \frac{dh_1}{dt} &= q(t) - \frac{h_1 - h_2}{R_1} \quad (1) \\ C_2 \frac{dh_2}{dt} &= \frac{h_1 - h_2}{R_1} - \frac{h_2 - h_3}{R_2} \quad (2) \\ C_3 \frac{dh_3}{dt} &= \frac{h_2 - h_3}{R_2} - \frac{h_3}{R_3} \quad (3) \end{aligned} \] --- ### **Step 2: Laplace transform** Assuming zero initial conditions: \[ \begin{aligned} C_1 s H_1(s) &= Q(s) - \frac{H_1(s) - H_2(s)}{R_1} \\ C_2 s H_2(s) &= \frac{H_1(s) - H_2(s)}{R_1} - \frac{H_2(s) - H_3(s)}{R_2} \\ C_3 s H_3(s) &= \frac{H_2(s) - H_3(s)}{R_2} - \frac{H_3(s)}{R_3} \end{aligned} \] Rearranged: \[ \begin{aligned} \left( C_1 s + \frac{1}{R_1} \right) H_1(s) - \frac{1}{R_1} H_2(s) &= Q(s) \quad (4) \\ -\frac{1}{R_1} H_1(s) + \left( C_2 s + \frac{1}{R_1} + \frac{1}{R_2} \right) H_2(s) - \frac{1}{R_2} H_3(s) &= 0 \quad (5) \\ -\frac{1}{R_2} H_2(s) + \left( C_3 s + \frac{1}{R_2} + \frac{1}{R_3} \right) H_3(s) &= 0 \quad (6) \end{aligned} \] --- ### **Step 3: Matrix form** Expressed as: \[ \mathbf{A} \cdot \mathbf{H}(s) = \mathbf{B} \cdot Q(s) \] where: \[ \mathbf{A} = \begin{bmatrix} A_{11} & A_{12} & 0 \\ A_{21} & A_{22} & A_{23} \\ 0 & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} C_1 s + \frac{1}{R_1} & - \frac{1}{R_1} & 0 \\ - \frac{1}{R_1} & C_2 s + \frac{1}{R_1} + \frac{1}{R_2} & - \frac{1}{R_2} \\ 0 & - \frac{1}{R_2} & C_3 s + \frac{1}{R_2} + \frac{1}{R_3} \end{bmatrix} \] and \[ \mathbf{H}(s) = \begin{bmatrix} H_1(s) \\ H_2(s) \\ H_3(s) \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \] The input vector: \[ \mathbf{X} = \mathbf{A}^{-1} \mathbf{B} Q(s) \] and since the output flow: \[ Q_o(s) = \frac{H_3(s)}{R_3} \] we focus on \( H_3(s) \). --- ### **Step 4: Find \( H_3(s) \) in terms of \( Q(s) \)** Using Cramer's rule: \[ H_3(s) = \frac{\det(\mathbf{A}_3)}{\det(\mathbf{A})} \] where \( \mathbf{A}_3 \) is the matrix formed by replacing the third column of \( \mathbf{A} \) with \( \mathbf{B} Q(s) \): \[ \mathbf{A}_3 = \begin{bmatrix} C_1 s + \frac{1}{R_1} & - \frac{1}{R_1} & 1 \\ - \frac{1}{R_1} & C_2 s + \frac{1}{R_1} + \frac{1}{R_2} & 0 \\ 0 & - \frac{1}{R_2} & 0 \end{bmatrix} \times Q(s) \] Thus, \[ H_3(s) = \frac{\det(\mathbf{A}_3)}{\det(\mathbf{A})} \times Q(s) \] and the transfer function: \[ \frac{Q_o(s)}{Q(s)} = \frac{H_3(s)}{Q(s)} \times \frac{1}{R_3} = \frac{1}{R_3} \times \frac{\det(\mathbf{A}_3)}{\det(\mathbf{A})} \] --- ### **Final expression:** \[ \boxed{ \frac{Q_o(s)}{Q(s)} = \frac{1}{R_3} \times \frac{\det(\mathbf{A}_3)}{\det(\mathbf{A})} } \] where both determinants are polynomial expressions in \( s \). This provides the **exact transfer function** of the system. --- ### **Conclusion:** Your approach—forming the equations, transforming into Laplace domain, expressing as a matrix, and calculating determinants—is correct and systematic. The key is to evaluate these determinants explicitly for the parameters involved, which yields the transfer function as a ratio of polynomials in \( s \). This method ensures a rigorous derivation consistent with control systems analysis.