Electrical Network Transfer Functions

Control 2.2

Imron Rosyadi

Learning Objectives

By the end of this section, you will be able to:

• Use Laplace-domain impedance and admittance to model R, L, C components.
• Derive transfer functions for simple RLC circuits using:
  • Differential equations
  • Mesh (loop) analysis
  • Nodal analysis
  • Voltage division
• Write mesh and node equations for multi-loop / multi-node circuits by inspection.
• Derive transfer functions for inverting and noninverting op-amp circuits.
• Recognize how these circuit transfer functions become building blocks in control systems (e.g., PID controllers).

Emphasize to students that this section is the “bridge” between basic circuit analysis and control systems.

We are not just solving circuits; we are systematically getting from physical circuits to transfer functions, which later plug directly into block diagrams and control design software.

Big Picture: Why We Care About Transfer Functions in ECE

• A transfer function tells us how an input signal \(U(s)\) affects an output signal \(Y(s)\) in the Laplace domain.
• For circuits, typical inputs: source voltages or currents.
• Typical outputs: voltages across elements, branch currents, etc.
• Once we have \(G(s) = \dfrac{Y(s)}{U(s)}\), we can:
  • Analyze stability and transient behavior.
  • Design controllers (like PID) around them.
  • Simulate entire systems in MATLAB/Simulink.

Tip

Think of the transfer function as a “black-box equation” describing your circuit’s behavior.

Later, multiple black boxes will be connected to form full control systems.

Connect this with what students have already seen: step response, eigenvalues, poles/zeros.

Also mention that RLC filters and op-amp circuits they saw in circuits courses are special cases of what they are now formalizing via transfer functions.

Fundamental R, L, C Relationships (Time Domain)

Variables and Units

• Voltage: \(v(t)\) [V]

• Current: \(i(t)\) [A]

• Charge: \(q(t)\) [C]

• Resistor: \(R\) [Ω]

• Conductance: \(G = 1/R\) [mho or S]

• Capacitor: \(C\) [F]

• Inductor: \(L\) [H]

Briefly review these as “known” from basic circuits, but stress we’ll now connect them systematically to Laplace-domain models.

Mention the zero-initial-condition assumption explicitly: this is crucial for using simple \(Z(s)\) expressions.

Table: R, L, C Relationships (Zero Initial Conditions)

Time-domain and impedance/admittance relationships

Component	Voltage–current	Current–voltage	Voltage–charge	Impedance \(Z(s) = \dfrac{V(s)}{I(s)}\)	Admittance \(Y(s) = \dfrac{I(s)}{V(s)}\)
Capacitor	\(v(t) = \dfrac{1}{C}\displaystyle\int_{0}^{t} i(\tau)\,d\tau\)	\(i(t) = C \dfrac{dv(t)}{dt}\)	\(v(t) = \dfrac{1}{C} q(t)\)	\(\dfrac{1}{Cs}\)	\(Cs\)
Resistor	\(v(t) = R\,i(t)\)	\(i(t) = \dfrac{1}{R} v(t)\)	\(v(t) = R \dfrac{dq(t)}{dt}\)	\(R\)	\(\dfrac{1}{R} = G\)
Inductor	\(v(t) = L \dfrac{di(t)}{dt}\)	\(i(t) = \dfrac{1}{L}\displaystyle\int_{0}^{t} v(\tau)\,d\tau\)	\(v(t) = L \dfrac{d^{2}q(t)}{dt^{2}}\)	\(Ls\)	\(\dfrac{1}{Ls}\)

Important

Zero initial conditions are assumed for the impedance formulas. If \(i(0^-)\) or \(v(0^-)\) is nonzero, extra terms appear in the Laplace transform.

Point out that the integral limits in the original text had typos (0 to 1); we correct to 0 to \(t\).

Stress: impedance expressions \(Z_C = 1/(Cs)\), \(Z_L = Ls\) are the key to building “transformed circuits” in the \(s\)-domain.

From Circuits to Transfer Functions: Overall Workflow

1. Choose input and output variables.
2. Use Kirchhoff’s laws (KVL or KCL) to write time-domain equations.
3. Convert to differential / integral equations if needed.
4. Apply Laplace transform (assume zero initial conditions).
5. Solve algebraically for \(\dfrac{\text{Output}(s)}{\text{Input}(s)}\).

Alternative shortcut:

• Replace elements by impedances \(Z(s)\) (or admittances \(Y(s)\)).
• Work entirely in the Laplace domain using KVL/KCL and circuit techniques (mesh, nodal, voltage division).

Explain that in simple circuits we can afford the “full” route via differential equations, but the impedance/admittance method is dramatically faster and generalizes to complex networks.

Set expectation: first example is “long way,” then we show multiple short ways leading to the same result.

Example 2.6 – Series RLC: Circuit and Goal

Problem

• Circuit: series RLC with input voltage \(v(t)\) and output \(v_C(t)\) across the capacitor.

• Find transfer function:

\[ \frac{V_C(s)}{V(s)} \]

Assume zero initial conditions.

Figure 2.3: Series RLC network

Ask students: “What other choices could we have made for the output?” (inductor voltage, resistor voltage, current).

Relate to control: \(v_C(t)\) could be a sensor voltage representing temperature, position, etc.

Example 2.6 – Step 1: KVL and Integro-Differential Equation

Using KVL around the loop:

\[ L\frac{di(t)}{dt} + R i(t) + \frac{1}{C}\int_{0}^{t} i(\tau)\,d\tau = v(t) \]

Use current–charge relationship:

\[ i(t) = \frac{dq(t)}{dt} \]

Substitute into KVL:

\[ L\frac{d^{2}q(t)}{dt^{2}} + R\frac{dq(t)}{dt} + \frac{1}{C}q(t) = v(t) \]

Highlight the appearance of a 2nd-order ODE: this is why RLC circuits are natural examples for 2nd-order system behavior (oscillation, damping, etc.).

We temporarily use \(q(t)\) just to leverage simpler capacitor relations.

Example 2.6 – Step 2: Express in Terms of Capacitor Voltage

For the capacitor, voltage–charge relationship:

\[ q(t) = C\,v_C(t) \]

Substitute into the ODE:

\[ LC\frac{d^{2}v_C(t)}{dt^{2}} + RC\frac{dv_C(t)}{dt} + v_C(t) = v(t) \]

Now both input and output are voltages, which is convenient for transfer functions.

Emphasize the modeling step: translating the loop equation into an ODE relating input \(v(t)\) and output \(v_C(t)\).

Students should see how choices of state variable (current, charge, voltage) are flexible as long as they are related.

Example 2.6 – Step 3: Laplace Transform and Transfer Function

Take Laplace transform (zero initial conditions):

\[ LC s^{2} V_C(s) + RC s V_C(s) + V_C(s) = V(s) \]

Factor \(V_C(s)\):

\[ (LC s^{2} + RC s + 1)V_C(s) = V(s) \]

Solve for transfer function:

\[ \frac{V_C(s)}{V(s)} = \frac{1}{LC s^{2} + RC s + 1} = \frac{\frac{1}{LC}}{s^{2} + \frac{R}{L}s + \frac{1}{LC}} \]

Block-diagram representation

Point out that the denominator looks like the standard 2nd-order form \(s^{2} + 2\zeta\omega_n s + \omega_n^{2}\).

Students can map: \(\omega_n^{2} = 1/(LC)\) and \(2\zeta\omega_n = R/L\).

This is a key connection between circuit parameters and dynamic behavior (damping, natural frequency).

Impedance in the Laplace Domain

Take Laplace transforms of time-domain voltage–current relations (zero initial conditions):

• Capacitor:

\[ v(t) = \frac{1}{C}\int_{0}^{t} i(\tau)\,d\tau \quad \Rightarrow \quad V(s) = \frac{1}{Cs}I(s) \]

• Resistor:

\[ v(t) = R i(t) \quad \Rightarrow \quad V(s) = R I(s) \]

• Inductor:

\[ v(t) = L \frac{di(t)}{dt} \quad \Rightarrow \quad V(s) = Ls I(s) \]

Define impedance:

\[ Z(s) \equiv \frac{V(s)}{I(s)} \]

Thus:

• \(Z_C(s) = \dfrac{1}{Cs}\)
• \(Z_R(s) = R\)
• \(Z_L(s) = Ls\)

Note

Impedance generalizes resistance to dynamic components (C and L). It captures frequency-dependent behavior in a single algebraic expression.

Relate to phasor analysis: \(Z(j\omega)\) for sinusoidal steady state is analogous, but here \(s\) is a complex variable (not restricted to \(j\omega\)).

Highlight that once we move to \(Z(s)\), we can solve circuits exactly like DC resistive networks, just with symbolic impedances.

Transforming the RLC Circuit to the \(s\)-Domain

Starting from the RLC loop equation in the \(s\)-domain:

\[ \left(Ls + R + \frac{1}{Cs}\right) I(s) = V(s) \]

This has the form:

\[ [\text{Sum of impedances}]\,I(s) = [\text{Sum of applied voltages}] \]

We can draw the transformed circuit by replacing each element with its impedance:

Figure 2.5: Laplace-transformed circuit

Steps to build the transformed circuit:

1. Replace all time functions \(v(t)\), \(i(t)\), etc. with \(V(s)\), \(I(s)\), etc.
2. Replace each component by its impedance \(Z(s)\).

Stress the payoff: we no longer have to write differential equations explicitly.

Once in the \(s\)-domain, we’re just doing algebra with impedances, exactly like resistive circuits.

Example 2.7 – Same RLC, Mesh (Loop) Analysis in \(s\)-Domain

From Figure 2.5:

KVL in \(s\)-domain:

\[ \left(Ls + R + \frac{1}{Cs}\right) I(s) = V(s) \]

Solve for current:

\[ \frac{I(s)}{V(s)} = \frac{1}{Ls + R + \frac{1}{Cs}} \]

Capacitor voltage:

\[ V_C(s) = I(s) \cdot Z_C(s) = I(s)\cdot \frac{1}{Cs} \]

Therefore:

\[ \frac{V_C(s)}{V(s)} = \frac{1}{Cs}\cdot \frac{1}{Ls + R + \frac{1}{Cs}} \]

Algebraic simplification recovers the same transfer function as in Example 2.6.

Tip

By working directly with \(Z(s)\), we bypass writing and transforming the differential equation. This becomes essential for more complex networks.

Ask: “Which method do you find conceptually clearer?” Some will like the time-domain ODE, others the algebra.

Encourage students to be fluent in both views but to solve exam / design problems using the transformed-circuit shortcut.

Example 2.8 – Single Node via Nodal Analysis

Use Figure 2.5 again, but now apply KCL at the node with voltage \(V_C(s)\).

Assume:

• Currents leaving the node are positive, entering are negative.

Currents:

• Through capacitor: \(\displaystyle I_C(s) = \frac{V_C(s)}{1/(Cs)} = Cs\,V_C(s)\)
• Through \(R\)–\(L\) series branch (from \(V_C\) to source node at \(V\)):

\[ I_{RL}(s) = \frac{V_C(s) - V(s)}{R + Ls} \]

KCL at node \(V_C(s)\) (sum of leaving currents = 0):

\[ \frac{V_C(s)}{1/Cs} + \frac{V_C(s) - V(s)}{R + Ls} = 0 \]

Solve this equation for \(\dfrac{V_C(s)}{V(s)}\); the result again matches Example 2.6.

Highlight that in nodal analysis, admittance \(Y = 1/Z\) is often more convenient than impedance.

Remind: this is purely an algebraic equation in \(s\)—no explicit derivatives or integrals now.

Example 2.9 – Voltage Division in the \(s\)-Domain

Same RLC transformed circuit: series combination of \(Z_L(s)\), \(Z_R(s)\), and \(Z_C(s)\).

Voltage division:

\[ V_C(s) = \frac{Z_C(s)}{Z_L(s) + Z_R(s) + Z_C(s)} \, V(s) = \frac{\dfrac{1}{Cs}}{Ls + R + \dfrac{1}{Cs}}\,V(s) \]

Thus:

\[ \frac{V_C(s)}{V(s)} = \frac{1/Cs}{Ls + R + 1/Cs} \]

Algebra gives the same transfer function as before.

Note

Three different methods (ODE, mesh, nodal/voltage division) all give the same \(G(s)\). Choosing the fastest correct method is a key engineering skill.

Use this as a checkpoint: quickly quiz students—“Which method would you use on an exam and why?”

Transition: simple circuits are fine with any of these; complex circuits demand a systematic mesh/nodal approach.

Moving to Complex Circuits – Mesh vs. Nodal

For multi-loop networks, direct ODE writing becomes painful.

Two scalable strategies:

1. Mesh analysis (KVL)
   • Unknowns: loop currents.
   • Best when fewer loops than nodes.
2. Nodal analysis (KCL)
   • Unknowns: node voltages.
   • Best when fewer nodes than loops, especially with many current sources.

Both reduce to solving simultaneous algebraic equations in \(s\).

Remind students of similar tradeoffs in basic circuit analysis.

The novelty here is simply that the “resistances” are now \(Z(s)\) or \(Y(s)\).

Example 2.10 – Two-Loop Transfer Function via Mesh Analysis

Given

• Network of Figure 2.6(a).
• Input: \(V(s)\).
• Output: loop 2 current \(I_2(s)\).

Goal

\[ G(s) = \frac{I_2(s)}{V(s)} \]

Figure 2.6a: Original two-loop network

Figure 2.6b: Transformed two-loop network

Figure 2.6c: Block diagram

Explain figure (b) carefully: all resistors and inductor now shown as \(R_1\), \(R_2\), \(Ls\), capacitor as \(1/(Cs)\).

Goal is to show how straightforward it is to write mesh equations in \(s\)-domain, then solve for \(I_2 / V\).

Example 2.10 – Mesh Equations

Assume mesh currents \(I_1(s)\) and \(I_2(s)\) as shown.

Mesh 1 (around left loop):

\[ R_1 I_1(s) + Ls I_1(s) - Ls I_2(s) = V(s) \]

Mesh 2 (around right loop):

\[ Ls I_2(s) + R_2 I_2(s) + \frac{1}{Cs} I_2(s) - Ls I_1(s) = 0 \]

Combine terms:

\[ (R_1 + Ls) I_1(s) - Ls I_2(s) = V(s) \]

\[ - Ls I_1(s) + \left(Ls + R_2 + \frac{1}{Cs}\right) I_2(s) = 0 \]

Ask students to identify:

• “Self-impedance” terms on the diagonal (sum of all impedances around each mesh).
• “Mutual impedance” terms (common elements with opposite sign).

This pattern leads to the inspection technique later.

Example 2.10 – Solving by Cramer’s Rule

Write the system in matrix form:

\[ \begin{bmatrix} R_1 + Ls & -Ls \\ -Ls & Ls + R_2 + \dfrac{1}{Cs} \end{bmatrix} \begin{bmatrix} I_1(s) \\ I_2(s) \end{bmatrix} = \begin{bmatrix} V(s) \\ 0 \end{bmatrix} \]

Let the determinant:

\[ \Delta = \begin{vmatrix} R_1 + Ls & -Ls \\ -Ls & Ls + R_2 + \dfrac{1}{Cs} \end{vmatrix} \]

For \(I_2(s)\), Cramer’s Rule:

\[ I_2(s) = \frac{ \begin{vmatrix} R_1 + Ls & V(s) \\ -Ls & 0 \end{vmatrix} }{\Delta} = \frac{Ls\,V(s)}{\Delta} \]

So:

\[ G(s) = \frac{I_2(s)}{V(s)} = \frac{Ls}{\Delta} \]

Carrying out determinant algebra yields:

\[ G(s) = \frac{LC s^{2}}{(R_1 + R_2)LC s^{2} + (R_1 R_2 C + L)s + R_1} \]

(as in Figure 2.6(c)).

You do not need students to memorize the exact polynomial; more important is that they understand how to get it.

Mention that Symbolic Math Toolbox in MATLAB can compute \(G(s)\) automatically from the matrix form.

Mesh-Equation Pattern (Two Loops)

The general form of the two-mesh equations observed:

\[ \left[\text{Sum of impedances around Mesh 1}\right] I_1(s) - \left[\text{Sum of common impedances}\right] I_2(s) = \left[\text{Sum of applied voltages around Mesh 1}\right] \]

\[ - \left[\text{Sum of common impedances}\right] I_1(s) + \left[\text{Sum of impedances around Mesh 2}\right] I_2(s) = \left[\text{Sum of applied voltages around Mesh 2}\right] \]

Recognizing this structure lets you write mesh equations by inspection without rederiving each term.

Tip

Inspection rule for mesh \(k\):

• Diagonal term: sum of all impedances in that mesh.
• Off-diagonal term: negative sum of impedances shared with the other mesh.

Prepare students for mechanical systems: inertia, springs, dampers produce the same matrix structure as RLC circuits once written in \(s\)-domain.

This “pattern recognition” is a huge time saver in modeling.

Nodal Analysis and Admittance

Define admittance:

\[ Y(s) = \frac{1}{Z(s)} = \frac{I(s)}{V(s)} \]

For basic elements:

• Capacitor: \(Y_C(s) = Cs\)
• Resistor: \(Y_R(s) = 1/R = G\)
• Inductor: \(Y_L(s) = 1/(Ls)\)

Using admittance can make KCL equations more compact, especially when multiple parallel branches are present.

Connect to DC nodal analysis: there you used conductances; here it’s the same but now frequency-dependent via \(s\).

Set up for Example 2.11 where the two-loop network is re-analyzed via nodes.

Example 2.11 – Transfer Function via Nodal Analysis

Use the transformed circuit of Figure 2.6(b).

Nodes:

• Node at inductor: \(V_L(s)\)
• Node at capacitor: \(V_C(s)\)

KCL at \(V_L(s)\): sum of currents leaving node = 0:

\[ \frac{V_L(s) - V(s)}{R_1} + \frac{V_L(s)}{Ls} + \frac{V_L(s) - V_C(s)}{R_2} = 0 \]

KCL at \(V_C(s)\):

\[ Cs\,V_C(s) + \frac{V_C(s) - V_L(s)}{R_2} = 0 \]

Express resistances as conductances \(G_1 = 1/R_1\), \(G_2 = 1/R_2\):

\[ \left(G_1 + G_2 + \frac{1}{Ls}\right) V_L(s) - G_2 V_C(s) = G_1 V(s) \]

\[ - G_2 V_L(s) + (G_2 + Cs) V_C(s) = 0 \]

Solve simultaneously for \(V_C(s)/V(s)\):

\[ \frac{V_C(s)}{V(s)} = \frac{\dfrac{G_1 G_2}{C} s}{ (G_1 + G_2) s^{2} + \dfrac{G_1 G_2 L + C}{LC} s + \dfrac{G_2}{LC} } \]

(as shown in Figure 2.7).

Block diagram from node voltage

Encourage students to compare this expression with the mesh-based one from Example 2.10.

Both represent the same physical network but with different input–output choices.

Point out how readily admittances appear in the coefficients when using nodal analysis.

Norton Equivalent and Current-Source View

Sometimes nodal analysis is simpler if we convert voltage sources in series with impedances into current sources in parallel with admittances (Norton’s theorem).

• Voltage source \(V(s)\) in series with \(Z_s(s)\) \(\Rightarrow\) equivalent current source \(\dfrac{V(s)}{Z_s(s)}\) in parallel with \(Z_s(s)\) (or \(Y_s = 1/Z_s\)).

This leads to nodal equations of the general form:

\[ \left[\text{Sum of admittances connected to Node 1}\right] V_1(s) - \left[\text{Common admittances}\right] V_2(s) = \left[\text{Sum of applied currents at Node 1}\right] \]

…and similarly for Node 2.

Use this to motivate Example 2.12 where the same two-node network is redrawn in a Norton form.

Stress that this is a modeling choice; physically both circuits are equivalent.

Example 2.12 – Nodal Analysis with Current Sources

Figure 2.8: Norton-equivalent nodal network

• Convert all impedances to admittances (\(G_1\), \(G_2\), \(Cs\), \(1/(Ls)\)).
• Convert series voltage sources to equivalent current sources.

KCL at \(V_L(s)\):

\[ G_1 V_L(s) + \frac{1}{Ls} V_L(s) + G_2\bigl[V_L(s) - V_C(s)\bigr] = G_1 V(s) \]

KCL at \(V_C(s)\):

\[ Cs\,V_C(s) + G_2\bigl[V_C(s) - V_L(s)\bigr] = 0 \]

These are algebraically identical to the equations in Example 2.11, thus yielding the same \(V_C(s)/V(s)\).

Tip

Pattern for node equations:

• Diagonal term (for node \(k\)): sum of admittances connected to node \(k\).
• Off-diagonal: negative sum of admittances common to nodes \(k\) and \(\ell\).

Reinforce the analogy to the mesh pattern: diagonals = self, off-diagonals = negative mutual.

Emphasize: once students see this pattern, they can write equations for large nodal networks very quickly.

Example 2.13 – Mesh Equations by Inspection (Three Loops)

Figure 2.9: Three-loop electrical network

Goal: Write, but do not solve, the mesh equations.

General structure for Mesh 1:

\[ \bigl[\text{Sum of impedances around Mesh 1}\bigr] I_1(s) - \bigl[\text{Common impedances (1–2)}\bigr] I_2(s) - \bigl[\text{Common impedances (1–3)}\bigr] I_3(s) = \bigl[\text{Sum of applied voltages around Mesh 1}\bigr] \]

Similarly for Meshes 2 and 3.

After substituting values from Figure 2.9, we obtain:

Mesh 1:

\[ (2s + 2) I_1(s) - (2s + 1) I_2(s) - I_3(s) = V(s) \]

Mesh 2:

\[ -(2s + 1) I_1(s) + (9s + 1) I_2(s) - 4s\,I_3(s) = 0 \]

Mesh 3:

\[ - I_1(s) - 4s\,I_2(s) + \left(4s + 1 + \frac{1}{s}\right) I_3(s) = 0 \]

These can be solved for any desired transfer function, e.g., \(I_3(s)/V(s)\).

Recommend that students practice identifying self and mutual impedances by visually circling each mesh and shared branches.

Mention the MATLAB snippet in the text (TryIt 2.8) which uses symbolic matrices to solve these equations.

Operational Amplifiers in Control Systems

An operational amplifier (op-amp) is used to implement transfer functions electronically.

Ideal op-amp assumptions:

1. Differential input: \(v_2(t) - v_1(t)\) drives the amplifier.
2. Input impedance \(Z_i \to \infty\) (no input current).
3. Output impedance \(Z_o \to 0\) (ideal voltage source behavior).
4. Open-loop gain \(A \to \infty\) (very large).

Basic relationship:

\[ v_o(t) = A \bigl(v_2(t) - v_1(t)\bigr) \]

Figure 2.10a: Op-amp symbol

Connect to prior circuits knowledge: they might know inverting/noninverting configurations already.

Key for control: op-amps let us realize transfer functions using RC networks—filters and controllers (P, PI, PID).

Inverting Op-Amp Configuration

If the non-inverting input is grounded (\(v_2 = 0\)), we get an inverting amplifier:

\[ v_o(t) = -A v_1(t) \]

With impedances \(Z_1(s)\) and \(Z_2(s)\) connected as in Figure 2.10(c):

Figure 2.10c: Inverting op-amp realization

Key consequences of ideal assumptions:

• Input current \(I_a(s) = 0\) (no current into op-amp input).
• Therefore, \(I_1(s) = -I_2(s)\) through \(Z_1\) and \(Z_2\).
• Large \(A\) forces \(v_1(t) \approx 0\) (virtual ground).

So:

\[ I_1(s) = \frac{V_i(s)}{Z_1(s)}, \quad I_2(s) = \frac{V_o(s)}{Z_2(s)} \]

From \(I_1 = -I_2\):

\[ \frac{V_o(s)}{Z_2(s)} = -\frac{V_i(s)}{Z_1(s)} \Rightarrow \frac{V_o(s)}{V_i(s)} = -\frac{Z_2(s)}{Z_1(s)} \]

Note

Inverting op-amp transfer function:

\[ G(s) = \frac{V_o(s)}{V_i(s)} = -\frac{Z_2(s)}{Z_1(s)} \]

Emphasize the generality: by choosing \(Z_1(s)\) and \(Z_2(s)\) as resistor, capacitor, or their combinations, we can realize many transfer functions.

Preview: Example 2.14 will show how this becomes a PID controller.

Example 2.14 – Inverting Op-Amp as a PID Controller

Figure 2.11: Inverting op-amp circuit

We know:

\[ \frac{V_o(s)}{V_i(s)} = -\frac{Z_2(s)}{Z_1(s)} \]

Compute \(Z_1(s)\): parallel \(R_1\) and \(C_1\):

• Admittances in parallel add: \(Y_1(s) = C_1 s + \dfrac{1}{R_1}\).
• Therefore:

\[ Z_1(s) = \frac{1}{C_1 s + \dfrac{1}{R_1}} = \frac{1}{C_1 s + G_1} \]

(Numerical combination gives: \(Z_1(s) = \dfrac{360\times 10^{3}}{2.016 s + 1}\) in the text.)

Compute \(Z_2(s)\): series \(R_2\) and \(C_2\):

\[ Z_2(s) = R_2 + \frac{1}{C_2 s} \]

Plug into \(-Z_2/Z_1\) and simplify:

\[ \frac{V_o(s)}{V_i(s)} = -1.232 \frac{s^{2} + 45.95 s + 22.55}{s} \]

This is of the form:

\[ K\left( s + \frac{\text{something}}{s} + 1 \right) \]

which corresponds to a PID (Proportional–Integral–Derivative) controller.

Emphasize:

• The \(s\) term behaves like a differentiator.
• The \(1/s\) term behaves like an integrator.
• Constant terms give proportional action.

This is a concrete example of how op-amp circuits implement control algorithms in analog hardware.

Noninverting Op-Amp Configuration

Figure 2.12: Noninverting op-amp

• Input \(V_i(s)\) is applied to the non-inverting terminal.
• Feedback network: \(Z_1(s)\)–\(Z_2(s)\) forming a voltage divider.

Op-amp equation:

\[ V_o(s) = A\bigl(V_i(s) - V_1(s)\bigr) \]

Noninverting Op-Amp Configuration

By voltage division:

\[ V_1(s) = \frac{Z_1(s)}{Z_1(s) + Z_2(s)} V_o(s) \]

Substitute into op-amp equation:

\[ V_o(s) = A\left[V_i(s) - \frac{Z_1}{Z_1 + Z_2} V_o(s)\right] \]

Rearrange and solve for \(V_o/V_i\):

\[ \frac{V_o(s)}{V_i(s)} = \frac{A}{1 + A Z_1(s)/(Z_1(s) + Z_2(s))} \]

For large \(A\), the “1” in the denominator is negligible:

\[ \frac{V_o(s)}{V_i(s)} \approx \frac{Z_1(s) + Z_2(s)}{Z_1(s)} \]

Note

Noninverting op-amp transfer function (large \(A\)):

\[ G(s) = \frac{V_o(s)}{V_i(s)} = \frac{Z_1(s) + Z_2(s)}{Z_1(s)} \]

Highlight difference vs inverting case: output is noninverting and gain expression differs.

Again, if \(Z_1\) and \(Z_2\) are frequency-dependent, we get frequency-selective amplifiers and control elements.

Example 2.15 – Noninverting Op-Amp with R–C Networks

Figure 2.13: Noninverting circuit

Given:

• \(Z_1(s) = R_1 + \dfrac{1}{C_1 s}\) (series \(R_1\) and \(C_1\)).
• \(Z_2(s) = \dfrac{R_2 \bigl(1/(C_2 s)\bigr)}{R_2 + 1/(C_2 s)}\) (parallel \(R_2\) and \(C_2\)).

Use:

\[ \frac{V_o(s)}{V_i(s)} = \frac{Z_1(s) + Z_2(s)}{Z_1(s)} \]

After substituting and simplifying:

\[ \frac{V_o(s)}{V_i(s)} = \frac{C_2 C_1 R_2 R_1 s^{2} + (C_2 R_2 + C_1 R_2 + C_1 R_1)s + 1} {C_2 C_1 R_2 R_1 s^{2} + (C_2 R_2 + C_1 R_1)s + 1} \]

Note that numerator and denominator are very similar; this suggests particular frequency-response characteristics (e.g., some zeros cancel near certain poles).

Students don’t need to memorize this expression; the key is the method: find \(Z_1\), \(Z_2\), use the general noninverting formula.

Skill-Assessment Example – Mesh vs. Nodal Consistency

Figure 2.14: RLC network for Skill-Assessment 2.6

Problem:

Find \(G(s) = V_L(s)/V(s)\) using:

1. Mesh analysis (KVL in \(s\)-domain).
2. Nodal analysis (KCL in \(s\)-domain).

Show both methods give:

\[ \frac{V_L(s)}{V(s)} = \frac{s^{2} + 2s + 1}{s^{2} + 5s + 2} \]

Tip

Try both methods yourself: - Mesh: define loop currents, write equations by inspection. - Nodal: select unknown node voltages, write KCL using admittances.

This is great practice for exam-style problems.

If time allows, outline one of the methods on the board and assign the other as an in-class exercise.

Talk through checking: compare final expressions term-by-term to verify equivalence.

Skill-Assessment Example – Simple Op-Amp Transfer Functions

Given:

• \(Z_1(s)\) = impedance of a 10 μF capacitor: \(Z_1(s) = \dfrac{1}{C s} = \dfrac{1}{10\times 10^{-6} s} = \dfrac{10^5}{s}\).
• \(Z_2(s)\) = impedance of a 100 kΩ resistor: \(Z_2(s) = 10^5\ \Omega\).

1. Inverting amplifier (Figure 2.10(c)):

\[ G(s) = \frac{V_o(s)}{V_i(s)} = -\frac{Z_2}{Z_1} = -\frac{10^5}{10^5/s} = -s \]

2. Noninverting amplifier (Figure 2.12):

\[ G(s) = \frac{Z_1 + Z_2}{Z_1} = \frac{\dfrac{10^5}{s} + 10^5}{\dfrac{10^5}{s}} = s + 1 \]

Note

Result:

• Inverting configuration behaves like a differentiator: \(G(s) = -s\).
• Noninverting configuration behaves like a differentiator plus proportional term: \(G(s) = s + 1\).

Mention that practical differentiators need high-frequency roll-off and noise considerations; ideal \(s\)-behavior is a starting point.

This illustrates how simple op-amp + RC networks already implement fundamental control operations (P, D, and later I).

Real-World ECE Application: RLC Filters and Sensors

• A series RLC circuit like Example 2.6 can be used as a band-pass filter in audio or communication systems.
• The transfer function \(V_C(s)/V(s)\) shapes how different frequencies are passed or attenuated.
• In control systems, the same RLC can be part of a sensor conditioning circuit, where \(v_C(t)\) represents a measured physical quantity (e.g., strain gauge + filter).
• Designing the resistor, inductor, and capacitor values sets the “dynamic response” (rise time, overshoot) of the measured signal feeding the controller.

Use this slide to connect the math back to hardware: filter design, sensor interfaces, analog front-ends.

Ask: “If we want a faster sensor, which parameter might we reduce — \(L\), \(C\), or \(R\)? What does that do to \(\omega_n\) and damping?”

Summary / Key Points

• R, L, C elements have simple Laplace-domain impedances: \(Z_R = R\), \(Z_L = Ls\), \(Z_C = 1/(Cs)\).
• We can model circuits in the \(s\)-domain by replacing each element with \(Z(s)\) and applying KVL/KCL just like in DC circuits.
• For simple circuits, transfer functions can be obtained via:
  • Differential equation → Laplace transform.
  • Mesh analysis (KVL in \(s\)-domain).
  • Nodal analysis (KCL in \(s\)-domain).
  • Voltage division using impedances.
• For multi-loop / multi-node circuits:
  • Mesh equations: diagonal terms = sum of mesh impedances; off-diagonals = negative common impedances.
  • Node equations: diagonal terms = sum of admittances at node; off-diagonals = negative common admittances.

• Op-amp circuits (inverting and noninverting) realize transfer functions:
  • Inverting: \(G(s) = -Z_2(s)/Z_1(s)\).
  • Noninverting (large \(A\)): \(G(s) = [Z_1(s)+Z_2(s)]/Z_1(s)\).
• These circuit transfer functions become building blocks for controllers (e.g., PID) and analog filters in control systems.

Reinforce the unifying theme: once we have \(G(s)\), we can treat circuits like any other dynamic system (mechanical, thermal, etc.), using the same control tools.

Preview: next, we’ll apply identical modeling ideas to mechanical systems via analogies (force–voltage, velocity–current).

Key Formulas Summary

R, L, C Impedances and Admittances

• Resistor:

\[ Z_R(s) = R, \quad Y_R(s) = \frac{1}{R} = G \]

• Capacitor:

\[ Z_C(s) = \frac{1}{Cs}, \quad Y_C(s) = Cs \]

• Inductor:

\[ Z_L(s) = Ls, \quad Y_L(s) = \frac{1}{Ls} \]

Generic Transfer Function Definition

• For input \(U(s)\) and output \(Y(s)\):

\[ G(s) = \frac{Y(s)}{U(s)} \]

Series RLC (capacitor output) – Example 2.6

\[ \frac{V_C(s)}{V(s)} = \frac{1/LC}{s^{2} + \dfrac{R}{L}s + \dfrac{1}{LC}} \]

Series Impedances in Loop Equation

• General KVL in \(s\)-domain:

\[ \bigl[\text{Sum of impedances}\bigr] I(s) = \bigl[\text{Sum of sources}\bigr] \]

Two-Mesh Pattern

\[ \begin{bmatrix} \sum Z_{\text{Mesh 1}} & -\sum Z_{\text{common}} \\ -\sum Z_{\text{common}} & \sum Z_{\text{Mesh 2}} \end{bmatrix} \begin{bmatrix} I_1(s) \\ I_2(s) \end{bmatrix} = \begin{bmatrix} \sum V_{\text{Mesh 1}} \\ \sum V_{\text{Mesh 2}} \end{bmatrix} \]

Two-Node Pattern (Admittances)

\[ \begin{bmatrix} \sum Y_{\text{Node 1}} & -\sum Y_{\text{common}} \\ -\sum Y_{\text{common}} & \sum Y_{\text{Node 2}} \end{bmatrix} \begin{bmatrix} V_1(s) \\ V_2(s) \end{bmatrix} = \begin{bmatrix} \sum I_{\text{Node 1}} \\ \sum I_{\text{Node 2}} \end{bmatrix} \]

Op-Amp Transfer Functions

• Inverting configuration:

\[ \frac{V_o(s)}{V_i(s)} = -\frac{Z_2(s)}{Z_1(s)} \]

• Noninverting configuration (large \(A\)):

\[ \frac{V_o(s)}{V_i(s)} = \frac{Z_1(s) + Z_2(s)}{Z_1(s)} \]

Encourage students to keep this formula slide handy as a reference while doing homework and labs, especially the \(Z(s)\) of R, L, C and the op-amp formulas.

This reduces algebra mistakes and lets them focus on modeling choices and interpretation of \(G(s)\).

Interactive Deck

Interactive Agenda

• Explore R, L, C impedances in Python (Pyodide).
• Experiment with pole locations of an RLC circuit and see their effect on \(G(s)\).
• Compare mesh vs. nodal analysis numerically.
• Interactively explore inverting and noninverting op-amp transfer functions.

This addendum is meant to be used after students have seen the main slides.

Encourage them to modify parameters, re-run cells, and observe how circuit behavior changes numerically and graphically.

Interactive: R, L, C Impedance Calculator

Use this live Python block to calculate impedances for R, L, C at a chosen value of \(s = \sigma + j\omega\).

Ask students to:

• Set \(\sigma = 0\) and vary \(\omega\) to see frequency-dependent behavior.
• Notice that \(|Z_L|\) increases with \(\omega\) and \(|Z_C|\) decreases.

Connect this to intuition from AC circuit analysis and Bode plots they’ll see later.

Reactive: Visualizing R, L, C Magnitudes vs Frequency

Use the sliders to change \(R\), \(L\), and \(C\) values, and see how the impedance magnitudes vary with frequency.

viewof R_val = Inputs.range([10, 5000], {step: 10, value: 1000, label: "R (Ω)"})
viewof L_val = Inputs.range([1e-4, 0.1], {step: 1e-4, value: 1e-2, label: "L (H)"})
viewof C_val = Inputs.range([1e-9, 1e-4], {step: 1e-9, value: 1e-6, label: "C (F)"})

Have students:

• Increase \(L\) and see \(|Z_L|\) grow faster with frequency.
• Increase \(C\) and see \(|Z_C|\) drop faster with frequency.

Link this to the behavior of filters and resonance in RLC circuits.

Interactive: Series RLC Transfer Function Coefficients

Recall for the series RLC with capacitor voltage output (Example 2.6):

\[ \frac{V_C(s)}{V(s)} = \frac{1/LC}{s^{2} + \dfrac{R}{L}s + \dfrac{1}{LC}} \]

Use Python to compute the denominator coefficients for your chosen \(R\), \(L\), \(C\).

Connect: - \(a_0 = \omega_n^2 = 1/(LC)\). - \(a_1 = 2 \zeta \omega_n = R/L\).

Students can change \(R\) and see how damping ratio changes qualitatively (although we don’t compute \(\zeta\) explicitly here).

Reactive: Poles of the Series RLC Transfer Function

Interactively vary \(R\), \(L\), and \(C\), and observe the pole locations of:

\[ G(s) = \frac{V_C(s)}{V(s)} = \frac{1/LC}{s^{2} + \frac{R}{L}s + \frac{1}{LC}} \]

viewof R_rlc = Inputs.range([0, 100], {step: 1, value: 10, label: "R (Ω)"})
viewof L_rlc = Inputs.range([1e-4, 1e-1], {step: 1e-4, value: 1e-3, label: "L (H)"})
viewof C_rlc = Inputs.range([1e-9, 1e-4], {step: 1e-9, value: 1e-6, label: "C (F)"})

Ask students:

• Reduce \(R\) towards 0. What happens to the poles? (Complex conjugate with small real part: oscillatory).
• Increase \(R\) large. Do the poles become real and move further left?

This visually ties component values to stability and damping behavior.

Interactive: Mesh vs. Nodal Equation Coefficients (Conceptual)

For the two-loop mesh system in Example 2.10, the equations:

\[ (R_1 + Ls) I_1(s) - Ls I_2(s) = V(s) \]

\[ - Ls I_1(s) + \left(Ls + R_2 + \frac{1}{Cs}\right) I_2(s) = 0 \]

Use Python to build the coefficient matrix symbolically for selected \(R_1\), \(R_2\), \(L\), \(C\) at a particular \(s\).

This reinforces the “pattern” of mesh equations and gives students a symbolic-algebra view.

They can change \(R_1, R_2, L, C\) and see how each term changes the matrix entries.

Reactive: Inverting Op-Amp – Frequency Response Magnitude

For an inverting op-amp:

\[ G(s) = -\frac{Z_2(s)}{Z_1(s)} \]

Here, choose:

• \(Z_1\) as a capacitor \(C_1\) (pure integrator).
• \(Z_2\) as a resistor \(R_2\).

Then:

\[ G(s) = -R_2 C_1 s \]

Magnitude grows linearly with \(\omega\). Explore this behavior interactively.

viewof R2_inv = Inputs.range([1e3, 1e6], {step: 1e3, value: 1e5, label: "R2 (Ω)"})
viewof C1_inv = Inputs.range([1e-8, 1e-4], {step: 1e-8, value: 1e-5, label: "C1 (F)"})

Students should see:

• Increasing \(R_2\) or \(C_1\) shifts the magnitude curve up.
• This illustrates how component choice sets differentiator gain.

You may also discuss why in practice high-frequency gain is limited to reduce noise amplification.

Interactive: Noninverting Op-Amp – DC Gain with Resistive Network

For large-\(A\) noninverting op-amp:

\[ G(s) = \frac{Z_1 + Z_2}{Z_1} \]

Let \(Z_1 = R_1\) and \(Z_2 = R_2\) (purely resistive):

\[ G = 1 + \frac{R_2}{R_1} \]

Use Python to compute the DC gain for selected resistor values.

Encourage students to try different resistor ratios and connect to amplifier design:

• For \(G = 2\), what ratio \(R_2/R_1\) do we need?
• For \(G = 10\)?

This also links to the special case used often in sensor amplifiers.

Reactive: Noninverting Op-Amp with Simple RC Feedback

Make \(Z_1 = R_1 + 1/(C_1 s)\) and \(Z_2 = R_2\) (for simplicity). Then approximately:

\[ G(s) = \frac{Z_1 + Z_2}{Z_1} = 1 + \frac{R_2}{R_1 + 1/(C_1 s)} \]

Interactively explore \(|G(j\omega)|\) as \(R_1\), \(R_2\), and \(C_1\) vary.

viewof R1_ni = Inputs.range([1e3, 1e6], {step: 1e3, value: 1e4, label: "R1 (Ω)"})
viewof R2_ni = Inputs.range([1e3, 1e6], {step: 1e3, value: 9e4, label: "R2 (Ω)"})
viewof C1_ni = Inputs.range([1e-9, 1e-4], {step: 1e-9, value: 1e-6, label: "C1 (F)"})

Students can see how adding a capacitor in series with \(R_1\) introduces frequency dependence into the gain.

This is a simple step towards understanding frequency-compensated amplifiers and lead/lag compensators in control.

Wrap-Up Interactive Check

Use the block below as a quick self-check: given \(R\), \(L\), \(C\) and the form

\[ G(s) = \frac{1/LC}{s^{2} + \frac{R}{L}s + \frac{1}{LC}}, \]

numerically evaluate \(|G(j\omega)|\) at a chosen frequency.

Encourage students to compare this numeric magnitude with what they see on the pole plot and Bode-like plots earlier.

This ties everything—component values, transfer function form, poles, and numerical frequency response—into a coherent picture.