---
title: "Block diagram and Vo/Vi of L-R ladder network"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:bd-l1-r1-l2-r2"
source: https://www.bctengineer.com/q/semester_3/control_systems/bd-l1-r1-l2-r2
site: BCT Engineering
---

# Block diagram and Vo/Vi of L-R ladder network

Control Systems · Chapter 2: Component Modeling

- Asked in: 2070 Chaitra (old course), 6 marks
- Video: [Problem on Transfer Function of Electrical Network](https://www.youtube.com/watch?v=5ii5WAfAGRs) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/bd-l1-r1-l2-r2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Draw the block diagram and reduce it to calculate Vo(s)/Vi(s) for the following network. [Figure: input Vi; series L1 to node 1; R1 from node 1 to ground; series L2 from node 1 to node 2; R2 from node 2 to ground; Vo across R2]

## Short answer

The block diagram is built from I1 = (Vi - V1)/(sL1), V1 = R1(I1 - I2), I2 = (V1 - Vo)/(sL2), Vo = R2 I2. Reduction gives Vo/Vi = R1 R2 / [L1 L2 s^2 + (L1 R1 + L1 R2 + L2 R1)s + R1 R2], a second-order low-pass filter.

## Answer

Let $I_1$ = current in $L_1$, $V_1$ = voltage across $R_1$ (node 1), $I_2$ = current in $L_2$ and $R_2$, $V_o$ = voltage across $R_2$.

#### Element equations (Laplace)
$$
\begin{aligned}
I_1 &= \frac{V_i - V_1}{sL_1} \\
V_1 &= R_1(I_1 - I_2) \\
I_2 &= \frac{V_1 - V_o}{sL_2} \\
V_o &= R_2I_2
\end{aligned}
$$

#### Block diagram
```text
 Vi +   ┌─────┐ I1 +   ┌────┐ V1 +   ┌─────┐ I2  ┌────┐ Vo
 ──►(Σ)►│1/sL1├──►(Σ)─►│ R1 ├─┬─►(Σ)►│1/sL2├──┬─►│ R2 ├─┬─►
    ▲-  └─────┘    ▲-  └────┘ │   ▲-  └─────┘  │  └────┘ │
    │              │          │   │            │         │
    │              └──────────┼───┼────────────┘ I2      │
    └────── V1 ──────────────┘   └──────── Vo ──────────┘
```

#### Reduction step by step
1. Last loop ($I_2 \to V_o$ fed back): forward $\dfrac{R_2}{sL_2}$, unity feedback:
$$
\frac{V_o}{V_1} = \frac{R_2/sL_2}{1 + R_2/sL_2} = \frac{R_2}{sL_2 + R_2}, \qquad \frac{I_2}{V_1} = \frac{1}{sL_2 + R_2}
$$
2. Middle loop: forward $R_1$ from $(I_1 - I_2)$ to $V_1$, feedback $\dfrac{1}{sL_2 + R_2}$:
$$
\frac{V_1}{I_1} = \frac{R_1}{1 + \frac{R_1}{sL_2 + R_2}} = \frac{R_1(sL_2 + R_2)}{sL_2 + R_1 + R_2}
$$
3. First loop: forward $\dfrac{1}{sL_1}\cdot\dfrac{V_1}{I_1}$, unity feedback:
$$
\frac{V_1}{V_i} = \frac{R_1(sL_2 + R_2)}{sL_1(sL_2 + R_1 + R_2) + R_1(sL_2 + R_2)}
$$
4. Multiply by $\dfrac{V_o}{V_1} = \dfrac{R_2}{sL_2 + R_2}$:

$$
\frac{V_o(s)}{V_i(s)} = \frac{R_1R_2}{L_1L_2s^2 + (L_1R_1 + L_1R_2 + L_2R_1)s + R_1R_2}
$$
At DC the gain is 1 (inductors act as shorts), so this is a second-order low-pass network.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which equation gives the current in L2?
   A. I2 = (V1 - Vo)/(s L2)
   B. I2 = V1/R1
   C. I2 = (Vi - V1)/(s L2)
   D. I2 = Vo/(s L2)
   Answer: A. The voltage across L2 is V1 - Vo, so its current is that divided by s L2. (Key point: Element equations)

2. Reducing the last loop (forward R2/(sL2), unity feedback) gives Vo/V1 equal to
   A. R2/(s L2)
   B. R2/(s L2 + R2)
   C. s L2/(s L2 + R2)
   D. 1/(s L2 + R2)
   Answer: B. G/(1+G) with G = R2/(sL2) gives R2/(sL2 + R2). (Key point: Inner loop reduction)

3. The final Vo/Vi has numerator
   A. R1 + R2
   B. R1 R2
   C. L1 L2
   D. s R1 R2
   Answer: B. Vo/Vi = R1R2 / [L1L2 s^2 + (L1R1 + L1R2 + L2R1)s + R1R2]. (Key point: Final transfer function)

4. Why is the DC gain of the network 1?
   A. Inductors act as shorts at DC, so Vo = Vi
   B. Inductors act as open circuits at DC
   C. R1 equals R2
   D. Capacitors charge fully
   Answer: A. At s = 0 the transfer function reduces to R1R2/R1R2 = 1, consistent with inductors being shorts. (Key point: DC gain)

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