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

# Block diagram model of R-L ladder network

Control Systems · Chapter 2: Component Modeling

- Asked in: 2071 Chaitra, 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-r1-l1-r2-l2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Develop block diagram model for the system below. [Figure: input Vin; series R1 to node 1; L1 from node 1 to ground; series R2 from node 1 to node 2; L2 from node 2 to ground; output VL across L2]

## Short answer

The block diagram uses summing junctions and blocks for 1/R1, sL1, 1/R2, sL2 with feedback loops. The overall transfer function is VL(s)/Vin(s) = L1L2 s^2 / [L1L2 s^2 + (L1R1 + L1R2 + L2R1)s + R1R2], a second-order high-pass filter.

## Answer

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

#### Element equations (Laplace)
$$
\begin{aligned}
I_1(s) &= \frac{V_{in}(s) - V_1(s)}{R_1} \\
V_1(s) &= sL_1\left[I_1(s) - I_2(s)\right] \\
I_2(s) &= \frac{V_1(s) - V_L(s)}{R_2} \\
V_L(s) &= sL_2\,I_2(s)
\end{aligned}
$$

#### Block diagram
```text
 Vin─►(Σ)►[1/R1]─►(Σ)►[sL1]─┬─►(Σ)►[1/R2]─┬─►[sL2]─┬─►VL
       ▲-          ▲-       │   ▲-        │        │
       │           └── I2 ──┼───┼─────────┘        │
       └── V1 ──────────────┘   └── VL ────────────┘
```

#### Reduction
Each block pair forms a simple loop; let
- loop $L_a = -\dfrac{sL_1}{R_1}$ (via $V_1$),
- loop $L_b = -\dfrac{sL_1}{R_2}$ (via $I_2$),
- loop $L_c = -\dfrac{sL_2}{R_2}$ (via $V_L$).

Only $L_a$ and $L_c$ do not touch. Forward path $P = \dfrac{1}{R_1}\cdot sL_1\cdot\dfrac{1}{R_2}\cdot sL_2$, with $\Delta_1 = 1$. Using Mason's rule (same result as stepwise reduction):
$$
\frac{V_L}{V_{in}} = \frac{\frac{s^2L_1L_2}{R_1R_2}}{1 + \frac{sL_1}{R_1} + \frac{sL_1}{R_2} + \frac{sL_2}{R_2} + \frac{s^2L_1L_2}{R_1R_2}}
$$
Multiplying by $R_1R_2$:
$$
\frac{V_L(s)}{V_{in}(s)} = \frac{L_1L_2s^2}{L_1L_2s^2 + (L_1R_1 + L_1R_2 + L_2R_1)s + R_1R_2}
$$
This is a second-order high-pass network: the output is zero at DC (inductors short) and approaches $V_{in}$ at high frequency.

## Check yourself

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

1. V1 across L1 is:
   A. sL1 (I1 - I2)
   B. (Vin - V1)/R1
   C. sL1 I1
   D. I2 R2
   Answer: A. L1 carries the difference I1 - I2. (Key point: Element equation)

2. Which loops do not touch?
   A. La and Lc
   B. Lb and Lc
   C. La and Lb
   D. All touch
   Answer: A. La (via V1) and Lc (via VL) share no node. (Key point: Mason non-touching)

3. The network is:
   A. Second-order high-pass, zero at DC
   B. Unity gain
   C. Low-pass
   D. Band-stop
   Answer: A. At DC inductors short, so VL=0; at high frequency VL approaches Vin. (Key point: Behaviour)

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