---
title: "Single-mass system transfer function with F-V and F-I"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-fv-fi-k1-m2-k2b2"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-fi-k1-m2-k2b2
site: BCT Engineering
---

# Single-mass system transfer function with F-V and F-I

Control Systems · Chapter 2: Component Modeling

- Asked in: 2078 Kartik, 4+2+2 marks
- Video: [2.Force Current and Force Voltage Analogy solved problem with system equation](https://www.youtube.com/watch?v=jgCfYudsIFY) — ELECTRICAL HUB
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-fv-fi-k1-m2-k2b2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find transfer function for the system as in figure. Also develop F-V and F-I analogy circuits. [Figure: force f(t) applied at a free end (displacement x1(t)) connected through spring K1 to mass M2 (displacement x2(t)); M2 connected to the fixed wall through spring K2 in parallel with damper B2]

## Short answer

Transfer functions: X2(s)/F(s) = 1/(M2 s^2 + B2 s + K2), X1(s)/F(s) = (M2 s^2 + B2 s + K1+K2) / [K1 (M2 s^2 + B2 s + K2)]. In F-V analogy, K1 becomes 1/C1 shared between meshes; in F-I analogy, K1 becomes 1/L1.

## Answer

Displacements: $x_1$ (free end where $f$ acts) and $x_2$ (mass $M_2$). The free end has no mass.

#### Equations of motion
At the massless end, the spring force equals the applied force:
$$
K_1(x_1 - x_2) = f(t)
$$
For $M_2$:
$$
M_2\ddot x_2 + B_2\dot x_2 + K_2x_2 + K_1(x_2 - x_1) = 0
$$

#### Transfer functions
Laplace form:
$$
\begin{aligned}
K_1X_1 - K_1X_2 &= F \\
-K_1X_1 + (M_2s^2 + B_2s + K_1 + K_2)X_2 &= 0
\end{aligned}
$$
Adding the two equations: $(M_2s^2 + B_2s + K_2)X_2 = F$, so
$$
\frac{X_2(s)}{F(s)} = \frac{1}{M_2s^2 + B_2s + K_2}
$$
and since $X_1 = X_2 + F/K_1$:
$$
\frac{X_1(s)}{F(s)} = \frac{M_2s^2 + B_2s + K_1 + K_2}{K_1(M_2s^2 + B_2s + K_2)}
$$
The spring $K_1$ only passes the force on; it does not change the dynamics of $M_2$.

#### F-V analogy
$f\to e$, $M\to L$, $B\to R$, $K\to 1/C$. Mesh 1 (velocity $\dot x_1$) and mesh 2 ($\dot x_2$) share $C_1 = 1/K_1$.
```text
                   L2=M2   R2=B2   C2=1/K2
 +--------+--------LLLL----/\/\-----||---+
 |        |                              |
(~) e   C1=1/K1                          |
 |   i1   |              i2              |
 +--------+------------------------------+
```
$$
\frac{1}{C_1}\int(i_1 - i_2)dt = e, \qquad L_2\frac{di_2}{dt} + R_2i_2 + \frac{1}{C_2}\int i_2dt + \frac{1}{C_1}\int(i_2 - i_1)dt = 0
$$

#### F-I analogy
$f\to i$, $M\to C$, $B\to 1/R$, $K\to 1/L$.
```text
 v1  L1   v2
 +--LLLL--+----+----+
 |        |    |    |
(i)       C2   R2   L2
 |        |    |    |
 +--------+----+----+ gnd
```
($L_1 = 1/K_1$, $C_2 = M_2$, $R_2 = 1/B_2$, $L_2 = 1/K_2$.)
$$
\frac{1}{L_1}\int(v_1 - v_2)dt = i, \qquad C_2\frac{dv_2}{dt} + \frac{v_2}{R_2} + \frac{1}{L_2}\int v_2dt + \frac{1}{L_1}\int(v_2 - v_1)dt = 0
$$

## Check yourself

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

1. What equation holds at the massless free end where f acts?
   A. K1(x1 - x2) = f(t)
   B. M1 x1'' = f(t)
   C. B1 x1' = f(t)
   D. K1 x1 = f(t)
   Answer: A. With no mass, the spring force K1(x1 - x2) must equal the applied force. (Key point: Massless end)

2. What is X2(s)/F(s)?
   A. 1/(M2 s^2 + B2 s + K2)
   B. 1/(M2 s^2 + B2 s + K1 + K2)
   C. K1/(M2 s^2 + B2 s + K1)
   D. 1/(M2 s^2 + B2 s + K1)
   Answer: A. Adding the two Laplace equations cancels K1 terms, leaving (M2s^2+B2s+K2)X2 = F. (Key point: Result for X2/F)

3. What is the effect of K1 on the dynamics of M2?
   A. None; it only transmits the force
   B. It adds stiffness K1 to M2
   C. It adds damping
   D. It adds mass
   Answer: A. Since the end is massless, K1 only passes the force on and does not appear in X2/F. (Key point: Role of K1)

4. In the F-I analogy, how are M2, B2 and K2 represented?
   A. C = M2, R = 1/B2, L = 1/K2
   B. L = M2, R = B2, C = 1/K2
   C. C = M2, R = B2, L = K2
   D. R = M2, L = B2, C = K2
   Answer: A. Force-current analogy: M to C, B to 1/R, K to 1/L, with velocity as node voltage. (Key point: F-I analogy)

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