Transfer Function of a Translational Mechanical System: \displaystyle Gs=\frac{V1s}{Fs} | AI Research

Transfer Function of a Translational Mechanical System: $\displaystyle G(s)=\frac{V_1(s)}{F(s)}$

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Sep 11, 2026

This section develops the transfer function between the applied force $F(s)$ and the velocity output $V_1(s)$ for the translational mechanical system shown in the figure.

The system contains a mass, viscous dampers, and springs. Because the diagram image does not expose readable component labels in the prompt, the derivation is presented for the standard interpretation of the figure: a mass with velocity $v_1(t)$ , connected to ground through a viscous damper and spring, with an additional damper connected to the applied-force node. Let

$m$ be the mass,
$b_1$ be the damper connected to ground,
$b_2$ be the damper associated with the force-input node,
$k$ be the spring stiffness,
$F(t)$ be the applied force,
$v_1(t)$ be the output velocity.

A transfer function is obtained by first writing the governing differential equation and then applying the Laplace transform. For a linear mass–damper–spring system, Newton’s second law gives the familiar force balance $m\ddot{x}+b\dot{x}+kx=F(t)$ . The force-to-displacement form is therefore $X(s)/F(s)=1/(ms^2+bs+k)$ .

Since velocity is the derivative of displacement,

v_1(t)=\dot{x}(t)

and, under zero initial conditions,

V_1(s)=sX(s).

Thus, once $X(s)/F(s)$ is known, the velocity transfer function follows by multiplying by $s$ .

The Mass-Damper-Spring System — Engineering LibreTexts - Derives Newton’s-law equation for a translational mass–damper–spring system. ↩
How to Find the Transfer Function of a System - Shows the derivation of $X(s)/F(s)=1/(ms^2+cs+k)$ . ↩

Translational Mechanical Systems: Solved Example

Physical modeling principles

The three ideal translational elements obey the following constitutive relations:

Element	Time-domain relation	Laplace-domain relation	Mechanical impedance
Mass	$F_m(t)=m\dot{v}(t)$	$F_m(s)=msV(s)$	$Z_m(s)=ms$
Damper	$F_b(t)=bv(t)$	$F_b(s)=bV(s)$	$Z_b(s)=b$
Spring	$F_k(t)=kx(t)$	$F_k(s)=\dfrac{k}{s}V(s)$	$Z_k(s)=\dfrac{k}{s}$

Here, mechanical impedance is defined as

Z(s)=\frac{F(s)}{V(s)}.

The mass stores kinetic energy, the spring stores potential energy, and the damper dissipates energy. These roles explain why the mass contributes a factor proportional to $s$ , the damper contributes a constant, and the spring contributes $1/s$ when velocity is used as the system variable.

For the standard arrangement, the force balance can be expressed using the equivalent damping

b_{\mathrm{eq}}=b_1+b_2.

The equation of motion is then

m\dot{v}_1(t)+b_{\mathrm{eq}}v_1(t)+kx(t)=F(t).

Because

v_1(t)=\dot{x}(t),

the spring displacement can be written as

x(t)=\int_0^t v_1(\tau)\,d\tau

for zero initial displacement.

Control Systems Engineering, Chapter 2 - Presents force–velocity relationships and mechanical impedances for masses, dampers, and springs. ↩

Key modeling shortcut

If the required output is velocity, use mechanical impedance directly: mass contributes ms, damper contributes b, and spring contributes k/s.

Derivation of $\u005cfrac{V_1(s)}{F(s)}$

1
Step 1
Take the direction of the applied force $F(t)$ as positive and define the mass velocity in this direction as $v_1(t)$ .
2
Step 2
The applied force is opposed by the inertial, viscous, and spring forces. Therefore, $F(t)=m\dot{v}_1(t)+b_{\mathrm{eq}}v_1(t)+kx(t)$ , where $b_{\mathrm{eq}}=b_1+b_2$ .
3
Step 3
Because $v_1(t)=\dot{x}(t)$ , the Laplace-domain relation under zero initial conditions is $X(s)=V_1(s)/s$ .
4
Step 4
Transforming the force balance gives $F(s)=msV_1(s)+b_{\mathrm{eq}}V_1(s)+kX(s)$ .
5
Step 5
Substitute $X(s)=V_1(s)/s$ to obtain $F(s)=\left(ms+b_{\mathrm{eq}}+\frac{k}{s}\right)V_1(s)$ .
6
Step 6
Rearranging yields $\frac{V_1(s)}{F(s)}=\frac{1}{ms+b_{\mathrm{eq}}+k/s}$ . Multiplying numerator and denominator by $s$ gives the polynomial form $\frac{V_1(s)}{F(s)}=\frac{s}{ms^2+b_{\mathrm{eq}}s+k}$ .

Final transfer function

With

b_{\mathrm{eq}}=b_1+b_2,

the transfer function is

\boxed{ G(s)=\frac{V_1(s)}{F(s)} =\frac{1}{ms+(b_1+b_2)+\dfrac{k}{s}} }

or, equivalently,

\boxed{ G(s)=\frac{V_1(s)}{F(s)} =\frac{s}{ms^2+(b_1+b_2)s+k} }

This is the required force-to-velocity transfer function for the standard translational mechanical arrangement.

The corresponding force-to-displacement transfer function is

\frac{X(s)}{F(s)} = \frac{1}{ms^2+(b_1+b_2)s+k}.

The velocity result follows immediately:

\frac{V_1(s)}{F(s)} = s\frac{X(s)}{F(s)} = \frac{s}{ms^2+(b_1+b_2)s+k}.

The numerator zero at $s=0$ is physically meaningful: a constant force applied to a stable spring–damper system eventually produces a static displacement, but the final velocity returns to zero.

Impedance interpretation

The same result can be obtained without explicitly writing displacement.

The mass impedance is

Z_m(s)=ms.

The damper impedances add when the dampers oppose the same velocity:

Z_b(s)=b_1+b_2.

The spring impedance is

Z_k(s)=\frac{k}{s}.

Therefore, the total mechanical impedance is

Z_{\mathrm{eq}}(s) = ms+(b_1+b_2)+\frac{k}{s}.

Since

Z_{\mathrm{eq}}(s)=\frac{F(s)}{V_1(s)},

the admittance, or force-to-velocity transfer function, is

G(s)=\frac{V_1(s)}{F(s)} = \frac{1}{Z_{\mathrm{eq}}(s)}.

Contribution of Each Mechanical Element

Terms appearing in the total force-to-velocity impedance

Dimensional and physical checks

A correct transfer function should pass several checks.

1. Units

Force has units of newtons, and velocity has units of meters per second. Therefore,

\left[\frac{V_1}{F}\right] = \frac{\mathrm{m/s}}{\mathrm{N}} = \frac{\mathrm{s}}{\mathrm{kg}}.

For the result

\frac{s}{ms^2+b_{\mathrm{eq}}s+k},

each denominator term has units of $\mathrm{kg/s^2}$ after treating $s$ as having units of $\mathrm{s^{-1}}$ , so the overall units are consistent.

2. Low-frequency behavior

As $s\to 0$ ,

G(s)=\frac{s}{ms^2+b_{\mathrm{eq}}s+k}\to 0.

A constant force does not produce a permanent velocity when a restoring spring is present.

3. High-frequency behavior

As $s\to\infty$ ,

G(s)\approx\frac{s}{ms^2}=\frac{1}{ms}.

At very high frequencies, inertia dominates and the system behaves approximately like a mass.

4. Pole locations

The poles satisfy

ms^2+b_{\mathrm{eq}}s+k=0.

Thus,

s_{1,2} = \frac{-b_{\mathrm{eq}}\pm \sqrt{b_{\mathrm{eq}}^2-4mk}}{2m}.

For positive $m$ , $b_{\mathrm{eq}}$ , and $k$ , the poles lie in the left half-plane, indicating asymptotic stability for the ideal linear model.

System Poles and Zeros — Engineering LibreTexts - Discusses the characteristic polynomial and pole locations of the mass–spring–damper transfer function. ↩

Common mistakes and edge cases

Do not confuse displacement and velocity outputs

The expressions 1/(ms² + bs + k) and s/(ms² + bs + k) describe different outputs. The first is X(s)/F(s); the second is V1(s)/F(s).

Generalization to a single equivalent damper

If the diagram uses a single damper coefficient $b$ rather than two dampers, substitute

b_{\mathrm{eq}}=b.

The result becomes

\boxed{ G(s)=\frac{V_1(s)}{F(s)} = \frac{s}{ms^2+bs+k} }

This is the standard mass–spring–damper force-to-velocity transfer function reported in control-system modeling references.3

If the figure labels the damping coefficients differently, the same procedure applies: identify every damper that experiences the mass velocity $v_1(t)$ , add those coefficients, and substitute their sum into the denominator.

The Mass-Damper-Spring System — Engineering LibreTexts - Derives Newton’s-law equation for a translational mass–damper–spring system. ↩
How to Find the Transfer Function of a System - Shows the derivation of $X(s)/F(s)=1/(ms^2+cs+k)$ . ↩
Mass Spring Damper System - Provides a transfer-function formulation and state-space interpretation of the mass–spring–damper model. ↩

Mechanical-System Transfer Function Review

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Question · Term

What is the impedance of a mass?

Click to reveal

Answer · Definition

$Z_m(s)=ms$ .

Knowledge Check

Question 1 of 4

Q1Single choice

For a mass–spring–damper system, what is the transfer function from applied force to displacement?

$\frac{1}{ms^2+bs+k}$

$\frac{s}{ms^2+bs+k}$

$ms+b+\frac{k}{s}$

$ms^2+bs+k$

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Transfer Function of a Translational Mechanical System: G(s)=V1(s)F(s)\displaystyle G(s)=\frac{V_1(s)}{F(s)}G(s)=F(s)V1​(s)​

Footnotes

Translational Mechanical Systems: Solved Example

Physical modeling principles

Footnotes

Key modeling shortcut

Derivation of $\u005cfrac{V_1(s)}{F(s)}$

Final transfer function

Impedance interpretation

Contribution of Each Mechanical Element

Dimensional and physical checks

1. Units

2. Low-frequency behavior

3. High-frequency behavior

4. Pole locations

Footnotes

Common mistakes and edge cases

Do not confuse displacement and velocity outputs

Generalization to a single equivalent damper

Footnotes

Mechanical-System Transfer Function Review

What is the impedance of a mass?

Knowledge Check

References

Explore Related Topics

Transfer Function of a Translational Mechanical System: $\displaystyle G(s)=\frac{V_1(s)}{F(s)}$