In Theory

Basics of suspension transfer function

Suspension plays a pivotal role in ensuring ride safety and comfort and hugely affects the driver’s confidence level and road feel. Since its invention a few centuries ago, there have been many different designs created by brilliant engineers and inventors. Some were mere evolutions of previous concepts, while others were poles apart.

Every suspension type was created with a particular goal in mind and was the result of a balancing act between cost, durability, performance, and other factors. But in informal conversations, you often hear absolutes, such as ‘stiffer suspension means better handling’, or ‘pushrod suspension is better’. If you ask why, though, the answer usually boils down to ‘it just is’. But rarely is our reality black or white, and suspension performance is no exception.

There are many aspects to suspension performance, but perhaps the most important one is its ability to absorb road irregularities. Not only does it hugely influence ride comfort and safety, but vehicle longevity and reliability as well.

Dynamic systems and transfer functions

So how do we go about the problem of characterizing the suspension’s ability to swallow up road imperfactions and offer that buttery smooth, magical carpet ride? Well, there are many ways to skin a cat, but the most common approach is to state the amplitude of the up-and-down motion of the vehicle if the wheel is undergoing a forced oscillating motion of a known frequency. By giving this relation for all possible frequencies, we arrive at the so-called suspension transfer function.

Suspension transfer function describes how the amplitude of vertical vehicle oscillations change in relation to the driving signal.

Transfer functions are a standard tool in control theory and analysis of dynamical systems. Their importance stems from the fact that input to the system (in our case wheel displacement at a given time) can be thought of as a sum of many oscillations of different frequencies. For linear systems, that enables us to compute the steady-state system output (in our case the vehicle vertical displacement), provided we also know the resulting phase shifts. But even for nonlinear systems, transfer functions can be used to at least estimate the magnitude of the output – a very valuable insight on its own right.

Transfer functions can be obtained either by theoretical considerations or direct measurement. They are thus useful in validating theoretical models and improving system design, provided we can find their form for a problem at hand. The standard approach to derive them is to use Laplace transforms, but for simple systems, such as some suspension designs, it is possible (and also instructive) to simply solve the underlying equations of motion – an approach we’ll use here.

Modelling suspension dynamics

A real vehicle is a three-dimensional object, and modeling its full dynamics involves at least six degrees of freedom: displacements along three mutually perpendicular axes and rotations around those. Even if we limit ourselves to vertical displacements, we are still left with three possible motions: up-and-down, back-and-forth, and side-to-side rocking. However, when thinking about the suspension transfer function, we focus on one-dimensional dynamics by replacing the load on the suspension with a single mass that is only allowed to move vertically – the so-called sprung mass.

In a way, this approach is equivalent to taking a car, cutting it into four independent quarters, and constraining the motion of each so that it can only move vertically. That allows us to analyze suspension independently of other vehicle characteristics and is so widespread that it has its own name: the quarter car model. It is not, however, limited to cars only – it can be used for other vehicles, such as motorcycles and bikes, just as well. Here, we’re gonna use it to derive the transfer function for models with one and two degrees of freedom.

One degree of freedom model

Let’s begin by considering one of the simplest suspension setups possible, that is one with only a single shock-absorber and a single damper. This setup is surprisingly common and can be found in all road and hardtail bicycles, some old and vintage-style motorcycles, and, more recently, electric scooters. In this case, the tire plays the role of both the spring and the damper, and we can model it in exactly this way: a sprung mass M attached to the road by a spring of stiffness k and damping coefficient c.

A simple suspension system can be modelled as a single mass connected to the ground with a string and a damper. Original picture by Dave_S licensed under cc-by-2.0.

To derive the equation of motion of the suspended mass, we need to identify all the forces acting on it. In this case, there are three: the spring force, the damping force, and the gravity force. The last one, however, can be disregarded if we measure the displacement of the mass, y(t), from its equilibrium position. If we denote by y_d(t) the vertical road displacement at time t, the Newton equation takes the following form:

    \[M \frac{d^2 y}{dt^2} = -k (y-y_d) - c \frac{d}{dt}(y - y_d)\]

Introducing the natural frequency \omega_0 = \sqrt{k/M}, and the damping ratio \zeta = c/(2 \sqrt{kM}), we can rearrangethe above equation into a more standard form:

    \[\frac{d^2y}{dt^2} + 2 \zeta \omega_0 \frac{dy}{dt} + \omega_0^2  y =\omega_0^2 y_d + 2 \zeta \omega_0 \frac{d y_d}{dt}\]

We’re interrested in response to a particular driving frequency, thus we take y_d as

    \[y_d(t) = A \sin{\omega t}\]

which would correspond to riding at speed v on a sinusoidal wave of length 2 \pi v / \omega.

The solution

The solution to the Newton equation can be found directly, and has the following form:

    \begin{align*}y(t) = & e^{-t \left(\sqrt{\left(\zeta ^2-1\right) \omega_0^2}+\zeta \omega_0\right)} \left(c_2 e^{2 t \sqrt{\left(\zeta ^2-1\right) \omega_0^2}}+c_1\right) + \\& A \frac{\omega_0 \left(\omega_0 \left(\left(4 \zeta ^2-1\right) \omega ^2+\omega_0^2\right) \sin (t \omega )-2 \zeta \omega ^3 \cos (t \omega )\right)}{2 \left(2 \zeta ^2-1\right) \omega ^2 \omega_0^2+\omega ^4+\omega_0^4}\end{align*}

As you can see, it consists of two parts. The first is a transient, exponentially decaying component of little interest to us (at least here). The second is a steady-state part and is an oscillating motion of the same frequency as the driving signal. It’s worth mentioning that this is a general property of damped oscillating systems, one that will come in handy later on.

To get the transfer function, we focus on the oscillating part and note the following identity:

    \begin{equation*}A_s \sin{x} + A_c \cos{x} = \sqrt{A_s^2 + A_c^2} \sin{x+\phi}\end{equation*}


    \begin{align*}\cos{\phi} &=  \frac{A_s}{\sqrt{A_s^2 + A_c^2}} \\\sin{\phi} &=  \frac{A_c}{\sqrt{A_s^2 + A_c^2}} \\\end{align*}

Using the above we can relate the amplitude of the oscillations of the suspended mass to the amplitude of the driving signal. After some straightforward manipulation we arrive at the following:

    \[T(n) = \sqrt{\frac{1+(2 n \zeta)^2}{(1-n^2)^2+(2 \zeta n)^2}}\]

where n=\omega/\omega_0 is the normilized frequency.


The transfer function depends on the normalized frequency as an argument, and on the damping ratio as a parameter. The latter can be tuned either by a different choice of the spring element, e.g. changing the tyre to a different one or altering the damper settings in more advanced systems. Let’s take a look at the plot of the suspension transfer function for various damping ratios:

Suspesion transfer function in the one degree of freedom model. A strong resonance develops around the natural frequency (i.e. n=1), but even light damping brings it to acceptable levels.

The first thing to note is the existence of a strong resonance around n=1 for low damping coefficients. It is quickly suppressed, though, as more damping is introduced in the suspension. It vanishes completely for an infinite damping ratio, but this corresponds to replacing the suspension with a solid block – as interesting as watching paint dry.

Second, note that regardless of the damping ratio, frequencies below the natural oscillation frequency are amplified, while those above it are suppressed. This imbalance is greater the higher the damping ratio, before being completely uniform for infinite damping. In real applications, though, damping ratios significantly bigger than 1 are uncommon. For \zeta=1 the transfer function drops quickly to 0: at n=20, it is already less than 0.1, and just shy of 0.04 at n=50.

This behaviour has an interesting physical interpretation: as the driving frequency grows, the force acting on the suspension approach infinity. However, the time it acts on the sprung mass goes rapidly to zero before reversing direction, which results in zero net movement of the mass.

Two degrees of freedom model

A far more common suspension setup involves two shock-absorbing elements and two dampers connected in series. Such an arrangement is what you’d find in most modern vehicles, including motorcycles, cars, trains, and even aircraft. Suspension forks in bicycles also fall into this category.

The two degrees of freedom model accounts for the influence of the unsprung mass on the dynamics of the suspension system. Original picture by Leon Wilson licensed under cc-by-2.0.

Including the second mass in the model allows for much richer dynamics, as we will see in a bit. We begin again by writing down the equations of motion for the sprung and unsprung masses in exactly the same way as before:

    \begin{align*}m_1 \frac{d^2 y_1}{dt^2} &= -k_1(y_1-y_2) - c_1 \frac{d}{dt} (y_1-y_2) \\m_2 \frac{d^2 y_2}{dt^2} &= k_1(y_1-y_2) + c_1 \frac{d}{dt} (y_1-y_2) - k_2(y_2 - y_d) -c_2\frac{d}{dt} (y_2-y_d)\end{align*}

where m_1 is the sprung mass, m_2 is the unsprung mass, k_1 and k_2 are spring rates for the ‘main suspension’ and the ‘tire’, respectively, and c_1 and c_2 are their corresponding damping coefficients. Just as before, y_d(t) is the elevation of the road at time t.

The above system can be represented in a matrix form and solved directly, but we’re only interested in the steady-state part of the solution. As before, we can simply assume it to be proportional to the driving signal with a possible phase shift. Furthermore, instead of using sines and cosines as the driving signal, we can move into the complex domain by taking y_d to have the form

    \[y_d(t) = A e^{i \omega t}\]

in this case the steady-state solutions can be assumed to be of the form

    \begin{align*}y_1(t) &= y_{1,0} e^{i (\omega t + \phi_1)} \\y_2(t) &= y_{2,0} e^{i (\omega t + \phi_2)}\end{align*}

where \phi_1 and \phi_2 are phase shifts.

The solution

Plugging the steady-state solutions into the equations of motion we arrive at a system of two linear equations over complex numbers with two unknowns, y_{1,0} and y_{2,0}. Obtaining the transfer functions is now as easy as taking the absolute values of those two unknowns and dividing them by A. Doing so, we obtain the following, rather convoluted, expressions:

    \begin{align*}&T_1(\omega) = \\&\frac{\sqrt{c_1^2 \omega ^2+k_1^2} \sqrt{c_2^2 \omega ^2+k_2^2}}{\sqrt{\omega ^2 \left(c_1 k_2+c_2 k_1 -\left(\omega ^2 (c_1 M+c_2 m_1)\right)\right)^2+\left(k_1 k_2+m_1 m_2 \omega ^4 -\omega ^2 (c_1 c_2+k_1 M+k_2 m_1)\right)^2}}\end{align*}


    \begin{align*}&T_2(\omega) = \\& \frac{\sqrt{c_2^2 \omega ^2+k_2^2} \sqrt{c_1^2 \omega ^2+\left(k_1-m_1 \omega ^2\right)^2}}{\sqrt{\omega ^2 \left(c_1 k_2+c_2 k_1 -\left(\omega ^2 (c_1 M+c_2 m_1)\right)\right)^2+\left(k_1 k_2+m_1 m_2 \omega ^4 -\omega ^2 (c_1 c_2+k_1 M+k_2 m_1) \right)^2}}\end{align*}

where M= m_1+m_2 is the total mass of the system.


The transfer functions in this model are indeed pretty nasty. We could use damping ratios and natural frequencies as in the one-dimensional model, but the resulting expressions would be even longer. We can, however, plot the transfer functions for particular cases of interest.

First let’s take a look at both transfer functions, both for the sprung and unsprung masses, without any damping, with frequency normalized to natural frequency of the sprung mass:

Without damping, the model displays resonances around natural freqencies of the suspension system. The second resonance at around n=9 was not present in the one degree of freedom model.

As you can see, we now experience two resonances, corresponding to two oscillation modes. This phenomenon was not present in the previous model, but it’s a general characteristic of oscillating systems: there are as many oscillations modes as there are oscillating bodies. In real systems, however, damping is always present, and the resonances are heavily supressed even for very small damping ratios. Let’s see how the system behaves for different damping ratios of the main suspension and a very modest value of \zeta_2 = 0.5:

Suspensions transfer function for a bicycle with very lightly damped tire (\zeta_2=0.5). It converges rapidly to 0, even for low damping.

As in the previous model, low frequencies are amplified, but higher are very quickly driven down. Compared to a one-degree of freedom model, the transmission drops to just shy of 0.02 for n=20 and about 0.005 for n=50. That’s about five-fold improvement!

Interestingly, all that happens at the expense of the unsprung mass, which acts as a mass damper. This can clearly be seen if we plot the transmission function for the unsprung mass:

The transfer function for the unsprung mass converges much slower and has a substantial value even for high driving frequencies.


The two models presented here cover a surprisingly wide range of real-world suspension systems. Obviously, they are a simplified version of reality and do not include certain effects, such as progressive spring rates, separate compression and rebound damping rates, and so on. But here is where the beauty of modelling reality lies: we can build models as simple and as complex as we want or need, depending on the level of insight that we need to achieve. There is also a trade-off involved: the more detailed the model, the more expansive it is to solve and analyze, and even the two cases presented here make it blatantly clear.

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