Overview
Sometimes engineering problems can best be assessed by looking at them in the frequency domain. This section will examine how Laplace Transforms allow us to do that.
Objective
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The student will examine how the Laplace Transform can be used to analyse a variety of engineering problems in the frequency domain.
Sometimes it is just as easy (or perhaps easier) to assess a particular engineering problem in the frequency domain rather than the time domain. Since the Laplace Transform allows us to enter the frequency domain, we have the tools which will allow us to achieve this.
Topic 1 - Complementary and Particular Solutions
Consider our old-familiar spring mass system. If it is at rest and we suddenly apply a sinusoidal forcing function, then there is a portion of the response that is controlled by the natural frequency and is due to the shock of the system being put into motion. If there is damping, this portion of the response dies out. This portion is called the transient response – resulting in what is known as the complementary solution.
Once the transient response dies out, we are left with the steady state response which is caused by the sinusoidal input and the response will be at the frequency of the sinusoidal forcing function, referred to as the input frequency or the forcing frequency. This portion called the steady state response – resulting in what is known as the particular solution.
So for our spring mass system, we have the equation:
where:
Divide by m and then Laplace Transform this and we have
Solve for X(s) and Inverse Laplace Transform
We can observe the following:
- Complementary Solution
- First term in the solution
- Vibration response at natural frequency
- Is a transient response
- Does not decay in this situation – but only because there is no damping – generally speaking it would decay due to damping
- Particular Solution
- Second term in the solution
- Vibration response at the forcing frequency
- Is a steady-state response and should not decay
- If the Forcing Frequency ω Increases from Zero
- Denominator k - mω2 becomes smaller
- Amplitudes become larger
- If the Forcing Frequency Equals Natural Frequency ω = ωn = (k/m)1/2
- Denominator k - mω2 becomes zero
- Resonance occurs
- Amplitude of vibration increases without bound (infinity)
- If the Forcing Frequency is Greater than Resonance
- Denominator k – mω2 becomes negative
- Response will be out of phase with the forcing function
- Denominator assumes increasingly larger values approaching negative infinity
- Amplitudes of vibration (at natural and forcing frequency) approach zero from negative side (e. starting at negative infinity when ω = ωn
- Phase Shift
- ω below resonance: the Part of the vibration at the forcing frequency is in phase with forcing sinusoid
- ω above resonance: the Part of the vibration at the forcing frequency is 180° out of phase
Topic 2 - Sinusoidal Transfer Function
If an input function p(t) causes a response function x(t), then the Transfer Function in the frequency domain is defined as G(s) = X(s)/P(s).
The Sinusoidal Transfer Function is defined as G(s) with s replaced by iω
If the transient response is needed, we must use G(s), because only it captures the decay with time. However, if only the steady state response is desired, then the solution process can be simplified by using this Sinusoidal Transfer Function.
If the forcing function is p(t) = Psinωt
Then x(t) = |G(iω)| Psin(ωt + Φ)
Where the term |G(iω)| is the amplitude of G(iω). To arrive at this amplitude, apply the Pythagorian Theorem as the square root of the sum of the squares of the imaginary part of G(iω) and the real part of G(iω)
And where the phase angle, Φ = -tan-1[imaginary part of G(iω) / real part of G(iω)]
Application 1
Consider a system with a transfer function of
If with a sinusoidal input of
Then the Sinusoidal Transfer Function will be with substituted for
The magnitude term is
And the phase angle is
So the response in the time domain, x(t) will be
Application 2
Consider a spring-mass-damper system with mass m, spring rate k, and damping coefficient b. Apply an input of p(t) = Asinωt and determine the steady state response using the sinusoidal transfer function.
The equation of motion will be
Transforming,
So the transfer function is
And the Sinusoidal Transfer Function will be
Magnitude is
And
So the response in the time domain is
Topic 3 - Uses of the Sinusoidal Transfer Function
The sinusoidal transfer function can be used to find the:
- Steady-state response of a system to a sinusoidal Input
- Frequency response characteristics
- Described by output-input amplitude ratio and phase angle between output and input sinusoids
- Can vary the frequency of input signal over wide range to study response of system
- Example: a shaker rig using a swept sine input varying in frequency where the output is used to characterize response of the test item
- Example: design of an isolation system
- Designed with the aim of minimizing or eliminating vibratory effects
- Reduce magnitude of force transmitted from a machine to its foundation
- Transmission Ratio = TR
- TR = transmitted force/exciting force
- Exciting force due to the unbalance of a rotating machine will be a sinusoidal force p(t) = Asinωt
- Reduce the magnitude of motion transmitted from a vibratory foundation to a machine
- Provides load support and also a means of energy dissipation
- Example of typical isolator: Spring and damper
- Example of simple isolator: Single element of synthetic rubber
- It can be shown that the TR for a force transmitted from a machine to the surroundings (or vice versa) is exactly the same as the TR for motion transmitted from a machine to the surroundings (or vice versa)
Application 1
Consider a machine with an unbalanced rotating element and determine its transmission ratio to the surroundings.
Force transmitted to foundation is sum of damper and spring forces:
Laplace Transforms of p(t) and f(t) assuming zero initial conditions:
The transfer functions are
Unbalance force to machine motion:
Machine to surroundings:
Combine these into one transfer function from the unbalance force to the surroundings
And convert to Sinusoidal Transfer Function
Divide all terms by m
Recall from vibration theory that the natural frequency so
And the frequency ratio is
And the damping ratio is
With these substitutions, we get
This is the sinusoidal transfer function from the unbalance force to the surroundings. That is by definition the transmission ratio to the surroundings that we were looking for. This shows that the transmission ratio depends on both damping ratio (ζ) and speed ratio (β) which is the proximity to resonance. This makes sense. From this we can plot curves of transmission ratio versus speed (β) for different values of damping (ζ).
Note that:
- All curves pass through a critical point where TR = 1 and β = 1.4 = square root of 2
- For β < 1.4, as the damping ratio increases the transmission at resonance decreases, so increasing the damping improves the vibration isolation
- For β > 1.4, as the damping ratio increases the transmission increases, so increasing the damping actually adversely affects the vibration
Application 2
A tuned vibration absorber is a spring-mass system added to a system experiencing vibration, which can actually reduce the vibration of a system. A 100 kg machine has a rotating component turning at a frequency of 10 hz. The machine has an unbalance which creates an unbalance force of F0 and can operate near the system’s natural frequency, which can result in significant vibration, transmitting significant force and motion to the surroundings. We wish to add a tuned vibration absorber to the system of mass ma and spring rate ka chosen to reduce the vibration that will be experienced.
Equation for the machine:
Equation for the added vibration absorber:
We would like to show that the vibration can be controlled by adding just the mass of the tuned absorber. To show this, let us consider all damping in the system as negligible. Thus these equations reduce to
And
Transform
And
We are only concerned about the motion of X, not of Y. So let us combine the equations to eliminate Y.
So
Changing to the sinusoidal transfer function
The amplitude of the response will be the magnitude of G(ωi) times F0. Thus, if we could make G(ωi) go to zero, there would be zero response at all. Looking at the equation above, we can see that if ka were to be equal to maω2 then
And we would have eliminated all vibration. We have ω = 10hz (2π) = 62.83 rad/sec
So for ka to be equal to maω2
Then THEORETICALLY the vibration of the machine would be zero. Pragmatically, this will not be true, however, the absorber would be expected to significantly reduce the vibration.
Topic 4 - Bode Diagram Representation of Systems in the Frequency Domain
Let us examine a representation known as a Bode Diagram. It is a means of plotting the sinusoidal transfer function as represented by two plots. The independent axis for both is frequency. Dependent axes are magnitude and phase angle.
The magnitude is usually expressed using the base 10 logarithm in the form 20log|G(jw)| with the units being decibels, or dB. The plots are then drawn on semi-log graph-paper, with the magnitude given on the vertical, linear scale and the frequency on the horizontal, logarithmic scale. Phase angle is likewise plotted on the vertical, linear scale and the frequency on the horizontal, log scale.
Consider this sinusoidal transfer function of a system governed by a first order differential equation (refer back to Topic 2 - Application 1)
The log magnitude of this sinusoidal transfer function is
Note that if ω is very small, then this essentially becomes -20log1 dB = 0 dB
Thus, for very low frequencies, the magnitude curve is flat near 0 dB
If we draw straight lines asymptotic to the two pieces of the curve, their intersection is a point called the corner frequency, which separates the low frequency range from the high frequency range. Phase angle at the corner frequency will always be 45o. We often approximate the curve as two straight lines using these asymptotes
(Ogata)
There is some error in this approximation, but not a large amount.
Consider the Bode Diagram of Second-order System
Then the sinusoidal transfer function is
From this we can plot the bode Diagram at different values of ζ.
Note from the red triangles that for different z values, there are slightly different peak frequencies. In a mechanical system, this would correspond to the fact that there is a slightly different damped natural frequency than undamped natural frequency.
Application
The test department has determined the experimental data below from a spring mass system, for which the values of m, k and c are unknown. They have given the data to you for evaluation. Can you determine the unknown quantities ?
Transfer Function:
Laplace Transform with zero initial conditions:
Transfer Function :
Sinusoidal Transfer Function:
From the Bode Diagram:
So
Corner frequency,
The peak occurs at -22 dB, which is 4 dB above the 0 dB asymptote. 4.38 dB corresponds to a damping ratio of 0.32, which is reasonably close. So use ζ = 0.32.
Summary
Some dynamic problems are more easily solved in the frequency domain, particularly if the response is known to be sinusoidal and only the steady state portion of the response (ignoring transients) is desired. The Bode Diagram is a useful means of visually displaying a system response in the frequency domain.
TASK 1
Tutorial problems will allow the student to practice the concepts shown in this discussion.
References and Bibliography
Ogata, K. (2004). System Dynamics, 4th edition. Upper Saddle River, USA: Pearson Prentice Hall.