Laplace transforms provide a method for representing and analyzing linear systems using algebraic methods. In systems that begin undeflected and at rest the Laplace 's' can directly replace the d/dt operator in differential equations. It is a superset of the phasor representation in that it has both a complex part, for the steady state response, but also a real part, representing the transient part. As with the other representations the Laplace s is related to the rate of change in the system.
Figure 18.1 The Laplace s
The basic definition of the Laplace transform is shown in Figure 18.2 The Laplace transform. The normal convention is to show the function of time with a lower case letter, while the same function in the s-domain is shown in upper case. Another useful observation is that the transform starts at t=0s. Examples of the application of the transform are shown in Figure 18.3 Proof of the step function transform for a step function and in Figure 18.4 Proof of the first order derivative transform for a first order derivative.
Figure 18.2 The Laplace transform
Figure 18.3 Proof of the step function transform
Figure 18.4 Proof of the first order derivative transform
The previous proofs were presented to establish the theoretical basis for this method, however tables of values will be presented in a later section for the most popular transforms.
The process of applying Laplace transforms to analyze a linear system involves the basic steps listed below.
1. Convert the system transfer function, or differential equation, to the s-domain by replacing 'D' with 's'. (Note: If any of the initial conditions are non-zero refer to the transform tables for extra terms.)
2. Convert the input function(s) to the s-domain using the transform tables.
3. Algebraically combine the input and transfer function to find an output function.
4. Use partial fractions to reduce the output function to simpler components.
5. Convert the output equation back to the time-domain using the tables.
Laplace transform tables are shown in Figure 18.5 Laplace transform tables, Figure 18.7 Laplace transform tables (continued) and Figure 18.8 Laplace transform tables (continued). These are commonly used when analyzing systems with Laplace transforms. The transforms shown in Figure 18.5 Laplace transform tables are general properties normally used for manipulating equations, and for converting them to/from the s-domain.
Figure 18.5 Laplace transform tables
Figure 18.6 Drill Problem: Converting a differential equation to s-domain
The Laplace transform tables shown in Figure 18.7 Laplace transform tables (continued) and Figure 18.8 Laplace transform tables (continued) are normally used for converting to/from the time/s-domain.
Figure 18.7 Laplace transform tables (continued)
Figure 18.8 Laplace transform tables (continued)
Figure 18.9 Drill Problem: Converting from the time to s-domain
Figure 18.10 Drill Problem: Converting from the s-domain to time domain
In previous chapters differential equations, and then transfer functions, were derived for mechanical and electrical systems. These can be converted to the s-domain, as shown in the mass-spring-damper example in Figure 18.11 A mass-spring-damper example. In this case we assume the system starts undeflected and at rest, so the 'D' operator may be directly replaced with the Laplace 's'. If the system did not start at rest and undeflected, the 'D' operator would be replaced with a more complex expression that includes the initial conditions.
Figure 18.11 A mass-spring-damper example
Impedances in the s-domain are shown in Figure 18.12 Impedances of electrical components. As before these assume that the system starts undeflected and at rest.
Figure 18.12 Impedances of electrical components
Figure 18.13 A circuit example shows an example of circuit analysis using Laplace transforms. The circuit is analyzed as a voltage divider, using the impedances of the devices. The switch that closes at t=0s ensures that the circuit starts at rest. The calculation result is a transfer function.
Figure 18.13 A circuit example
At this point two transfer functions have been derived. To state the obvious, these relate an output and an input. To find an output response, an input is needed.
Figure 18.14 Transform table examples
An input to a system is normally expressed as a function of time that can be converted to the s-domain. An example of this conversion for a step function is shown in Figure 18.15 An input function.
Figure 18.15 An input function
In the previous section we converted differential equations, for systems, to transfer functions in the s-domain. These transfer functions are a ratio of output divided by input. If the transfer function is multiplied by the input function, both in the s-domain, the result is the system output in the s-domain.
Figure 18.16 A transfer function multiplied by the input function
Output functions normally have complex forms that are not found directly in transform tables. It is often necessary to simplify the output function before it can be converted back to the time domain. Partial fraction methods allow the functions to be broken into smaller, simpler components. The previous example in Figure 18.16 A transfer function multiplied by the input function is continued in Figure 18.17 Partial fractions to reduce an output function using a partial fraction expansion. In this example the roots of the third order denominator polynomial, are calculated. These provide three partial fraction terms. The residues (numerators) of the partial fraction terms must still be calculated. The example shows a method for finding resides by multiplying the output function by a root term, and then finding the limit as s approaches the root.
Figure 18.17 Partial fractions to reduce an output function
After simplification with partial fraction expansion, the output function is easily converted back to a function of time as shown in Figure 18.18 Partial fractions to reduce an output function (continued).
Figure 18.18 Partial fractions to reduce an output function (continued)
The flowchart in Figure 18.19 The methodology for doing an inverse transform of an output function shows the general procedure for converting a function from the s-domain to a function of time. In some cases the function is simple enough to immediately use a transfer function table. Otherwise, partial fraction expansion is normally used to reduce the complexity of the function.
Figure 18.19 The methodology for doing an inverse transform of an output function
Figure 18.20 The methodology for solving partial fractions shows the basic procedure for partial fraction expansion. In cases where the numerator is greater than the denominator, the overall order of the expression can be reduced by long division. After this the denominator can be reduced from a polynomial to multiplied roots. Calculators or computers are normally used when the order of the polynomial is greater than second order. This results in a number of terms with unknown residues that can be found using a limit or algebra based technique.
Figure 18.20 The methodology for solving partial fractions
Figure 18.21 Partial fractions when the numerator is larger than the denominator shows an example where the order of the numerator is greater than the denominator. Long division of the numerator is used to reduce the order of the term until it is low enough to apply partial fraction techniques. This method is used infrequently because this type of output function normally occurs in systems with extremely fast response rates that are infeasible in practice.
Figure 18.21 Partial fractions when the numerator is larger than the denominator
Partial fraction expansion of a third order polynomial is shown in Figure 18.22 A partial fraction example. The s-squared term requires special treatment. Here it produces two partial fraction terms divided by s and s-squared. This pattern is used whenever there is a root to an exponent.
Figure 18.22 A partial fraction example
Figure 18.23 Partial fractions with repeated roots shows another example with a root to an exponent. In this case each of the repeated roots is given with the highest order exponent, down to the lowest order exponent. The reader will note that the order of the denominator is fifth order, so the resulting partial fraction expansion has five first order terms.
Figure 18.23 Partial fractions with repeated roots
Algebra techniques are a reasonable alternative for finding partial fraction residues. The example in Figure 18.24 Solving partial fractions algebraically extends the example begun in Figure 18.23 Partial fractions with repeated roots. The equivalent forms are simplified algebraically, until the point where an inverse matrix solution is used to find the residues.
Figure 18.24 Solving partial fractions algebraically
For contrast, the example in Figure 18.24 Solving partial fractions algebraically is redone in Figure 18.25 Solving partial fractions with limits using the limit techniques. In this case the use of repeated roots required the differentiation of the output function. In these cases the algebra techniques become more attractive, despite the need to solve simultaneous equations.
Figure 18.25 Solving partial fractions with limits
An inductive proof for the limit method of solving partial fractions is shown in Figure 18.26 A proof of the need for differentiation for repeated roots.
Figure 18.26 A proof of the need for differentiation for repeated roots
Figure 18.27 Finding Partial Fractions in Scilab
A mass-spring-damper system is shown in Figure 18.28 A mass-spring-damper example with a sinusoidal input.
Figure 18.28 A mass-spring-damper example
The residues for the partial fraction in Figure 18.28 A mass-spring-damper example are calculated and converted to a function of time in Figure 18.29 A mass-spring-damper example (continued). In this case the roots of the denominator are complex, so the result has a sinusoidal component.
Figure 18.29 A mass-spring-damper example (continued)
It is not necessary to develop a transfer functions for a system. The equation for the voltage divider is shown in Figure 18.30 A circuit example. Impedance values and the input voltage are converted to the s-domain and written in the equation. The resulting output function is manipulated into partial fraction form and the residues calculated. An inverse Laplace transform is used to convert the equation into a function of time using the tables.
Figure 18.30 A circuit example
Figure 18.31 A circuit example (continued)
In some cases a system input function is comprised of many different functions, as shown in Figure 18.32 Switching on and off function parts. The step function can be used to switch function on and off to create a piecewise function. This is easily converted to the s-domain using the e-to-the-s functions.
Figure 18.32 Switching on and off function parts
The initial and final values an output function can be calculated using the theorems shown in Figure 18.33 Final and initial values theorems.
Figure 18.33 Final and initial values theorems
· The convolution integral can be difficult to deal with because of the time shift. But, the Laplace transform for the convolution integral turns it into a simple multiplication.
Figure 18.34 The convolution integral in the Laplace s-domain
- the inverse Laplace transform provides the impulse response function.
- consider the example of a system. Note that the transfer function is the same as the output response for a unit impulse input.
- this is the most useful when trying to find the response to an arbitrary input.
The following map is to be used to generally identify the use of the various topics covered in the course.
Figure 18.35 A map of Laplace analysis techniques
· Transfer and input functions can be converted to the s-domain
· Output functions can be calculated using input and transfer functions
· Output functions can be converted back to the time domain using partial fractions.
1. Convert the following functions from time to laplace functions using the tables.
2. Convert the following functions below from the laplace to time domains using the tables.
3. Convert the following functions below from the laplace to time domains using partial fractions and the tables.
4. Convert the output function below Y(s) to the time domain Y(t) using the tables.
5. Convert the following differential equations to transfer functions.
6. Given the transfer function, G(s), determine the time response output Y(t) to a step input X(t).
7. Given the following input functions and transfer functions, find the response in time.
8. Do the following conversions as indicated.
9. Convert the output function to functions of time.
10. Solve the differential equation using Laplace transforms. Assume the system starts undeflected and at rest.
11. For the following control system select a controller transfer function, Gc, that will make the overall system performance match the desired transfer function.
12. Write a C program for an ATMega microcontroller to implement the control system in the dashed line below with an update time of 10ms.
13. A feedback control system is shown below. The system incorporates a PID controller. The closed loop transfer function is given.
a) Develop the transfer function for the system.
b) Select controller values that will result in a response that includes a natural frequency of 2 rad/sec and damping factor of 0.5. Verify that the controller will be stable.
c) If the values of Kp = Ki = Kd = 1 find the response to an input of 5sin(10t) as a function of time using the LaPlace Transforms.
d) Find the response in part c) using numerical methods. Show the results as a table and graph. The results should show the region(s) of greatest interest.
e) Find the system response to an input of X = 5sin(100t + 1) using phasor transforms.
1. Prove the following relationships.
2. The applied force `f' is the input to the system, and the output is the displacement `x'.
b) What is the steady state response for an applied force F(t) = 10cos(t + 1) N ?
c) Give the transfer function if `x' is the input.
d) Find x(t), given F(t) = 10N for t >= 0 seconds using Laplace methods.
3. The following differential equation is supplied, with initial conditions.
a) Solve the differential equation using calculus techniques.
b) Write the equation in state variable form and solve it numerically.
c) Find the frequency response (gain and phase) for the transfer function using the phasor transform. Roughly sketch the bode plots.
d) Convert the differential equation to the Laplace domain, including initial conditions. Solve to find the time response.
4. Given the transfer functions and input functions, F, use Laplace transforms to find the output of the system as a function of time. Indicate the transient and steady state parts of the solution.
5. Select a controller transfer function, Gc, that will reduce the system to a first order system with a time constant of 0.5s, as shown below.
6. a) The block diagram below is for an angular positioning system. The setpoint is a desired angle, which is converted to a desired voltage. This is compared to a feedback voltage from a potentiometer. A PID controller is used to generate an output voltage to drive a DC motor. Simplify the block diagram.
b) Given the transfer function below, select values for Kp, Ki and Kd that will result in a second order response that has a damping factor of 0.125 and a natural frequency of 10rad/s. (Hint: eliminate Ki).
c) The function below has a step input of magnitude 1.0. Find the output as a function of time using numerical methods. Give the results in a table OR graph.
d) The function below has a step input of magnitude 1. Find the output as a function of time by integrating the differential equation (i.e., using the homogeneous and particular solutions).
e) The function below has a step input of magnitude 1. Find the output as a function of time using Laplace transforms.
f) Given the transfer function below; 1) apply a phasor/Fourier transform and express the gain and phase angle as a function of frequency, 2) calculate a set of values and present them in a table, 3) use the values calculated in step 4) to develop a frequency response plot on semi-log paper, 5) draw a straight line approximation of the Bode plot on semi-log paper.
7. a) Simplify the block diagram as far as possible.
b) Given the transfer function below, select values for Kp and tau that will include a second order response that has a damping factor of 0.125 and a natural frequency of 10rad/s.
c) The function below has a step input of magnitude 1.0. Find the output as a function of time using numerical methods. Give the results in a table OR graph.
d) The function below has a step input of magnitude 1. Find the output as a function of time by integrating the differential equation (i.e., using the homogeneous and particular solutions).
e) The function below has a step input of magnitude 1. Find the output as a function of time using Laplace transforms.
f) Given the transfer function below; 1) apply a phasor/Fourier transform and express the gain and phase angle as a function of frequency, 2) calculate a set of values and present them in a table, 3) use the values calculated in step 4) to develop a frequency response plot on semi-log paper, 5) draw a straight line approximation of the Bode plot on semi-log paper.
g) Select a controller transfer function, Gc, that will reduce the system to a first order system with a time constant of 0.5s, as shown below.
8. Write a C program for an ATMega 32 microcontroller to implement the control system in the dashed line below with an update time of 10ms.
9. a) Given the experimental Bode (Frequency Response Function) plot below, find a transfer function to model a positioning system. The input is a voltage `V' and the output is a displacement `x'. (Hint: after calculating the function develop a Bode plot to verify the system performance.)
b) The transfer function found in step a) will be used for the positioning system in the block diagram below. Find the overall transfer function for the system.
c) Find a new controller transfer function (V/e) that would give the overall response below.
d) Write a subroutine that implements the control function (V/e) found in step c). (Hint: Convert it to state equations first.)
e) The controller function is replaced with the function below. Draw a Bode plot using the straight line approximation. Compare it to the Bode plot in part a).
10. a) Develop differential and state equations describing the following system. The input is Vd and the output is the motor shaft speed. Assume all components are ideal. The motors are identical with a resistance of 20 ohms. With an input voltage of 5V the motors spin at 2000RPM, and have a time constant of 0.25s.
b) Find the system response to a unit step input explicitly (i.e. homogenous and particular).
c) Find the system response to a unit step input using Laplace transforms.
d) Solve the problem numerically and report the results using a table of values.
e) Use phasors to find the steady state output response to an input of,
11. a) Using Laplace transforms find the system response for the transfer function and input function (given as a function of time).
b) Use numerical methods to find the response of the system described with a transfer function. Report the results using a table of values.
c) Use phasors to find the steady state output response to the input given below.
12. Convert the following differential equations to physical systems. Show the method.
13. For the following control system select a controller transfer function, Gc, that will make the overall system performance match the desired transfer function.
14. Write a C program for an ATMega microcontroller to implement the control system in the dashed line below with an update time of 10ms.
15. A feedback control system is shown below. The system incorporates a PID controller. The closed loop transfer function is given.
a) Develop the transfer function for the system.
b) Select controller values that will result in a response that includes a natural frequency of 2 rad/sec and damping factor of 0.5. Verify that the controller will be stable.
c) If the values of Kp = Ki = Kd = 1 find the response to a unit ramp input as a function of time using LaPlace Transforms.
d) Find the response in part c) using numerical methods.
e) Find the system response to an input of X = 5sin(100t + 1)
f) If the input X is a trapezoidal motion profile with an acceleration time of 2 seconds, and a maximum velocity of 5, what would the response Y look like.
16. For the following control system select a controller transfer function, Gc, that will make the overall system performance match the desired transfer function.
17. A feedback control system is shown below. The system incorporates a PID controller. The closed loop transfer function is given.
a) Develop the transfer function for the system.
b) Select controller values that will result in a response that includes a natural frequency of 2 rad/sec and damping factor of 0.5. Verify that the controller will be stable.
c) If the values of K1 = K2= 1 find the response to a unit ramp input as a function of time using LaPlace Transforms.
d) Find the response in part c) using numerical methods.
e) Find the system response to an input of X = 5sin(10t + 1) using phasor transforms.
f) If the input X is a trapezoidal motion profile with an acceleration time of 2 seconds, and a maximum velocity of 5, calculate the response Y.
Irwin, J.D., and Graf, E.R., Industrial Noise and Vibration Control, Prentice Hall Publishers, 1979.
Close, C.M. and Frederick, D.K., "Modeling and Analysis of Dynamic Systems, second edition, John Wiley and Sons, Inc., 1995.