ASEN 3200 Orbital Mechanics and Attitude Dynamics
Spring 2006
Laplace Transform Solution of Linear Differential Equations
1 General Form for Linear Time-Invariant Systems
A single-input single-output (SISO) linear time-invariant (LTI) linear dynamical system can be represented
in the time domain with an nth order ordinary differential equation (ODE) with constant
coefficients (ai , bi ) :
dn
d n 1
y
t

a
y t  


n 1
dt n
dt n 1
 a0 y  t   bm
dm
u t  
dt m
 b0u (t )
(1)
Here, y (t ) is the output (dependent variable) and u (t ) is the input (independent variable). Solving this
differential equation for y (t ) given a specific u (t ) is difficult, and this prevents a comprehensive
understanding of system behavior for a variety of inputs. In addition, this difficulty complicates the design
process, where system characteristics are specified or modified with feedback control to achieve desired
system behavior. This difficulty has motivated a large amount of theoretical work in understanding these
systems, and the development of a variety of analytical and numerical tools to aid in the design of feedback
control systems. The resulting Theory of Linear Dynamical Systems (studied in more depth in ASEN
4114/5114 and ASEN 5014) is very powerful and enables the behavior of these systems to be understood
and specified with surprising ease. In ASEN 3200 we will touch on some of the fundamental insights and
tools of this theory.
2 Laplace Transforms
Formal and complete derivations of the Laplace Transform and its properties are beyond the scope of these
notes. There are many good references on the subject, a few of which are listed in the references. The
objective of these notes is to provide enough background to use Laplace Transforms as a tool for solving
second order linear ODE's and for the design and analysis of simple control laws.
bg
The Laplace Transform is an integral transformation that maps a function of time f t , into a function of a
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complex variable, F s . The single-sided Laplace Transform is defined as follows:
L  f  t   F  s  

 f t  e
 st
dt
(2)
0
where s is a complex number defined by,
s   +j
with

the real part of
and j 
s , designated by Re( s ) , and  the imaginary part of s , designed by Im( s ) ,
bg
1. Note that  and  are real numbers. F s is a complex number for each s .
bg
bg
The inverse Laplace Transform is another integral transform that maps F s back to f t . It is given by,
f t   L
1
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 F  s   
c  j

F  s  e st ds
(3)
c  j
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Spring 2006
These transforms are rarely calculated directly. Instead, tables of transform pairs are used to quickly solve
for solutions to (1) for given “standard” inputs. Often, lower case letters are used to represent time domain
functions and the corresponding capital letters to represent their Laplace Transforms. It is also common to
use the same letter for both the Laplace and time functions, using the argument s or t to designate which
form of the function is being used.
The main purpose of the Laplace Transform and its inverse is that they enable direct solution of ODE's
using algebraic operations and function table lookup. In order to use it for this purpose we must understand
two key properties.
Linearity: The Laplace Transform is a linear transformation between functions in the t domain to functions
in the s domain. That means that the Laplace Transform of the sum of two or more functions is the sum of
their transforms, and the Laplace Transform of a function scaled by a constant is the constant times the
Laplace Transform of the original function.
L  a1 f1  t   a2 f 2  t    a1F1  s   a2 F2  s 
(4)
Derivatives: The Laplace Transform of the time derivative of a function is s times the Laplace Transform
of the function minus the initial value of the function. Higher order derivatives can be handled by iterating
this rule. For example, the Laplace Transform of the second derivative is s2 times the Laplace Transform of
the function minus s times the initial value of the function, minus the initial value of the time derivative of
the function.
df (t ) 
L 
 sL  f (t )   f  t  0   sF  s   f  t  0 
 dt 
2
L  d dtf 2(t )   s 2 L  f (t )   sf  t  0   f  t  0   s 2 F  s   sf  t  0   f  t  0 


(5)
Use of these properties are illustrated in the following section, which applies the Laplace Transform to
second order LTI ODEs.
2.1
Laplace Transform of Second Order ODE's
Applying the derivative and linearity properties to the dynamical system represented by the ODE
d2
d
y  t   2n y  t   n2 y  t   n2u  t 
2
dt
dt
(6)
we find the Laplace Transform of this system to be
 d2

d
L  2 y  t   2n y  t   n2 y  t    L n2u  t  
dt
 dt

 s 2Y ( s)  sy (0)  y (0)   2n Y ( s)  y (0)   n2 X  s   n2U ( s)
(7)
 s 2  2n s  n2  Y( s)  n2U ( s)   s  2n  y (0)  y (0)
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which can be rearranged to solve for the (transform of the) output Y ( s ) due to the (transform of the) input
U ( s ) , and the initial conditions as follows:
Y ( s) 
n2
( s  2n ) y(t  0)  y(t  0)
U (s) 
2
2
 s  2n s  n 
 s 2  2n s  n2 
(8)
The first term in this expression for the response Y ( s ) is a product of U(s) and the transfer function from
the input U(s) to the output Y(s), given by:
n2
Y ( s)
H ( s) 

U ( s)  s 2  2n s  n2 
(9)
The second term in (8) is due only to initial conditions (initial value and initial rate of change of y (t ) ).
To solve for y (t ) , a particular u (t ) is selected, transformed to obtain U ( s ) , then Y ( s ) is found from
(8) with specific initial conditions. Then Y ( s ) is inverse transformed, either from a table of transforms, or
split up first using a partial fraction expansion of Y ( s ) into simple terms that can be inverse transformed
by inspection.
The partial fraction expansion also provides important insight into the type of solutions that can be
obtained, which is very useful for design. Assuming that there are no repeated roots of the denominator of
Y ( s ) , the partial fraction expansion has the form
Y ( s) 
A1
A2
B
B
C1
C2

 1  2 

s  p1 s  p2 s  q1 s  q2 s  p1 s  p2
(10)
pi are the roots of the denominator of the transfer function (called the poles of the system), and qi
are the roots of the denominator of the Laplace expression for U ( s ) (called the poles of the input).
where
Together these poles produce the poles of the output. Note that the initial condition terms in (8) have
partial fraction terms in the same form as for the transfer function (same poles pi ).
Using the transform pair
e t 1(t ) 
where 1(t ) is the unit step function (zero before
solution
1
s 
(11)
t  0 and 1 thereafter), we obtain the time domain
y (t )  ( A1  C1 )e p1t  ( A2  C2 )e p2t  B1e q1t  B2e q2t  1(t )
Observe that the shapes (or character) of the time response terms depend only on the poles
(12)
pi and qi . The
amplitudes of the response terms depend on the amplitude of the input, as well as the size of the initial
conditions. Importantly, the shape of the input carries through to the output via the input poles qi . This
portion of the response is called the forced response. The portion of the response due to the system
poles pi has a shape that is purely a function of the characteristics of the system (ODE) relating input to
output. This part of the response is called the natural response. These system poles are discussed in more
detail below.
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2.2
Spring 2006
Roots of the Characteristic Equation
The characteristic equation is obtained by setting the denominator polynomial of the transfer function to
zero. The roots of the characteristic equation are therefore the poles of the system. (As an aside, the roots
of the numerator polynomial of H ( s ) are called the zeros of the transfer function.) Knowing the poles of
a transfer function tells us what the natural response looks like, and this is an important part of the overall
response to initial conditions and inputs.
Im(s)
A second-order system (from an n=2 order ODE) has two poles, from
the two roots of its second-order characteristic polynomial. The roots of
polynomials with real coefficients may be real, imaginary or complex,
but any complex roots must occur in complex-conjugate pairs. Pole
locations in the complex “s-plane” are often indicated by an “x” as
shown in Figure 1.
x
n
For the general form of the 2nd order ODE, the poles are given by:
s1 , s2   n   n   1
2
where

is called the “damping ratio”, and
n
d

Re(s)
(13)
is called the “natural
x
frequency”.
For   1 both poles are real and negative. From (12) we see that the
corresponding response terms are decaying exponentials, with no
oscillation. This is termed an overdamped system.
Figure 1: Pole locations in the
s-plane
For   1 the roots are negative, real, and repeated. This is termed “critical damping” and again no
oscillations occur.
s1 , s2  n
This is a special case (poles are not distinct) and the transform pair (11) does not provide the correct
response. Instead, these two poles produce a response in the form te
shape.
 n t
, which has a non-exponential
For 0    1 the roots are complex conjugates with negative real parts. In this case, the system is said to
be “underdamped” and the system poles are given by
p1 , p2    jd
where the real part   n and the imaginary part d  n 1   2
(14)
Since the poles are again distinct, (11) applies so that the time response corresponding to the two system
poles is given by
A1e(  jd )t  A2e(  jd )t  e t ( A1e jd t  A2e jd t )
(15)
Observe that the exponential “envelope” of this response is determined by the real part of the poles
  n . When this is negative, these response terms decay to zero. When  is zero, the envelope is
constant. When

is positive, the response grows exponentially. The remaining complex exponentials in
(15) are oscillation terms, as seen by applying the Euler formula
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e jz  cos( z )  j sin( z ) . This makes it
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ASEN 3200 Orbital Mechanics and Attitude Dynamics
clear that whenever there is an imaginary part
an oscillation at the frequency
d
Spring 2006
d in a pole, the corresponding time response term contains
[rad/sec].
A quantitative estimate of the decay rate of the envelope in (15) can be found by calculating the settling
time
ts where the envelope is a certain percentage of its initial value, e.g. 5%. Since the initial value of e t
is 1, this requires that
e ts  0.05  ts 
ln(0.05)


3
(16)

Hence, settling time is a function of the real part of the system poles. In a system with many system poles,
the composite settling time can be a complex function, but for design purposes it is usually specified that
the settling time of (the envelopes of) each of the natural response terms satisfy a settling time requirement
like (16).
Performance objectives for a control system are often stated in terms of required settling time, which is a
speed of response measure, and on oscillation frequency or damping ratio, which relates to the “overshoot”
and “ringing” obtained tracking a reference or command pointing trajectory. The above relationships
between pole locations and corresponding time response terms provide a theory that enables a control
system to be designed “in the frequency domain” (i.e. by placing system poles in “good” locations).
3 Initial Conditions and Forced Responses
In addition to the natural response of a system, we are also interested in understanding how it responds to
input. For example, what is the motion of the spacecraft attitude in response to repeated gas jet firing or to
slowly varying aerodynamic drag, or to sudden retargeting commands in a feedback control system? This
constitutes the analysis of the forced response of a system. Of course the form of the forcing input can take
a wide variety of forms. Fortunately, it can frequently be described by the sum of some rather simple
components. The ones we will consider here are the impulse, step, and sinusoid. For now we will ignore
other unbounded input such as a ramp input. It is interesting to note that any periodic input can always be
described by the sum of a sinusoidal inputs (by Fourier analysis).
Examples:
Impulse
An input that occurs over a very short period of time is described as an impulse with obvious analogy to
linear and angular impulses that we have considered in ASEN2003. Gas jets used on spacecraft are often
treated as impulses, due to their short on-time (tens of milliseconds) relative to the settling time of the
spacecraft natural response.
Step
An input that begins at a specific time and then persists is described as a step input. Step type inputs can be
created by sudden changes in pointing angle commands. Often, these are used to measure the behavior of a
control system, since they enable both steady state tracking errors and transient natural responses to be
measured.
Sinusoid
A harmonic forcing function is an input that varies sinusoidally at a fixed frequency. Many spacecraft have
reaction wheels, momentum wheels, or control moment gyros which have spinning masses. Imbalance in
these masses causes sinusoidal vibration forces which can disturb spacecraft line-of-sight stability, inducing
jitter or smear in imaging instruments. This is also a common test signal, where a system is subjected to
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sinusoidal forcing functions at various frequencies to measure its frequency response, and thereby provide
more accurate models of its dynamics (particularly for identifying structural resonance modes).
The Laplace Transforms of these and other input functions can be found on the class web site in the
Handouts section. Each of these functions has specific poles, which describe the character (or shape) of the
time function. This carries though to the system output, since the input poles qi appear also as poles of the
output. Hence the output will contain terms that have the same character as the inputs. Exceptions to this
only occur in the case of the impulse input (which has no poles), and when input and system poles coincide,
which produces repeated roots and new types of functions on the output, not found in either the input or in
the system natural modes (natural response). In this case, the functions can be found in Laplace Transform
tables. Note overlapping input and system poles are rather difficult to obtain, since this requires precise
matching between system and input. Hence the insight obtained from the above analysis of the distinct
roots case is used extensively for system understanding and design efforts to achieve prescribed
performance objectives.
The following sections explore the outputs of the specific system (6) to various inputs. It is instructive
to compare these responses to the shapes of the response predicted by the poles of the intput and the poles
of the system.
4 Response of an Undamped 2nd Order System
  0 , and system poles at s   jd   jn . All responses are valid for t  0 , and are
zero for t  0 .
This case has
4.1
Free Response - Initial condition only
y (t )  y  0  cos nt 
4.2
y  0
n
sin nt
(17)
Unit Step Response (with zero initial conditions)
With u(t) = 0 for t<0 and u(t)=1 for t>0.
y(t )  1  cos nt
4.3
With
(18)
Sinusoidal Input Response (with zero initial conditions)
u  t   a cos t ,   n ,
y (t ) 
a
cos t  cos nt 
1   2 
(19)
where    n is the input-system frequency ratio
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5 Response of an Underdamped 2nd Order System
This case assumes that 1    1 .
5.1
Free Response - Initial condition

y (t )  y  0  ent cos d t 

1
2

sin d t 
y  0
d
e nt sin d t
(20)
Notice that the decay envelope is determined by the real part of the roots (n) and the frequency of
oscillation is determined by the imaginary part of the roots ( d).
5.2
Unit Step Response (with zero initial conditions)
With u(t) = 0 for t<0 and u(t)=1 for t>0.

y(t )  1  ent cos d t 

1 2
sin d t

(21a)
or
y (t )  1 
e nt
1  2
sin d t   
(21b)
where tan  
5.3
With
1  2

Sinusoidal Input Response (with zero initial conditions)
u  t   a cos t ,
y (t ) 
a cos t   
1      2 
2 2
2

 
2
  2  1 cos d t    2 1 
 

1  2

n t 

 ae

2

1   2    2 2


where    n is the input-system frequency ratio

 

  1 sin d t 


 




(22)
 2 
and   tan 1 
2 
 1  
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The so-called amplification factor is given by:
AF 
1
1   2    2 
2
(23)
2
To find this peak amplification we take the derivative with respect to  and set it to zero. This leads to two
possibilities:
For  
For  
1
2
1
2
, the peak occurs at   0, and has a peak value of 1
, the peak occurs at   1  2 2 , and has a peak value 
For very small  the peak value 
1
2 1   2
1
2
(24)
Figure 2 – Amplification factor as a function of
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 for various damping ratios  .
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6 Time domain performance measures
Performance of a second order system is often specified in terms of the characteristics of the step response.
Figure 3 illustrates the step response for the 2 nd order system (6) with damping ratio ranging from 0.2 to
1.2. The horizontal axis is a normalized time =nt so these graphs apply to all systems defined by the
form given in Equation 6. The vertical axis is the output y (t ) .
6.1
Rise Time
The rise time is generally defined as the time required for a unit step response to go from 0.1 to 0.9 or for a
more generic step response to go from 10% to 90% of the input value. It can be easily seen on the step
response graph.
For the unit step it can be found analytically by setting y(tstart) = 0.1 in Equation 21b and solving for the
first time at which this occurs; then setting x(tstop) = 0.9 and solving for the first time this occurs. The
difference in the times is the rise time tr=tstop – tstart.
An approximation for underdamped 2nd order responses with damping ratio,  near 0.5 is:
tr 
1.8
(25)
n
A closely related time is the time required to go from 0 to the first crossing of 1 is given by:
t zero _ to _1 
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 
d
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ASEN 3200 Orbital Mechanics and Attitude Dynamics
6.2
Spring 2006
Peak Time
The peak time is the time required for a unit step response to reach is maximum value. To find it
analytically set the time derivative of x(t) to zero and find the first time this occurs. Using Equation 18b,
you find:
 t
 t
d
e np
e np
y (t )  0  n
sin d t p     d
cos d t p   
dt
1  2
1  2
n sin d t p     d cos d t p   
tan d t p    
1  2
d

 tan 
n

so
d t p  
6.3
and
tp 

d
(26)
Maximum Overshoot
The maximum overshoot is how much above 1 the step response goes. To find it analytically compute
x(tp)-1.
Mp  
e n / d
1 2

Mp e
1 2
For small 
sin     
e  n /  d
1 2
sin 

(27)
M p  e 
Mp can also be represented as a percent overshoot.
6.4
Settling Time
The settling time is the time required for the system to almost reach steady state. It is defined as the time
required for the response to settle within some percent of its final value; which can be read off a plot of the
response. Analytically it is approximated by the time at which the exponential decay envelope drops with
some percent of its final value. So you set the envelope to 0.01, 0.02 or 0.05 and solve for the time.
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for 1%
e nts  0.01
ts 
for 2%
e nts  0.02
ts 
for 5%
e nts  0.05
ts 
4.6
 n
3.9
 n
(28)
3.0
 n
7 References
Blanchard, P., R. L. Devaney, and G.R. Hall, Differential Equations, Chapter 6, Brooks/Cole Publishing,
1997.
DiStefano, Stubberud, Williams, Feedback and Control Systems, Shaum's Outline Series.
Dorf, R.C., Modern Control Systems, Addison-Wesley, 1994.
Franklin, G.F., Powell, J.D., and A. Emami-Naeini, Feedback Control of Dynamic Systems, Fourth Edition,
Prentice Hall, 2002.
Woods, R. L., and K.L. Lawrence, Modeling and Simulation of Dynamic Systems, Appendices E&F,
Prentice Hall, 1997.
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