3.1
Time-Domain Analysis of Continuous-Time Systems:
Linear time-invariant continuous-time systems (LTIC) are modeled by linear
differential equations:
n −1
n
m −1
m
d y
d y
dy
d f
d f
df
+a
+L+ a
+a y =b
+b
+L+b
+b f
dt
dt
dt
dt
dt
dt
n
n −1
n −1
1
0
m
m
m −1
m −1
1
0
where f(t) and y(t) are system inputs and outputs, respectively, and a0 .. an-1, b0 .. bm
are constants. Input-output behavior of any LTIC system is characterized by these
constants.
A more compact notation will be used to represent the above system, where Dk is a
derivative operator:
( D + a D +L+ a D + a ) y (t ) = (b D + b D +L+b D + b ) f (t )
n −1
n
n −1
m −1
m
1
0
m
m −1
1
0
3.2
The Zero-Input Response:
Consider systems excited or driven only by initial conditions (i.e. stored energy):
( D + a D +L+a D + a ) y (t ) = 0
n −1
n
n −1
1
0
0
Solution of first order equation:
( D + a ) y (t ) = 0
0
0
Show solution is: y (t ) = c exp( − a t ) , where c is an arbitrary constant. If initial
condition given at t = t0, then y (t ) = y (t ) exp( − a t ) for t ≥t0
0
0
0
0
0
For a general order, assume solution is of the form c exp(λi t), substitute into zeroinput response differential equation to show that a system can be characterizes by
its characteristic polynomial: Q( λ ) = λ + a λ +L+a λ + a
and the solution can be obtained based on the roots of the characteristic equation:
Q(λ ) = 0
Let the roots of the characteristic equation of an n-th order differential equation be
denoted as λ , λ ,L, λ
n −1
n
n −1
1
2
n
1
0
3.3
Modes of the Zero-Input Response
The component of the system’s zero-input response that corresponds to a root of the
characteristic equation is referred to a mode or natural frequency.
a) If roots are real and distinct, each system mode is an exponential and the total
response is described by: y0 (t ) = c1 exp(λ1t ) + c2 exp(λ2 t ) + L + cn exp(λn t )
To view examples of how various combination of these modes can behave, try
running the following script in Matlab:
t = [0:.1:10]; % Define Time Axis for function evaluation
% Loop through 10 different functions
for k=1:10
c = 5*(rand(1,10) - .5); % Randomly pick 10 coefficients
subplot(3,1,1), plot(c) % Plot them out
ylabel('Coefficients')
lam = -2*rand(1,10);
% Randomly pick 10 time constants
subplot(3,1,2), plot(lam) % Plot them out
ylabel('Time Constants')
subplot(3,1,3), plot(t, c*exp(lam'*t)) % Evaluate and plot function
ylabel('Function')
xlabel('seconds')
pause
% Hold figure until Key is struck
end
3.4
Real and Distinct Roots Example:
Coefficents
2
0
Time Constants
-2
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
0
-1
-2
Function
1.5
1
0.5
0
0
1
2
3
4
5
seconds
6
7
8
9
10
3.5
b) If roots are real and repeated, each system mode is the product of an exponential
and the time variable raised to a power. The total response is described by:
y ( t ) = (c + c t +L+ c t ) exp( λ t )
To view examples of how various combination of these modes can behave, try
running the following script in Matlab:
n −1
0
1
2
n
1
t = [0:.1:20]; % Set up Time Axis matrix for function evaluation
tm =[];
for k=1:10
tm = [tm; t.^(k-1)]; % Each row is t raised to the next power
end
% Loop through 10 different functions
for k=1:10
c = 5*(rand(1,10) - .5); % Randomly pick 10 coefficients
subplot(2,1,1), plot(c) % Plot them out
ylabel('Coefficients')
lam = -2*rand(1,1) -1
% Randomly pick 1 time constant (between -1 and -3) and print it out
subplot(2,1,2), plot(t, (c*tm).*exp(lam*t)) % Evaluate and plot function
ylabel('Function'); xlabel('seconds')
pause
% Hold figure until Key is struck
end
3.6
Real and Repeated Roots Example (time constant = 1/1.4279)
Coefficents
4
2
0
-2
-4
1
2
3
4
5
6
7
8
9
10
18
20
60
Function
40
20
0
-20
-40
0
2
4
6
8
10
seconds
12
14
16
3.7
If roots are complex, each system mode is an exponentially weighted sinusoid. The
total response is described by (let λ = α ± jβ ):
i1
y (t ) = ∑ exp(α
i=1
y (t ) = ∑ c exp( λ
t )[ c
n /2
0
2 *1− 1
i
n /2
0
i =1
2 *i −1
2 *i −1
cos( β
i
i
t ) + c sin( β
i
2 *i
t ) + c exp( λ t ) , where c
2 *i
2*i
i
t )] , or
= c and λ
*
2 *i −1
2 *i
2 *i − 1
=λ
*
2 *i
To view examples of how various combinations of these modes can behave, try
running the following script in Matlab:
t = [0:.1:10]; % Define Time Axis for function evaluation
for k=1:10 % Loop through 10 different functions
cr = 5*(rand(1,5) - .5); % Randomly pick 5 complex coefficients
ci = 5*(rand(1,5) - .5);
c = [(cr+j*ci), (cr-j*ci)];
subplot(3,1,1), plot(abs(c)); ylabel('Coefficient Magnitudes') % Plot out magnitudes
lamr = -2*rand(1,5);
% Randomly pick 10 complex frequencies
lami = 1 + 3*rand(1,5); lam = [(lamr+j*lami), (lamr-j*lami)];
subplot(3,1,2), plot(abs(lam)) % Plot out magnitudes
ylabel('Time Constant Magnitudes')
subplot(3,1,3), plot(t, real(c*exp(lam'*t))) % Evaluate and plot function
ylabel('Function'); xlabel('seconds')
pause
% Hold figure until Key is struck
end
3.8
Time Constant Magnitudes
Coefficent Magnitudes
Complex Roots Example:
2.5
2
1.5
1
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
3.5
3
2.5
2
Function
10
5
0
-5
-10
0
1
2
3
4
5
seconds
6
7
8
9
10
3.9
Example: Identify the modes of the system described below.
( D + 2 D + 10D + 7 D + D + 1) y(t ) = f (t )
5
4
3
2
Find roots of ( D + 2 D + 10D + 7 D + D + 1) = 0
5
4
3
Use Matlab:
» lam = roots([1, 2, 10, 7, 1, 1])
lam =
-0.6224 + 2.9286i
-0.6224 - 2.9286i
-0.8073
0.0261 + 0.3708i
0.0261 - 0.3708i
2
3.10
Solutions to Zero-Input Systems:
To have a nontrivial solution, the system must have some non-zero initial conditions.
In addition, to obtain a unique solution, the initial conditions on y0(t), and all its
derivatives must be specified up to the order of the differential equation minus 1.
Example: Consider ( D + 2 D + 10D + 7 D + D + 1) y(t ) = f ( t ) , assume that
y ( 0) = 0, y& ( 0) = 10, &&y ( 0) = −1, &&&
y ( 0) = 0, &&&
y& ( 0) = 0 . Find y0(t) for t ≥0.
5
0
0
4
0
3
0
2
0
Show solution is:
coefficients
0.0085 - 0.0009i
0.0085 + 0.0009i
-1.6741 - 0.0000i
0.8286 +11.6242i
0.8286 -11.6242i
which implies y0(t) =
0.0085exp( −.6224t ) cos(2.93t − 01
. ) + 1.67 exp( −0.81t ) + 1156
. exp( 0.03t ) cos(.37t + 2.5)
for t ≥0
3.11
The Unit Impulse (Dirac Delta) Function:
The unit impulse function δ(t) can be defined in the following equation:
for a < t < b
⎧ f (t )
∫ f ( t )δ (t − t )dt = ⎨
elsewhere
⎩ 0
This is sometimes referred to as the sampling property of the impulse function.
Since this function is defined only in terms of another function, it is referred to as a
generalized function.
b
0
0
0
a
Other important properties of the unit Impulse function:
The integral of δ(t) over any t interval that includes the point where the argument
becomes zero is one:
t 0+
∫ δ (t − t )dt = 1
t 0−
0
Although not a good definition, the impulse function can also be described as:
⎧ 0 for t < 0
d
d
⎪
r (t ) = u(t ) = δ (t ) = ⎨∞ for t = 0
dt
dt
⎪ 0 for t > 0
⎩
The impulse function can be consider as a limit of another function:
2
3.12
1⎡
ε
ε ⎤
δ (t ) = ⎢u⎛⎜ t + ⎞⎟ − u⎛⎜ t − ⎞⎟ ⎥
ε ⎣ ⎝ 2⎠ ⎝ 2⎠ ⎦
What is the limit for δε(t) as ε→0?
For any ε, what is the area under δε(t)?
δε(t)
ε
1
ε
t
Examples:
Evaluate the integral
+∞
⎛ π ⎞
δ
(
t
−
4
)
exp
⎜ j t ⎟ dt
∫
⎝ 4 ⎠
−∞
Show
+a
∫ u&(t ) f (t )dt = f (0)
−a
for a > 0
−ε
2
show by using integration by parts
ε
2
3.13
The Unit Impulse Response:
If the input to a LTIC system is an impulse, energy is injected into the system in a
moment of time. This creates a jump in the stored energy, and therefore an
instantaneous change in the initial conditions.
The effect of the impulse is to energize all modes of the system, whose behavior can
be observed in the impulse response.
Finding the impulse response:
The impulse response is similar to the zero-input response, since for t > 0 δ(t) = 0.
For a general differential Equation:
( D + a D +L+ a D + a ) y (t ) = (b D + b D +L+b D + b ) f (t )
Q( D) y ( t ) = P ( D) f ( t )
The impulse response is:
h ( t ) = b δ ( t ) + [ P ( D) y ( t )] u ( t ) where if m<n, bn=0, and y0(t) is the zero-input
response where the initial conditions were:
y ( 0) = 1, y ( 0) = 0, y ( 0) = 0, K y ( 0) = 0,
Examples: Find the impulse response for the given systems.
n −1
n
n −1
n
m −1
m
1
0
m
0
( n −1 )
( n− 2 )
( n− 3)
0
0
0
0
m −1
1
0
3.14
(D
+ 10D + 1) y ( t ) = Df ( t )
.
exp( −9.9t ) − 0.0102 exp( −01
. t )] u( t )
Ans: h(t ) = [10102
2
( D + 40) y (t ) = ( D + 2) f (t )
Ans: h( t ) = δ (t ) + [ − 40 exp( −40t ) + 2 exp( −40t )] u(t )
( D + 4 D + 4) y ( t ) = 3 f ( t )
Ans: h ( t ) = 3t exp( −2 t ) u ( t )
2