55
Problems
ub(t)
3.4 A fixed, linear system has the unit-step response a(t) te l l(t).
a) Find the impulse response h(t) = Scfw_8(t).
b) Find the response m2(t) to the unit ramp function /x2(0c) Find the response y(t) to the function u(t) shown below.
3.
3
Principles of time-domain analysis
3.1 Introduction
As we indicated earlier, in the indirect approach to the analysis of linear systems,
when we seek the response of some linear system S to a general input function
u(t), we express u(t) as a linear comb
3.6 Graphical interpretation of the convolution integral
49
Of course, this is just a graphical interpretation of the process of convolution. To
evaluate, we actually integrate the functions inside the integral.
Example 3.1
A fixed linear system has the s
3.6 Graphical interpretation of the convolution integral
47
For fixed systems
pOO
y(t)=
pOQ
u(k)h(t-k)dk=
Joo
ucfw_t-t-)hcfw_t-)dl-
(3.23)
u(t-%)a(%)d%
(3.24)
Joo
or
poo
y(t) =
poo
u(k)a(t - k) dk =
Joo
Joo
and
poo
a(t-r)=
/
h(t-X)dX.
(3.25)
JX
For r = 0
40
2 Finite dimensional linear systems
2.8 Find the state differential equations and the A, B, C, and D matrices for
each of the mechanical systems in Fig. P 2.8. The input is fit) and the
output is z\(t) for both systems,
a)
K
D
| _
2(0
b)
D
Kt)
l_l
U
Z2
52
3 Principles of time-domain analysis
where h(k,i) = the response observed at time k when a discrete delta function is
applied at time i. This is called the discrete delta response function or just the delta
response.
As before for fixed systems
oo
y(k)
48
3 Principles of time-domain analysis
Suppose
and cfw_/2(A.) are as shown in the figure.
1.5
0
0
Then if we plot cfw_f2(k) against A., this looks like
l
Now, shift by an amount t\, so 1/2(^1 A) is
and the product function cfw_/i(A.) f2(t\ X) versus X is
Problems
39
2.5 In the circuit shown below, the inputs are the applied voltages e\cfw_t) and
e2(t). The output is the voltage v(t).
Write the state equations and output equations for this circuit.
2.6 For the circuit shown below, write the state different
2.7 Systems of multiple dynamic equations
35
Example 2.10
h + y\ + h + yi = u
y\ + y\ - h - h = u - u + u.
Procedure 1
Begin by placing the highest order derivative of a different output variable on the
left in each of the equations
D2ycfw_ = -D(yi + y2)
Problems
41
2.11 Find the simulation diagram and the A, B, C, and D matrices for the analog
systems:
a) y'(t) + 2y(t) - y(t) + 3y(t) = AU(t) + u(t)
b) 5t2y(t) + (t- l)y(t) + t2y(t) = u(t) + tu(t).
2.12 For the system described by the differential equation
54
3 Principles of time-domain analysis
I with all other inputs zero. Then
y(k)= V H(k,l)u(l)
(3.40)
where H(k, I) is the m x r matrix with elements htj(k, I), and it is called the delta
response matrix.
Problems
3.1 For the function shown below:
1.51-
0.
44
3 Principles of time-domain analysis
3.3 Impulses and the impulse response
The unit impulse or 8-function can be defined by the two properties
F
/
f0
fiX)8iX
- x)dX = <
for t < x
r/
x
(3.3)
r
7-00
1 / ( T ) forr > r
and
8it) = 8i-t)
foralH.
From this,
42
2 Finite dimensional linear systems
2.20 Write the state difference equations and the A, B, C, and D matrices for
the system
1
ycfw_k + 2) + k2y(k) = u(k + 2) + ku(k + 1) + 2u(k).
k +l
2.21 Find the simulation diagram and the A, B, C, and D matrices of
36
2 Finite dimensional linear systems
u(t) o-
Fig. 2.15.
Procedure 2
We can frequently get a clue to the order by trying to write a single equation in
terms of one of the output variables. In Example 2.10 we write
D2yx + Dyx + Dy2 + y2 = u
+y\-
D2y2 - Dy
46
3 Principles of time-domain analysis
which is the integral over the second variable of mt (t, rj) or
rmit, r) = mi+iV>T\
(3 15)
Now, if a system is causal then
h(t,x)=Q
for^<r
a n d a(t,x) = 0 f o r f < r
(3.16)
since the response to an impulse or step
3.8 Systems with multiple inputs and outputs
53
so the discrete step response can be found from the discrete delta response.
Note: It is not always easy to find a closed form expression from this.
3.8 Systems with multiple inputs and outputs
When a linear
3.5 Relation between the step and impulse response
45
Thus, for t > t0,
rco
y(t) = u(to)a(t,to)+
iicfw_k)a(t,k)dk
(3.9)
Jto
where a(t, z) = the zero-state response observed at time t to a unit step that starts
at time r. This is called the unit-step respo
38
2 Finite dimensional linear systems
2.2 Write the state differential equations for the circuit below.
R
e(t)l
)
2
Cx
2.3 For the circuit shown below, e(t) is the input voltage and the outputs are
the voltages vR(t) and vc(t). Write the state differenti
50
3 Principles of time-domain analysis
Example 3.2
Suppose we want to find the zero-state response of the system of Example 3.1 to
the input u(t) = 5 1(0 5 \(t 1). Since the system is linear,
Scfw_5 1(0 - 5 \cfw_t - 1) = 55(1(0 - 5Scfw_l(t - 1)
and since
34
2 Finite dimensional linear systems
Formally we can obtain the lowest order simulation for a fixed system with one
input and one output by canceling common factors of the polynomials on each side
of the equation. Note, however, that because of the pres
26
2 Finite dimensional linear systems
We see therefore that the choice of state variables is fairly wide. For circuits,
it is most convenient to choose inductor currents and capacitor voltages (for fixed
networks) or inductor fluxes and capacitor charges
20
2 Finite dimensional linear systems
algebraic equations for z. When Fn is singular (i.e., it has no inverse) then
the equations are degenerate and cannot be put in normal form. (This occurs
when the circuit contains a loop of capacitors or a node conne
2.2 State differential equations of circuits
17
or
iL(t)
(2.6)
L(t)- = vL(t)
dt
dt
can be used to describe the basic terminal relation for an inductor. When the inductor
value does not change with time (i.e., L is a constant) then L = 0, so the terminal
r
16
2 Finite dimensional linear systems
2.2 State differential equations of circuits
We start by showing how passive electrical circuits can be described by state differential equations in normal form. The basic elements of such systems are resistors,
capa
24
2 Finite dimensional linear systems
The equations of motion are
at
=
_
K[Z2(t)
(2.15)
(2.16)
- f(t) - K[Z2(t) - D2v2cfw_t) at
where the velocity vt (t) is the derivative of displacement position zt (t) so vt (t) =
M2
Zi(f).
We choose the natural state
2.2 State differential equations of circuits
19
a For resistive branches (Rk) the equation is vt Vj = ikRk.
b For inductive branches (Lk) the equation is vt Vj = Lk(dik/dt).
c For capacitative branches (Ck) vt Vj = vc (where v( is at the +
terminal),
d Fo
30
2 Finite dimensional linear systems
Xi
x2
Xn
0
1
0
0
0
1
0
0
0
0
X\
-a0
x2
x3
-a2
1 -_!
+
xn_
A(t)
= [o o
b0 -aobn
bx-axbn
b2 -a2bn
A-.-a ^b _
n
n
Bit)
(2.26)
= bn.
Note: If bn = 0 then D = 0. This means that all forward paths from the
input to output
2.6 Systems governed by a single dynamic equation
29
where y(j) denotes the jth derivative, and the coefficient of the highest derivative
of y(t) has been normalized to 1. Furthermore, notice that the highest order derivative of u(t) cannot exceed the hig