EL625
Solutions to Problem Set 1
1.1
a) The system is not causal since we can see that y (t) depends on u(t + ) for 0 < 1.
b) This could be confusing because of the way the variable t appears. To avoid this
confusion, let = t + so d = d and = 1 gives = t

Solutions to Problem Set 2
EL625
2.1
a) Let x1 = iL1 , x2 = iL2 and x3 = vC be the state variables. We have then
dx2
vL
= 2;
dt
L2
dx1
vL
= 1;
dt
L1
dx3
x1 + x2
=
dt
C
Applying Kirchos voltage law around the rst loop gives e1 + R1 x1 + vL1 + x3 = 0. Then

Chapter 6 solutions
6.1
a) - This Z -transform does not exist because f (k ) is not bounded at k = 1.
b), c), e) and h) - These Z -transforms exist since the f (k ) are bounded for all k 0 and
go to zero as k so they are of exponential order with A = 0.
d

Solutions to Chapter 5
EL625
5.1 For this system, we see that
s1
0
2 s + 3
[sI A] =
Since
[sI A]1 =
so
1
s1
2
(s1)(s+3)
0
1
s+3
.
2
1
1
1
=
,
(s 1)(s + 3)
2 s1 s+3
we nd that
et
0
1
t
3t
3t
(e e ) e
2
(t) =
and
H (s) = C [sI A]1 B + D =
5.3 We apply Lever

Solutions to Problem Set 5
EL625
4.12 We have
1 1
0
0
1
A=0
0 2 3
so
+ 1 1
0
1
det[I A] = det 0
0
2 +3
= ( + 1)(2 + 3 + 2) = ( + 1)2 ( + 3) = 0.
Thus the eigenvalues are 1 = 1 with multiplicity 2 and 2 = 2 with multiplicity 1. By
Method 2 with the trial

Solutions to Problem Set 4
EL625
4.1
a) The current entering C (t) equals the rate of change of charge on C (t). Thus
dq (t)
d[C (t)vC (t)]
=
= C (t)vC (t) + C (t)vC (t) = iC (t).
dt
dt
Now
v (t) vC (t) vC (t)
= v (t) vC (t) 2vC (t) = v (t) 3vC (t)
R1
R2

EL625
Solutions to Problem Set 3
2.22
a) For the system:
D2 y1 + 4Dy1 + 3y2 = u1 + u2 D2 y2 + 5Dy2 + Dy1 + y1 = u2 ,
we have
y1 = D1 4y1 + D1 (3y2 + u1 + u2 )
y2 = D1 5y2 y1 + D1 (y1 + u2 )
Using x1 = y1 , x3 = y2 , x2 = 3y2 + u1 + u2 , and x4 = y1 + u2 w

EL 625 Lecture 11
1
EL 625 Lecture 11 Frequency-domain analysis of discrete-time systems Theorem: If S is a xed linear discrete-time system, the zero-state response, S cfw_z k = H (z )z k Proof: S cfw_z k =
S cfw_z k
(1)
zk
H (z,k )
.z k = H (z, k

Recipes for Pole Placement Task: Consider an nth order linear controllable single-input (i.e., u being scalar) system x = Ax + Bu. With the linear state feedback u = Kx, the closed-loop system is x = (A + BK )x. Given a set of n desired eigenvalues p1 , .

EL 625 Lecture 10
1
EL 625 Lecture 10 Pole Placement and Observer Design Pole Placement Consider the system x = Ax The solution to this system is x(t) = eAtx(0) (2) (1)
If the eigenvalues of A all lie in the open left half plane, x(t) asymptotically appro

EL 625 Lecture 9
1
EL 625 Lecture 9 Observability of linear systems Denition of observability: A system is called (completely) observable if for any initial time, t0, any initial state, x(t0) can be determined from observation of the output y [t0, t] over

EL 625 Lecture 8
1
EL 625 Lecture 8 Controllability and observability of linear systems Denition of controllability: A system is called completely controllable if for any pair of arbitrary initial and nal states, an input can be found to transfer the syst

EL 625 Lecture 7
1
EL 625 Lecture 7 Frequency domain analysis of time-invariant systems Theorem: If S is a linear time-invariant continuous-time system, S cfw_est = H (s)est Proof:
S cfw_est st st S cfw_e = st .e = H (t, s)est e
H (t,s)
S cfw_es(t ) =

EL 625 Lecture 5
1
EL 625 Lecture 5 Cayley-Hamilton Theorem:
Every square matrix satises its own characteristic equation If the characteristic equation is p() = det(I A) = n + an1n1 + an2n2 + . . . + a1 + a0 From Cayley-Hamilton theorem, p(A) = An + an1An

EL 625 Lecture 4
1
EL 625 Lecture 4 Solution of the dynamic state equations
x(t) = A(t)x(t) + B (t)u(t) y(t) = C (t)x(t) + D(t)u(t) Homogeneous equation (Unforced system): x = A(t)x (2) (1)
Consider the matrix dierential equation, Q(t) = A(t)Q(t). If th

EL 625 Lecture 3
1
EL 625 Lecture 3 Principles of time-domain analysis Singularity functions:
t
i+1(t ) =
i( )d
i ( t ) E
E
i+1 (t )
0(t) = (t) 1(t) = 1(t) Unit impulse or -function:
T
0
E
t
t
f () ( )d =
0
for t < for t >
f ( )
EL 625 Lecture 3
2
(

EL 625 Lecture 2
1
EL 625 Lecture 2 State equations of nite dimensional linear systems Continuous-time: x(t) = A(t)x(t) + B (t)u(t) y(t) = C (t)x(t) + D(t)u(t) Discrete-time: x(tk+1) = A(tk )x(tk ) + B (tk )u(tk ) y(tk ) = C (tk )x(tk ) + D(tk )u(tk ) sta

EL 625 Lecture 1
1
EL 625 Lecture 1 Denition: A physical system is an interconnection of physical components that perform a specic function. These components may be electrical, mechanical,hydraulic,thermal and so forth.
Inputs or excitations
u1 u2 ur
E E