LTI System Review

Black-box model: Mathematical descriptions of the system and the input and
output signals x and y, respectively, are used to represent a physical system:
x

Model
y
o Mathematical model:
 Internal description: Differential/difference equations
 External description: Input/output terminals relationship
o Data required for analysis: Input and initial conditions
Systems - Modify signals
o Linear / Nonlinear systems: a linear system satisfies the superposition
property, which can be expressed into two separate parts:
 The additivity property:
If x1  y1 and x2  y2, then x1 + x2  y1 + y2
 The homogeneity (scaling) property:
If x  y, then ax  ay (a is a constant)
 Combining we have:
If x1  y1 and x2  y2, then ax1 + bx2  ay1 + by2
(a and b are constants)
 Responses:
 Zero-input response
 Zero-state response
 Total response = Zero-input response + Zero-state response
o Time-invariant / varying systems: A time invariant system’s properties do
not change with time – If x(t)  y(t), then x(t-)  y(t-)
o Instantaneous / dynamic system: An instantaneous system responds to
present input only. A dynamic system’s response also depends on the
history of the system as well as past input.
o Causal / noncausal systems: A causal system’s response does not depend
on future inputs. On the other hand, a noncausal system’s response does
depend on future inputs.
o Lumped / distributed systems: A lumped system’s physical dimensions are
much smaller than the wavelength of the signal. A distributed system’s
physical dimensions are comparable to the wavelength of the signal.
o Continuous- / discrete-time systems: Time axis
o Analog / digital systems: Vertical axis
o Number of inputs and outputs
 Single-input single-output (SISO) systems
 Single-input multi-output (SIMO) systems
 Multi-input single-output (MISO) systems
 Multi-input multi-output (MIMO) systems
o Invertible / noninvertible systems: An invertible system’s input can be
recovered from the output by some operations.
o Stable / unstable systems: A stable system observes the BIBO (boundedinput bounded-output) rule.
1
o Uncontrollable systems: Example

S1
S2
o Unobservable systems: Example
S1
S2

Signals - V or I
o Size: For convenience, the load is assumed to be 1 ohm.
 Amplitude and duration: Peak value and length of the signal
 Signal energy:

E

f
2
(t ) dt 




Real


2
f (t ) dt



Complex
Signal power (Periodic signals):
 T2

1
2 
if E   P  lim 
 f (t ) dt 
T  T T


2


For Periodic signals with period T :
T
1
P
T
2

2
f (t ) dt
T
2
o Types of signals
 Energy/ power signals (or neither: ramp function)
 Periodic signals: f(t) = f(t + t0) where t0 < 
 Random / deterministic signals
 Casual signals
1
1
 Even / odd signals: f (t )   f (t )  f (t )   f (t )  f (t )









2
2 
even
o Signal operations:
 Time-shifting (delay / advance)
 Time scaling
 Time-inversion (reversal)
2
odd
o Basic signals:
 Unit step and ramp: u(t) and r(t)
 Using these operations one can combine basic waveforms to obtain
other more complex waveforms.
o Impulse: (t-t0)
 (t-t0) = 0 for t  t0
 (t-t0) is undefined for t = t0
h
h
w0
w
Area  h  w  1
The impulse function can only be defined under integrations.

t2
  (t  t0 ) dt  1
if t1  t0 and t 2  t0
t1

Sampling property:
f (t )  (t )  f (0)  (t ) 

 f (t )  (t ) dt  f (0)

f (t )  (t  t 0 )  f (t 0 )  (t  t 0 ) 

 f (t )  (t  t0 ) dt  f (t0 )


Relationship with u(t) and r(t)
t
  ( ) d  u (t ),

d u (t )
  (t )
dt
t
 u ( ) d  r (t ),

d r (t )
 u (t )
dt
 Often used as a test signal (impulse response)
o Exponential function: ex

A natural solution for DEs:

For a signal:
de x
 ex
dx
f (t )  A e st  A e (  j )t
 - real and negative: Decaying exponential (stable)
 - real and positive: Growing exponential (unstable)
 - nonzero: oscillating

Response of an LTI (Linear time-invariant) system:
1
1
1
2
1
1
1
2
1
3
2
3
Question (Lathi, p. 101): For an LTI system with input f(t), output y(t), and two
initial conditions x1(0) and x2(0), the following measurements were made:
f(t)
x1(0)
x2(0)
y(t)
0
1
-1
e-t u(t)
0
u(t)
2
-1
e-t (3t + 2) u(t)
2 u(t)
1
-1
Determine y(t) when both initial conditions are zero and the input is as follow:
1
-5
0
5
t
Solution:
1. Decompose f(t): f(t) = u(t + 5) – u(t – 5)
2. Eliminate x2(0):
f(t)
x1(0)
x2(0)
0
1
-1
y(t)
e-t u(t)
+)
0
2
1
e-t (3t + 2) u(t)
0
3
0
e-t (3t + 3) u(t)
0
1
0
e-t (t + 1) u(t)
3. Eliminate x1(0):
f(t)
x1(0)
0
1
x2(0)
-1
y(t)
e-t u(t)
-)
or
or
0
1
0
e-t (t + 1) u(t)
0
0
-1
e-t (-t) u(t)
0
0
1
e-t (t) u(t)
4. Combining the above two steps:
f(t)
x1(0)
x2(0)
0
1
0
y(t)
e-t (t + 1) u(t)
+)
0
0
1
e-t (t) u(t)
0
1
1
e-t (2t + 1) u(t)
x2(0)
1
-1
0
y(t)
e-t (2t + 1) u(t)
2 u(t)
[e-t (2t + 1) + 2] u(t)
5. Eliminate both initial conditions:
f(t)
x1(0)
0
1
+)
u(t)
-1
u(t)
0
4
6. Now consider the time-shift:
u(t + 5)
0
0
[e-(t+5) (2t + 11) + 2] u(t + 5)
u(t – 5)
[e-(t–5) (2t – 9) + 2] u(t – 5)
0
0
7. Final result:
y(t) = [e-(t+5) (2t + 11) + 2] u(t + 5) – [e-(t–5) (2t – 9) + 2] u(t – 5)

System analysis techniques:
o Classical technique: Solving the differential/difference equations
 Undetermined coefficients
 Tedious
 Difficult to interpret the physical meaning
 Must repeat the effort for different input
o Convolution: Find the impulse response h(t) and convolve it with the input
signal to obtain the output signal.
 Obtain the impulse response h(t) by solving the DEs with no input
and a special set of initial conditions.
 The output is obtained by performing the convolution integral
between the input and h(t).
 The DEs are easier to solve.
 Same h(t) is used for any input.
 Impulse response can be obtained experimentally or numerically.
 Zero-state and zero-input responses can be identified separately.
 Convolutions are difficult.
 Table look-up
o Transform techniques: Take the Laplace/Fourier or z-transform and solve
the algebraic equations in the transformed domain. Use the inverse
transform to recover the time domain solutions.
 Fourier series: Decomposing periodic or finite-duration signals
 Fourier transform:

Aperiodic signals

Steady-state response (no initial conditions)

Frequency response

Direct integration or table look-up
 Laplace / z-transform:

Total response (initial conditions)

Frequency response

Direct integration or table look-up

Inverse transform difficult (partial-fraction expansion +
table look-up)
o State-variable technique (time-domain): EEE-241
5
Circuit Review

Component models in the time domain:
i(t)
+
v(t)
-
Component
Resistor
Capacitor
v(t ) 
Voltage: v(t )  R i (t )

1
C
Inductor
t
 i(t ) dt

t
v(t )  L
1
i (t ) dt  i (0  )
C
di (t )
dt
0
Current:
i (t ) 
v (t )
R
i (t )  C
1
i (t ) 
L
dv (t )
dt

t
 v(t ) dt

t
1
v(t ) dt  v(0  )
L
0

Formulating circuit equations
i1
i3
i4
i2
C
o Choose a ground node (node 0)
o Label other nodes
o Write nodal equations (KCL):
KCL at node 2: i1 + i2 + i3 = 0
KCL at node 3: (-i3) + i4 = 0
o Represent the currents in terms of voltages and component models (for
simplicity, initial conditions are ignored for now):
v = vi – vj
vi
vj
Component
i
6
1
i1 
L
t
 v2 (t )  v1 (t )dt

v (t )
i2  2
R1
v (t )  v3 (t )
i3  2
R2
v (t )  v2 (t )
  i3  3
R2
d
v3 (t )
dt
o Write nodal equations (KCL) in terms of node voltages:
i4  C
1
L
t
 v2 (t )  v1(t )dt 

v2 (t ) v2 (t )  v3 (t )

0
R1
R2
v3 (t )  v2 (t )
d
 C v3 (t )  0
R2
dt
o Removing the integral:
v2 (t )  v1 (t ) 1 d
v2 (t )  1 d v2 (t )  v3 (t )  0

L
R1 dt
R2 dt
o Substituting vin for v1, vout for v3, and eliminating v2 from the two
equations we have the following differential for the above circuit:
 R  d2
 L
d

LC 1  2 
vout (t )    R2C  vout (t )  vout (t )  vin (t )
R1  dt 2

 R1
 dt

Impulse response: The impulse response of an LTI system can be obtained from
the zero-input response of the system. A specific set of initial condition must be
included in the solution procedures.
Example: y”(t) + 3y’(t) + 2y(t) = 0
To obtain h(t) we need to set y(0) = 0 and y’(0) = 1
(see Lathi P.116 for the
general formula)
2  3  2  0    1,2  h(t )  C1e t  C2 e 2t
y (0)  0  C1  C2  0

y ' (0)  1   C1  2C2  1
 C1  1 C2  1 h(t )  e  t  e  2t
Total response: The response for a specific input can be obtained from the
convolution of the input and the impulse response.
Example: The impulse response for a series R-C network is: h(t )  e t / t 0 u (t ) ,
where t0 is the R-C time constant. The input is a rectangular pulse with amplitude
equals to 1 and width equals to T. Find the output of the system.
t
y (t )   h(t   ) x( ) d  h(t )  x(t )
0
x(t )  u (t )  u (t  T ) 
y (t )  h(t )  u (t )  h(t )  u (t  T )
7
Using a convolution table we have:
y1 (t )  e
t / t0
y2 (t )  e


1  e t / t 0
u (t )  u (t ) 
u (t )  t0 1  e  t / t 0 u (t )
1 / t0
t / t 0
1  e  t T  / t 0
u (t )  u (t  T ) 
u (t  T )  t0 1  e  t T  / t 0 u (t  T )
1 / t0


 y (t )  y1 (t )  y2 (t )

We will join the textbook at this point. In order to make the transition we need to
clarify the difference in notations:
o xi(t) represent the internal states of the system.
o u(t) is the input. The step function is replaced by specifying the duration of
the function such as (t > 0, …).
o g(t) is the impulse response.
L.T.
o The Laplace transform of a function is represented by g (t ) 
 gˆ (s) .
However, the ĝ notation is difficult to produce. As a result, I will use the
L.T.

conventional notation for Laplace transform: g (t )  G ( s ) .
For SISO linear systems:

 g (t , )u ( ) d
y (t ) 


where g is a function for the time at which the impulse is applied to the system.
For SISO casual LTI systems:
t
t
0
0
y(t )   g (t   )u ( ) d   g ( )u (t   ) d  g (t )  u (t )  u (t )  g (t )
8

For MIMO (p inputs and q outputs) casual LTI systems:
g1 p (t ) 
 y1 (t ) 
 g11 (t ) g12 (t )
 u1 (t ) 
 y (t ) 
 g (t ) g (t )

 u (t ) 
g 2 p (t )
22
 2 
 21
 2 
 u(t )   . 
y (t )   .  g(t )  






 . 


 . 
 y q (t )
 g q1 (t ) g q 2 (t )
u p (t )
g qp (t ) 





t
y (t )   g (t   )u( ) d
0

Laplace transform

L.T.
f (t )  F ( s) : F ( s) 
 f (t ) e
 st
dt
0
we have:
SISO :
MIMO :
Y (s)  G (s) U (s)
Y( s)  G ( s) U( s)
 Y1 ( s ) 
Y ( s ) 
 2 
Y(s)   . 


 . 
Yq ( s )


 G11 ( s )
G ( s )
 21
G (s)  


Gq1 ( s )


 U1 ( s ) 
U ( s ) 
 2 
U(s)   . 


 . 
U p ( s )


G12 ( s )
G1 p ( s ) 
G22 ( s )
G2 p ( s )



Gq 2 ( s )
Gqp ( s ) 
G(s) [G(s)] is called the transfer function [matrix] of the system.
o The transfer functions are irrational functions of s  G(s) = N(s) / D(s),
where N(s) and D(s) are polynomials of s (poles and zeros).
a s m  a s m1    a1 s  a0
G ( s)  m n m1 n1
bn s  bn1 s    b1 s  b0
o The concepts of proper, strictly proper, biproper, and improper transfer
functions.
o For physical systems deg D(s)  deg N(s).
o Zero/pole gain form of G(s):
s  z1 s  z 2 s  z m 
G( s)  k
s  p1 s  p2 s  pn 
9
o Coprime: N(s) and D(s) have no common factors of degree 1 or higher.
o Component models in the s-domain (for simplicity, initial conditions are
ignored for now):
Resistor
Voltage: V ( s )  R I ( s )
Current: I ( s ) 
V (s)
R
Capacitor
1
V (s) 
I (s)
sC
Inductor
I ( s )  sC V ( s )
I ( s) 
V ( s )  sL I ( s )
1
V (s)
sL
Formulating the same circuit equations using Laplace transform we have:
s
1/sC

s 2 LC 1 

 L

R2 
Vout ( s)  s  R2C Vout ( s)  Vout ( s)  Vin ( s)
R1 
 R1

Y ( s) Vout ( s)


U ( s) Vin ( s)
1

 L

R 
LC 1  2  s 2    R2C  s  1
R1 

 R1

o G(s) fully describes the characteristics of the system (no initial
conditions).
o Stability: No RHP poles
o The effect of feedback
 G ( s )
+

X(s)
E(s)
H(s)
Y(s)
Y(s)G(s)
G(s)
E ( s)  X ( s)  Y ( s)G ( s)
and
Y ( s)  E ( s) H ( s)
 Y ( s)  X ( s)  Y ( s)G ( s)H ( s)  H ( s) X ( s)  H ( s)Y ( s)G ( s)
 Y ( s)1  H ( s)G ( s)  H ( s) X ( s) 
Y (s)
H (s)

 H FB ( s)
X ( s) 1  H ( s)G ( s)
(New transfer function)
10

State-space equations: For a lumped, LTI system we can use the following set of
equations:
x (t )  A x(t )  B u(t )
y (t )  C x(t )  D u(t )
o The vector x(t) represents the internal states of the system.
o A, B, C, and D are constant matrices.
o For a system with p inputs and q outputs, the dimensions of the constant
matrices are n  n, n  p, q  n, and q  p, respectively.
o Applying Laplace transform we can convert the DEs to simple algebraic
equations:
sX( s )  x(0)  A X( s )  B U( s )
Y ( s )  C X( s )  D U ( s )
o The internal states and the output can be expressed as:
X( s)  ( sI  A) 1 x(0)  ( sI  A) 1 B U( s)
Y( s)  C( sI  A) 1 x(0)  C( sI  A) 1 B U( s)  D U( s)
o The zero-state response and the system transfer function/matrix are:
Y( s)  C( sI  A) 1 B  D U( s)


G( s)  C( sI  A) 1 B  D
11