CHAPTER
Time-Domain Representations of LTI Systems
2.11 Characteristics of Systems Described by Differential
and Difference Equations
Complete solution: y = y (n) + y (f)
y (n) = natural response, y (f) = forced response
2.11.1 The Natural Response
Example 2.24 RC Circuit (continued): Natural Response
The system In Example 2.17 is described by the differential equation
d
y  t   RC y  t   x  t 
dt
Find the natural response of the this system, assuming that y(0) = 2 V, R = 1 
and C = 1 F.
<Sol.>
h
y    t   c1e t V
1. Homogeneous sol.:
2. I.C.: y(0) = 2 V
y (n) (0) = 2 V
3. Natural Response:
c1 = 2
y
n
 t   2e  t
V
1
CHAPTER
Time-Domain Representations of LTI Systems
Example 2.25 First-Order Recursive System (Continued): Natural Response
The system in Example 2.21 is described by the difference equation
y n 
1
y  n  1  x  n 
4
Find the natural response of this system.
n
<Sol.>
1

h
1. Homogeneous sol.: y  n   c1  
4
2. I.C.: y[ 1] = 8
1
8  c1  
4
1
c1 = 2
3. Natural Response:
n
1
y  n   2   , n  1
4
n
The forced response is valid only
for t  0 or n  0
2.11.2 The Forced Response
The forced response is the system output due to the input signal assuming
zero initial conditions.
2
CHAPTER
Time-Domain Representations of LTI Systems
 The at-rest conditions for a discrete-time system, y[ N] = 0, …, y[ 1] = 0,
must be translated forward to times n = 0, 1, …, N  1 before solving for the
undetermined coefficients, such as when one is determining the complete
solution.
Example 2.26 First-Order Recursive System (Continued): Forced Response
The system in Example 2.21 is described by the difference equation
y n 
1
y  n  1  x  n 
4
Find the forced response of this system if the input is x[n] = (1/2)n u[n].
<Sol.>
n
n
1
1


1. Complete solution: y  n   2    c1   , n  0
2
4
2. I.C.: Translate the at-rest condition y[ 1] to time n = 0
1
y 0  x 0  y  1
4
3. Finding c1:
y[0] = 1 + (1/4)  0 =1
0
1
1
1  2    c1  
2
4
0
c1 =  1
3
CHAPTER
Time-Domain Representations of LTI Systems
4. Forced response:
n
n
1 1
f
y   n  2      , n  0
2 4
Example 2.27 RC Circuit (continued): Forced Response
The system In Example 2.17 is described by the differential equation
y  t   RC
d
y t   x t 
dt
Find the forced response of the this system, assuming that x(t) = cos(t)u(t) V, R
= 1  and C = 1 F.
From Example 2.22
<Sol.>
1
1
t
1. Complete solution: y  t   ce  cos t  sin t V
2
2
2. I.C.:
y(0) = y(0+) = 0
c =  1/2
3. Forced response:
1
1
1
y  f   t    e t  cos t  sin t V
2
2
2
4
CHAPTER
Time-Domain Representations of LTI Systems
2.11.3 The Impulse Response
 Relation between step response and impulse response
2. Discrete-time case:
1. Continuous-time case:
h( t ) 
h  n  s[n]  s[n  1]
d
s(t )
dt
2.11.4 Linearity and Time Invariance
 Forced response  Linearity
 Natural response  Linearity
Input
Forced response
Initial Cond.
Natural response
x1
y1(f)
I1
y1(n)
x2
y2(f)
I2
y2(n)
 x1 +  x2
y1(f) +  y2 (f)
 I1 +  I2
y1(n) +  y2 (n)
 The complete response of an LTI system is not time invariant.
Response due to initial condition will not shift with a time shift of the
input.
2.11.5 Roots of the Characteristic Equation
5
CHAPTER
Time-Domain Representations of LTI Systems
 Roots of characteristic equation
Forced response, natural response, stability, and response time.
★ BIBO Stable:
1. Discrete-time case:
ri n
bounded
rt
2. Continuous-time case: e i
bounded
ri  1,
for all i
eri   0
ri  1 and eri   0  The system is said to be on the verge of instability.
2.12 Block Diagram Representations
 A block diagram is an interconnection of the elementary operations that act
on the input signal.
 Three elementary operations for block diagram:
1. Scalar multiplication: y(t) = cx(t) or y[n] = cx[n], where c is a scalar.
2. Addition: y(t) = x(t) + w(t) or y[n] = x[n] + w[n].
t
3. Integration for continuous-time LTI system:
y (t ) 
x( )d

; and a time shift for discrete-time LTI
system: y[n] = x[n  1].
Fig. 2.32.

6
CHAPTER
Time-Domain Representations of LTI Systems

t
y(t )   x( )d

(a)
(c)
Figure 2.32 (p. 162)
Symbols for elementary operations in block diagram
(b)
descriptions of systems. (a) Scalar multiplication. (b)
Addition. (c) Integration for continuous-time systems and
time shifting for discrete-time systems.
Ex. A discrete-time LTI system: Fig. 2.33.
 Direct Form I:
1. In dashed box:
w[n]  b0 x[n]  b1x[n  1]  b2 x[n  2] (2.49)
2. y[n] in terms of w[n]:
7
CHAPTER
Time-Domain Representations of LTI Systems
y[n]  w[n]  a1y[n  1]  a2 y[n  2] (2.50)
3. System output y[n] in terms of input x[n]:
y[n]  a1y[n  1]  a2 y[n  2]  b0 x[n]  b1x[n  1]  b2 x[n  2]
y[n]  a1y[n  1]  a2 y[n  2]  b0 x[n]  b1x[n  1]  b2 x[n  2]
Cascade Form
(Direct Form I)
(2.51)
Figure 2.33
(p. 162)
Block diagram
representation of
a discrete-time
LTI system
described by a
second-order
difference
equation.
8
CHAPTER
Time-Domain Representations of LTI Systems
 Direct Form II:
1. Interchange the order of Direct Form I. h1 (t ) * h2 (t )  h2 (t ) * h1 (t ) 
2. Denote the output of the new first system as f[n].
Input: x[n]
f[n]  a f[n  1]  a f[n  2]  x[n] (2.52)
1
2
3. The signal is also the input to the second system. The output of the second
system is
y[n]  b0 f[n]  b1f[n  1]  b2f[n  2] (2.53)
Fig. 2.35.
Figure 2.35 (p. 164)
Direct form II representation of an LTI
system described by a second-order
difference equation.
9
CHAPTER
Time-Domain Representations of LTI Systems
 Block diagram representation for continuous-time LTI system:
1. Differential Eq.:
M
dk
dk
ak k y(t)  bk k x(t) (2.54)

dt
dt
k 0
k 0
2. Let v(0)(t) = v(t) be an arbitrary signal, and set
N

n
 t    n1  d ,
t
n  1, 2, 3, ...
v(n)(t) is the n-fold integral of v(t) with respect to time
3. Integrator with initial condition:

n
 t   0 n1  d  n  0 ,
t
n  1, 2, 3, ...
d n
  t     n 1  t  , t  0 and n  1, 2, 3, ...
dt
Integrate N times to eq. (2.54)
N
 ak y
k 0
(Nk )
M
(t)  bk x
k 0
(Nk )
(t)
(2.55)
10
CHAPTER
Time-Domain Representations of LTI Systems
Ex. Second-order system:
y(t)  a1y(1) (t)  a0 y(2) (t)  b2x(t)  b1x(1) (t)  b0 x(2)(t)
(2.56)






(a)
(b)
Figure 2.37 (p. 166)
Block diagram representations of a continuous-time LTI system described by a
11
second-order integral equation. (a) Direct form I. (b) Direct form II.
CHAPTER
Time-Domain Representations of LTI Systems
2.13 State-Variable Description of LTI Systems
 The state of a system may be defined as a minimal set of signals that
represent the system’s entire memory of the past.
Given initial point ni (or ti) and the input for time n  ni (or t  ti), we can
determine the output for all times n  ni (or t  ti).
2.13.1 The State-Variable Description
1. Direct form II of a second-order LTI system: Fig. 2.39.
2. Choose state variables: q1[n] and q2[n].
3. State equation:
q1[n  1]  a1q1[n]  a2q2 [n]  x[n] (2.57)
q2 [n  1]  q1[n]
(2.58)
4. Output equation:
y[n]  x[n]  a1q1[n]  a2q2 [n]  b1q1[n]  b2q2 [n],
y[n]  (b1  a1 )q1[n]  (b2  a2 )q2 [n]  x[n]
(2.59)
5. Matrix Form of state equation:  q [n  1]   a a   q [n]   1
1
1
1
2
q [n  1]   1

  0  x[n]

q
[n]
0
 2   
 2
 
(2.60)
12
CHAPTER
Time-Domain Representations of LTI Systems
Figure 2.39 (p. 167)
Direct form II representation of a second-order discrete-time LTI system
depicting state variables q1[n] and q2[n].
13
CHAPTER
Time-Domain Representations of LTI Systems
6. Matrix form of output equation:
 q1[n] 
y[n]  [b1  a1 b2  a2 ] 
  [1]x[n] (2.61)
q
[n]
 2 
Define state vector as the column vector
 q [n] 
q[n]   1 
q2 [n]
We can rewrite Eqs. (2.60) and (2.61) as
q[n  1]  Aq[n]  bx[n] (2.62)
y[n]  cq[n]  Dx[n]
(2.63)
where matrix A, vectors b and c, and scalar D are given by
 a a2 
A 1
0 
 1
1
b 
0 
c  b1  a1 b2  a2 
D 1
Example 2.28 State-Variable Description of a Second-Order System
Find the state-variable description corresponding to the system depicted in
Fig. 2.40 by choosing the state variable to be the outputs of the unit delays.
<Sol.>
14
CHAPTER
Time-Domain Representations of LTI Systems
Figure 2.40 (p. 169)
Block diagram of LTI system for Example 2.28.
1. State equation:
q1  n  1   q1  n   1 x  n 
q2  n  1   q1  n    q2  n    2 x  n 
15
CHAPTER
Time-Domain Representations of LTI Systems
2. Output equation:
y  n   1q1 n   2q2 n 
3. Define state vector as
 q1  n  
q n   

q
n


 2 
In standard form of dynamic equation:
q[n  1]  Aq[n]  bx[n] (2.62)
y[n]  cq[n]  Dx[n]

A

0
 
(2.63)
 
b   1
 2 
c  1 2 
D   2
 State-variable description for continuous-time systems:
d
q(t)  Aq(t)  bx(t) (2.64)
dt
y(t)  cq(t)  Dx(t) (2.65)
16
CHAPTER
Time-Domain Representations of LTI Systems
Example 2.29 State-Variable Description of an Electrical Circuit
Consider the electrical circuit depicted in Fig. 2.42. Derive a state-variable
description of this system if the input is the applied voltage x(t) and the output
is the current y(t) through the resistor.
<Sol.>
Figure 2.42 (p. 171)
Circuit diagram of LTI
system for Example 2.29.
1. State variables: The voltage across
each capacitor.
2. KVL Eq. for the loop involving x(t), R1, and C1:
x  t   y  t  R1  q1  t 
y(t)  
1
1
q1(t) 
x(t)
R1
R1
Output equation
(2.66)
3. KVL Eq. for the loop involving C1, R2, and C2:
q1  t   R2i2 (t )  q2 (t )
17
CHAPTER
i2 (t) 
Time-Domain Representations of LTI Systems
1
1
q1(t) 
q2 (t)
R2
R2
(2.67)
Use Eq. (2.67) to eliminate i2(t)
4. The current i2(t) through R2:
d
i2  t   C2 q2 (t )
dt
d
1
1
q2  t  
q1 (t ) 
q2 (t )
dt
C2 R2
C2 R2
(2.68)
5. KCL Eq. between R1 and R2:
y  t   i1  t   i2  t 
where
i1  t   C1
Current through C1 = i1(t)
d
q1  t 
dt
 1
d
1 
1
1
q1  t    

q
(
t
)

q
(
t
)

x (t )
 1
2
dt
C1R2
C1R1
 C1R1 C2 R2 
(2.69)
◆ Eqs. (2.66), (2.68), and (2.69) = State-Variable Description.
d
q(t)  Aq(t)  bx(t) (2.64)
dt
y(t)  cq(t)  Dx(t) (2.65)
18
CHAPTER
Time-Domain Representations of LTI Systems
  1
1 
1 


 1 
 


C
R
C
R
C
R
 1

1 1
1 2 
1 2 



A
, b  C1R1
c  




1
1
 R1
0







C
R
C
R

2 2
2 2
1

D

0 , and
R1

Example 2.30 State-Variable Description from a Block Diagram
Determine the state-variable description corresponding to the block diagram in
Fig. 2.44. The choice of the state variables is indicated on the diagram.

<Sol.>

Figure 2.44 (p. 172)
Block diagram of LTI system for Example 2.30.
19
CHAPTER
Time-Domain Representations of LTI Systems
1. State equation:
d
q1  t   2q1  t   q2  t   x  t 
dt
d
q2  t   q1  t 
dt
3. State-variable description:
 2 1
A
,

1 0 
1 
b   ,
0 
c  3 1,
D  0
2. Output equation:
y  t   3q1  t   q2  t 
2.13.2 Transformations of The State
The transformation is accomplished by defining a new set of state variables
that are a weighted sum of the original ones.
The input-output characteristic of the system is not changed.
1. Original state-variable description:
q  Aq  bx
y  cq  Dx
(2.70)
(2.71)
2. Transformation:
q’ = Tq
T = state-transformation matrix
q = T1 q’
20
CHAPTER
Time-Domain Representations of LTI Systems
3. New state-variable description:
1) State equation: q  TAq  Tbx.
q  TAT1q  Tbx.
q  Tq
q = T1 q’
2) Output equation:
y  cT1q  Dx.
3) If we set
A  TAT1, b  Tb, c  cT1, and D  D
then
q  Aq  bx
and
y  cq  Dx
Ex. Consider Example 2.30 again. Let us define new states
q2 (t )  q1 (t ) and q1(t )  q2 (t )
Find the state-variable description.
<Sol.>
d
q1 (t )  2q1 (t )  q2 (t )  x(t ) 
1. State equation:
dt
old description
d
q2 (t )  2q2 (t )  q1(t )  x(t )
dt
New description
21
CHAPTER
Time-Domain Representations of LTI Systems
3. State-variable description:
d
d
q2 (t )  q1 (t ) 
q1(t )  q2 (t )
dt
dt
Old description
New description
2. Output equation:
y  3q1 (t )  q2 (t )  y  3q2 (t )  q1(t )
Old description
 0 1
A  
,

 1 2 
c  1 3,
0 
b    ,
1 
D  0.
New description
Example 2.31 Transforming The State
A discrete-time system has the state-variable description
1  1 4 
A 
,

10  4 1
2
b   ,
4
c
1
1 1,
2
and
D  2.
Find the state-variable description A, b, c, D corresponding to the new states
q1[n]   12 q1[n]  12 q2[n] and q2 [n]  12 q1[n]  12 q2 [n]
<Sol.>
1. Transformation: q = Tq, where
1  1 1
T 
.

2  1 1
22
CHAPTER
Time-Domain Representations of LTI Systems
 1 1
T 
.
1
1


1
2. New state-variable description:
  12
A  
 0
0
,
3
10 

1 
b    ,
3
c  0 1,
and
D  2.
 This choice for T results in A being a diagonal matrix and thus separates
the state update into the two decoupled first-order difference equations
1
q1  n  1   q1  n   x  n 
2
and
q2  n  1 
3
q2  n   3x  n 
10
2.14 Exploring Concepts with MATLAB
 Two limitations:
1. MATLAB is not easily used in the continuous-time case.
2. Finite memory or storage capacity and nonzero computation times.
 Both the MATLAB Signal Processing Toolbox and Control System Toolbox
are use in this section.
23
CHAPTER
Time-Domain Representations of LTI Systems
2.14.1 Convolution
x and h are signal vectors.
1. MATLAB command: y = conv(x, h)
2. The number of elements in y is given by the sum of the number of elements in
x and h, minus one.
<pf.>
1) Elements in vector x: from time n = kx to n = lx
2) Elements in vector h: from time n = kh to n = lh
3) Elements in vector y: from time n = ky = kx + kh to n = ly = lx + ly
4) The length of x[n] and h[n] are Lx = lx  kx + 1 and Lh = lh  kh +1
5) The length of y[n] is Ly = Lx + Lh  1
Ex. Repeat Example 2.1
Impulse and Input : From time n = kh = kx = 0 to n = lh = 1 and n = lx =2
Convolution sum: From time n = ky = kx + kh = 0 to n = ly = lx + lh = 3
The length of convolution sum: Ly = ly – ky + 1 = 4
MATLAB Program: >> h = [1, 0.5];
y=
>> x = [2, 4, -2];
>> y = conv(x,h)
2
5
0
-1
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CHAPTER
Time-Domain Representations of LTI Systems
Repeat Example 2.3
Given
Impulse response
h  n   1 4   u  n   u  n  4
and
Input
x  n  u  n  u n 10.
Find the convolution sum x[n]  h[n].
<Sol.>
1. In this case, kh = 0, lh = 3, kx = 0 and lx = 9
2. y starts at time n = ky = 0, ends at time n = ly =12, and has length Ly = 13.
3. Generation for vector h with MATLAB:
>> h = 0.25*ones(1, 4);
>> x = ones(1, 10);
4. Output and its plot:
>> n = 0:12;
>> y = conv(x, h);
>> stem(n, y); xlabel('n'); ylabel('y[n]')
Fig. 2.45.
25
CHAPTER
Time-Domain Representations of LTI Systems
Figure 2.45 (p. 177)
Convolution sum computed using MATLAB.
26
CHAPTER
Time-Domain Representations of LTI Systems
2.14.2 The Step Response
1. Step response = the output of a system in response to a step input
2. In general, step response is infinite in duration.
3. We can evaluate the first p values of the step response using the conv
function if h[n] = 0 for n < kh by convolving the first p values of h[n] with a
finite-duration step of length p.
1) Vector h = the first p nonzero values of the impulse response.
2) Define step: u = ones(1, p).
3) convolution: s = conv(u, h).
Ex. Repeat Problem 2.12
Determine the first 50 values of the step response of the system with impulse
response given by
h  n     u  n
n
with  =  0.9, by using MATLAB program.
<Sol.>
1. MATLAB Commands:
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CHAPTER
Time-Domain Representations of LTI Systems
>> h = (-0.9).^[0:49];
>> u = ones(1, 50);
>> s = conv(u, h);
>> stem([0:49], s(1:50))
2. Step response: Fig. 2.47.
Figure 2.47 (p. 178)
Step response
computed using
MATLAB.
2.14.3 Simulating Difference equations
N
M
1. Difference equation:
 ak y[n  k]  bk x[n  k]
k 0
k 0
(2.36)
Command:
filter
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CHAPTER
Time-Domain Representations of LTI Systems
2. Procedure:
1) Define vectors a = [a0, a1, …, aN] and b =[b0, b1, …, bM] representing the
coefficients of Eq. (2.36).
2) Input vector: x
3) y = filter(b, a, x) results in a vector y representing the output of the system
for zero initial conditions.
4) y = filter(b, a, x, zi) results in a vector y representing the output of the
system for nonzero initial conditions zi.
 The initial conditions used by filter are not the past values of the output.
 Command zi = filtic(b, a, yi), where yi is a vector containing the initial
conditions in the order [y[1], y[2], …, y[N]], generates the initial
conditions obtained from the knowledge of the past outputs.
Ex. Repeat Example 2.16
The system of interest is described by the difference equation
y n  1.143 y n  1  0.4128 y n  2  0.0675x n   0.1349x n  1  0.675x n  2
Determine the output in response to zero input and initial condition
y[1] = 1 and y[2] = 2.
<Sol.>
(2.73)
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CHAPTER
Time-Domain Representations of LTI Systems
1. MATLAB Program:
>> a = [1, -1.143, 0.4128]; b = [0.0675, 0.1349, 0.675];
>> x = zeros(1, 50);
0.5
>> zi = filtic(b, a, [1, 2]);
>> y = filter(b, a, x, zi);
0.4
>> stem(y)
0.3
2. Output: Fig. 2.28(b).
3. System response to an
0.2
input consisting of the
Intel stock price data
0.1
Intc:
>> load Intc;
>> filtintc = filter(b, a, Intc);
 We have assume that the
Intel stock price data are
in the file Intc.mat.
0
-0.1
-0.2
0
10
20
30
40
50
 The command [h, t] = impz(b, a, n) evaluates n values of the impulse response
of a system described by a different equation.
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CHAPTER
Time-Domain Representations of LTI Systems
2.14.4 State-Variable Descriptions
Representing the matices A,b,c, and D.
 MATLAB command: ss
1. Input MATLAB arrays: a, b, c, d
2. Command: sys = ss(a, b, c, d, -1) produces an LTI object sys that represents
the discrete-time system in state-variable form.
★ Continuous-time case: sys = ss(a, b, c, d)
No  1
 System manipulation:
1. sys = sys1 + sys2
Parallel combination of sys1 and sys2.
2. sys = sys1  sys2
Cascade combination of sys1 and sys2.
 MATLAB command: lsim
1. Command form: y = lsim(sys, x)
2. Output = y, input = x.
 MATLAB command: impulse
1. Command form: h = impulse(sys, N)
2. This command places the first N values of the impulse response in h.
 MATLAB routine: ss2ss
Perform the state transformation
1. Command form: sysT = ss2ss(sys, T), where T = Transformation matrix
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CHAPTER
Time-Domain Representations of LTI Systems
Ex. Repeat Example 2.31.
1. Original state-variable description:
A
1  1 4 
,


10  4 1
2
b   ,
4
1
c  1 1,
2
and
D  2,
2. State-transformation matrix:
1  1 1
T 
.

2  1 1
3. MATLAB Program:
>> a = [-0.1, 0.4; 0.4, -0.1]; b = [2; 4];
>> c = [0.5, 0.5]; d = 2;
>> sys = ss(a, b, c, d, -1); % define the state-space object sys
>> T = 0.5*[-1, 1; 1, 1];
>> sysT = ss2ss(sys, T)
4. Result:
32
CHAPTER
a=
x1 x2
x1 -0.5 0
x2 0 0.3
Time-Domain Representations of LTI Systems
b=
u1
x1 1
x2 3
c=
x1 x2
y1 0 1
d=
u1
y1 2
Sampling time: unspecified
Discrete-time model.
Ex. Verify that the two systems represented by sys and sysT have identical
input-output characteristic by comparing their impulse responses .
<Sol.>
1. MATLAB Program: >> h = impulse(sys, 10); hT = impulse(sysT, 10);
>> subplot(2, 1, 1)
>> stem([0:9], h)
>> title ('Original System Impulse Response');
>> xlabel('Time'); ylabel('Amplitude')
>> subplot(2, 1, 2)
>> stem([0:9], hT)
>> title('Transformed System Impulse Response');
>> xlabel('Time'); ylabel('Amplitude')
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CHAPTER
Time-Domain Representations of LTI Systems
Original System Impulse Response
2. Simulation results:
Fig. 2.48.
Amplitude
 We may verify that
the original and
transformed systems
have the (numerically)
identical impulse
response by computing
the error, err = h – hT.
3
2
1
0
0
2
4
6
8
10
Time
Transformed System Impulse Response
Figure 2.48
(p. 181)
Impulse responses
associated with the
original and transformed
state-variable
descriptions computer
using MATLAB.
Amplitude
3
2
1
0
0
2
4
6
8
10
Time
34
CHAPTER
Time-Domain Representations of LTI Systems
1
x 10
-17
0
Amplitude err
-1
-2
-3
-4
-5
-6
0
2
4
6
8
10
Time
Plot for err = h  hT
35