sig and sys

1

Time-Domain Representations of LTI SystemsCHAPTER

2.11 Characteristics of Systems Described by Differential

and Difference EquationsComplete 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

Find the natural response of the this system, assuming that y(0) = 2 V, R = 1

and C = 1 F.

The system In Example 2.17 is described by the differential equation

d

y t RC y t x tdt

<Sol.>

1. Homogeneous sol.: 1 Vh ty t c e

2. I.C.: y(0) = 2 V

y (n) (0) = 2 V c1 = 2

3. Natural Response: 2 Vn ty t e

2


Example 2.25 First-Order Recursive System (Continued): Natural Response

The system in Example 2.21 is described by the difference equation

1

14

y n y n x n

Find the natural response of this system.

<Sol.>

1. Homogeneous sol.: 1

1

4

n

hy n c

2. I.C.: y[ 1] = 8

1

1

18

4c

c1 = 2

3. Natural Response:

1

2 , 14

n

ny n n

2.11.2 The Forced Response

The forced response is the system output due to the input signal assuming

zero initial conditions.

The forced response is valid only

for t 0 or n 0

3


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

1

14

y n y n x n

Find the forced response of this system if the input is x[n] = (1/2)n u[n].

<Sol.>

1. Complete solution:

2. I.C.: Translate the at-rest condition y[ 1] to time n = 0

1

0 0 14

y x y

1

1 12 , 0

2 4

n n

y n c n

y[0] = 1 + (1/4) 0 =1

3. Finding c1: 0 0

1

1 11 2

2 4c

c1 = 1

4


4. Forced response:

1 1

2 , 02 4

n n

fy n n

Example 2.27 RC Circuit (continued): Forced Response

Find the forced response of the this system, assuming that x(t) = cos(t)u(t) V, R= 1 and C = 1 F.

The system In Example 2.17 is described by the differential equation

d

y t RC y t x tdt

<Sol.>1. Complete solution:

1 1cos sin V

2 2

ty t ce t t

From Example 2.22

2. I.C.:

y(0) = y(0+) = 0 c = 1/2

3. Forced response:

1 1 1

cos sin V2 2 2

f ty t e t t

5


2.11.3 The Impulse Response

Relation between step response and impulse response

1. Continuous-time case:

( ) ( )d

h t s tdt

2. Discrete-time case:

[ ] [ 1]h n s n s n

2.11.4 Linearity and Time Invariance

Input Forced response

x1 y1(f)

x2 y2(f)

x1 + x2 y1(f) + y2

(f)

Forced response Linearity

Initial Cond. Natural response

I1 y1(n)

I2 y2(n)

I1 + I2 y1(n) + y2

(n)

Natural response Linearity

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

6


Roots of characteristic equation

Forced response, natural response, stability, and response time.

★ BIBO Stable:

1. Discrete-time case: boundedn

ir 1, for allir i

2. Continuous-time case: boundedir te 0ie r

and1ir 0ie r 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].

3. Integration for continuous-time LTI system: ( ) ( )

t

y t x d

; and a time shift for discrete-time LTI

system: y[n] = x[n 1].Fig. 2.32.

7

(a)

(b)

( ) ( )t

y t x d

(c)

Figure 2.32 (p. 162)Symbols for elementary operations in block diagram

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.

1. In dashed box:

0 1 2w[n] b x[n] b x[n 1] b x[n 2] (2.49)

2. y[n] in terms of w[n]:

Direct Form I:

8


1 2y[n] w[n] a y[n 1] a y[n 2] (2.50)

3. System output y[n] in terms of input x[n]:

1 2 0 1 2y[n] a y[n 1] a y[n 2] b x[n] b x[n 1] b x[n 2]

1 2 0 1 2y[n] a y[n 1] a y[n 2] b x[n] b x[n 1] b x[n 2] (2.51)

Figure 2.33

(p. 162)

Block diagram

representation of

a discrete-time

LTI system

described by a

second-order

difference

equation.

Cascade Form

(Direct Form I)

9


Direct Form II:

1. Interchange the order of Direct Form I.

2. Denote the output of the new first system as f[n].

1 2f[n] a f[n 1] a f[n 2] x[n] (2.52) Input: x[n]

3. The signal is also the input to the second system. The output of the second

system is

0 1 2y[n] b f[n] b f[n 1] b f[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.

)(*)()(*)(1221

thththth

10


Block diagram representation for continuous-time LTI system:

1. Differential Eq.:

k kN M

k kk kk 0 k 0

d da y(t) b x(t)

dt dt

(2.54)

2. Let v(0)(t) = v(t) be an arbitrary signal, and set

1, 1, 2, 3, ...

tn n

t d n

v(n)(t) is the n-fold integral of v(t) with respect to time

3. Integrator with initial condition:

1, 0 and 1, 2, 3, ...

n ndt t t n

dt

1

00 , 1, 2, 3, ...

tn n n

t d n

(N k )N M

(N k)

k k

k 0 k 0

a y (t) b x (t)

(2.55)

Integrate N times to eq. (2.54)

11


Figure 2.37 (p. 166)

Block diagram representations of a continuous-time LTI system described by a

second-order integral equation. (a) Direct form I. (b) Direct form II.

(a)

(b)

Ex. Second-order system:(1) (2) (1) (2)

1 0 2 1 0y(t) a y (t) a y (t) b x(t) b x (t) b x (t)

(2.56)

12


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:

1 1 1 2 2q [n 1] a q [n] a q [n] x[n] (2.57)

2 1q [n 1] q [n] (2.58)

4. Output equation:

1 1 1 2 2 2y[n] (b a )q [n] (b a )q [n] x[n] (2.59)

1 11 2

2 2

q [n 1] q [n]a a 1x[n]

q [n 1] q [n]1 0 0

(2.60)

1 1 2 2 1 1 2 2y[n] x[n] a q [n] a q [n] b q [n] b q [n],

5. Matrix Form of state equation:

13


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].

14


6. Matrix form of output equation:

1

1 1 2 2

2

q [n]y[n] [b a b a ] [1]x[n]

q [n]

(2.61)

Define state vector as the column vector

1

2

q [n][n]

q [n]q

We can rewrite Eqs. (2.60) and (2.61) as

[n 1] [n] x[n] q Aq b (2.62)

y[n] [n] Dx[n] cq (2.63)

where matrix A, vectors b and c, and scalar D are given by

1 2

1 0A

a a

1

0b

1 1 2 2c b a b a 1D

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.>

15


Figure 2.40 (p. 169)

Block diagram of LTI system for Example 2.28.1. State equation:

1 1 11q n q n x n

2 1 2 21q n q n q n x n

16


2. Output equation:

1 1 2 2y n q n q n

3. Define state vector as

1

2

qq n

nq n

In standard form of dynamic equation:

[n 1] [n] x[n]q Aq b

y[n] [n] Dx[n]cq (2.63)

(2.62)

0A

1

2

b

1 2c 2D

State-variable description for continuous-time systems:

d(t) (t) x(t)

dt q Aq b (2.64)

y(t) (t) Dx(t) cq (2.65)

17


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:

1 1x t y t R q t

1

1 1

1 1y(t) q (t) x(t)

R R (2.66)

Output equation

3. KVL Eq. for the loop involving C1, R2, and C2:

1 2 2 2( ) ( )q t R i t q t

18


2 1 2

2 2

1 1i (t) q (t) q (t)

R R (2.67)

4. The current i2(t) through R2:

2 2 2( )d

i t C q tdt

2 1 2

2 2 2 2

1 1( ) ( )

dq t q t q t

dt C R C R (2.68)

Use Eq. (2.67) to eliminate i2(t)

5. KCL Eq. between R1 and R2:

1 2y t i t i t Current through C1 = i1(t)

where 1 1 1

di t C q t

dt

1 1 2

1 1 2 2 1 2 1 1

1 1 1 1( ) ( ) ( )

dq t q t q t x t

dt C R C R C R C R

(2.69)

◆ Eqs. (2.66), (2.68), and (2.69) = State-Variable Description.

d

(t) (t) x(t)dt

q Aq b (2.64)

y(t) (t) Dx(t)cq (2.65)

19


1 1 1 2 1 2

2 2 2 2

1 1 1

,1 1

AC R C R C R

C R C R

1 1

1

0

b C R

1

10 ,c

R

1

1D

Rand

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.

Figure 2.44 (p. 172)

Block diagram of LTI system for Example 2.30.

<Sol.>

20


1. State equation:

1 1 22d

q t q t q t x tdt

2 1

dq t q t

dt

2. Output equation:

1 23y t q t q t

3. State-variable description:

2 1,

1 0A

1,

0b

3 1 ,c 0D

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 (2.70)

cqy Dx (2.71)

2. Transformation: q’ = Tq

T = state-transformation matrix

q = T1 q’

21


3. New state-variable description:

.q TAq Tbx 1) State equation:q Tq

1 .q TAT q Tbx q = T1 q’

2) Output equation:

1 .cT qy Dx

3) If we set

1 1, , , andA TAT b Tb c cT D D

then

q A q b x and c qy D x

Ex. Consider Example 2.30 again. Let us define new states

2 1 1 2( ) ( ) and ( ) ( )q t q t q t q t

Find the state-variable description.<Sol.>

1. State equation: 1 1 2 2 2 1( ) 2 ( ) ( ) ( ) ( ) 2 ( ) ( ) ( )

old description New description

d dq t q t q t x t q t q t q t x t

dt dt

22


2 1 1 2( ) ( ) ( ) ( )

Old description New description

d dq t q t q t q t

dt dt

1 2 2 13 ( ) ( ) 3 ( ) ( )

Old description New description

y q t q t y q t q t

2. Output equation:

3. State-variable description:

0 1,

1 2A

0,

1b

1 3 ,c 0 .D

Example 2.31 Transforming The State

A discrete-time system has the state-variable description

1 41A ,

4 110

2,

4b

1

1 1 ,2

c 2.D and

Find the state-variable description A, b, c, D corresponding to the new states 1 1

1 1 22 2[ ] [ ] [ ]q n q n q n

1 1

2 1 22 2[ ] [ ] [ ]q n q n q n and

<Sol.>

1. Transformation: q = Tq, where1 11

.1 12

T

23


11 1

.1 1

T

2. New state-variable description:

1

2

3

10

0,

0A

1,

3b

0 1 ,c 2.D and

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 1

11

2q n q n x n 2 2

31 3

10q n q n x n and

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.

24


2.14.1 Convolution

1. MATLAB command: y = conv(x, h)x and h are signal vectors.

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 = lx2) Elements in vector h: from time n = kh to n = lh3) Elements in vector y: from time n = ky = kx + kh to n = ly = lx + ly4) 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];

>> x = [2, 4, -2];

>> y = conv(x,h)

y =

2 5 0 -1

25


Repeat Example 2.3

Given 1 4 4h n u n u n 10 .x n u n u n and

Impulse response Input

Find the convolution sum x[n] h[n].

1. In this case, kh = 0, lh = 3, kx = 0 and lx = 9

<Sol.>

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.

26


Figure 2.45 (p. 177)

Convolution sum computed using MATLAB.

27


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

n

h n u n

with = 0.9, by using MATLAB program.

<Sol.>

1. MATLAB Commands:

28


>> 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

1. Difference equation:

N M

k k

k 0 k 0

a y[n k] b x[n k] (2.36) Command:

filter

29


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

1.143 1 0.4128 2 0.0675 0.1349 1 0.675 2y n y n y n x n x n x n

(2.73)Determine the output in response to zero input and initial condition

y[1] = 1 and y[2] = 2.

<Sol.>

30


1. MATLAB Program:

>> a = [1, -1.143, 0.4128]; b = [0.0675, 0.1349, 0.675];

>> x = zeros(1, 50);

>> zi = filtic(b, a, [1, 2]);

>> y = filter(b, a, x, zi);

>> stem(y)

2. Output: Fig. 2.28(b).

0 10 20 30 40 50-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

3. System response to an

input consisting of the

Intel stock price data

Intc:

>> load Intc;

>> filtintc = filter(b, a, Intc);

We have assume that the

Intel stock price data are

in the file Intc.mat.

The command [h, t] = impz(b, a, n) evaluates n values of the impulse response

of a system described by a different equation.

31


2.14.4 State-Variable Descriptions

MATLAB command: ss

1. Input MATLAB arrays: a, b, c, d

Representing the matices A,b,c, and 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

2. This command places the first N values of the impulse response in h.

1. Command form: h = impulse(sys, N)

MATLAB routine: ss2ss Perform the state transformation

1. Command form: sysT = ss2ss(sys, T), where T = Transformation matrix

32


Ex. Repeat Example 2.31.

1. Original state-variable description:

1 41,

4 110A

2,

4b

1

1 1 ,2

c 2,D and

2. State-transformation matrix:

1 11.

1 12T

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:

33


a =

x1 x2

x1 -0.5 0

x2 0 0.3

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')

34


2. Simulation results:

Fig. 2.48.

0 2 4 6 8 100

1

2

3Original System Impulse Response

TimeA

mplitu

de

0 2 4 6 8 100

1

2

3Transformed System Impulse Response

Time

Am

plitu

de

Figure 2.48

(p. 181)

Impulse responses

associated with the

original and transformed

state-variable

descriptions computer

using MATLAB.

We may verify that

the original and

transformed systems

have the (numerically)

identical impulse

response by computing

the error, err = h – hT.

35


0 2 4 6 8 10-6

-5

-4

-3

-2

-1

0

1x 10

-17

Time

Am

plitu

de

err

Plot for err = h hT

sig and sys

Documents

time n

y n y

y n y n x n

response time

n ny n c n y0

h n s n s n

n nexample

times n