1 time-domain representations of lti systems chapter 2.6 interconnection of lti systems 2.6.1...

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Time-Domain Representations of LTI SystemsTime-Domain Representations of LTI Systems CHAPTER

2.6 Interconnection of LTI Systems2.6 Interconnection of LTI Systems2.6.1 Parallel Connection of LTI Systems2.6.1 Parallel Connection of LTI Systems

1. Two LTI systems: Fig. 2.18(a).Fig. 2.18(a).Figure 2.18 (p. 128)

Interconnection of two LTI systems. (a) Parallel connection of two

systems. (b) Equivalent system.

2. Output:1 2

1 2

( ) ( ) ( )

( ) ( ) ( ) ( )

y t y t y t

x t h t x t h t

1 2( ) ( ) ( ) ( ) ( )y t x h t d x h t d

1 2( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

y t x h t h t d

x h t d x t h t

where h(t) = h1(t) + h2(t)

Fig. 2.18(b)Fig. 2.18(b)

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Distributive property for Continuous-time case:

1 2 1 2x(t) h (t) x(t) h (t) x(t) {h (t) h (t)} (2.15)

Distributive property for Discrete-time case:

1 2 1 2x[n] h [n] x[n] h [n] x[n] {h [n] h [n]} (2.16)

2.6.2 Cascade Connection of LTI Systems2.6.2 Cascade Connection of LTI Systems

1. Two LTI systems: Fig. 2.19(a).Fig. 2.19(a).

Figure 2.19 (p. 128)Interconnection of two LTI systems. (a) Cascade connection of two

systems. (b) Equivalent system. (c) Equivalent system: Interchange system order.

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2. The output is expressed in terms of z(t) as

2y(t) z(t) h (t) (2.17)

2-y(t) z( )h (t )d

(2.18)

Since z(t) is the output of the first system, so it can be expressed as

1 1-z( ) x( ) h ( ) x( )h ( )d

(2.19)

Substituting Eq. (2.19) for z(t) into Eq. (2.18) gives

1 2( ) ( ) ( ) ( )y t x v h v h t dvd

1 2-y(t) x( ) h ( )h (t )d d

(2.20)

Change of variable =

Define h(t) = h1(t) h2(t), then

1 2( ) ( ) ( )h t v h h t v d

-y(t) x( )h(t )d x(t) h(t)

(2.21)

3. Associative property for continuous-time case:

Fig. 2.19(b).Fig. 2.19(b).

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1 2 1 2{x(t) h (t)} h (t) x(t) {h (t) h (t)} (2.22)

4. Commutative property:

Write h(t) = h1(t) h2(t) as the integral

1 2( ) ( ) ( )h t h h t d

1 2 2 1-h(t) h (t )h ( )d h (t) h (t)

(2.23)

Change of variable = t

Fig. 2.19(c).Fig. 2.19(c).

Interchanging the order of the LTI systems in the cascade without affecting the result:

1 2 2 1( ) ( ) ( ) ( ) ( ) ( ) ,x t h t h t x t h t h t

Commutative property for continuous-time case:

1 2 2 1h (t) h (t) h (t) h (t) (2.24)

5. Associative property for discrete-time case:

1 2 1 2{x[n] h [n]} h [n] x[n] {h [n] h [n]} (2.25) Commutative property for discrete-time case:

1 2 2 1h [n] h [n] h [n] h [n] (2.26)

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Example 2.11Example 2.11 Equivalent System to Four Interconnected SystemsConsider the interconnection of four LTI systems, as depicted in Fig. 2.20Fig. 2.20. The impulse responses of the systems are

1[ ] [ ],h n u n 2[ ] [ 2] [ ],h n u n u n 3[ ] [ 2],h n n 4[ ] [ ].nh n u nand

Find the impulse response h[n] of the overall system.

<Sol.><Sol.>

1. Parallel combination of h1[n] and h2[n]:h12[n] = h1[n] + h2[n] Fig. 2.21 (a).Fig. 2.21 (a).

Figure 2.20 Figure 2.20 (p. 131)(p. 131)

Interconnection of systems for

Example 2.11.

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Figure 2.21 (p. 131)(a) Reduction of parallel combination of LTI systems in upper branch of Fig. 2.20. (b) Reduction of cascade of systems in upper branch of Fig. 2.21(a). (c) Reduction of parallel combination of systems in Fig. 2.21(b) to obtain an equivalent system for Fig. 2.20.

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2. h12[n] is in series with h3[n]:

h123[n] = h12[n] h3[n]

h123[n] = (h1[n] + h2[n]) h3[n] Fig. 2.21 (b).

3. h123[n] is in parallel with h4[n]:

h[n] = h123[n] h4[n]

1 2 3 4[ ] ( [ ] [ ]) [ ] [ ],h n h n h n h n h n Fig. 2.21 (c).

Thus, substitute the specified forms of h1[n] and h2[n] to obtain

12[ ] [ ] [ 2] [ ]

[ 2]

h n u n u n u n

u n

Convolving h12[n] with h3[n] gives

123[ ] [ 2] [ 2]

[ ]

h n u n n

u n

[ ] 1 [ ].nh n u n

Table 2.1 summarizes the interconnection properties presented in this section.Table 2.1 summarizes the interconnection properties presented in this section.

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2.7 Relation Between LTI System Properties and the2.7 Relation Between LTI System Properties and the Impulse ResponseImpulse Response2.7.1 Memoryless LTI Systems2.7.1 Memoryless LTI Systems

1. The output of a discrete-time LTI system:

[ ] [ ] [ ] [ ] [ ]k

y n h n x n h k x n k

y[n] h[ 2]x[n 2] h[ 1]x[n 1] h[0]x[n] h[1]x[n 1] h[2]x[n 2]

(2.27)

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2. To be memoryless, y[n] must depend only on x[n] and therefore cannot depend on x[n k] for k 0.

A discrete-time LTI system is memoryless if and only if

[ ] [ ]h k c k c is an arbitrary constant

Continuous-time system:

1. Output:

( ) ( ) ( ) ,y t h x t d

2. A continuous-time LTI system is memoryless if and only if

( ) ( )h c c is an arbitrary constant

2.7.2 Causal LTI Systems2.7.2 Causal LTI Systems

Discrete-time system:

The output of a causal LTI system depends only on past or present values of the input.

1. Convolution sum: [ ] [ 2] [ 2] [ 1] [ 1] [0] [ ]

[1] [ 1] [2] [ 2] .

y n h x n h x n h x n

h x n h x n

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2. For a discrete-time causalcausal LTI system,

[ ] 0 for 0h k k

3. Convolution sum in new form:

0

[ ] [ ] [ ].k

y n h k x n k

Continuous-time system:1. Convolution integral:

( ) ( ) ( ) .y t h x t d

2. For a continuous-time causalcausal LTI system,

( ) 0 for 0h

3. Convolution integral in new form:

0( ) ( ) ( ) .y t h x t d

2.7.3 Stable LTI Systems2.7.3 Stable LTI Systems

A system is BIBO stable if the output is guaranteed to be bounded for every bounded input.

Discrete-time case: Input [ ] xx n M [ ] yy n M Output:

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1. The magnitude of output:

[ ] [ ] [ ] [ ] [ ]k

y n h n x n h k x n k

[ ] [ ] [ ]k

y n h k x n k

a b a b

[ ] [ ] [ ]k

y n h k x n k

ab a b

2. Assume that the input is bounded, i.e.,

[ ] xx n M [ ] xx n k M

and it follows that

xk

y[n] M h[k]

(2.28)

Hence, the output is bounded, or y[n] ≤ for all n, provided that the impulse response of the system is absolutely summable.

3. Condition for impulse response of a stable discrete-time LTI system:

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[ ] .k

h k

Continuous-time case:

Condition for impulse response of a stable continuous-time LTI system:

Example 2.12Example 2.12 Properties of the First-Order Recursive SystemThe first-order system is described by the difference equation

[ ] [ 1] [ ]y n y n x n and has the impulse response

[ ] [ ]nh n u nIs this system causal, memoryless, and BIBO stable?<Sol.><Sol.>1. The system is causal, since h[n] = 0 for n < 0.2. The system is not memoryless, since h[n] 0 for n > 0.3. Stability: Checking whether the impulse response is absolutely summable?

dh )(

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0 0

[ ]kk

k k k

h k

if and only if < 1

◆ Note:A system can be unstable even though the impulse response has a finite value.

1. Ideal integrator:t

y(t) x( )d

(2.29)

Input: x() = (), then the output is y(t) = h(t) = u(t).

h(t) is not absolutely integrable

Ideal integrator is not stable!

2. Ideal accumulator:

[ ] [ ]n

k

y n x k

Impulse response: h[n] = u[n]

h[n] is not absolutely summable

Ideal accumulator is not stable!

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2.7.4 Invertible Systems and Deconvolution2.7.4 Invertible Systems and Deconvolution

A system is invertible if the input to the system can be recovered from the output except for a constant scale factor.

1. h(t) = impulse response of LTI system, 2. hinv(t) = impulse response of LTI inverse system

Fig. 2.24.Fig. 2.24.

Figure 2.24 (p. 137)Cascade of LTI system with impulse response h(t) and inverse system with

impulse response h-1(t).

3. The process of recovering x(t) from h(t) x(t) is termed deconvolution.4. An inverse system performs deconvolution.

in( ) ( ( ) ( )) ( ).vx t h t h t x t

( ) ( ) ( )invh t h t t (2.30)

Continuous-time case

5. Discrete-time case: invh[n] h [n] [n] (2.31)

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Example 2.13Example 2.13 Multipath Communication Channels: Compensation by means of an Inverse SystemConsider designing a discrete-time inverse system to eliminate the distortion associated with multipath propagation in a data transmission problem. Assume that a discrete-time model for a two-path communication channel is

[ ] [ ] [ 1].y n x n ax n Find a causal inverse system that recovers x[n] from y[n]. Check whether this inverse system is stable.<Sol.><Sol.>1. Impulse response:

1, 0

[ ] , 1

0, otherwise

n

h n a n

2. The inverse system hinv[n] must satisfy h[n] hinv[n] = [n].

[ ] [ 1] [ ].inv invh n ah n n (2.32)

1) For n < 0, we must have hinv[n] = 0 in order to obtain a causal inverse system

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2) For n = 0, [n] = 1, and eq. (2.32) implies that

[ ] [ 1] 0,inv invh n ah n

(2.33)

3. Since hinv[0] = 1, Eq. (2.33) implies that hinv[1] = a, hinv[2] = a2, hinv[3] = a3, and so on.

The inverse system has the impulse response

[ ] ( ) [ ]inv nh n a u n

inv invh [n] ah [n 1]

4. To check for stability, we determine whether hinv[n] is absolutely summable, which will be the case if

[ ]kinv

k k

h k a

is finite.

For For aa < 1, the system is stable. < 1, the system is stable.

★ Table 2.2 summarizes the relation between LTI system properties and impulse response characteristics.

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2.8 Step Response2.8 Step Response1. The step response is defined as the output due to a unit step input signal.

2. Discrete-time LTI system:

Let h[n] = impulse response and s[n] = step response.

[ ] [ ]* [ ] [ ] [ ].k

s n h n u n h k u n k

3. Since u[n k] = 0 for k > n and u[n k] = 1 for k ≤ n, we have

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[ ] [ ].n

k

s n h k

The step response is the running sum of the impulse response.

Continuous-time LTI system:t

s(t) h( )d

(2.34)

The step response s(t) is the running integral of the impulse response h(t). ◆ Express the impulse response in terms of the step response as

[ ] [ ] [ 1]h n s n s n ( ) ( )d

h t s tdt

and

Example 2.14Example 2.14 RC Circuit: Step ResponseThe impulse response of the RC circuit depicted in Fig. 2.12Fig. 2.12 is

1( ) ( )

t

RCh t e u tRC

Find the step response of the circuit.<Sol.><Sol.>

Figure 2.12 (p. 119)RC circuit system with the voltage source x(t) as input and the voltage measured across the capacitor y(t), as output.

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1. Step response:

1( ) ( ) .

tRCs t e u d

RC

0, 0

( ) 1( ) 0

tRC

t

s te u d t

RC

0

0, 0

( ) 10

0, 0

1 , 0

tRC

t

RC

t

s te d t

RC

t

e t

Fig. 2.25Fig. 2.25

Figure 2.25 (p. 140)RC circuit step response for RC = 1 s.

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2.9 Differential and Difference Equation Representations of 2.9 Differential and Difference Equation Representations of LTILTI Systems Systems1. Linear constant-coefficient differential equation:

k kN M

k kk kk 0 k 0

d da y(t) b x(t)

dt dt

(2.35) Input = x(t), output = y(t)

2. Linear constant-coefficient difference equation:N M

k kk 0 k 0

a y[n k] b x[n k]

(2.36) Input = x[n], output = y[n]

The order of the differential or difference equation is (N, M), representing the number of energy storage devices in the system.

Ex. RLC circuit depicted in Fig. 2.26Fig. 2.26.

2. KVL Eq.:

1. Input = voltage source x(t), output = loop current

1 tdRy t L y t y d x t

dt C

Often, N M, and the order is described

using only N.

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Figure 2.26 (p. 141)Example of an RLC circuit described by a differential equation.

2

2

1 d d dy t R y t L y t x t

C dt dt dt N = 2

Ex. Accelerator modeled in Section 1.10:

2

22

nn

d dy t y t y t x t

Q dt dt

where y(t) = the position of the proof mass, x(t) = external acceleration.

N = 2

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Ex. Second-order difference equation:

1y[n] y[n 1] y[n 2] x[n] 2x[n 1]

4 (2.37) N = 2

Difference equations are easily rearranged to obtain recursive formulas for computing the current output of the system from the input signal and the past outputs.

Ex. Eq. (2.36) can be rewritten as

0 10 0

1 1M N

k kk k

y n b x n k a y n ka a

Ex. Consider computing y[n] for n 0 from x[n] for the second-order difference equation (2.37).<Sol.><Sol.>

1. Eq. (2.37) can be rewritten as

1

y[n] x[n] 2x[n 1] y[n 1] y[n 2]4

(2.38)

2. Computing y[n] for n 0:

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1y[0] x[0] 2x[ 1] y[ 1] y[ 2]

4 (2.39)

1y[1] x[1] 2x[0] y[0] y[ 1]

4 (2.40)

12 2 2 1 1 0

4y x x y y

13 3 2 2 2 1

4y x x y y

Initial conditions: y[ 1] and y[ 2].

◆ The initial conditions for Nth-order difference equation are the N values

, 1 ,..., 1 ,y N y N y

◆ The initial conditions for Nth-order differential equation are the N values

2 1

0 , 2 10 00

, , ...,N

t Nt tt

d d dy t y t y t y t

dt dt dt

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Example 2.15Example 2.15 Recursive Evaluation of a Difference Equation

Find the first two output values y[0] and y[1] for the system described by Eq. (2.38), assuming that the input is x[n] = (1/2)nu[n] and the initial conditions are y[ 1] = 1 and y[ 2] = 2.

<Sol.><Sol.>1. Substitute the appropriate values into Eq. (2.39) to obtain

1 10 1 2 0 1 2

4 2y

2. Substitute for y[0] in Eq. (2.40) to find

1 1 1 31 2 1 1 1

2 2 4 4y

Example 2.16Example 2.16 Evaluation of a Difference Equation by means of a Computer

A system is described y 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

Write a recursive formula that computes the present output from the past outputs and the current inputs. Use a computer to determine the step response

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of the system, the system output when the input is zero and the initial conditions are y[ 1] = 1 and y[ 2] = 2, and the output in response to the

sinusoidal inputs x1[n] = cos(n/10), x2[n] = cos(n/5), and x3[n] = cos(7n/10),

assuming zero initial conditions. Last, find the output of the system if the input is the weekly closing price of Intel stock depicted in Fig. 2.27Fig. 2.27, assuming zero initial conditions.<Sol.><Sol.>1. Recursive formula for y[n]:

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. Step response: Fig. 2.28 (a).Fig. 2.28 (a).3. Zero input response: Fig. 2.28 (b).Fig. 2.28 (b).

4. The outputs due to the sinusoidal inputs x1[n], x2[n], and x3[n]: Fig. 2.28 Fig. 2.28

(c),(c), (d), and (e).(d), and (e).5. Fig. 2.28(f)Fig. 2.28(f) shows the system output for the Intel stock price unit.

A comparison of peaks in Figs. 2.27Figs. 2.27 and 2.28 (f)2.28 (f) Slightly delay!Slightly delay!

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Figure 2.27 (p. 144)Weekly

closing price of Intel stock.

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Fig. 2.28 (a).Fig. 2.28 (a).

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Fig. 2.28 (b).Fig. 2.28 (b).

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Fig. 2.28 (d).Fig. 2.28 (d).

Fig. 2.28 (c).Fig. 2.28 (c).

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Figure 2.28a (p. 145)Illustration of the solution to Example 2.16.

(a) Step response of system. (b) Output due to nonzero initial conditions with zero input.

(c) Output due to x1[n] = cos (1/10n). (d) Output due to x2[n] = cos (1/5n). (e) Output due to x3[n] = cos(7/10n).

Fig. 2.28 (e).Fig. 2.28 (e).

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Figure 2.28f (p. 146)Output associated with the weekly closing price of Intel stock.

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2.10 Solving Differential and Difference Equations2.10 Solving Differential and Difference EquationsComplete solution: y = y

(h) + y (p)

y (h) = homogeneous solution, y

(p) = particular solution

2.10.1 The Homogeneous Solution2.10.1 The Homogeneous Solution Continuous-time case:

1. Homogeneous differential equation: 0

0kN

hk k

k

da y tdt

2. Homogeneous solution:

i

Nrt(h)

ii 0

y (t) c e

(2.41)

3. Characteristic eq.: N

kk

k 0

a r 0

(2.42)

Discrete-time case:1. Homogeneous differential equation:

2. Homogeneous solution:

0

0N

hk

k

a y n k

N

(h) ni i

i 1

y [n] c r

(2.43)

3. Characteristic eq.: N

N kk

k 0

a r 0

(2.44)

Coefficients ci is determined by I.C.

Coefficients ci is determined by I.C.

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If a root rj is repeated p times in characteristic eqs., the corresponding

solutions are1, , ...,j j jr r rt t p te te t e

1, , ...,n n p nj j jr nr n r

Continuous-time case:

Discrete-time case:

Example 2.17Example 2.17 RC Circuit: Homogeneous SolutionThe RC circuit depicted in Fig. 2.30Fig. 2.30 is described by the differential equation

dy t RC y t x t

dt

Determine the homogeneous solution of this equation.<Sol.><Sol.>1. Homogeneous Eq.: 0

dy t RC y t

dt

2. Homo. Sol.: 11 Vh r ty t c e

3. Characteristic eq.: 11 0RCr r1 = 1/RC

4. Homogeneous solution: 1 V

th RCy t c e

Figure 2.30 (p. 148)RC circuit.

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Example 2.18Example 2.18 First-Order Recursive System: Homogeneous Solution

Find the homogeneous solution for the first-order recursive system described by the difference equation

1y n y n x n

<Sol.><Sol.>1. Homogeneous Eq.: 1 0y n y n 2. Homo. Sol.:

1 1h ny n c r

3. Characteristic eq.: 1 0r r1 =

4. Homogeneous solution: 1h ny n c

2.10.2 The Particular Solution2.10.2 The Particular Solution

A particular solution is usually obtained by assuming an output of the same general form as the input.

Table 2.3Table 2.3

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Example 2.19Example 2.19 First-Order Recursive System (Continued): Particular SolutionFind a particular solution for the first-order recursive system described by the difference equation

1y n y n x n

if the input is x[n] = (1/2)n.<Sol.><Sol.>1. Particular solution form: 12[ ]

nppy n c

2. Substituting y(p)[n] and x[n] into the given difference Eq.:

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11 1 1

2 2 2

n n n

p pc c

pc (1 2 ) 1 (2.45)

Both sides of above eq. are multiplied by (1/2) n

3. Particular solution:

1 1

1 2 2

npy n

Example 2.20Example 2.20 RC Circuit (continued): Particular SolutionConsider the RC circuit of Example 2.17Example 2.17 and depicted in Fig. 2.30Fig. 2.30. Find a particular solution for this system with an input x(t) = cos(0t).

Figure 2.30 (p. 148)

RC circuit.

<Sol.><Sol.>

1. Differential equation: dy t RC y t x t

dt

2. Particular solution form:( )

1 2( ) cos( ) sin( )py t c t c t

3. Substituting y(p)(t) and x(t) = cos(0t) into the given differential Eq.:

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1 2 0 1 0 0 2 0 0cos( ) sin( ) sin( ) cos( ) cos( )c t c t RC c t RC c t t

1 0 2 1c RC c

0 1 2 0RC c c

4. Coefficients c1 and c2:

1 2

0

1

1c

RC

02 2

01

RCc

RC

and


00 02 2

0 0

1cos sin V

1 1

p RCy t t t

RC RC

2.10.3 The Complete Solution2.10.3 The Complete SolutionComplete solution: y = y

(h) + y (p)

y (h) = homogeneous solution, y

(p) = particular solution

The procedure for finding complete solution of differential or difference equations is summarized as follows:

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Procedure 2.3:Procedure 2.3: Solving a Differential or Difference equation1. Find the form of the homogeneous solution y(h) from the roots of the characteristic equation.2. Find a particular solution y(p) by assuming that it is of the same form as the input, yet is independent of all terms in the homogeneous solution.3. Determine the coefficients in the homogeneous solution so that the complete solution y = y(h) + y(p) satisfies the initial conditions.

★ ★ Note that the initial translation is needed in some cases.Note that the initial translation is needed in some cases.

Example 2.21Example 2.21 First-Order Recursive System (Continued): Complete SolutionFind the complete solution for the first-order recursive system described by the difference equation

1y[n] y[n 1] x[n]

4 (2.46)

if the input is x[n] = (1/2)n u[n] and the initial condition is y[ 1] = 8.<Sol.><Sol.>

1. Homogeneous sol.: 11 4

nhy n c

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122

npy n

3. Complete solution:

n n1

1 1y[n] 2( ) c ( )

2 4 (2.47)

4. Coefficient c1 determined by I.C.:

I.C.: 0 0 1 4 1y x y 0 0 (1 4) 8 3y x

We substitute y[0] = 3 into Eq. (2.47), yielding0 0

1

1 13 2

2 4c

c1 = 1

5. Final solution:

1 122 4

n n

y n

for n 0

Example 2.22Example 2.22 RC Circuit (continued): Complete Response

Find the complete response of the RC circuit depicted in Fig. 2.30Fig. 2.30 to an input x(t) = cos(t)u(t) V, assuming normalized values R = 1 and C = 1 F and assuming that the initial voltage across the capacitor is y(0) = 2 V.

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<Sol.><Sol.>1. Homogeneous sol.: V

th RCy t ce


2 2

1cos sin V

1 1

p RCy t t t

RC RC

4. Coefficient c1 determined by I.C.:


1 1cos sin V2 2

ty t ce t t

0 = 1

R = 1 , C = 1 F

y(0y(0) = y(0) = y(0++))

0 1 1 12 cos0 sin 0

2 2 2ce c

c = 3/2

5. Final solution:

3 1 1cos sin V

2 2 2ty t e t t

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Example 2.23Example 2.23 Financial Computations: Loan Repayment

The following difference equation describes the balance of a loan if x[n] < 0 represents the principal and interest payment made at the beginning of each period and y[n] is the balance after the principal and interest payment is credited. As before, if r % is the interest rate per period, then = 1 + r/100.

1y n y n x n

Use the complete response of the first-order difference equation to find the payment required to pay off a $20,000 loan in 10 periods. Assume equal payments and a 10% interest rate.<Sol.><Sol.>1. We have = 1.1 and y[ 1] = 20,000, and we assume that x[n] = b is the payment each period.

2. The first payment is made when n = 0. The loan balance is to be zero after 10 payments, thus we seek the payment b for which y[9] = 0.3. Homogeneous sol.:

1.1nh

hy n c


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ppy n c

Substituting y(p)[n] = cp and x[n] = b into the difference equation y[n] 1.1y[n 1] = x[n], we obtain

10pc b


nhy[n] c (1.1) 10b, n 0 (2.48)

4. Coefficient ch determined by I.C.:

I.C.: 0 1.1 1 0 22,000y y x b

022,000 1.1 10hb c b

22,000 11hc b

22,000 11 1.1 10n

y n b b

5. Payment b: By setting y[9] = 0, we have

90 22,000 11 1.1 10b b

9

9

22,000 1.13,254.91

11 1.1 10b

Fig. 2.31.Fig. 2.31.

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Figure 2.31 (p. 155)Balance on a $20,000 loan for Example 2.23. Assuming 10% interest per

period, the loan is paid off with 10 payments of $3,254.91.

1 time-domain representations of lti systems chapter 2.6 interconnection of lti systems 2.6.1...

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