Difference Equations

DIFFERENCE EQUATIONS

Consider the difference equation,

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Here x(n-k) are the inputs and y(n-k) are the outputs.

The difference equation can be solved to obtain an expression for y(n). This expression for y(n) is made up of two components.

5.1.1 Natural Response (Zero Input Response)

The natural response is the output of the system with zero input. This response is obtained only with initial conditions. It is denoted by y⁽ⁿ⁾(n).

5.1.2 Forced Response (Zero State Response)

The forced response is the output of the system for given input and zero initial conditions. Thus forced response is obtained only with given input, it is denoted by y^(f)(n).

The complete response of the system is given as the sum of natural response and forced response.

Natural Response y⁽ⁿ⁾(n):

The natural response is obtained for zero input. Hence equation (1) becomes,

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The above equation can also be written as,

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The above difference equation is also called homogeneous difference equation. The solution of the above equation has the following forms:

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Here r_i are real and distinct.

Here r_i are the N roots of the discrete time system's characteristics equation. This characteristic equation is given as,

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If r_i is complex conjugate ie r_i e ^{j Ω}. Then,

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If r_i is repeated ie r_i^p+1. Then,

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Forced Response y^(f)(n):

The forced response consists of sum of the two components:

(i) The term similar to natural response, y⁽ⁿ⁾(n)

(ii) Particular solution, y^(p)(n).

The particular solution is solution of difference equation for given input. The particular solution has same form as that of input.

For example, if x(n) = aⁿ, then particular solution is of the form y^(p)(n) = C αⁿ. Here 'C' is some constant. The table below lists the form of particular solutions for commonly applied inputs.

 

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Complete Response y(n)

The complete response of the system is equal to sum of natural response and forced response. Thus input as well as initial conditions are taken care of in complete response.

Example:

Problems Based on Difference Equations

Problem Using Natural Response:

Problem 1:

Determine the natural response (zero input response) for the following system:

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Solution:

With zero inputs, the given equation becomes,

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Step 1: Roots of characteristics equation

Here N = 2, hence the characteristic equation of equation (4) becomes.

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From equation (1), Comparatively we can get

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Roots of the above equation are,

Step 2: Form of natural response

The natural response is given by equation (3),

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For N = 2 above equation becomes,

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Putting value of r₁ = 1/2 and r₂ = -1/4 in above equation,

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Step 3: Values of constants in y⁽ⁿ⁾(n).

With n = 0, above equation becomes,

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With n = 1 equation (2) becomes,

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Now let us put n = 0 in equation (1),

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We know that y(-1) = 0 and y(-2) = 1, hence above equation becomes,

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This should be equal to y⁽ⁿ⁾(0) of equation (3). Putting y⁽ⁿ⁾(0) = 1/8 in equation (3) we get,

C₁ + C₂ = 1/8

Now let us put n = 1 in equation (1),

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We know that y(−1) = 0 and y(0) = 1/8. Hence the above equation becomes,

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This should be equal to y⁽ⁿ⁾(1) of equation (4) putting y⁽ⁿ⁾(1) = 1/32 in equation (4)

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Solving above equation and equation (3) for C₁ and C₂, we get,

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Putting these values of C₁ and C₂ in equation (2),

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This is the natural response of the system.

Problem based on Forced Response y^(f)(n):

Problem 2:

Determine the forced response for the following system,

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Assume zero initial conditions.

Solution:

Step 1: Roots of characteristic equations

Let us first find out characteristic equation. For this we have to make input zero in the given equation. Then we get,

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Here N = 2, hence the characteristic equation of equation (4) becomes.

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From equation (1) a₀ = 1, a₁ = -1/4 and a₂ = -1/8, hence above equation becomes,

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Roots of the above equation are,

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Step 2: Form of natural response.

Natural response is given by equation (3) as,

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For N = 2, above equation becomes,

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Putting values of r₁ = 1/2 and r₂ = -1/4 in above equation,

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Step 3: Form of a particular solution.

Here x(n) = (1/8)ⁿ u(n). Hence from the table of the particular solution has the form of

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Step 4: Values of constants in y^(p)(n)

Putting y^(p)(n) = y(n) in given system equation and input we get,

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For n ≥ 2, all the terms in above equation will be present. Hence we will obtain value of 'k' for n ≥ 2. Then we can drop all unit step functions in above equation ie.

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Hence particular solution of equation (3) becomes,

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The forced response of the system is equal to sum of natural response and particular solution.

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Putting values in above equation from equation (2) and equation (4),

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Step 5: Values of constants in y^(f)(n) with initial conditions.

The given system equation can be written as,

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With n = 0 above equation becomes,

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In above equation y(-1), y(-2) are zero. And x(-1) = 0 since eeeeeeeeeee

Hence the above equation becomes.

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With n = 1 in equation (6)

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Here y(-1) = 0 and y(0) = 1. And x(1) = (1/8)¹, x(0) = (1/8)⁰. The above equation becomes.

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Putting n = 0 in equation (5),

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This value of y^(f)(0) must be equal to y(0) = 1 of equation (7). Hence above equation becomes,

C₁ + C₂ -1 = 1 ⇒ C₁ + C₂ = 2 ..........(9)

Similarly putting n = 1 in equation (5),

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This value of y^(f)(1) must be equal to

y(1) = 11/8 of equation (8), ie.

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Solving equation (9) and above equation for C₁ and C₂ we get,

C₁ = 8/3 and C₂ =-2/3

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This is the forced response of the system for zero initial conditions.

Problems based on Complete Response y(n):

Problem 3:

Determine the step response of the difference equation,

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Solution:

Step 1: Roots of characteristic equation

The given system equation is,

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Since we have to obtain step response, the input is,

x(n) = u(n) ..............(2)

To obtain characteristic equation make inputs equal to zero in equation (1), then we get,

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Here N = 2, hence characteristic equation of equation (4) becomes,

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From equation (3), a₀ = 1, a₁ = 0 and a₂ = -1/9, hence above equation becomes,

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Step 2: Form of a natural response

Natural response is given by equation (3) as,

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Putting r₁ and r₂ in above equation

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Step 3: Form of a particular solution

Here x(n) = u(n). Hence from the table the particular solution has the form of

y^(p)(n) = k u(n)

Step 4: Value of k in y^(p)(n)

Putting y^(p)(n) in system equation of equation (1) and inputs,

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For n ≥ 2 all the terms in above equation will be present. Hence we will obtain value of k for n ≥ 2 ie,

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Hence particular solution of equation (5) becomes,

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Forced response of the system is equal to sum of natural response and particular solution i.e,

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Putting values in above equation from equation (4) and equation (6),

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Step 5: Values of C₁ and C₂ with initial conditions

With n = 0 in system equation of equation (1),

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Since x(n) = u(n), x(-1) = 0 and x(-2) = 0

y(0) = 0 ..............(8)

With n = 1 in system equation of equation (1)

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Here x(0) = 1 and y(-1) = 1, hence we get,

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Putting n = 0 in equation (7),

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This value of y^(f)(0) must be equal to y(0) = 0 of equation (8). Hence above equation becomes,

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Similarly putting n = 1 in equation (7),

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This value of y^(f)(1) must be equal to y(1) = 10/9 of equation (9). Then above equation becomes.

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On solving above equation and equation (10) for C₁ and C₂ we get

C₁ = -7/12, C₂ = -13/24

 

Putting these values in equation (7),

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This is the complete response of the system. It includes natural response as well as forced response.

Here we consider initial conditions as well as input. Hence step response of the system is

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Problem 4:

Determine the response y(n), n ≥ 0 of the system (by finding the homogeneous and particular solutions) described by the second order difference equation y(n) – 3 y(n-1) – 4 y(n-2) = x(n) + 2 x(n-1) when the input sequencer is x(n) = 4ⁿ u(n).

Solution:

(i) Roots of Characteristic Equation

Characteristic equation can be written as,

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Here a₀ = 1, a₁ = -3 and a₂ = −4. Putting values,

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(ii) Form of a Natural Response

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(iii) Form of a Particular Solution

Here x(n) = 4ⁿ u(n). Hence particular solution will have following form.

y^(p)(n) = k (4)ⁿ u(n).

Above particular solution is already included in natural response given by equation (1), so we must select the particular solution, which is linearly independent of terms in natural response. Hence let us select the particular solution for multiple root. ie,

y^(p)(n) = k (4)ⁿ u(n) .............(2)

(iv) Value of k in y^(p)(n)

Putting this value of y^(p)(n) for y(n) and x(n) = 4ⁿ u(n) in a given difference equation,

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For n ≥ 2 above equation will be,

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Dividing by 4ⁿ.

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Let n = 2 for simplicity.

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Hence equation (2) becomes,

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The response of the system is equal to sum of natural response and particular solution ie

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v) Value of C₁ and C₂ with initial conditions

The given difference equation is,

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With n = 0 in above equation

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Then above equation will be

y(0) - 3 × 0 - 4 × 0 = 1 ⇒ y(0) = 1

Similarly putting n = 1 in equation (5)

y(1) - 3 y(0) - 4(-1) = 41 + 2 x 40

Here y(0) = 1 and y(-1) = 0,

y(1) − 3 = 4 + 2 ⇒ y(1) = 9.

Now let us evaluate equation (5) for n = 0 and n = 1

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Solving above equations,

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Putting these values of C₁ and C₂ in equation (5),

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This is the required response.