Question

Determine the natural response of the system: Difference equation is

(a) -y(n-1)-2y(n-2)=x(n)  and y(-1) = 1; y(-2) = 0

I had been asked this question in semester exam.

Asked question is from Discrete-Time Systems in the Time-Domain topic in portion Signals and Systems Basics of Signals and Systems

Answer 1

To determine the natural response of the system, we need to analyze the homogeneous solution (i.e., when \( x(n) = 0 \)) of the difference equation:

\[

y(n) + y(n-1) + 2y(n-2) = 0

\]

Given initial conditions:

- \( y(-1) = 1 \)

- \( y(-2) = 0 \)

### Step 1: Find the Characteristic Equation

Rewrite the difference equation as:

\[

y(n) + y(n-1) + 2y(n-2) = 0

\]

Assume a solution of the form \( y(n) = r^n \). Substituting \( y(n) = r^n \), \( y(n-1) = r^{n-1} \), and \( y(n-2) = r^{n-2} \) into the equation:

\[

r^n + r^{n-1} + 2r^{n-2} = 0

\]

Dividing by \( r^{n-2} \):

\[

r^2 + r + 2 = 0

\]

This is the characteristic equation.

### Step 2: Solve the Characteristic Equation

The characteristic equation is:

\[

r^2 + r + 2 = 0

\]

Using the quadratic formula:

\[

r = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2}

\]

This gives complex roots:

\[

r = \frac{-1 \pm j\sqrt{7}}{2}

\]

Let \( r = \alpha \pm j\beta \), where \( \alpha = -\frac{1}{2} \) and \( \beta = \frac{\sqrt{7}}{2} \). The general solution for the homogeneous equation is:

\[

y(n) = A \left( e^{\alpha n} \cos(\beta n) \right) + B \left( e^{\alpha n} \sin(\beta n) \right)

\]

Substituting \( \alpha = -\frac{1}{2} \) and \( \beta = \frac{\sqrt{7}}{2} \):

\[

y(n) = A \left( e^{-\frac{n}{2}} \cos\left(\frac{\sqrt{7}}{2} n\right) \right) + B \left( e^{-\frac{n}{2}} \sin\left(\frac{\sqrt{7}}{2} n\right) \right)

\]

### Step 3: Apply Initial Conditions

Using \( y(-1) = 1 \) and \( y(-2) = 0 \), substitute these values to solve for \( A \) and \( B \). Once \( A \) and \( B \) are determined, you’ll have the complete natural response \( y(n) \).