
A DT LSI system is described by the LCCDE

y[n]−0.5y[n−1]=x[n]+x[n−1]
 

The input signal is  x[n]=3^(−n)u[n]
  and  y[−1]=4
 .. The output signal takes the form

y[n]=(Aa^(−n)+Bb^(−n))u[n]

To find the output signal \( y[n] \) of the given discrete-time linear shift-invariant (DT LSI) system, we need to solve the linear constant-coefficient difference equation (LCCDE) and apply the initial condition. Here's a step-by-step solution:

### Given:
1. **Difference Equation (LCCDE):**

   \[
   y[n] - 0.5y[n-1] = x[n] + x[n-1]
   \]

2. **Input Signal:** 

   \[
   x[n] = 3^{-n}u[n]
   \]

3. **Initial Condition:** 

   \[
   y[-1] = 4
   \]

4. **Output Signal Form:** 

   \[
   y[n] = (A a^{-n} + B b^{-n})u[n]
   \]

### Step 1: Find the Homogeneous Solution

The homogeneous part of the difference equation is:

\[
y_h[n] - 0.5y_h[n-1] = 0
\]

The characteristic equation for this homogeneous equation is:

\[
r - 0.5 = 0
\]

Solving for \( r \), we get:

\[
r = 0.5
\]

So, the homogeneous solution \( y_h[n] \) is:

\[
y_h[n] = C (0.5)^n
\]

where \( C \) is a constant to be determined.

### Step 2: Find the Particular Solution

Given that the input \( x[n] = 3^{-n}u[n] \), we look for a particular solution of the form:

\[
y_p[n] = K \cdot 3^{-n}
\]

Substitute \( y_p[n] \) into the LCCDE:

\[
y_p[n] - 0.5y_p[n-1] = x[n] + x[n-1]
\]

Substituting for \( y_p[n] = K \cdot 3^{-n} \):

\[
K \cdot 3^{-n} - 0.5K \cdot 3^{-(n-1)} = 3^{-n} + 3^{-(n-1)}
\]

Simplify the terms:

\[
K \cdot 3^{-n} - 0.5K \cdot 3 \cdot 3^{-n} = 3^{-n} + \frac{1}{3} \cdot 3^{-n}
\]

\[
K \cdot 3^{-n} - \frac{3}{2}K \cdot 3^{-n} = 3^{-n} + \frac{1}{3} \cdot 3^{-n}
\]

Combine like terms:

\[
(K - \frac{3}{2}K) \cdot 3^{-n} = (1 + \frac{1}{3}) \cdot 3^{-n}
\]

\[
-\frac{1}{2}K \cdot 3^{-n} = \frac{4}{3} \cdot 3^{-n}
\]

Divide both sides by \( 3^{-n} \):

\[
-\frac{1}{2}K = \frac{4}{3}
\]

Solve for \( K \):

\[
K = -\frac{8}{3}
\]

Thus, the particular solution is:

\[
y_p[n] = -\frac{8}{3} \cdot 3^{-n}
\]

### Step 3: General Solution

The general solution \( y[n] \) is the sum of the homogeneous and particular solutions:

\[
y[n] = y_h[n] + y_p[n]
\]

\[
y[n] = C(0.5)^n - \frac{8}{3} \cdot 3^{-n}
\]

### Step 4: Apply Initial Condition

To determine \( C \), use the initial condition \( y[-1] = 4 \).

Substitute \( n = -1 \) into the general solution:

\[
y[-1] = C(0.5)^{-1} - \frac{8}{3} \cdot 3^{1}
\]

\[
4 = 2C - 8
\]

\[
2C = 12
\]

\[
C = 6
\]

### Step 5: Final Output Expression

Substitute \( C = 6 \) back into the general solution:

\[
y[n] = 6(0.5)^n - \frac{8}{3} \cdot 3^{-n}
\]

Since the output is defined for \( n \geq 0 \) (due to \( u[n] \)), we multiply by \( u[n] \):

\[
y[n] = \left(6(0.5)^n - \frac{8}{3} \cdot 3^{-n}\right) u[n]
\]

### Conclusion:

The output signal \( y[n] \) is:

\[
y[n] = \left(6(0.5)^n - \frac{8}{3} \cdot 3^{-n}\right) u[n]
\]

This matches the form given in the problem statement with \( A = 6 \), \( a = 0.5 \), \( B = -\frac{8}{3} \), and \( b = 3 \).

Would you like more details on any specific step, or do you have any questions?

### Related Questions:

1. What is the significance of the unit step function \( u[n] \) in discrete-time signals?
2. How do you find the zero-input response of a DT LSI system?
3. What is the role of the characteristic equation in solving LCCDEs?
4. How does the initial condition affect the output of a discrete-time system?
5. Can you explain the difference between homogeneous and particular solutions in difference equations?

### Tip:

Always check the stability of a discrete-time system by examining the roots of the characteristic equation. If all roots lie within the unit circle, the system is stable.

Learn how to solve a discrete-time linear shift-invariant (DT LSI) system described by a linear constant-coefficient difference equation (LCCDE) y[n] - 0.5y[n-1] = x[n] + x[n-1]. Includes step-by-step solution, initial conditions, and output signal form derivation. Suitable for advanced undergraduate level.

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