𝑦[𝑛]=𝑥[𝑛]+0.23∙𝑥[𝑛−1]+𝑥[𝑛−2]−(-1.19)∙𝑦[𝑛−1]−0.79∙𝑦[𝑛−2] Derive an expression of the output y[n] in steady state when fsignal equals 100 Hz. The steady state is defined here as the state when all transients have died out.

To derive the steady-state output expression \( y[n] \) for the given difference equation, we need to consider the following difference equation:

\[
y[n] = x[n] + 0.23 \cdot x[n-1] + x[n-2] - (-1.19) \cdot y[n-1] - 0.79 \cdot y[n-2]
\]

First, we can rewrite this equation as:

\[
y[n] + 1.19 \cdot y[n-1] + 0.79 \cdot y[n-2] = x[n] + 0.23 \cdot x[n-1] + x[n-2]
\]

### Steady-State Response

In steady state, we assume all transients have died out, which means the system response will be due solely to the input signal. If we assume the input signal is a sinusoidal signal of frequency 100 Hz, say:

\[
x[n] = A \cos(2\pi f n T_s + \phi)
\]

where \( A \) is the amplitude, \( f = 100 \) Hz is the frequency, \( T_s \) is the sampling period, and \( \phi \) is the phase.

For steady-state analysis, the output \( y[n] \) will have the same frequency as the input but generally a different amplitude and phase. So, we can assume:

\[
y[n] = B \cos(2\pi f n T_s + \theta)
\]

where \( B \) and \( \theta \) are the steady-state amplitude and phase shift, respectively.

### Substituting into the Difference Equation

Substitute \( x[n] = A \cos(2\pi f n T_s + \phi) \) and \( y[n] = B \cos(2\pi f n T_s + \theta) \) into the difference equation:

\[
B \cos(2\pi f n T_s + \theta) + 1.19 \cdot B \cos(2\pi f (n-1) T_s + \theta) + 0.79 \cdot B \cos(2\pi f (n-2) T_s + \theta)
\]
\[
= A \cos(2\pi f n T_s + \phi) + 0.23 \cdot A \cos(2\pi f (n-1) T_s + \phi) + A \cos(2\pi f (n-2) T_s + \phi)
\]

We know the trigonometric identity:

\[
\cos(\theta + \alpha) = \cos(\theta)\cos(\alpha) - \sin(\theta)\sin(\alpha)
\]

We can use this to simplify the expressions, and since we are looking for the steady-state solution, we equate the coefficients of the corresponding cosine terms.

### Solving for B and θ

This process involves equating the coefficients of \(\cos(2\pi f n T_s + \phi)\) and \(\sin(2\pi f n T_s + \phi)\) from both sides of the equation. The resulting system of equations can be solved to find \( B \) and \( \theta \).

However, this is a complex process requiring solving two equations for two unknowns (B and θ). The general process is as follows:

1. Equate the coefficients of \(\cos(2\pi f n T_s)\) on both sides to solve for B.
2. Equate the coefficients of \(\sin(2\pi f n T_s)\) on both sides to solve for θ.
3. Express the final steady-state response \( y[n] \) in terms of \( A \), \( B \), and \( \theta \).

Given the complexity of these equations, the solution typically involves the use of phasor analysis or Laplace transform methods to obtain \( B \) and \( \theta \).

Do you want to see the detailed steps involved in solving for \( B \) and \( \theta \), or have any other questions? Here are some related questions:

1. What is the difference between transient and steady-state response in a system?
2. How do you perform phasor analysis for solving difference equations?
3. What role does the sampling frequency \( T_s \) play in discrete-time signal processing?
4. How does the stability of a system affect its steady-state response?
5. Can this difference equation be solved using the Z-transform?
6. What are the conditions for a system to reach steady-state?
7. How does the input signal frequency affect the steady-state response of a system?
8. What happens to the steady-state response if the input signal frequency is not 100 Hz?

**Tip:** The steady-state response of a linear system to a sinusoidal input is a sinusoid at the same frequency as the input but generally with a different amplitude and phase.

Learn how to derive the steady-state output y[n] from a difference equation involving cosine functions and trigonometric identities. This problem explores steady-state analysis in discrete-time systems, suitable for advanced undergraduate students.

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