The image contains a system of equations and problems related to signal processing in the time domain. Let’s break this down step by step:

1. Given System:

   • The system $S$ takes input $x(t)$ and gives output $y(t)$ with different responses for $e^{j2t}$ and $e^{-j2t}$.

     For $x(t) = e^{j2t}$, $y(t) = e^{j3t}$.

     For $x(t) = e^{-j2t}$, $y(t) = e^{-j3t}$.

2. Tasks:

   • $(a)$ Compute the output $y(t)$ corresponding to $x_1(t) = \cos(2t)$.
   • $(b)$ Compute the output $y(t)$ corresponding to $x_2(t) = \cos\left(2\left(t - \frac{1}{2}\right)\right)$.

Solution:

(a) For $x_1(t) = \cos(2t)$:

• Using Euler's formula: $\cos(2t) = \frac{1}{2}\left(e^{j2t} + e^{-j2t}\right)$.
• The system is linear, so the output $y_1(t)$ is the sum of the outputs for $e^{j2t}$ and $e^{-j2t}$: $y_1(t) = \frac{1}{2}\left(S(e^{j2t}) + S(e^{-j2t})\right)$
• Substitute the responses: $S(e^{j2t}) = e^{j3t}, \quad S(e^{-j2t}) = e^{-j3t}.$
• Thus: $y_1(t) = \frac{1}{2}\left(e^{j3t} + e^{-j3t}\right) = \cos(3t).$
• Final result for (a): $y_1(t) = \cos(3t)$.

(b) For $x_2(t) = \cos\left(2\left(t - \frac{1}{2}\right)\right)$:

• Using the time-shifting property: $\cos\left(2\left(t - \frac{1}{2}\right)\right) = \frac{1}{2}\left(e^{j2(t - 1/2)} + e^{-j2(t - 1/2)}\right)$.
• Expanding: $$$$

### Solution Continued:

Expanding for $x_2(t) = \cos\left(2\left(t - \frac{1}{2}\right)\right)$:

(b) Deriving $y_2(t)$:

Using the time-shifted input:

[ x_2(t) = \cos\left(2\left(t - \frac{1}{2}\right)\right) = \frac{1}{2}\left(e^{j2t}e^{-j} + e^{-j2t}e^{j}\right).$$

1. Apply the linearity of the system $S$:

   • The input $e^{j2t}e^{-j}$ has the response $S(e^{j2t}) = e^{j3t}$, so: $S(e^{j2t}e^{-j}) = e^{j3t}e^{-j} = e^{j(3t - 1)}.$
   • Similarly, the input $e^{-j2t}e^{j}$ has the response $S(e^{-j2t}) = e^{-j3t}$, so: $S(e^{-j2t}e^{j}) = e^{-j3t}e^{j} = e^{-j(3t - 1)}.$

2. Combine the results:

   • The output is: $y_2(t) = \frac{1}{2}\left(e^{j(3t - 1)} + e^{-j(3t - 1)}\right).$

3. Simplify using Euler's formula: $y_2(t) = \cos\left(3\left(t - \frac{1}{3}\right)\right).$

Final Results:

• (a) $y_1(t) = \cos(3t)$.
• (b) $y_2(t) = \cos\left(3\left(t - \frac{1}{3}\right)\right)$.

Do you have any further questions or would like detailed explanations on specific steps? Here are 5 related questions to consider:

1. How does the system's linearity affect the computation of outputs for superimposed signals?
2. What is the significance of the time-shifting property in solving (b)?
3. Can you explain the difference between $e^{j\omega t}$ and $\cos(\omega t)$ in this context?
4. How would the solution change if the system's response were nonlinear?
5. What role does Euler's formula play in solving these problems efficiently?

Tip: When working with signals in the frequency domain, always verify the linearity and time-invariance properties to simplify the calculations.