To calculate $y(t)$ based on the given integral and functions, let's break down the problem step by step. We are given two functions:

1. $y(t) = \int_{-\infty}^{t} g(\xi) \, d\xi$
2. ( g(t) = \begin{cases} \cos\left( \frac{\pi t}{T_{\text{per}}} \right) & \text{if } |t| \leq \frac{T}{2}, \ 0 & \text{elsewhere}. \end{cases} )

Steps to calculate $y(t)$:

1. Understand the function $g(t)$:

   • $g(t) = \cos\left( \frac{\pi t}{T_{\text{per}}} \right)$ only when $|t| \leq \frac{T}{2}$, i.e., the function is a cosine wave restricted to the interval $-\frac{T}{2} \leq t \leq \frac{T}{2}$.
   • For $t$ outside this range, $g(t) = 0$.

2. Integrating $g(\xi)$: The integral $y(t) = \int_{-\infty}^{t} g(\xi) \, d\xi$ can be split into cases depending on the value of $t$.

3. Case 1: $t < -\frac{T}{2}$:

   • In this range, $g(\xi) = 0$ for all $\xi$, since the cosine function is zero outside $|t| \leq \frac{T}{2}$.
   • Therefore, $y(t) = 0$ for $t < -\frac{T}{2}$.

4. Case 2: $-\frac{T}{2} \leq t \leq \frac{T}{2}$:

   • In this range, we need to integrate the function $g(\xi) = \cos\left( \frac{\pi \xi}{T_{\text{per}}} \right)$ from $-\frac{T}{2}$ to $t$.

   • The integral is: $y(t) = \int_{-\frac{T}{2}}^{t} \cos\left( \frac{\pi \xi}{T_{\text{per}}} \right) d\xi$

     The antiderivative of $\cos\left( \frac{\pi \xi}{T_{\text{per}}} \right)$ is: $\int \cos\left( \frac{\pi \xi}{T_{\text{per}}} \right) d\xi = \frac{T_{\text{per}}}{\pi} \sin\left( \frac{\pi \xi}{T_{\text{per}}} \right)$

     Applying the limits of integration, we get: $y(t) = \frac{T_{\text{per}}}{\pi} \left[ \sin\left( \frac{\pi t}{T_{\text{per}}} \right) - \sin\left( -\frac{\pi}{2} \right) \right]$

     Since $\sin\left( -\frac{\pi}{2} \right) = -1$, we have: $y(t) = \frac{T_{\text{per}}}{\pi} \left[ \sin\left( \frac{\pi t}{T_{\text{per}}} \right) + 1 \right]$

5. Case 3: $t > \frac{T}{2}$:

   • For $t > \frac{T}{2}$, the function $g(\xi)$ is zero outside the interval $|t| \leq \frac{T}{2}$. Therefore, the integral is the same as the value at $t = \frac{T}{2}$.

   • So, $y(t) = y\left( \frac{T}{2} \right)$.

     This is just the result from Case 2 evaluated at $t = \frac{T}{2}$: $y\left( \frac{T}{2} \right) = \frac{T_{\text{per}}}{\pi} (1 + 1) = \frac{2 T_{\text{per}}}{\pi}$

Final result for $y(t)$:

• For $t < -\frac{T}{2}$: $y(t) = 0$
• For $-\frac{T}{2} \leq t \leq \frac{T}{2}$: $y(t) = \frac{T_{\text{per}}}{\pi} \left[ \sin\left( \frac{\pi t}{T_{\text{per}}} \right) + 1 \right]$
• For $t > \frac{T}{2}$: $y(t) = \frac{2 T_{\text{per}}}{\pi}$

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Would you like me to provide more details or clarify any part of the solution?

Related Questions:

1. How would the solution change if the periodic function was sine instead of cosine?
2. What happens to $y(t)$ if the period $T_{\text{per}}$ becomes very large?
3. Can you explain how to compute the integral for different types of periodic functions?
4. What is the physical interpretation of integrating a periodic signal over time?
5. How does the result change if the interval of nonzero $g(t)$ is shifted?

Tip: When integrating trigonometric functions, always remember to adjust the antiderivative to account for the coefficient inside the argument of the sine or cosine function.