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The problem asks for simplifications of various signals that involve Dirac delta functions and other functions such as the sinc function and cosine.

Let’s go through each part:

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### (l) \( x_7(t) = \text{sinc}(t)\delta(t) \)
The Dirac delta function \( \delta(t) \) has the property that it picks the value of the function it multiplies at \( t = 0 \). Since \( \text{sinc}(0) = 1 \), we get:

\[
x_7(t) = \text{sinc}(t) \delta(t) = \delta(t)
\]

---

### (m) \( x_8(t) = \text{sinc}(t) \delta(2t - 3) \)
To simplify this, recognize that the delta function \( \delta(2t - 3) \) picks the value of \( \text{sinc}(t) \) when \( 2t - 3 = 0 \), i.e., when \( t = \frac{3}{2} \). Therefore, it becomes:

\[
x_8(t) = \text{sinc}\left(\frac{3}{2}\right) \delta(2t - 3)
\]

---

### (n) \( x_9(t) = \cos(2\pi t + \frac{\pi}{3})\delta(3t) \)
The delta function \( \delta(3t) \) picks the value of the cosine function when \( 3t = 0 \), i.e., at \( t = 0 \). So, we evaluate the cosine at \( t = 0 \):

\[
\cos\left(2\pi(0) + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
\]

Thus, the expression simplifies to:

\[
x_9(t) = \frac{1}{2} \delta(3t)
\]

---

### (o) \( x_{10}(t) = \int_0^\infty \Lambda(2t)\delta(3t - 1) dt \)

The delta function \( \delta(3t - 1) \) selects the value of \( \Lambda(2t) \) at the point where \( 3t - 1 = 0 \), or \( t = \frac{1}{3} \). So we substitute this into \( \Lambda(2t) \), yielding:

\[
\Lambda\left(2 \times \frac{1}{3}\right) = \Lambda\left(\frac{2}{3}\right)
\]

Thus, the integral reduces to:

\[
x_{10}(t) = \Lambda\left(\frac{2}{3}\right)
\]

---

### (p) \( x_{11}(t) = \int_2^\infty \cos(2\pi - \pi)\delta(t - 1) dt \)

In this case, the delta function \( \delta(t - 1) \) picks the value of the cosine function when \( t = 1 \). However, since the limits of integration are from \( t = 2 \) to infinity, the delta function has no effect because its argument is outside the bounds of integration. Therefore, the integral evaluates to zero:

\[
x_{11}(t) = 0
\]

---

Would you like more details or explanations on any of these steps?

### 5 Relative Questions:
1. What are the key properties of the Dirac delta function?
2. How does the sinc function behave and where is it commonly used?
3. Why does the cosine term in (n) simplify to a specific value when combined with the delta function?
4. How does the integration of functions involving the delta function typically work?
5. What happens when the argument of a delta function falls outside the bounds of integration?

### Tip:
When dealing with Dirac delta functions, always identify the point where the delta function forces the argument to zero and evaluate the surrounding function at that point.

應用脈衝函數之化簡運算，求出下列訊號之最簡表示式：
(l) x₇(t) = sinc(t)δ(t)
(m) x₈(t) = sinc(t)δ(2t - 3)
(n) x₉(t) = cos(2πt + π/3)δ(3t)
(o) x₁₀(t) = ∫₀⁰∞ Λ(2t)δ(3t - 1) dt
(p) x₁₁(t) = ∫₂⁰∞ cos(2π - π)δ(t - 1) dt

Learn how to simplify complex signals involving impulse and delta functions, including sinc functions and cosine terms. This problem involves key properties of the delta function and demonstrates how to evaluate integrals with δ(t). Ideal for students in Electrical Engineering or Applied Mathematics.

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