Let's solve the problem step-by-step. --- ## **Given** Signal: \[ h(t) = 12 \cdot \sin^2(2\pi f_c t - \pi/3) \] where \( f_c = 40\, \text{Hz} \). --- ## **Step 1: Formula for Energy of a Signal** The **energy** of a signal \( h(t) \) is: \[ E = \int_{-\infty}^{\infty} |h(t)|^2 dt \] --- ## **Step 2: Substitute the Given Signal** \[ |h(t)|^2 = \left[12 \cdot \sin^2(2\pi f_c t - \pi/3)\right]^2 = 144 \cdot \sin^4(2\pi f_c t - \pi/3) \] So, \[ E = \int_{-\infty}^{\infty} 144 \cdot \sin^4(2\pi f_c t - \pi/3) \, dt \] --- ## **Step 3: Energy of a Periodic Signal** For a **periodic signal**, the energy over infinite time is infinite. Instead, we usually compute the **power**, or the energy in one period: \[ E_T = \int_{}^{T} |h(t)|^2 dt \] where \( T = \frac{1}{f_c} \) is the period. --- ## **Step 4: Simplify \( \sin^4(x) \)** Use the identity: \[ \sin^4(x) = \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x) \] So, \[ E_T = 144 \int_{}^{T} \left[ \frac{3}{8} - \frac{1}{2}\cos(2\theta) + \frac{1}{8}\cos(4\theta) \right] dt \] where \(\theta = 2\pi f_c t - \pi/3\). But cosine terms integrate to zero over a period, so: \[ \int_^T \cos(2\theta) \, dt = , \quad \int_^T \cos(4\theta) \, dt = \] Thus, \[ E_T = 144 \cdot \frac{3}{8} \int_^T dt = 144 \cdot \frac{3}{8} \cdot T \] --- ## **Step 5: Substitute \( T \) and Calculate** \[ T = \frac{1}{f_c} = \frac{1}{40} \text{ s} \] \[ E_T = 144 \cdot \frac{3}{8} \cdot \frac{1}{40} \] Calculate: - \( 144 \cdot \frac{3}{8} = 144 \cdot .375 = 54 \) - \( 54 \cdot \frac{1}{40} = 1.35 \) --- ## **Step 6: Final Answer** \[ \boxed{E_T = 1.35} \] **So, the energy of \( h(t) \) in one period is \( 1.35 \) (units depend on the signal, e.g., Joules if it's a voltage/current signal).** --- ### **Summary** - The energy in one period is \( 1.35 \). - For a periodic signal, total energy is infinite, so we calculate energy per period or power.

## **Explanation of Steps** 1. **Formula**: Energy of a signal is calculated using \( E = \int_{-\infty}^{\infty} |h(t)|^2 dt \). 2. **Substitution**: Substitute the given signal into the energy formula. 3. **Periodic Signal**: For periodic signals, use energy over one period instead of total energy. 4. **Simplification**: Use trigonometric identities to simplify \( \sin^4(x) \). 5. **Calculation**: Substitute the period into the simplified energy expression and compute the final energy value.

## **Step 1: Formula for Energy of a Signal** The energy of a signal quantifies its total power over time. It is calculated using the integral \( E = \int_{-\infty}^{\infty} |h(t)|^2 dt \). This formula is fundamental in signal processing to analyze signal properties and behavior over infinite time. --- ## **Step 2: Substitute the Given Signal** To find the energy of a specific signal, we substitute it into the energy formula. For the signal \( h(t) = 12 \cdot \sin^2(2\pi f_c t - \pi/3) \), we compute \( |h(t)|^2 \) to prepare for integration. This step is crucial for evaluating energy accurately. --- ## **Step 3: Energy of a Periodic Signal** For periodic signals, total energy over infinite time is infinite. Instead, we compute the energy in one period, denoted as \( E_T \). The period \( T \) is given by \( T = \frac{1}{f_c} \). This approach simplifies calculations and provides meaningful insights into signal behavior. --- ## **Step 4: Simplify \( \sin^4(x) \)** Using trigonometric identities, \( \sin^4(x) \) can be expressed as a combination of cosine functions. This simplification allows us to break down the integral into manageable parts, making it easier to evaluate energy over one period. It's a critical technique in signal processing. --- ## **Step 5: Substitute \( T \) and Calculate** After simplifying, we substitute the period \( T \) into the energy equation. We then compute the energy by integrating the simplified expression. This step combines all previous calculations and leads to the final energy value, providing a clear measure of the signal's energy. --- ## **Step 6: Final Answer** The final step involves presenting the calculated energy value clearly. In this case, the energy over one period is determined to be \( 1.35 \). This final answer encapsulates the total energy of the signal within its periodicity, offering insights into its characteristics and behavior.