https://file.aiquickdraw.com/mathbot/file/17272823376586b2jg8kv.jpg

The image you've uploaded contains a worked example regarding the Fourier series for a half-wave rectified sinusoidal voltage. The solution is broken into various steps, and I'll walk through the key calculations step by step.

### Problem Setup:
The voltage \( u(t) \) is given as:
\[
u(t) = 
\begin{cases} 
0 & \text{if} \ -L < t < 0, \\
E \sin(\omega t) & \text{if} \ 0 < t < L.
\end{cases}
\]
Where \( L = \frac{\pi}{\omega} \) and \( p = 2L = \frac{2\pi}{\omega} \) gives the period of the function.

We need to find the Fourier series for the resulting periodic function \( u(t) \). 

### Step 1: Compute \( a_0 \) (DC component)
The constant term \( a_0 \) is the average value of \( u(t) \) over a period \( p \). This can be computed as:
\[
a_0 = \frac{\omega}{2\pi} \int_0^{\frac{\pi}{\omega}} E \sin(\omega t) \, dt.
\]

Solving this integral:
\[
a_0 = \frac{\omega}{2\pi} \cdot \left[ -\frac{E}{\omega} \cos(\omega t) \right]_0^{\frac{\pi}{\omega}} = \frac{E}{\pi}.
\]

### Step 2: Compute \( a_n \) (Fourier coefficients for cosine terms)
For \( n \geq 1 \), the Fourier coefficients for cosine terms are given by:
\[
a_n = \frac{\omega E}{2\pi} \int_0^{\frac{\pi}{\omega}} \sin(\omega t) \cos(n \omega t) \, dt.
\]

We use a standard trigonometric identity:
\[
\sin(\omega t) \cos(n \omega t) = \frac{1}{2} [\sin((1+n)\omega t) + \sin((1-n)\omega t)].
\]
Substituting this into the integral, we get:
\[
a_n = \frac{\omega E}{2\pi} \cdot \frac{1}{2} \int_0^{\frac{\pi}{\omega}} \left[ \sin((1+n)\omega t) + \sin((1-n)\omega t) \right] dt.
\]

Solving this integral:
- For \( n = 1 \), the integral evaluates to zero because the sine functions vanish over the integration limits.
- For \( n \neq 1 \), the result simplifies to:
\[
a_n = \frac{E}{\pi} \cdot \left( \frac{(-1)^n}{1-n^2} \right).
\]

### Step 3: Fourier Coefficients for Even \( n \)
When \( n \) is even, say \( n = 2, 4, 6, \dots \), a similar pattern emerges, and the general formula simplifies to:
\[
a_n = \frac{2E}{(1 - n^2)\pi}.
\]

### Summary:
- The DC component \( a_0 = \frac{E}{\pi} \).
- The Fourier coefficients \( a_n \) are zero for odd \( n \), and for even \( n \), we have:
\[
a_n = \frac{2E}{(1 - n^2)\pi}.
\]

Would you like further clarifications on any specific step or more details on trigonometric identities and integration techniques used here?

### 5 Related Questions:
1. How does a half-wave rectifier alter the original sinusoidal wave?
2. What is the significance of the Fourier series in analyzing waveforms?
3. How do we derive the Fourier coefficients for more complex periodic functions?
4. What would change if we had a full-wave rectifier instead?
5. How does the period of the function affect the Fourier series components?

### Tip:
When calculating Fourier series, it's essential to clearly distinguish between odd and even terms as their integration behavior can differ significantly, simplifying the overall computation.

Learn how to calculate the Fourier series of a half-wave rectified sinusoidal voltage. This example covers key signal processing concepts such as Fourier coefficients, integration techniques, and trigonometric identities, suitable for university-level engineering and physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation