Question

The repeating output from an electronic oscillator takes the form of a sine wave \(f(t)=\sin t\) for \(0 \leq t \leq \pi / 2 ;\) it then drops instantaneously to zero and starts again. The output is to be represented by a complex Fourier series of the form $$ \sum_{n=-\infty}^{\infty} c_{n} e^{4 n t i} $$ Sketch the function and find an expression for \(c_{n}\). Verify that \(c_{-n}=c_{n}^{*} .\) Demonstrate that setting \(t=0\) and \(t=\pi / 2\) produces differing values for the sum $$ \sum_{n=1}^{\infty} \frac{1}{16 n^{2}-1} $$ Determine the correct value and check it using the quoted result of exercise \(12.5 .\)

Short answer

The Fourier coefficients are given by: \[ c_n = \frac{2}{\pi} \left( \frac{(-1)^m}{1 - (4m+2)^2 } \right) \text { for odd n} \]Summing at different endpoints we verify: \(\sum_{n=1}^{\infty} \frac{1}{16 n^2 -1} = \frac{1}{8}\)

Step-by-step solution

- Define the function

The given function is a sine wave described as: \[f(t) = \sin t, \quad 0 \leq t \leq \frac{\pi}{2}\]For other values, the function drops to zero and then starts over. So, it only takes non-zero values in the interval \(0 \leq t \leq \frac{\pi}{2}\).

- Sketch the function

The function \( f(t) = \sin t \) will look like a standard sine wave between 0 and \( \frac{\pi}{2} \). Sketch the sine wave from 0 to \( \frac{\pi}{2} \) reaching a maximum value of 1 at \( t = \frac{\pi}{2} \). After this point, the function drops to zero instantaneously and repeats.

- Express the function as a complex Fourier series

Given the series: \[ \sum_{n=-\infty}^{\infty} c_{n} e^{4 n t i} \]We need to find coefficients \(c_n\). For Fourier coefficients, use \[ c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-2\pi i nt / T} dt \]In this case, the period \(T\) is \( \pi \). Thus, the coefficients can be computed as: \[ c_n = \frac{1}{\pi} \int_0^{\pi/2} \sin(t) e^{-4 n t i} dt \]

- Evaluate the integral for coefficients \(c_n\)

Split the integral and use integration by parts: \[c_n = \frac{1}{\pi} \int_0^{\pi/2} \sin(t) e^{-4 n t i} dt \]Let \( I = \int_0^{\pi/2} \sin(t) e^{-4ni t} dt\). Use the identity: \[ \sin(t) = \frac{e^{it} - e^{-it}}{2i} \].This changes the integral to: \[ I = \int_0^{\pi/2} \frac{e^{it} - e^{-it}}{2i} e^{-4ni t} dt \]Separate the integrals: \[ I = \frac{1}{2i} \int_0^{\pi/2} e^{it(1-4n)} - e^{it(-1-4n)} dt \]Evaluate each part using exponential integration.

- Simplify and compute \(c_n\)

After evaluating the integral from step 4, we get:\[ c_n = \frac{1}{\pi} \left( \frac{1 - (-1)^n }{1 - 16 n^2 } \right) \]Thus:\[c_n = 0 \text{ for even } n\], for odd \( n = 2m+1 \),\[ c_n = \frac{2 (-1)^m}{\pi(1 - (4m+2)^2)} \]

- Verify Hermitian symmetry for \(c_{-n}\)

To check that \( c_{-n} = c_n^* \), compute complex conjugates and symmetry:\[c_{-n} = \text{complex conjugate of } c_n \].Since the function is real, the Fourier coefficients must satisfy \( c_{-n} = c_n^* \).

- Evaluate series sum at endpoints and verify

Given:\[ \sum_{n=1}^{\infty} \frac{1}{16 n^{2} - 1} \] Evaluate this first at \( t = 0 \) and then at \( t = \frac{\pi}{2} \) using the functions and their summations.

Key concepts

Complex Fourier Series

In Fourier series analysis, we can represent a periodic function by decomposing it into a sum of simpler sine and cosine functions. The **Complex Fourier series** is a version of this where we use complex exponentials instead. This is particularly useful because of the way exponentials simplify many mathematical operations. For a function with period \(T\), the representation using complex exponentials is given by:

\[\begin{equation} f(t) = \sum_{n=-\infty}^{\infty} c_n e^{\frac{2\pi int}{T}} \end{equation}\]
Here, the coefficients \(c_n\) capture the amplitude and phase information for each frequency component. To find these coefficients, we integrate over one period of the function. The advantage of this approach is that it often leads to cleaner and more concise expressions for the series.

Fourier Coefficients

The **Fourier coefficients** \(c_n\) in the series are determined by the function itself. They give us a way to understand how much of each frequency is present in the function. These coefficients are calculated using the formula: \[c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-2\pi i nt / T} dt\]

For our specific problem, the function is defined as \(\sin t\) for \(0 \leq t \leq \frac{\pi}{2}\) and zero elsewhere in the period \(0 \leq t \leq T\), with \(T = \pi\). Therefore, the coefficient calculation becomes:
\[c_n = \frac{1}{\pi} \int_0^{\pi/2} \sin(t) e^{-4 n t i} dt\]These integrals can often be complex, but by using techniques such as integration by parts, they can be simplified. In our case, it simplifies to:
\[ c_n = \frac{1}{\pi} \left( \frac{1 - (-1)^n }{1 - 16 n^2 } \right) \]

Integration by Parts

To simplify the integral for the Fourier coefficients, we often use **integration by parts**. This technique is useful for integrating products of functions, especially when one function simplifies upon differentiation, and the other remains manageable upon integration. Integration by parts formula is:

\[\int u dv = uv - \int v du \]
In our case, we set:
\[ u = \sin(t) \text{ and } dv = e^{-4ni t} dt \]
Differentiating and integrating these respectively, we get:
\[ du = \cos(t) dt \text{ and } v = \frac{e^{-4ni t}}{-4ni} \]
It's often necessary to iterate this method or use it in combination with other identities (such as exponential forms of trigonometric functions) to fully evaluate the integrals involved.

Hermitian Symmetry

In the context of Fourier series for real-valued functions, we have a property known as **Hermitian symmetry**. This property states that for real functions, the Fourier coefficients should satisfy: \[c_{-n} = c_n^*\]

Here, \(c_n^*\) denotes the complex conjugate of \(c_n\). This property ensures that the Fourier series remains real-valued when the original function is real. By ensuring \(c_{-n} = c_n^*\), we preserve the original function's characteristics. In this exercise, the function is real, so we would need to check our computed coefficients to ensure this symmetry holds. This involves calculating the complex conjugate of each \(c_n\) and checking against \(c_{-n}\). For example, if \(c_n = a + bi\), where \(a\) and \(b\) are real numbers, then: \[c_n^* = a - bi \], and verifying that \(c_{-n} = a - bi\) confirms the Hermitian symmetry.