Question

Find all values \(\omega\) for which \(\int_{-\infty}^{\infty} \frac{\sin ^{2} 5 x}{x^{2}} \cos \omega x d x=0\)

Short answer

The values of \(\omega\) are \(\pm 10n\) for any integer \(n\).

Step-by-step solution

Recognizing the integral's form

The given integral involves sinusoidal functions. We recognize that it resembles a Fourier transform of a function over the real line, specifically focusing on \[ \int_{-\infty}^{\infty} \left(\frac{\sin 5x}{x}\right)^{2} \cos(\omega x) \, dx = 0. \] This expression suggests we are seeking values of \( \omega \) where the cosine component orthogonally cancels out the integrated function.

Analyze the function inside the integral

Notice that \( \frac{\sin^2(5x)}{x^2} \) is even because \( \sin^2 \) is even, and \( \cos(\omega x) \) is also even. Therefore, this integral involves the orthogonality of an even function with \( \cos(\omega x) \).

Relate to known integral property

Recall that the integral \[ \int_{-\infty}^{\infty} \frac{\sin^2(ax)}{x^2} e^{i\omega x} \, dx \] has solutions where its real part (the cosine transform) is zero. This typically occurs at specific values of \( \omega \) determined by the characteristic integrand frequencies.

Express in terms of Fourier transform

Focusing on the \[ \int_{-\infty}^{\infty} sinc^2(5x) \cos(\omega x) \, dx = 0 \] allows us to analyze when the Fourier cosine transform of \( sinc^2(5x) \) is zero. Recognize \( sinc(x) = \frac{\sin \pi x}{\pi x} \), hence \( sinc^2(5x) \) is its square.

Determine the zeros and solve for \( \omega \)

The zeros for such transforms often lie in rational harmony with the base frequency. Given that \( sinc \) functions drop to zero on non-zero integer multiples of their frequency scale, explore solutions like \( \omega = \pm 10n \).

Finalize the general solution

Combine insights to express the general condition. The frequency response function \( sinc^2(ax) \), due to even harmonics and integration symmetry, is zero at twice the frequency, leading to \[ \omega = \pm 10n, \quad n = 1, 2, 3, \ldots \] as precise zero-crossing points for the cosine component.

Key concepts

Fourier Series

The Fourier series is a way to represent a function as the sum of simple sine and cosine waves. This is incredibly useful when analyzing periodic functions. Each component in the series is like a building block that, when added together, perfectly express the original function. The beauty of the Fourier series lies in how it breaks down complex periodic signals into understandable wave patterns.

• This decomposition helps analyze signal frequencies.
• Each frequency in the series is a harmonic of the base or fundamental frequency.
• It provides a link between the time domain representation and the frequency domain.

When we talk about solving integrals like the one in the exercise, we are essentially utilizing the idea that any function can be decomposed into these harmonics. The integral represents how well these harmonics fit together. In essence, Fourier series is about understanding complex periodic functions in a simpler way through their frequency components.

Orthogonality

Orthogonality is a principle that comes from linear algebra, and it’s a core feature of the Fourier series and transforms. Two functions are orthogonal if their inner product (a kind of multiplication) is zero. In the realm of trigonometric functions, such as sines and cosines, orthogonality is crucial.

• It means that the functions don't "interfere" with each other.
• This property is what allows us to separate signals into distinct frequency components.

In our integral, the orthogonality comes into play because the cosine component cancels out the integrated function when the equation equals zero. This occurs at values of \( \omega \) where the function inside the integral and the cosine are orthogonal. This concept of orthogonality helps significantly simplify and solve many kinds of integrals in analysis.

Sinc Function

The sinc function, expressed as \( \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \), is a critical concept in signal processing and Fourier transforms. It describes a completely smooth and bell-like function that tapers off to zero as \( x \) increases.

• It has important properties that make it the "ideal" low pass filter in the frequency domain.
• In the original integral, the factor \( \left(\frac{\sin(5x)}{x}\right)^2\) resembles a squared sinc function.

This makes its analysis dependent on understanding the zero-crossings, which are symmetrically placed away from the origin. These zero-crossings are key to understanding the spectral characteristics and behavior of the sinc function in Fourier transforms. Understanding these halts help solve integrals where the function and cosine component are orthogonal.

Cosine Transform

The Fourier cosine transform is similar to the ordinary Fourier transform but specifically uses only cosine functions to build the frequency domain representation. This type of transform is particularly useful for even functions that are symmetric around the origin.

• Even functions are those where \( f(x) = f(-x) \).
• The cosine transform efficiently represents such functions because cosines are themselves even.

Within the given context of the exercise, we utilize the property of the cosine transform to determine when our integral equals zero. The exercise demonstrates how the zeros of the cosine transform correspond to frequencies at which the transform is zero. This understanding allows us to identify the zero-crossing points like \( \omega = \pm 10n \), providing a clear solution to the integral problem. Leveraging the cosine transform simplifies solving problems involving even symmetry in functions.