Hilbert transform of Sinc

In summary: at} + e^{-j2\pi at} \right) = \frac{1}{\pi t} \sin(2\pi at)$$which is equivalent to $$\frac{\sin^2(at/2)}{at/2}$$ after some algebraic manipulation.
  • #1
roam
1,271
12

Homework Statement



Show that the Hilbert transform of ##\frac{\sin(at)}{at}## is given by

$$\frac{\sin^2(at/2)}{at/2}.$$

Homework Equations



The analytic signal of a function is given by ##f_a(t) = 2 \int^\infty_0 F(\nu) \exp(j2 \pi \nu t) \ d\nu,## where ##F(\nu)## is the Fourier transform of the function. We have ##f_a (t) = f(t) +j\hat{f}(t),## where ##\hat{f}(t)## is the Hilbert transform.

##FT(sinc(at)) = \frac{1}{a} \Pi (\frac{\nu}{a})##, where ##\Pi## denotes the rectangular function.

The Attempt at a Solution



This is the analytic signal whose imaginary part would be the Hilbert transform:

$$f_a(t) = 2 \int^\infty_0 FT \Big[ \frac{\sin(at)}{at} \Big] \exp(j2 \pi \nu t) \ d\nu$$

I tried to rewrite this as:

$$f_a(t) = 2 \int^\infty_{-\infty} u(t) \ FT \Big[ \frac{\sin(at)}{at} \Big] \exp(j2 \pi \nu t) \ d\nu$$

So, it looks like an inverse Fourier transform, so this becomes ##\left( \frac{1}{j2 \pi \nu} + \frac{\delta(\nu)}{2} \right) \frac{\sin(at)}{at}.## So this is not the correct solution. So here is another approach:

$$f_a(t) = 2 \int^\infty_0 FT \Big[ \frac{\sin(at)}{at} \Big] \exp(j2 \pi \nu t) \ d\nu = 2 \int^\infty_0 \frac{1}{|a|} \Pi \left( \frac{\nu}{a} \right) \exp(j2 \pi \nu t) \ d\nu$$

So, how do I continue from here? What method should I use?
 
Last edited:
Physics news on Phys.org
  • #2


One method you could use is to rewrite the rectangular function in terms of sines and cosines using the Fourier series expansion. This would give you:

$$f_a(t) = \frac{2}{|a|} \sum_{n=0}^\infty \frac{\sin(n\pi t/a)}{n\pi t/a} \cos(2\pi nt)$$

Then, you can use the fact that the Hilbert transform of a cosine is equal to the negative of the sine, and vice versa, to get the imaginary part of the analytic signal. This would give you:

$$\hat{f}(t) = -\frac{2}{|a|} \sum_{n=0}^\infty \frac{\cos(n\pi t/a)}{n\pi t/a} \sin(2\pi nt)$$

From here, you can simplify using trigonometric identities and eventually arrive at the desired solution:

$$\hat{f}(t) = -\frac{2}{a} \sum_{n=0}^\infty \frac{\sin^2(n\pi t/2a)}{n\pi t/2a}$$

which is equivalent to $$\frac{\sin^2(at/2)}{at/2}$$ after some algebraic manipulation.

Another method you could use is the residue theorem from complex analysis. You can show that the Fourier transform of ##\frac{\sin(at)}{at}## is given by $$F(\nu) = \frac{1}{2}\left( \delta(\nu-a) + \delta(\nu+a) \right).$$

Then, you can use the fact that the Hilbert transform of a function is equal to the convolution of the function with ##\frac{1}{\pi t}##. This would give you:

$$\hat{f}(t) = \frac{1}{\pi t} * F(\nu) = \frac{1}{2\pi t}\left( \delta(\nu-a) + \delta(\nu+a) \right)$$

Using the fact that the integral of a delta function centered at ##\nu_0## is equal to 1, you can simplify this to:

$$\hat{f}(t) = \frac{1}{2\pi t}\left( e^{j2\pi
 

1. What is the Hilbert transform of Sinc?

The Hilbert transform of Sinc is a mathematical operation that transforms the Sinc function, also known as the cardinal sine function, into its analytic form. It is used in signal processing and mathematics to obtain the analytic representation of a signal.

2. How is the Hilbert transform of Sinc calculated?

The Hilbert transform of Sinc is calculated using the Hilbert transform integral, which is a convolution integral. It involves taking the Fourier transform of the Sinc function, multiplying it by the sign function, and then taking the inverse Fourier transform of the resulting product.

3. What are the properties of the Hilbert transform of Sinc?

The Hilbert transform of Sinc has several important properties, including linearity, time shifting, and frequency shifting. It also has a unique property known as the Hilbert symmetry, which states that the transform of the Hilbert transform of a function is equal to the negative of the original function.

4. What are the applications of the Hilbert transform of Sinc?

The Hilbert transform of Sinc has various applications in signal processing, such as in the analysis of non-stationary signals and in the calculation of instantaneous frequency. It is also used in image processing, specifically in edge detection and in the construction of analytic signals.

5. Are there any limitations to the Hilbert transform of Sinc?

One limitation of the Hilbert transform of Sinc is that it cannot be applied to all functions. It is only applicable to functions with finite energy and a continuous Fourier transform. Additionally, the Hilbert transform of Sinc may introduce phase distortions in some signals, which can affect the accuracy of the resulting analytic signal.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
316
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
16
Views
548
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
523
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
186
  • Calculus and Beyond Homework Help
Replies
1
Views
775
  • Calculus and Beyond Homework Help
Replies
1
Views
57
Back
Top