Inverse Fourier Transform of a squared sinc function

Question

I have here a squared sinc function, which is the Fourier Transform of some triangular pulse:

$$\mathrm H(f)= 2\mathrm A\mathrm T_\mathrm o \frac{\sin^2(2\pi f \mathrm T_\mathrm o)}{(2\pi f\mathrm T_\mathrm o)^2}$$

As an excercise, I would like to go back to the original time domain triangular pulse, using the inverse Fourier Transform.

I feel like I'm very close to achieving it, however, I stumbled upon this integral:

$$\mathrm h(t)=\frac {\mathrm A} {2\pi ^2 \mathrm T_\mathrm o}\int_{0}^{\infty}\frac{(1-\cos(4\pi f\mathrm T_\mathrm o))\cos(2\pi ft)}{f^2}df$$

I'm not sure if there is already an expression which defines the value of this integral, or if this integral is of a well-known type, with a well-defined value. The tables of integrals that I have right now are basic and don't cover integrals of such type.

Does someone know if there is a well-known expression for the value of integrals of this type?

Or, would you be so kind as to recommend a good table of integrals for Fourier Analysis?

Thank you very much!

Answer 1

For $T_o>0$, this integral can be tackled with a Schwinger parameterization and Fubini's Theorem. (The equivalent of the intermediate integral you specifically asked about is at the 7th equal sign below, with the trig functions written in terms of complex exponentials.)

$$\begin{align*}h(t) &= 2AT_o\int_{-\infty}^{\infty} \frac{\sin^2(2\pi fT_o)}{(2\pi fT_o)^2}e^{2\pi ift} \; df\\ \\ &= 2AT_o\int_{-\infty}^{\infty} \frac{\sin^2(2\pi fT_o)}{(2\pi fT_o)^2}\cos(2\pi ft) \; df\\ \\ &= 4AT_o\int_{0 }^{\infty} \frac{\sin^2(2\pi fT_o)}{(2\pi fT_o)^2}\cos(2\pi ft) \; df\\ \\ &= 4AT_o\int_{0}^{\infty} \sin^2(2\pi fT_o)\cos(2\pi ft)\int_0^{\infty}ue^{-u2\pi fT_o} \; du\; df \\ \\ &= 4AT_o\int_{0}^{\infty} \int_{0}^{\infty} u\sin^2(2\pi fT_o)\cos(2\pi ft)e^{-u2\pi fT_o} \; df\; du \\ \\ &= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{i2\pi fT_o}-e^{-i2\pi fT_o}\right)^2\left(e^{i2\pi ft}+e^{-i2\pi ft}\right)e^{-u2\pi fT_o} \; df\; du \\ \\ &= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{i2\pi f 2T_o}-2+e^{-i2\pi f 2T_o}\right)\left(e^{i2\pi ft}+e^{-i2\pi ft}\right)e^{-u2\pi fT_o} \; df\; du \\ \\ &= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{i2\pi f (2T_o+t)}+e^{i2\pi f (2T_o-t)}-2\left[e^{i2\pi ft}+e^{-i2\pi ft}\right]+e^{-i2\pi f (2T_o-t)}+e^{-i2\pi f (2T_o+t)}\right)e^{-u2\pi fT_o} \; df\; du \\ \\ &= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{-f(2\pi uT_o - i2\pi [2T_o+t])}+e^{-f(2\pi uT_o - i2\pi [2T_o-t])}-2\left[e^{-f(2\pi uT_o -i2\pi t)}+e^{-f(2\pi u T_o+i2\pi t)}\right]+e^{-f(2\pi uT_o + i2\pi [2T_o+t])}+e^{-f(2\pi uT_o + i2\pi [2T_o-t])}\right) \; df\; du \\ \\ &= -\dfrac{AT_o}{2}\int_{0}^{\infty} u\left(\dfrac{1}{2\pi uT_o - i2\pi [2T_o+t]}+\dfrac{1}{2\pi uT_o - i2\pi [2T_o-t]}-\dfrac{2}{2\pi uT_o -i2\pi t}-\dfrac{2}{2\pi u T_o+i2\pi t}+\dfrac{1}{2\pi uT_o + i2\pi [2T_o+t]}+\dfrac{1}{2\pi uT_o + i2\pi [2T_o-t]}\right) \; du \\ \\ &= -\dfrac{A}{4\pi}\int_{0}^{\infty} \dfrac{u}{u - i\left[2+\frac{t}{T_o}\right]}+\dfrac{u}{u - i\left[2-\frac{t}{T_o}\right]}-\dfrac{2u}{u -i\frac{t}{T_o}}-\dfrac{2u}{u+i\frac{t}{T_o}}+\dfrac{u}{u+ i\left[2+\frac{t}{T_o}\right]}+\dfrac{u}{u + i\left[2-\frac{t}{T_o}\right]} \; du \\ &= -\dfrac{A}{2\pi}\int_{0}^{\infty} \dfrac{u^2}{u^2 +\left[2+\frac{t}{T_o}\right]^2}+\dfrac{u^2}{u^2 + \left[2-\frac{t}{T_o}\right]^2}-\dfrac{2u^2}{u^2 +\left[\frac{t}{T_o}\right]^2}\; du \\ \\ &= -\dfrac{A}{2\pi}\int_{0}^{\infty} \dfrac{u^2}{u^2 +\left[\frac{t}{T_o}+2\right]^2}+\dfrac{u^2}{u^2 + \left[\frac{t}{T_o}-2\right]^2}-\dfrac{2u^2}{u^2 +\left[\frac{t}{T_o}\right]^2}\; du \\ \\ &= -\dfrac{A}{2\pi} \left(u - \left[\frac{t}{T_o}+2\right] \tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}+2}\right]+u - \left[\frac{t}{T_o}-2\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}-2}\right]-2u +2\left[\frac{t}{T_o}\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}}\right]\right)\bigg{|}_0^\infty\\ \\ &= \dfrac{A}{2\pi} \left(\left[\frac{t}{T_o}+2\right] \tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}+2}\right]+\left[\frac{t}{T_o}-2\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}-2}\right]- 2\left[\frac{t}{T_o}\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}}\right]\right)\bigg{|}_0^\infty\\ \\ &= \begin{cases} \dfrac{A}{2\pi} \left(-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]+2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right) & \dfrac{t}{T_o} \le -2 \\ \\ \dfrac{A}{2\pi} \left(\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]+2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right) &-2 \le \dfrac{t}{T_o}\le 0 \\ \\ \dfrac{A}{2\pi} \left(\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]-2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right)&0 \le\dfrac{t}{T_o}\le 2 \\ \\ \dfrac{A}{2\pi} \left(\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]+\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]-2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right)&\dfrac{t}{T_o}\ge 2\\ \end{cases} \\ \\ h(t) &= \begin{cases} 0 & \dfrac{t}{T_o} \le -2 \\ \\ A\left(1+\dfrac{1}{2}\cdot\dfrac{t}{T_o}\right) &-2 \le \dfrac{t}{T_o}\le 0 \\ \\ A\left(1-\dfrac{1}{2}\cdot\dfrac{t}{T_o}\right)&0 \le\dfrac{t}{T_o}\le 2 \\ \\ 0 &\dfrac{t}{T_o}\ge 2\\ \end{cases}\\ \\ \end{align*}$$

Answer 2

Per Wikipedia* the integral of the sinc function is, $$\int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2}$$ The integrand is an even function, so $$\int_{-\infty}^\infty \frac{\sin x}{x} dx = \pi$$ Consider this the end of a string. It's a target point.

Your integrand is $$\frac{(1-\cos(4\pi f\mathrm T_\mathrm o))\cos(2\pi ft)}{f^2}$$ With suitable substitutions, this can be expanded into three fractions of the form $$\frac{\cos x}{x^2}$$

So the problem boils down to finding $$\int_{-\infty}^\infty \frac{\cos x}{x^2} dx$$

Integrating by parts, with $dv = \frac{1}{x^2} dx$, $u = \cos x$, we first get $v = -\frac{2}{x}$ and $du = -\sin x\ dx$. Substituting, we get $$\int_{-\infty}^\infty \frac{\cos x}{x^2} dx = \left . 2\frac{\cos x}{x} \right |_{-\infty}^\infty - 2\int_{-\infty}^\infty \frac{\sin x}{x} dx$$

Assuming no stupid math errors (I found one going from my notes to this answer), you should be able to do the rest.

* However, solving this integral requires solving the Dirchlet integral. But, again per Wikipedia: "The value of the integral (in the Riemann or Henstock sense) can be derived using various ways, including the Laplace transform, double integration, differentiating under the integral sign, contour integration, and the Dirichlet kernel. " So you're back tables of Laplace transforms (which you don't want to do), or contour integration, or other advanced calculus methods.

Answer 3

While searching on the Internet for a more specialized table of integrals, I found this source:

Erdélyi, A. et al., "Tables of Integral Transforms", vols. I and II. McGraw Hill, New York, 1954.

On page 20 of volume I, I found this integral (number 16):

For $y>0,\,a>0$,

\begin{align} \ g(y)=\int_0^\infty x^{-2}(1-\cos(ax))\cos(xy)dx & = \frac{\pi}{2}(a-y)\quad & y<a \\ & = 0\quad & a<y \\ \end{align} So, with $a=4\pi\mathrm T _\mathrm o$ and $y=2\pi t$,

\begin{align} \ h(t) & = \mathrm A\left(1-\frac{t}{2\mathrm T_\mathrm o}\right)\quad & t<2\mathrm T_\mathrm o \\ & = 0\quad & t>2\mathrm T_\mathrm o \\ \end{align}

Since $y$ must be positive and since $t$ varies from $-\infty$ to $\infty$, we can finally write the inverse Fourier Transform of our squared sinc function as:

\begin{equation*} h(t) = \left\{ \begin{array}{l} \mathrm A\left(1-\frac{|t|}{2\mathrm T_\mathrm o}\right) & |t|<2\mathrm T_\mathrm o\\ 0 & |t|>2\mathrm T_\mathrm o\\ \end{array} \right. \end{equation*}