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I want to find the Fourier transform of the raised cosine with $\alpha=1$, i.e.

$g(t) = \text{sinc}(t/T)\frac{\cos(\pi t/T)}{1-4t^2/T^2}$

We can with substitution $u=t/T$, a trig identity and partial fractions expansion rewrite it as

$\frac{1}{4\pi u}\Big(\frac{\sin(2\pi u)}{2u+1} - \frac{\sin(2\pi u)}{2u-1}\Big)$. I also tried various variants of $\cos(u) = \sin(\pi/2-u)$, obtaining

$\frac{\pi}{2}\frac{\text{sinc}(u)}{2}\Big(\text{sinc}(2u+1)+\text{sinc}(2u-1)\Big)$. But from here I'm not sure how to proceed.

Edit: actually got as far as

$= \frac{\sin(2\pi u)}{4\pi}\Big(\frac{1}{u}+\frac{1}{1-2u}-\frac{1}{1+2u}\Big)$ now, if I haven't made any mistakes...

Benjamin Lindqvist
  • 3,112

2 Answers2

1

The last line is wrong, it's supposed to be: $$\frac{\sin(2\pi u)}{2\pi}\Big(\frac{1}{u}-\frac{1}{1-2u}-\frac{1}{1+2u}\Big)$$ I'll work out the first term, the rest are similar: $$F(\omega)=\mathcal{F}_u\bigg[\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)\bigg]=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du$$

Keep in mind that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Hence: $$F(\omega)=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du=\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega-2\pi) u}du-\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega+2\pi) u}du$$

Using two facts $$\mathcal{F}_t[e^{i\omega_0t}\cdot f(t)]=F(\omega-\omega_0)$$ where $G(\omega)=\mathcal{F}_t[f(t)]$ and $$\mathcal{F}_t\bigg[\frac 1t\bigg]=-i\pi\text{sgn} (\omega)$$

You get $$F(\omega)=-\frac 1{4} \text{sgn}(\omega-2\pi)+\frac 1{4} \text{sgn}(2 \pi+\omega)$$

For the other parts, using the following property (very easy to prove): $$\mathcal{F}_t(f(t-t_0))=e^{-it_0\omega}F(\omega)$$ $$F_1(\omega)=\frac 12 e^{-i\omega/2}F(\omega)$$ $$F_2(\omega)=\frac 12 e^{i\omega/2}F(\omega)$$

The transform $G_u(\omega)$ of the function $f(u)$ is: \begin{align*} G_u(\omega)&=F(\omega)+F_1(\omega)+F_2(\omega)\\ &=(1+\cos(w/2)F(\omega)\\ &=2\cos^2(\omega/4)F(\omega) \end{align*}

So that $$\mathcal{F}_t(g(t))=G(\omega)=\frac T2 \cos^2(\frac{T\omega} 4) \left(\text{sgn}(2 \pi - T\omega) + \text{sgn}(2 \pi + T\omega)\right)$$

entrelac
  • 4,681
0

Note that we can write

$$\frac{\text{sinc}(t/T)\cos(\pi t/T)}{1-4t^2/T^2}=\frac{\sin(2\pi t/T)}{(2\pi t/T)(1-4t^2/T^2)}$$

Then, letting $t=(T/2)z$, we have

\begin{align} \int_{-\infty}^{\infty}\frac{\sin(2\pi t/T)e^{i\omega t}}{(2\pi t/T)(1-4t^2/T^2)}\,dt&=\frac{1}{2i}\frac{T}{2\pi}\left(\int_{-\infty}^{\infty}\frac{e^{i(\omega T/2+\pi)z}}{z(1-z)(1+z)}\,dz\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\int_{-\infty}^{\infty}\frac{e^{i(\omega T/2-\pi)z}}{z(1-z)(1+z)}\,dz\right) \tag 1 \end{align}

We will evaluate the right-hand side of $(1)$ using the residue theorem.

CASE 1: $\omega >2\pi/T$

For $\omega >2\pi/T$, we can close the contour of both integrals on the right-hand side of $(1)$ in the upper-half of the complex $z$ plane. Note, that the poles at $z=0$, and $z=\pm 1$ are on the real axis. Therefore, we have

$$\int_{-\infty}^{\infty}\frac{\sin(2\pi t/T)}{(2\pi t/T)(1-4t^2/T^2)}\,e^{i\omega t}\,dt=\pi i \text{Res}\left(\frac{T}{2\pi}\frac{e^{i(\omega Tz/2)}\sin (\pi z)}{z(1-z)(1+z)}\right)=0$$

CASE 2: $\omega <-2\pi/T$

Similarly, for $\omega <-2\pi/T$, we close the contours of both integrals on the right-hand side of $(1)$ in the lower-half $z$ plane and find

$$\int_{-\infty}^{\infty}\frac{\sin(2\pi t/T)}{(2\pi t/T)(1-4t^2/T^2)}\,e^{i\omega t}\,dt=0$$

CASE 3: $-2\pi/T<\omega<2\pi/T$

For the case $-2\pi/T<\omega<2\pi/T$, we close the first integral in the upper-half plane and the second integral in the lower-half plane and find

$$\begin{align} \int_{-\infty}^{\infty}\frac{\sin(2\pi t/T)}{(2\pi t/T)(1-4t^2/T^2)}\,e^{i\omega t}\,dt&=\pi i \text{Res}\left(-i\frac{T}{2\pi}\frac{e^{i(\omega Tz/2)}\cos (\pi z)}{z(1-z)(1+z)}\right)\\\\ &=\frac{T}{2}\left(1+\frac12 e^{i\omega T/2}+\frac12 e^{-i\omega T/2}\right) \\\\ &=\frac{T}{2}\left(1+\cos(\omega T/2)\right) \\\\ &=T\cos^2 (\omega T/4) \end{align}$$

Putting it all together yields

$$ \int_{-\infty}^{\infty}\frac{\sin(2\pi t/T)}{(2\pi t/T)(1-4t^2/T^2)}\,e^{i\omega t}\,dt= \begin{cases} T\cos^2 (\omega T/4)&,|\omega|<2\pi/T\\\\ 0&,|\omega|>2\pi/T \end{cases} $$

Mark Viola
  • 179,405