correct answer for Fourier Transform of $\cos(at+b)$

Question

I'm trying to find out the (continuous time) Fourier transform of $\cos(at+b)$. Now I've managed to derive some properties shown in the book:

if $\mathcal F\{x(t)\} = X(jw) = \int_{-\infty}^\infty x(t)e^{-jwt}dt$ , then

$\mathcal F\{x(at)\} = \frac{1}{|a|}X(\frac{jw}{a})$ and $\mathcal F\{x(t-t_0)\} = e^{-jwt_{0}}X(jw)$

Also, $\mathcal F\{cos(t)\} = \pi [\delta(w-1) + \delta(w+1)]$ was derived pretty easily.

Now I came up with three ways to find $F\{cos(at+b)\}$:

1. Find $\mathcal F\{cos(at)\}$, then time-shift by $b/a$.

2. Find $\mathcal F\{cos(t+b)\}$ and scale $t$ by $a$.

3. Use the fact that $cos(at+b) = cos(at)cos(b) - sin(at)sin(b) = \frac{e^{jb}+e^{-jb}}{2}cos(at) - \frac{e^{jb}-e^{-jb}}{2}sin(at)$

   and use the linearity of Fourier transform.

Let's assume $a$ is positive.

Method #1:

$\mathcal F\{cos(t)\} = \pi [\delta(w-1) + \delta(w+1)]$

$\mathcal F\{cos(at)\} = \pi\frac{1}{a} [\delta(\frac{w}{a}-1) + \delta(\frac{w}{a}+1)] = \pi [\delta(w-a) + \delta(w+a)]$

Thus, $\mathcal F\{cos(a(t+\frac{b}{a}))\} = e^{j\frac{b}{a}}\mathcal F\{cos(at)\} = \pi e^{j\frac{b}{a}}[\delta(w-a) + \delta(w+a)]$

Method #2:

$\mathcal F\{cos(t)\} = \pi [\delta(w-1) + \delta(w+1)]$

$\mathcal F\{cos(t+b)\} = \pi e^{jb}[\delta(w-1) + \delta(w+1)]$

$\mathcal F\{cos(at+b)\} = \pi e^{jb}\frac{1}{a}[\delta(\frac{w}{a}-1) + \delta(\frac{w}{a}+1)] = \pi e^{jb}[\delta(w-a) + \delta(w+a)]$

Method #3:

$\mathcal F\{cos(at+b)\} = \mathcal F\{cos(at)cos(b) - sin(at)sin(b)\} = \mathcal F\{\frac{e^{jb}+e^{-jb}}{2}cos(at) - \frac{e^{jb}-e^{-jb}}{2j}sin(at)\} = \frac{e^{jb}+e^{-jb}}{2}\mathcal F\{cos(at)\} - \frac{e^{jb}-e^{-jb}}{2j}\mathcal F\{sin(at)\}$ by linearity of Fourier transforms.

Thus, $\mathcal F\{cos(at+b)\} = \frac{e^{jb}+e^{-jb}}{2}\pi[\delta(w-a) + \delta(w+a)] - \frac{e^{jb}-e^{-jb}}{2j}\frac{\pi}{j}[\delta(w-a) - \delta(w+a)] = \pi[e^{jb}\delta(w-a) + e^{-jb}\delta(w+a)]$

I see no errors in any of the three methods, but they all yield different results, so I'm really confused.

Answer 1

You seem to forget to put $\omega$ in the exponents. For example, in the first and second cases you should have

Method $1$

$$\mathcal{F}\left\{\cos\left(t\right)\right\}=\pi\left[\delta\left(\omega-1\right)+\delta\left(\omega+1\right)\right]$$

$$\mathcal{F}\left\{\cos\left(at\right)\right\}=\pi\frac{1}{a}\left[\delta\left(\frac{\omega}{a}-1\right)+\delta\left(\frac{\omega}{a}+1\right)\right]=\pi\left[\delta\left(\omega-a\right)+\delta\left(\omega+a\right)\right]$$

$$\mathcal{F}\left\{\cos\left(a\left(t+\frac{b}{a}\right)\right)\right\}=e^{j\frac{b}{a}\color{red}{\omega}}\mathcal{F}\left\{\cos\left(at\right)\right\}=\pi e^{j\frac{b}{a}\color{red}{\omega}}[\delta\left(\omega-a\right)+\delta\left(\omega+a\right)]$$

Method $2$

$$\mathcal{F}\left\{\cos\left(t\right)\right\}=\pi\left[\delta\left(\omega-1\right)+\delta\left(\omega+1\right)\right]$$

$$\mathcal{F}\left\{\cos\left(t+b\right)\right\}=\pi e^{jb\color{red}{\omega}}\left[\delta\left(\omega-1\right)+\delta\left(\omega+1\right)\right]$$

$$\mathcal{F}\left\{\cos\left(at+b\right)\right\}=\pi e^{jb\color{red}{\frac{\omega}{a}}}\frac{1}{a}\left[\delta\left(\frac{\omega}{a}-1\right)+\delta\left(\frac{\omega}{a}+1\right)\right]=\pi e^{j\frac{b}{\color{red}{a}}\color{red}{\omega}}\left[\delta\left(\omega-a\right)+\delta\left(\omega+a\right)\right]$$

Method $3$

$$\mathcal{F}\left\{\cos\left(at+b\right)\right\}=\pi\left[e^{jb}\delta\left(\omega-a\right)+e^{-jb}\delta\left(\omega+a\right)\right]=\pi e^{j\frac{b}{\color{red}{a}}\color{red}{\omega}}\left[\delta\left(\omega-a\right)+\delta\left(\omega+a\right)\right]$$

by using the fact that $f\left(x\right)\delta\left(x-a\right)=f\left(a\right)\delta\left(x-a\right)$.