I've met some problems when calculating the Fourier transform of $\cos(at+b)$. I want to use the shifting and scaling properties to solve this problem. First, when I look up in the book and some online resources, I saw a post about the Fourier transform of the signal $x(3t-6)$, where it tells me that $x(3t-6)$ is scaling $x(t)$ by $3$ and shift it by $2$, which I will get the answer that $$F\{x(3t-6)\} = \frac13 X(j\omega/3)exp(-j2\omega)$$ This indicates that the Fourier transform of $x(at+b)$ is $$\frac{1}{|a|}X(j\omega/a)\cdot \exp(j\omega b/a)$$

Thus, I apply the conclusion to the Fourier transform of $cos(at+b)$ is

$$\frac{1}{|a|}\exp(j\omega b/a)\,\pi\, [\delta(\omega/a−1)+\delta(\omega/a+1)]$$

How do I proceed to the right answer which is

$$\pi [\exp(jb)\delta(\omega-b)+\exp(-jb)\delta(\omega+b)]$$ ? I appreciate your help.

(I'm new here so I haven't figure out how to type the fractions or exponentials, sorry for the bad typing)


1 Answer 1


Using the shifting and scaling properties of the Fourier transform is a rather complicated way of computing the Fourier transform of $\cos(at+b)$. A more straightforward way is to realize that


From the Fourier transform pair

$$e^{jat}\Longleftrightarrow 2\pi\delta(\omega-a)\tag{2}$$

we immediately obtain

$$\cos(at+b)\Longleftrightarrow \pi\left[e^{jb}\delta(\omega-a)+e^{-jb}\delta(\omega+a)\right]\tag{3}$$

But having said that, your result is correct. You just need to simplify it appropriately. In order to do so, you need to know two things. First,


and, second,


for any $f(\omega)$ that is continuous at $\omega_0$.

From $(4)$ you get


And from $(5)$ you get

$$e^{j\omega b/a}\delta(\omega-a)=e^{jb}\delta(\omega-a)\tag{7}$$

Applying similar transformations to the second term in your result leads to the form given in $(3)$.


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