# Is there an easy way to translate a Fourier transform table from angular frequency $\omega$ to Hertz $f$?

I have a table with transform operations, e.g. scaling:

$$\begin{equation} \tag{0} \label{0} x(at) \iff \frac{1}{a} X(\frac{\omega}{a}) \end{equation}$$

or frequency shifting:

$$\begin{equation}\tag{1} \label{1} x(t)e^{jw_ot} \iff X(\omega - \omega_o) \end{equation}$$

I also have a table with transform pairs, for example:

$$\begin{equation}\tag{2} \label{2} e^{-a|t|} \iff \frac{2a}{a^2+\omega^2} \end{equation}$$

Now, I would like to use these tables but with hertz $$f$$ instead. Note that there are many more operations and transform pairs in the tables than I have mentioned. I hope those can be translated similary as the examples I have given. Is there an easy way to do this?

For instance, I have tried to find the inverse transformation of $$X(f - f_o)$$ using equation $$\eqref{0}$$ and $$\eqref{1}$$

$$\begin{equation} X(f - f_o) =^? X(\frac{\omega - \omega_o}{2\pi}) \iff //eq. \eqref{0}// \iff 2\pi \cdot x(2\pi t)\cdot e^{j2\pi w_ot} = \end{equation}$$

$$\begin{equation}\tag{3} \label{3} = 2\pi \cdot x(2\pi t)\cdot e^{j4\pi^2 f_ot} \end{equation}$$

but equation $$\eqref{3}$$ gives $$\begin{equation}\tag{4} \label{4} \frac{1}{2}(X(f - f_o) + X(f + f_o)) \iff 2\pi \cdot x(2\pi t)\cdot \frac{e^{j4\pi^2 f_ot} + e^{-j4\pi^2 f_ot}}{2} \end{equation}$$

which simplifies to

$$\begin{equation}\tag{5} \label{5} \frac{1}{2}(X(f - f_o) + X(f + f_o)) \iff 2\pi \cdot x(2\pi t)\cdot \cos(j4\pi^2 f_ot) \end{equation}$$

which doesn't equal the correct expression

$$\begin{equation}\tag{6} \label{6} \frac{1}{2}(X(f - f_o) + X(f + f_o)) \iff x(t)\cdot \cos(j2\pi f_ot) \end{equation}$$.

Even if equation $$\eqref{5}$$ is correct, there is non-trivial calculation involved. I would like a faster way similar to "replace $$\omega$$ with $$2\pi f$$"

Thanks

## 2 Answers

Your confusion comes from the fact that you use $$X(\cdot)$$ for denoting both functions, the function of $$\omega$$ and the function of $$f$$, but they are really two different functions, because

$$X(\omega)=\hat{X}(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt,\quad\omega=2\pi f\tag{1}$$

That's why in all the correspondences you mentioned you can just replace $$\omega$$ by $$f$$ inside the function argument (and note that the two frequency domain functions are actually different).

Note, however, that outside the function you have to replace $$\omega$$ by $$2\pi f$$, as for instance in the following correspondence:

$$\frac{d^nx(t)}{dt^n}\Longleftrightarrow (j\omega)^nX(\omega)=(j2\pi f)^n\hat{X}(f)\tag{2}$$

There is one more complication that arises due to the transform as defined by $$(1)$$ being unitary if $$\hat{X}(f)$$ is used, but being non-unitary when using $$X(\omega)$$. This can be seen by the factor $$1/(2\pi)$$ for the inverse transform when using $$X(\omega)$$. This implies, for instance, that you need a factor $$1/(2\pi)$$ in the following correspondence when using $$X(\omega)$$, but not when you use $$\hat{X}(f)$$:

$$x(t)y(t)\Longleftarrow \frac{1}{2\pi}(X\star Y)(\omega)=(\hat{X}\star\hat{Y})(f)\tag{3}$$

Also be careful when a Dirac delta impulse is involved, because clearly $$\delta(\omega)\neq\delta(f)$$. The correct correspondence is

$$\delta(f)=2\pi\delta(\omega)\tag{2}$$

Consequently,

$$\mathcal{F}\{1\}=2\pi\delta(\omega)=\delta(f)\tag{3}$$

and

$$\mathcal{F}\{e^{j\omega_0t}\}=\mathcal{F}\{e^{j2\pi f_0t}\}=2\pi \delta(\omega-\omega_0)=\delta(f-f_0)\tag{4}$$

from which the Fourier transforms of sine and cosine follow easily.

• i would have defined the default convention the other way (because of the duality property) using $$\hat{X}(\omega)=X(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt,\quad\omega=2\pi f\tag{1}$$ – robert bristow-johnson Dec 12 '20 at 7:40

The properties that you cite are for actually changing the frequency components, by stretching/compressing or shifting, not for just changing the units on the axis. When you do the conversion, $$\omega=2\pi f$$, you are doing a mapping from radians per second to Hertz but the frequencies in the signal are not changing. When you do the frequency stretching/compressing or shifting, you are truly changing the frequency components of the signal. The best comparison I can think of is to consider loading up barbell with 110 pounds, but then a friend comes over who insists on using kilograms as a measure instead so you take off the 100 pounds and put on 55 kilograms (55 kilograms = 110 pounds). You haven't changed the weights, only the units that it is measured in (this is what is happening when changing from $$\omega \rightarrow f$$). However, imagine if you actually had loaded with 55 pounds, now you've truly changed something! You've scaled the amount of weight (this is what is happening when changing from $$\omega \rightarrow \frac{\omega}{a}$$). To more directly answer your question: If you call the Fourier transform of $$x(t)$$ using radians as $$X_{\omega}(\omega)=\int x(t) e^{-j\omega t} dt$$ and the Fourier transform using $$\text{Hz}$$ as $$X_f(f)=\int x(t) e^{-j 2\pi ft} dt$$, then your question boils down to showing that $$X_{\omega}(\omega)=X_f(f)$$ for each property. First replace $$f$$ with $$\frac{\omega}{2\pi}$$, then simplify.

\begin{align} X_{\omega}(\omega-\omega_0) &= X_f(f-f_0) \\ &= X_f(\frac{\omega-\omega_0}{2\pi}) \\ &= \int x(t) e^{-j2\pi \big(\frac{\omega-\omega_0}{2\pi}\big) t} dt \\ &= \int x(t) e^{-j (\omega-\omega_0) t} dt \\ &= X_{\omega}(\omega-\omega_0) \end{align}

And you can do a similar thing for the scaling property.

\begin{align} \frac{1}{a}X_{\omega}\bigg(\frac{\omega}{a}\bigg) &= \frac{1}{a}X_f\bigg(\frac{f}{a}\bigg) \\ &= \frac{1}{a}X_f\bigg(\frac{\omega}{2\pi a}\bigg) \\ &= \frac{1}{a}\int x(t) e^{-j2\pi \big(\frac{\omega}{2\pi a}\big) t} dt \\ &= \frac{1}{a}\int x(t) e^{-j \big(\frac{\omega}{a}\big) t} dt \\ &= \frac{1}{a}X_{\omega}\bigg(\frac{\omega}{a}\bigg) \end{align}