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 $2\pi f$"



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





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

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

  • $\begingroup$ i would have defined the default convention the other way (because of the duality property) using $$\hat{X}(\omega)=X(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt,\quad\omega=2\pi f\tag{1}$$ $\endgroup$ – 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.

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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 $\omega \rightarrow \frac{\omega}{a}$).

enter image description here

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.

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


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