# Deducting Fourier Transform pair from existing pair

I need to determine Fourier Transform of $\delta(2\pi f) = \frac{1}{2 \pi} \delta (f)$ [Note: $\delta$ is Unit impulse function]

Answer is written in my book as follows :

\begin{align} F^{-1}[\delta(2\pi f)] & = \int_{-\infty}^\infty \delta(2\pi f)e^{j2\pi ft} \, df = \frac{1}{2\pi}\int_{-\infty}^\infty \delta(2\pi f)e^{j2\pi ft} \, d(2\pi f) \\[10pt] & =\frac{1}{2\pi} \end{align}

Therefore,

$$\frac{1}{2\pi}\Longleftrightarrow \delta(2 \pi f)$$

$$1 \Longleftrightarrow \delta(f)$$

Here, the left part is in time domain and the right part is in frequency domain. I understood inverse fourier integral part. But couldn't understand how we could deduct the last line from it's upper one. I know that $\int_{-\infty}^\infty \delta (f) e^{j2\pi ft} \, df =1$. I just need to know if there is any way of deducting further Fourier transform pair from an existing one by multiplication or division as it seems to be happened here.

• For functions, you have $$\mathcal{F}^{-1} \big(f(Cx)\big)(\xi) = \int f(Cx) e^{ix\xi} \,dx = [u=Cx]= \frac{1}{C}\int_{} f(u) e^{iu\xi/C} \, du = \frac{1}{C} \mathcal{F}^{-1} (f)(\xi/C).$$ A similar thing happens with distributions.
– MSDG
Aug 29, 2018 at 18:21
• Why is $\delta(2\pi f) = \tfrac 1{2\pi}\delta(f)$. The symbol $\delta$ here is a distribution that maps a function $f$ to a value, namely to $f(0)$. So, $\delta(2\pi f) = 2\pi f(0) = 2\pi\delta(f)$. Aug 29, 2018 at 18:29
• @amsmath The $f$ is not the function here, but the argument of the Fourier transform. The distribution usually written $\delta(ax)$ is indeed the same as $\frac1{|a|} \delta(x)$ (you want formal integration by parts to work correctly, and give the same result as for a sequence of functions converging to a delta). Aug 29, 2018 at 19:17
• @Kusma What is $\delta(ax)$ here? Aug 29, 2018 at 19:22
• $\delta(af)$ is defined (for $a>0$) via change of variables by $\int g(f) \delta(af) df= \int g(\frac s a) \delta(s) \frac 1a ds = \frac1a g(0)$. Aug 29, 2018 at 19:26

\begin{align} g_1(t)\Longleftrightarrow G_1(f) ~ ~~~~~and ~~~~~~ g_2(t)\Longleftrightarrow G_2(f) \end{align} then, \begin{align} \sum_{k}^{}a_kg_k(t) \Longleftrightarrow \sum_{k}^{}a_kG_k(f) \end{align} For any constants $\{a_k\}$ and signal $\{g_k(t)\}$.
If we take the signal $g_1(t)$ n times then corresponding fourier tansform $G_1(f)$ must be taken n times too.
\begin{align} ng_1(t) \Longleftrightarrow nG_1(f) \end{align} Correspondingly if, $$\frac{1}{2\pi}\Longleftrightarrow \delta(2 \pi f)$$ then using linearity property, (The following line was skipped in the book) $$1 \Longleftrightarrow 2\pi \delta(2 \pi f)$$ But according to question, $\delta(2\pi f) = \frac{1}{2 \pi} \delta (f)$ $$1 \Longleftrightarrow \delta(f)$$