Sign in

username:

password:



Not a member?

Search Online Books



Search tips

Free Online Books

Sponsor

TI's Lowest Power DSPs: TMS320VC5505 and TMS320VC5504 Industry's best combination of standby and active power for longer battery life
Click Here!

Chapters

See Also

Embedded Systems
Chapter Contents:

Search Mathematics of the DFT

  

Book Index | Global Index


Would you like to be notified by email when Julius Orion Smith III publishes a new entry into his blog?

  

Differentiation Theorem

Let $ x(t)$ denote a function differentiable for all $ t$ such that $ x(\pm\infty)=0$ and the Fourier Transforms (FT) of both $ x(t)$ and $ x^\prime(t)$ exist, where $ x^\prime(t)$ denotes the time derivative of $ x(t)$. Then we have

$\displaystyle \zbox {x^\prime(t) \;\longleftrightarrow\;j\omega X(\omega)}
$

where $ X(\omega)$ denotes the Fourier transform of $ x(t)$. In operator notation:

$\displaystyle \zbox {\hbox{\sc FT}_{\omega}(x^\prime) = j\omega X(\omega)}
$



Proof: This follows immediately from integration by parts:

\begin{eqnarray*}
\hbox{\sc FT}_{\omega}(x^\prime)
&\isdef & \int_{-\infty}^\in...
...\infty x(t) (-j\omega)e^{-j\omega t} dt\\
&=& j\omega X(\omega)
\end{eqnarray*}

since $ x(\pm\infty)=0$.

The differentiation theorem is implicitly used in §E.6 to show that audio signals are perceptually equivalent to bandlimited signals which are infinitely differentiable for all time.


Order a Hardcopy of Mathematics of the DFT

Previous: Selected Continuous-Time Fourier Theorems
Next: Scaling Theorem

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


Comments


No comments yet for this page


Add a Comment
You need to login before you can post a comment (best way to prevent spam). ( Not a member? )