Generally, everyone knows that "the part of signal with high-slope is the high frequency part."

What is the definition or theory behind it?
How to prove this well-known thing?
perhaps, I miss something apparent.

  • $\begingroup$ Are you talking about the slope in time domain? If so you might be referring to the fact that the derivative of a sinusoidal signal is proportional to its frequency, thus high slope can either be achieved with high gain or high frequency. $\endgroup$
    – fibonatic
    Commented Mar 28, 2015 at 13:23
  • $\begingroup$ Yes, I am. I mean what you mean and I need the mathematical proof of that fact. don't know where should I start by. Any suggestion would be helpful. Thanks. $\endgroup$
    – pakornosky
    Commented Mar 28, 2015 at 15:11

1 Answer 1


The Fourier transform of the derivative of a signal $x(t)$ is given by

$$\dot{x}(t)\Longleftrightarrow j\omega X(\omega)\tag{1}$$

where $\dot{x}(t)$ denotes the derivative of $x(t)$, and $X(\omega)$ is the Fourier transform of $x(t)$. From (1) we have

$$\dot{x}(t)=\frac{j}{2\pi}\int_{-\infty}^{\infty}\omega X(\omega)e^{j\omega t}\,d\omega\tag{2}$$

If we now assume that $x(t)$ is band-limited with cut-off frequency $\omega_c$ (in radians), then (2) becomes

$$\dot{x}(t)=\frac{j}{2\pi}\int_{-\omega_c}^{\omega_c}\omega X(\omega)e^{j\omega t}\,d\omega\tag{3}$$

From (3) we can determine an upper bound for $|\dot{x}(t)|$:

$$|\dot{x}(t)|\le \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}|\omega| |X(\omega)|\,d\omega \le\frac{\omega_c}{2\pi}\int_{-\omega_c}^{\omega_c}|X(\omega)|\,d\omega\tag{4}$$

The upper bound (4) on the magnitude of the derivative of $x(t)$ increases with increasing $\omega_c$, so you can expect faster changes of $x(t)$ with increasing values of the cut-off frequency $\omega_c$.

  • $\begingroup$ Thanks @Matt. This is what I exactly want. I would like to ask one more question. This proof is conventional or you solve it by yourself? By the way, you have my thanks. :) $\endgroup$
    – pakornosky
    Commented Mar 29, 2015 at 14:44
  • $\begingroup$ @pakornosky: This is a standard way of deriving an upper bound for the magnitude of a signal, by writing it as an integral, bounding it by the integral over the magnitude of the integrand, and pulling out (part of) the integrand by multiplying by its maximum. I'm sure you can find this somewhere else, but I don't know any reference right now. $\endgroup$
    – Matt L.
    Commented Mar 29, 2015 at 15:07
  • $\begingroup$ really appreciate your answer and comments. Thx again. $\endgroup$
    – pakornosky
    Commented Mar 29, 2015 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.