0
\$\begingroup\$

I was studying the Laplace transform and the Fourier transform, and I can't figure out if there is some intuitive way of finding out how a signal looks in the time domain if we know its frequency domain. For example, are there some rules like:

If there is a frequency whose amplitude is approaching infinity then.... (like when there are poles on the imaginary axis in the s plane, like this: enter image description here

If the frequency graph is decaying towards 0 then....

Because from what I'm reading about the Laplace transform, the position of poles and zeros gives me a better understanding of how the frequency response looks like, but what do I care what it looks like if I don't know what the time domain will look like?

\$\endgroup\$
  • 1
    \$\begingroup\$ The frequency domain is often a far better way of evaluating signals (particularly in communications systems where the eye diagram is not necessarily going to give all the information you need). Given a frequency domain response plot I can certainly predict the eye diagram. That said, these are two different ways of viewing a signal, each with its own strengths and weaknesses. \$\endgroup\$ – Peter Smith Aug 10 at 15:59
  • \$\begingroup\$ Since the data-eye is what counts, that becomes a 3rd view into the proper use of signal energy. The masks imposed on GMSK ( for proper GSM cellphone energy) are crafted to set adjacent-channel energy bleedover for low InterSymbolInterference. To visualize what -40dBc versus -30dBc will cause in the dataeye ---- is nie impossible. \$\endgroup\$ – analogsystemsrf Aug 10 at 16:27
  • \$\begingroup\$ A frequency response can often give information about resonances that can't be seen on a time domain (e.g.step) response. It's a sinusoidal steady state measurement containing information on time domain steady state and transient characteristics. \$\endgroup\$ – Chu Aug 10 at 16:28
  • 2
    \$\begingroup\$ Your picture of the bode plot makes no sense to what a signal will look like because that picture is a spectral plot of the transfer function of the “filter” and not the output signal hence, it has no relevance in this question. \$\endgroup\$ – Andy aka Aug 10 at 19:13
  • \$\begingroup\$ @analogsystemsrf Did I miss something in the question or why did you start going into GSM and data-eye? I can't imagine the data-eye is "what counts" when designing filters for analog signals etc. \$\endgroup\$ – pipe Aug 10 at 23:19
0
\$\begingroup\$

One of the best properties of both Fourier and Laplace transforms is that, for a signal \$x\$ resulting of a convolution of two signals in time \$a \star b\$, the transform of \$x\$ (let's use Laplace) is equal to the multiplication of the transforms for each signal. So, \$L\{x\}= L\{a\} \cdot L\{b\}\$. That way, You can calculate the TF in the frequency domain then the transform for the input signal, the multiplication of the two will be the resulting signal (though representing it in time might be difficult, you usually rearrange the terms in the equation until you find one or more know transform terms).

For example, you have the TF, \$ B(s)=\frac{s+2}{s+1}\$ and step signal \$ A(s)=\frac{1}{s}\$, the result of both signals will be \$ X(s)=\frac{s+2}{s(s+1)}\$.

Which may seem to not give any insight into the time response, but it is known that: \$L\{e^{-at}\}=\frac{s}{s+a}\$, so we transform the equation for \$ X(s)\$, using the Heaviside cover-up method, into \$ X(s)=\frac{-1}{s+1}+\frac{2}{s}\$. Which is the sum of two of the exponential terms mentioned above, so:

\$x(t) = -e^{-1t}+2\$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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