Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

(signal processing)Time domain vs Frequency domain, totally confused!

  1. Sep 7, 2010 #1
    Hey guys,
    I am totally lost in the idea why we have to change function in time domain to frequency domain to ease with our calculations.
    Let me break my question into serveral smaller ones.

    1)What is the physical implication of transferring a signal to frequency domain?
    (I know that in frequency domain you have the information of the gain and phase)

    2)For example, in time domain a step function x(t) in piecewise, after transformation, in freq. domain it will become 1/s. It doesn't make sense to me that 1/s implies as frequency increases the gain will decrease. By looking at the time domain, it seems not to convey with such kind of information regarding the responses of other frequencies. Seriously, what does 1/s really mean? how does it relate to the time domain function?

    3) Why determining the transfer function of an open loop can help to know the stability of the closed loop?

    Please help.
    I am totally lost.
  2. jcsd
  3. Sep 7, 2010 #2
    Though I dont know much I can tell what I know about frequency domain calculations
    1) It is just a mathematical technique and has no physical implications. The signal whether represented in time or frequency domain has no change in itself and only the way it is viewed is different. So both are one and same on physical level.
    2) It shows that step functions contains all frequencies and (1/s) shows that higher frequencies contribution to step function is less compared to low frequencies(the magnitude of higher frequencies present in a step function is lesser than that of the lower frequencies). Meaning of 1/s is the magnitude of frequency component (in step function) is inversely proportional to the frequency.

    Hope this helps you a little bit
  4. Sep 7, 2010 #3
    Do you mean that y(t),for example, is composed of frequency components?( y(t)= Sum(1/s1+1/s2+1/s3....)?
  5. Sep 7, 2010 #4
    There are two main reasons for changing from the time domain to the frequency domain.
    1) It allows easy analysis of the frequency response of the system (if it's a filter, say)
    2) Calculations are often required when passing an input signal through and LTI system. This is a convolution in the time domain. A convolution in the time domain is the same as multiplication in the frequency domain and vice versa. Obviously multiplication is easier to do than a convolution.

    n.karthik is correct regarding step input/physical differences.

    With regards to stability, this comes down to your understanding of Nyquist plots and the Nyquist criterion. This is a means of checking closed loop stability given the open loop transfer function. A quick google search will provide you with more in depth info on this. Also, the poles of the transfer function need to be in the negative plane.
  6. Sep 8, 2010 #5
    No, Y(s)=1/s
    If you want y(t) you have to take inverse transformation of Y(s)
  7. Sep 13, 2010 #6
    ddarvil is right on. primarily its used to make life easier. Circuit analysis comes down to differential equations (inductors and caps are functions of varying parameters) and its easier to analyse all that in a different domain. Algebra is easy.

    In todays world there are tons of programs and stuff so it really doesnt even matter. (granted you need to know what you're doing and why you're doing it)
  8. Feb 13, 2012 #7
    It pays to send frequency modulated signal over the air (FM over AM) as it is much less vulnerable to distortion.
  9. Feb 13, 2012 #8
    because zeros of OLTF are same as poles of CLTF
    so if there are any zeros of CLTF it right side of s plane that means that poles of CLTF are on right side of s plane so CL system is unstable.
  10. Feb 13, 2012 #9

    jim hardy

    User Avatar
    Science Advisor
    Gold Member


    i went through school without ever forming a physical understanding of that Laplace transform. I accepted it was for the intelligent people to understand but for me to just use.

    for now just take it as a tool that cracks tough caluculs problems into manageable algebra problems.

    With repetition you will become fluent in its use.
    you will get to the point you can build a circuit just from looking at the desired transfer function.

    If you figure it out in simple terms please post.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook