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Electr. Engineering - Digital Sig. Processing

  1. Sep 4, 2007 #1
    Determine if the CT systems are 1) casual or uncasual 2) memory or memoryless.

    Definitions:

    Casual: If for any time t1, the output response y(t1) at time t1 resulting from input x(t) does not depend on the values of the input x(t) for t > t1.
    Memory: If the output at time t1 depends in general on the past values of the input x(t) for some range of values of t up to t=t1.


    [tex]x(t)[/tex] is random input and [tex] y(t)[/tex] is the output of [tex]x(t)[/tex]

    For Eq1:

    [tex]y(t) = |x(t)| = \left\{ \begin{array}{l}
    x(t)\; \mathrm{if}\, x(t) \geq 0 \\
    -x(t)\; \mathrm{if}\, x(t) < 0
    \end{array}\right.
    [/tex]

    I said this system is CASUAL and MEMORYLESS.
    • Casual - because at time t, y(t) will depend only t from the input function x(t), not some other arbitrary t value.
    • Memoryless - because the outputs at time t do not depend on previous inputs.

    For Eq 2:


    [tex]y(t) = \int_0^t\lambda x(\lambda)d\lambda[/tex]

    I said this system is CASUAL and has MEMORY.
    • Casual - because at time t, it doesn't really depend on the future. It only depends on the past, so I'm guessing casual. *This I'm not too sure about*
    • Memory - because the outputs at time t do depend on previous inputs since youre taking the integral from 0 to time t. *I'm almost sure about this one*
     
    Last edited: Sep 4, 2007
  2. jcsd
  3. Sep 5, 2007 #2

    learningphysics

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    Homework Helper

    Both your answers look good to me. By the way, it's "causal" not "casual". :wink:
     
  4. Sep 5, 2007 #3
    Hahha, I just realized that. Wow.

    Thanks tho.
     
  5. Sep 5, 2007 #4
    Now, Eq1 is obviously linear, but when I graph eq2, it seems to be nonlinear... does that make sense?
     
    Last edited: Sep 5, 2007
  6. Sep 5, 2007 #5

    chroot

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    Staff Emeritus
    Science Advisor
    Gold Member

    Yes.

    - Warren
     
  7. Sep 5, 2007 #6

    is it because when you take derivatives and integrals, the terms will become nonlinear

    also, in the one DefualtName posted, for equation 2, would that be a time varying or time invariant one. i would say varying because the actual output will be different from the input
     
    Last edited: Sep 5, 2007
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