Linearity and Time Invariance

Discussion in 'Math' started by RdAdr, Sep 14, 2015.

  1. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    Something is bothering me.

    So, if I input the signal u(n) to a discrete-time sytem I can calculate the output like this:

    s(n) = k(1-a^(n+1))*u(n). So, this is the model of the system when I apply the step input.

    Now, I want to input the signal x(n) = u(n)-u(n-5).

    So, y(n) = s(n) - s(n-5) = k(1-a^(n+1))*u(n) - k(1-a^(n-5+1))*u(n-5)

    So, I used linearity and time invariance to arrive to this result.

    But, if I take this route instead:

    y(n) = k(1-a^(n+1))x(n) = k(1-a^(n+1))(u(n)-u(n-5)) = k(1-a^(n+1))*u(n) - k(1-a^(n+1))*u(n-5)

    Now it is wrong. The last term is k(1-a^(n+1))*u(n-5) instead of k(1-a^(n-5+1))*u(n-5).

    So, why can't I write like this:

    y(n) = k(1-a^(n+1))*x(n)

    and find the answer?

    What is happening?

    I think it has something to do with operators, transformations. I don't know exactly. I've heard of these terms but I did not study linear algebra very hard. Maybe I try to apply the function to the input x(n), but I must apply the operator or transformation that characterizes the system.

    If someone could point me into the right direction with this, what to search further on the internet, I will appreciate.
     
    Last edited: Sep 14, 2015
  2. RdAdr

    Thread Starter Member

    May 19, 2013
    214
    1
    I figured it out.
     
  3. amilton542

    Active Member

    Nov 13, 2010
    494
    64
    Why do z transforms constitute a linear time invariant system?
     
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