# Linearity and Time Invariance

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

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)

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