# LTI, causal, discrete time system output

Consider a discrete time LTI causal system $$S: y = S(u)$$, with its impulse response $$h:{Z} \rightarrow R:h(n)=3^{n+1}{H(n)}$$ with $$H$$ the Heaviside function. We know the values of input:

$$u(0) = 2$$ $$u(1) = 1$$ $$u(2) = 1$$ $$u(3) = 1$$ $$u(4) = 2$$

Also the $$y(2) = 66$$. I'm trying to find the value of $$y(3)$$.

I may say, for a LTI, causal system the output is the convolution of the input with the impulse response:

$$y(n) = \sum^\infty_{k=0}3^{k+1}u(n-k)$$

Then

$$y(3) = 3u(3)+9u(2)+27u(1)+81u(0) = 201$$

Which is the correct answer! But I don't understand, why should I stop at $$u(0)$$ or why the negative values of the input $$u(n), n< 0$$ are zero here ?

Alternative solution:

recursion gives us this form:

$$y(n) = 3y(n-1)+3u(n)$$

• I think that usually for such questions, the input for negative time values is defined as zero. – Gideon Genadi Kogan Dec 15 '19 at 7:51

Unless explicitly stated, you shall not assume that the input is zero for $$n<0$$, hence the convolution sum should be

$$y(n) = \sum^\infty_{k=-\infty}3^{k+1}u(n-k)=\sum^\infty_{k=0}3^{k+1}u(n-k).$$

In this situation, without making any assumption on $$u[n]$$ for $$n<0$$ you can still compute $$y[n]$$ for $$n \geq 0$$ based on the recursive relation as you have outlined :

$$y(n) = 3y(n-1)+3u(n) \implies y[3] = 3y[2] + 3u[3] = 201$$

What can be said about $$u[n]$$ for $$n<0$$ ? Actually nothing but, given that $$y[2] = 66$$ we may try the folowing :

\begin{align} y[2] &= \sum^2_{k=0}3^{k+1}u(2-k) + \sum^\infty_{k=3}3^{k+1}u(2-k) \\ &= 66 + \sum^\infty_{k=3}3^{k+1}u(2-k) \\ 66 &= 66 + \sum^\infty_{k=3}3^{k+1}u(2-k) \\ 0 &= \sum^\infty_{k=3}3^{k+1}u(2-k) \end{align}

That the sum in the parenthesis equals zero, has a trivial solution of $$x[n] = 0$$ for $$n <0$$.

However, we may also ask whether another, nonzero, solution is also possible that the sum be zero? Consider, for example, the input $$u[n] = (-1)^{2-n} ~~3^{n-3}$$ for $$n<0$$ which yields for the sum

$$\sum^\infty_{k=3}3^{k+1}u(2-k) = 1-1+1-1+1-...$$

This is an alternating sequence and it does not converge. However, if it's possible to restrict the length of the input to be even for $$n<0$$, then the sum will be zero too...

Nevertheless, in order to uniquely specify $$u[n]$$ for $$n<0$$ a single sum is not enough. We must know values of output $$y[n]$$ for all $$n<0$$ for that. Hence you cannot conlcude that $$x[n]=0$$ for $$n<0$$ under the given situation.