So for n less than 0, so we accumulate no terms in the sum.
In particular, if we first consider what will turn out to be the simple case namely, n less than 0 for n less than 0, this unit step is 0 for k greater than n.
With n less than 0, that means that this unit step never is non-zero for k positive.
Second, the impulse response is 0 for n less than 0. That implies that the system is causal.
Note that the calculation used to predict bound nuclei in this chart is limited to those with N less than 160.
Here is the unit step u[n-k], and for n less than 0, so that this point comes before this point, the product of these two is equal to 0. That means there is no overlap between these two terms, and so it says that y[n], the output, is 0 for n less than 0. Well, that was an easy one.
That's for n greater than 0. We had found out previously there it was 0 for n less than 0. Finally, if we were to plot this, what we would get is the graph that I indicate here.
In particular, a necessary and sufficient condition for causality in the case of linear time-invariant systems is that the impulse response be 0, for t less than 0 in the continuous-time case, or for n less than 0 in the discrete-time case.
So it's 0 for n less than or equal to 0. And clearly, if we subtract the delayed step from the original unit step, everything subtracts out except at n = 0, at which point the difference is equal to 1.
Crisnatol had no effect at k less than 30 (n less than or equal to 1); was growth inhibitory at k = 30-1000 (n = 1), cytostatic at 1500, and cytotoxic at k greater than 2000 microMn-h (n = 2).
