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Using the count() method:
This following loop creates a unit ramp input : for t= 0:0.1:5 if(t<=1) y(count) = t; end if(t>1) y(count) = 1; end Rz(count) = tf(y(count)*T,[1 -2 1],T); Rs(count) = d2c(Rz(count)); end % T = 0.005; this is for a discrete system
yes u count by 1, 2, and three
That depends on how fast you can count.
if you count from one to one hundred how many ones will there be?