# Confusion regarding overshoot?

As shown underlined in attached photo Overshoot is mentioned as 9% I am confused in understanding it It is 9% of error or size of jump(1)?

It is the error with respect to size of jump. It has nothing to do with error at discontinuity. As per the explanation given, suppose the perfect square is having values $$0$$ and $$1$$. Just after the discontinuity of a rise time, the perfect square would take value $$1$$, while the reconstructed wave would take the value $$1.09$$ at the peak of overshoot. Here the size of discontinuity is difference between perfect low value $$0$$ and perfect high value $$1$$ of the square wave. In general if it were $$A$$ and $$B$$, the overshoot percentage would have been $$\frac{(Ovs_{peak} - (B-A))}{(B-A)} \times 100$$
The error of $$0.5$$ in the error graph is not related to overshoot. For perfect square wave of $$0$$ and $$1$$, at discontinuity it is either $$0$$ or $$1$$, but the reconstructed wave will be midway $$0.5$$ ($$|0.5−0|$$ or $$|0.5−1|$$). So max error will be at discontinuity and its absolute value will be $$0.5$$.
• The error of 0.5 in the error graph is not related to overshoot. For perfect square wave of $0$ and $1$, at discontinuity it is either $0$ or $1$, but the reconstructed wave will be midway $0.5$ ($|0.5-0|$ or $|0.5-1|$). So max error will be at discontinuity and its absolute value will be $0.5$. In summary, max error is measured at discontinuity, while overshoot is measured with respect to how much it exceeds the step size. – jithin Apr 4 '20 at 9:58