Question The conditions for the signal generated from the Q0.0 output of an S7-1200 class PLC whose input-output connections are shown in the figure on the right are given as follows:  The system will be activated when the S1 button is pressed, and will be deactivated when the S0 button is pressed.  When the system is active and the piece is moving to the right, the transition time (t1) of the piece will be determined by PS1 sensor.  Q0.0 = 1 immediately after t1 time the part is detected by PS2 sensor.  As soon as the PS1 sensor detects a new part, Q0.0 = 0. PLC main program block (OB1) average cycle time is 35ms, t1 time varies between 25ms and 500ms. Explain the PLC program structure that meets these conditions, the related programs and the operation by indicating the reasons. so PS1 $1 I PS2 M 1M .0.1 2 .3 S7-1200 CPU 3L+ 3M .0 .1 .2 3 + 9 우 07 8 PS1(10.0) PS2(10.1) Q0.0 ti ti

(j x+)_{\text {now }}=j 0.1 \times \frac{100}{100} \times\left(\frac{11}{11}\right)^{2}=j 0.1 \mathrm{p4}Total lemith of T L=0.5 \mathrm{2} / \mathrm{km} \times 100 \mathrm{~km}=50 \OmegaZ_{p u}(\text { new })=Z_{p u(0 i d)} \times \frac{S_{b}(\text { now })}{S_{b}(o \& d)}(352)(This is used when rated voltage is base voltage, system should Operate at rated valul]\text { i). } \begin{aligned}I_{f} & =\frac{1 \angle 0^{\circ}}{j 0.4}=-j 2.5 \mathrm{Pu} \\I_{2} & =\frac{1 \angle 0^{\circ}}{j 0.5}=-12 \mathrm{pu}\end{aligned}\begin{array}{l} \rightarrow\left(I_{f}\right)_{K A}=2.5 \times \frac{\text { Sbase }}{\sqrt{3} V_{B M S E}} \quad I_{F_{3}}=\frac{1 \angle 0^{\circ}}{j 0.28}=-11.28 P 4 \\=2.5 \times \frac{100 \mathrm{MUA}}{\sqrt{3} \times 11 \mathrm{KV}}=13.12 \mathrm{kA} . \\\left(I_{f}\right)_{2 K A}=2 \times \frac{100 \mathrm{MMA}}{\sqrt{3 \times 132 \mathrm{KV}}=0.874 \mathrm{KA}} \\\left(I_{f}\right)_{2 K A}=2 \times 1.28 \times \frac{100 \mathrm{MVA}}{\sqrt{3} \times 13.2 \mathrm{kA}}=0 . \sqrt{6} \mathrm{kA} \\\left.V_{1}=(-j 1.28) \times 1 j 0.38\right) \times 11 \mathrm{KV}=5.28 \mathrm{kV}\end{array} ...