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.

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(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} ...