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, S0 It will be deactivated when the button is pressed.  When the system is active and the part is moving to the right The transition time (t1) of the piece will be determined with the 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 will be. PLC main program block (OB1) average cycle time 35ms, The t1 time ranges from 25ms to 500ms. Explain the PLC program structure that meets these conditions, the related programs and the operation by indicating the reasons. SO PS1 I PS2 هابية M 1M . .1 .2 S7-1200 CPU 3L+ 3M .0 .1 .2 بية |8 20 PS1(10.0) PS2(10.1) Q0.0 ti ti

Answer : An::-\Rightarrow V_{s}-V_{f}= iffrential signalN_{s}+v_{f}= cannot be zeroV_{G c} is only valid for Negative.feedback systemInverting Amplifier:-\begin{array}{c}\Rightarrow R_{i n}=\frac{V_{s}}{I_{i n}}=\frac{V_{i n}}{\operatorname{Iin}_{i n} \rightarrow 0} \\R_{i n}=2 M \Omega\left(I_{c}-I_{e 1}\right) \\\left.R_{\text {in }}=\infty \text { (ideal }\right)\end{array}tig: Negetive feedback amplifierIdcal op-Amp:-\begin{array}{ll}A_{O_{L}}=\infty & V_{G C}=f\left(A_{O L}\right) \\I_{\text {in }}=\infty & I_{\text {in }}=f\left(R_{\text {in }}\right)\end{array}Apply K_{C} inverting terminal\begin{array}{c}\Rightarrow \frac{0-V_{S}}{R_{1}}+\frac{0-V_{\text {out }}}{R_{2}}+0=0 \\\Rightarrow \frac{-V_{S}}{R_{1}}+\frac{V_{\text {out }}}{R_{2}}\end{array}It Deceaser the gain \Rightarrow ctabify \uparrowA_{O_{L}}=\frac{V_{\text {out }}}{\left(V_{d}\right)=0} \infty \quad \infty \quad \frac{V_{\text {out }}}{V_{S}}=A c_{2}=\frac{-R_{2}}{R_{1}}internal gain lopenclosed loop lgain total loop gaingain | externalgainApplying K_{C L} at inverting terminal.\begin{array}{l}\frac{V_{I}-V_{C}}{10}+\frac{V_{I}-V_{\text {out }}}{100} \neq 0=0 \\V_{I}=\frac{10}{11} V_{S}+\frac{1}{11} V_{\text {out }} \\V_{\text {out }}=-10\left[\frac{10}{11} V_{S}+\frac{1}{11} V_{\text {out }}\right] \\\frac{V_{\text {out }}}{V_{S}}=-\frac{100}{21}=-4.76\end{array}\begin{array}{l}V_{G C} \text { Not Valid }\left(V_{N I} \neq V_{S}\right) \quad \quad A C_{L}=-4.76 \\\frac{V_{\text {out }}}{V_{S}}=? \\\text { Vout }=A_{O L}\left(V_{N I}-V_{I}\right) \\V_{\text {out }}=10\left(-V_{I}\right) \longrightarrow(1)\end{array}Non- inverting Amplifier:-Applying K C_{L} at inverting terminal\begin{array}{l}\frac{V_{S}-0}{R_{1}}+\frac{V_{S}-V_{\text {out }}}{R_{2}}=0 \\V_{S}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)=\frac{V_{\text {out }}}{R_{2}} \\A_{C_{L}} \frac{V_{\text {out }}}{V_{S}}=\frac{1+R_{2}}{R_{1}}\end{array}     .................................................................................... I provided you complete and correct answers for your question.So, please encourage me by giving an Upvote. If u have any doubt or if u want anything else, please drop a comment here. Thank you:) ...