% latex diagram produces diagram for the following question% Obtain the closed-loop pulse transfer function of the system shown in the figure below.% Final Answer:%% \begin{align*}% & C^\ast(s) = \frac{G_2^\ast(s)G_1^\ast(s)R^\ast(s)}{1 +G_2^\ast(s)+G_1^\ast(s)%  \left[G_2H(s)\right]^\ast} \quad  C(z) =  \frac{R(z)G_2(z) G_1(z)}{1 +G_2(z)+G_1(z)%  \left[G_2H(z)\right]} %\\% \end{align*}\documentclass{standalone}\usepackage{tikz}\usetikzlibrary{positioning}\usetikzlibrary{shapes,arrows,calc}\usetikzlibrary{decorations.text}\tikzset{add/.style n args={4}{minimum width=6mm,path picture={\draw[black](path picture bounding box.south east) -- (path picture bounding box.north west)(path picture bounding box.south west) -- (path picture bounding box.north east);\node at ($(path picture bounding box.south)+(0,0.13)$)     {\tiny #1};\node at ($(path picture bounding box.west)+(0.13,0)$)      {\tiny #2};\node at ($(path picture bounding box.north)+(0,-0.13)$)        {\tiny #3};\node at ($(path picture bounding box.east)+(-0.13,0)$)     {\tiny #4};}}}\begin{document}%\begin{figure}%\centering\tikzstyle{block} = [draw, fill=blue!20, rectangle, minimum height=3em, minimum width=4em]\tikzstyle{controller} = [draw, fill=red!20, rectangle, minimum height=3em, minimum width=4em]\tikzstyle{sum} = [draw, fill=blue!20, circle, node distance=1cm]\tikzstyle{input} = [coordinate]\tikzstyle{output} = [coordinate]\tikzstyle{sampleSP} = [coordinate]\tikzstyle{sampleEP} = [coordinate]\tikzstyle{otherPoint} = [coordinate]\begin{tikzpicture}[auto, >=latex']% Nodes\node [input] (input) {};%\node [sum, right = 1cm of input] (sum) {};\node[draw,circle,add={--}{+}{}{},right of= input](sum){};\node [sampleSP, right = 1cm of sum] (sumSP) {};\node [sampleEP, right = 1cm of sumSP] (sumEP) {};\node [sampleEP, above = 1cm of sumEP] (sumEPTOP) {};\node [controller, right = 2cm of sumEP] (system) {$G(s)$};\node [otherPoint,right = 1cm of system] (branchPoint) {};\node [otherPoint,below = 1 cm of system] (belowsystem) {};%{$\frac{1}{Ts+1}$};%\node [block, right = 1cm of system] (system2) {$\frac{1}{Ts+1}$};\node [output, right = 2cm of branchPoint] (output) {};%\node [input, below = 0.5cm of system] (m) {};\node [block, right = 0.75 of belowsystem] (systemH) {$H_1(s)$};% Second Sampler Point\node [sampleSP, left = 1cm of systemH] (sysHSP2) {};\node [sampleEP, left = 1cm of sysHSP2] (sysHEP2) {};\node [sampleEP, above = 1cm of sysHEP2] (sumEPTOP2) {};% Second block\node [block, left = 1.5cm of sysHEP2] (systemH2) {$H_2(s)$};% Arrows\draw [draw,->] (input) -- node {$R$} (sum);% Arrows for first sampler\draw [-] (sum) -- node {$E(s)$} (sumSP);\draw [-,thick] (sumEPTOP) -- node {$\delta_\tau$} (sumSP);\draw [->] (sumEP) -- node {$E^\ast(s)$} (system);%Arrows for second sampler (bottom)\draw [-]  (sysHSP2)-- node {$M(s)$} (systemH);\draw [-,thick] (sysHSP2) -- node {$\delta_\tau$} (sumEPTOP2);%\draw [->] (sumEP) -- node {$M^\ast(s)$} (systemH);% \draw [->] (system) --  (system2);\draw [-] (system) --  (branchPoint);\draw [->] (branchPoint) -- node (y) {$C(s)$}(output);\draw [-] (y) |- (systemH) {};%\draw [-] (y) |- (m) {} ;\draw [<-] (systemH2) -- node {$M^\ast(s)$} (sysHEP2);\draw [->] (systemH2) -| (sum);%{$-$}  node [near end] {} (sum);%\draw [->] (m) -| node[pos=0.99] {} node [near end] {} (sum); %{$-$}  node [near end] {} (sum);\end{tikzpicture}%\end{figure}\end{document}