Let's derive the transfer function formula from the state-space equations step by step. Step 1: Start with the standard state-space equations The state equation is: x ˙ ( t ) = A x ( t ) + B u ( t ) \dot{x}(t)=Ax(t)+Bu(t) The output equation is: y ( t ) = C x ( t ) + D u ( t ) y(t)=Cx(t)+Du(t) Where: A A = System matrix B B = Input matrix C C = Output matrix D D = Feedforward matrix Step 2: Take the Laplace transform Taking the Laplace transform of: x ˙ ( t ) = A x ( t ) + B u ( t ) \dot{x}(t)=Ax(t)+Bu(t) We know that: L { x ˙ ( t ) } = s X ( s ) − x ( 0 ) \mathcal{L}\{\dot{x}(t)\}=sX(s)-x(0) Therefore, s X ( s ) − x ( 0 ) = A X ( s ) + B U ( s ) sX(s)-x(0)=AX(s)+BU(s) For zero initial conditions: x ( 0 ) = 0 x(0)=0 Hence, s X ( s ) = A X ( s ) + B U ( s ) sX(s)=AX(s)+BU(s) Step 3: Rearrange the equation Move A X ( s ) AX(s) to the left side: s X ( s ) − A X ( s ) = B U ( s ) sX(s)-AX(s)=BU(s) Taking X ( s ) X(s) commo...