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By Galal Rabie

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Conventionally, the symbol G(s) is used for the transfer function, but if the element appears in the feedback path, the symbol H(s) is used. The linear, single-input single-output, system can be described by a single linear differential equation of zero initial conditions. This differential equation (DE), can be transformed into Laplace domain by replacing the term (d/dt) by (s). 2) For real systems, the order of P(s) should not exceed that of Q(s); n  m . 1 Consider the linear system, of input x(t) and output y(t), described by the following differential equation of zero initial conditions.

17) 1 s(s  2) The inverse Laplace transform of this expression can be obtained after performing a partial fraction expansion as follows. e. Repeated real roots of the polynomial of the denominator). , n  1   s a 1 ds i! 10 Find the partial fractions of Y(s)  Y( s )  s4 (s  1)(s  2) 2  (s  1)Y(s)  3 k 3  lim ( s  2 ) 2 Y ( s )  2 s 2 (s  1)(s  2) 2 k3 k1 k  2  s  1 s  2 ( s  2) 2 k 1  lim s  1 s4  (s  1)  (s  4)   d  s  4  d  (s  2) 2 Y(s)   lim    lim  k 2  lim     3        s 2 ds (s  1) 2   s 2  ds  s  1  s 2    Y( s )   3 3 2   s  1 s  2 ( s  2) 2 Or, calculate k1 and k3 as shown above then calculate k2 as follows.

Output node (Sink): is a node, which has only ingoing branches X5. Forward path: is a path from input node to output node along which no node is encountered more than once. Feedback loop: is a path, which originates and terminates at the same node, along which no node is encountered more than once. Path gain: is the product of all gains (transfer functions) along the forward path. 1. M1  t 12 t 23 t 34 t 45 M2  t12 t 24 t 45 M3  t12 t 25 Loop gain: is the product of the gains of branches forming the loop.

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