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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition B 3 Copyright © Orchard Publications Simulink and its Relation to MATLAB (B.7) Solution of (B.7) yields of and and with these values (B.6) is written as (B.8) The forced component is found from (B.5), i.e., (B.9) Since the right side of (B.9) is a constant, the forced response will also be a constant and we denote it as . By substitution into (B.9) we obtain or (B.10) Substitution of this value into (B.8), yields the total solution as (B.11) The constants and will be evaluated from the initial conditions. First, using and evaluating (B.11) at , we obtain (B.12) Also, and (B.13) Next, we differentiate (B.11), we evaluate it at , and equate it with (B.13). Thus, (B.14) By equating the right sides of (B.13) and (B.14) we obtain s 2 4s 3 ++ 0 = s 1 1 = s 2 3 = v c t () k 1 e t k 2 e 3 t v cf t = v t d 2 v C dt 2 ----------- 4 dv C -------- 3v C 3 = t0 > v Cf k 3 = 003 k 3 3 = v k 3 1 == v C t v Cn t v + = k 1 e t k 2 e 3 t 1 = k 1 k 2 v C 0 0.5 V = = v C 0 k 1 e 0 k 2 e 0 1 0.5 k 1 k 2 + 0.5 = i L i C C C C i L C ---- = , C = i L 0 C ------------ 0 C --- 0 = = C = k 1 3k 2 =

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Introduction to Simulink B 4 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications (B.15) Simultaneous solution of (B.12) and (B.15), gives and . By substitution into (B.8), we obtain the total solution as (B.16) Check with MATLAB: syms t % Define symbolic variable t y0= 0.75*exp( t)+0.25*exp( 3*t)+1; % The total solution y(t), for our example, vc(t) y1=diff(y0) % The first derivative of y(t) y1 = 3/4*exp(-t)-3/4*exp(-3*t) y2=diff(y0,2) % The second derivative of y(t) y2 = -3/4*exp(-t)+9/4*exp(-3*t) y=y2+4*y1+3*y0 % Summation of y and its derivatives y = 3 Thus, the solution has been verified by MATLAB. Using the expression for in (B.16), we find the expression for the current as (B.17) Second Method Using the Laplace Transformation The transformed circuit is shown in Figure B.2. Figure B.2. Transformed Circuit for Example B.1 k 1 3k 2 0 = k 1 0.75 = k 2 0.25 = v C t () 0.75 e t 0.25e 3 t 1 ++ u 0 t = v C t ii L = i C C dv C dt --------- 4 3 -- 3 4 --e t 3 4 e 3t   e t e A == + R L + C 1 V s s 1s = V C s Is 0.25s 34 s + V C 0 0.5 s
Signals and Systems with MATLAB Computing and Simulink

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