Taking the Laplace transform of the both sides and assuming that all initial conditions are zero,

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1 The transfer function Let s begin with a general nth-order, linear, time-invariant differential equation, d n a n dt nc(t)... a d dt c(t) a 0c(t) d m = b m dt mr(t)... a d dt r(t) b 0r(t) () where c(t) is the output, r(t) the input, and a i s, b i s are coefficients in the differential equation. Taking the Laplace transform of the both sides and assuming that all initial conditions are zero, (a n s n... a s a 0 )C(s) = (b m s m... b s b 0 )R(s) (2)

2 or G(s) = C(s) R(s) = b ms m... b s b 0 a n s n... a s a 0 (3) We call this ratio, G(s), the transfer function. The transfer function can be represented as a block diagram as above. Note that the denominator of the transfer function is just the characteristic polynomial of the differential equation. Also, we can find out the output C(S) by using C(s) = R(s)G(s) 2

3 Example: (i) Find the transfer function represented by d c(t) 2c(t) = r(t) (4) dt (ii) Find the response, c(t), to an input, r(t) = u(t), a unit step, assuming zero initial conditions. Solution: (i) Taking the Laplace transform of both sides, assuming zero initial conditions, we have sc(s) 2C(s) = R(s) (5) The transfer function, G(s), is G(s) = C(s) R(s) = s 2 (6) (ii) Since r(t) = u(t) R(s) = /s. C(s) = R(s)G(s) =. Expanding by partial fractions, we get C(s) = /2 s s(s2) /2 s2 Finally, taking the inverse Laplace transform of each side yields c(t) = 2 2 e 2t 3

4 Electric network transfer functions We now combine electrical components into circuits, decide on the input and output, and find the transfer function. Example: Find the transfer function relating the capacitor voltage, V c (s), to the input voltage, V (s), in the Figure below: L R v(t) _ C v c (t) i(t) 4

5 Solution: Summing the voltages around the loop, assuming zero initial conditions, yields the integrodifferential equations for this network as L d dt i(t) Ri(t) i(τ)dτ = v(t) (7) C 0 Changing the variable from current to charge using i(t) = dt d q(t) yields dt 2q(t) R d dt q(t) q(t) = v(t) (8) C The voltage-charge relationship for a capacitor is L d2 t q(t) = Cv c (t) (9) Substituting above to (8) yields LC d2 dt 2v c(t) RC d dt v c(t) v c (t) = v(t) (0) Taking the Laplace transform assuming zero initial conditions (LCs 2 RCs )V c (S) = V (s) () 5

6 so G(s) = V c(s) V (s) = /LC (s 2 R L s LC ) (2) A technique for simplifying the solution. Redraw the original network showing all time variables, such as v(t), i(t) and v c (t), as Laplace transforms V (s), I(s) an V c (s) respectively. 2. Replace the component values with their impedance values. for the capacitor V (s) = Cs I(s) for the inductor V (s) = RI(s) for the resistor V (s) = LsI(s) 3. Finally we bypass the writing of differential equation, and applying kirchhof f s voltage law to the transform circuits. 6

7 Example: Repeat the same example. Redrawn L R Ls R v(t) _ C v c (t) V(s) _ Cs V c (s) i(t) I(s) We obtain (Ls Rs )I(s) = V (s) (3) Cs But the voltage across the capacitor, V c (s), is V c (s) = I(s) Cs (4) Hence we have the same result. (LCs 2 RCs )V c (s) = V (s) (5) 7

8 Complex circuit via mesh analysis. Replace passive elements values with their impedances. 2. Replace all sources and time variables with their Laplace transform. 3. Assume a transform current and a current direction in each mesh. 4. Write Kirchhoff s voltage law around each mesh. 5. Solve the simultaneous equations for the output. 6. Form the transfer function. 8

9 Example: Given the network of the Figure (a) below, find the transfer function, I 2 (s)/v (s). R R 2 v(t) _ L C i (t) i (t) 2 Figure (a) The first step is to convert the network into Laplace transforms for impedances and circuits variables. The result is shown in the Figure (b). 9

10 R R 2 V(s) _ Ls Cs I (s) I (s) 2 Figure (b) Around Mesh, where I (s) flows, R I (s) LsI (s) LsI 2 (s) = V (s) (6) Around Mesh 2, where I 2 (s) flows, LsI 2 (s)r 2 I 2 (s) Cs I 2(s) LsI (s) = 0 (7) Rearrange the two equations above, [ ] [ ] R Ls Ls I (s) Ls Ls R 2 Cs = I 2 (s) [ ] V (s) 0 (8) 0

11 Using Cramer s rule I 2 (s) = R Ls Ls R Ls V (s) Ls 0 Ls Ls R 2 Cs (9) = LsV (s) (R Ls)(Ls R 2 Cs ) L2 s 2 (20) Forming the transfer function,g(s) G(s) = I 2(s) V (s) = Ls (R Ls)(Ls R 2 Cs ) L2 s 2 = LCs 2 (R R 2 )LCs 2 (R R 2 C L)s R

12 Operational amplifiers An operational amplifier is an electronic amplifier used as building block to implement transfer functions. It has the following characteristics; (i)differential two inputs; v 2 (t) v (t); (ii)high input impedance, Z i = (ideal); (iii)low output impedance, Z o = 0; ; and (iv)high constant gain amplification, A = (ideal) The output v o (t) = A(v 2 (t) v (t)). 2

13 Inverting operational amplifier If v 2 (t) is grounded, the amplifier is called an inverting operational amplifier. We have (i) I a (s) = 0 and I (s) = I 2 (s); (ii)v (s) 0 I (s) = V i(s) Z (s) = V o(s) Z 2 (s) = I 2(s) Hence, the transfer function is V o (s) V i (s) = Z 2(s) Z (s) (2) 3

14 Example: Find the transfer function, V o(s) V i (s), for the circuit given below 4

15 Solution: The transfer function of the circuit is given by (2). Since the admittances of parallel components add, Z (s) is the reciprocal of the sum of the admittances, or Z (s) = = C s R = s s For Z 2 (s), the impedances add, or Z 2 (s) = R 2 C 2 s = s The transfer function is V o (s) V i (s) = Z 2(s) Z (s) =.232 s s s The resulting circuit is called a PID controller and can be used to improve the performance of a control system. 5

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