Unit 2: Modeling in the Frequency Domain. Unit 2, Part 4: Modeling Electrical Systems. First Example: Via DE. Resistors, Inductors, and Capacitors

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1 Unit 2: Modeling in the Frequency Domain Part 4: Modeling Electrical Systems Engineering 582: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 20, 200 Inverting Op-Amp The following table gives the relevant relationships between voltage, current, and charge for resistors, inductors, and capacitors: e.g. Find the transfer function for the circuit below. Consider the output to be the capacitor voltage and the input to be v(t), Voltage-current Current-voltage Voltage-charge Resistor v(t) Ri(t) i(t) R v(t) dq v(t) R Inductor Capacit. v(t) L di(t) v(t) t dv(t) C 0 i(τ)dτ i(t) C i(t) t L 0 v(τ)dτ v(t) L d2 q(t) 2 v(t) C q(t) These components are considered both passive and linear. Passive because they involve no internal source of energy (although inductors and capacitors can store energy). We consider them linear because their behavior is well-described using linear DE s. Apply KVL around the loop: L di(t) + Ri(t) + vc (t) v(t) We need to express i(t) in terms of vc (t) so that there are only two variables in the equation. This is achieved by the relationship between current and voltage in a capacitor: Hence, dvc (t) i(t) C d 2 vc (t) dvc (t) 2 + RC + vc (t) v(t)

2 d 2 vc (t) dvc (t) 2 + RC + vc (t) v(t) We can now apply the LT, assuming zero initial conditions, L 2 + R + V (s) Solve for the transfer function /V (s), V (s) L 2 + R + / s 2 + R L s + The approach above is perfectly fine. However, we can arrive at the final answer a bit quicker if we can actually state the problem directly in the frequency domain. This can be done by applying the LT to our table of laws for resistors, inductors, and capacitors, v-i i-v Imped. Admit. Resistor V (s) RI (s) I (s) R V (s) R R Inductor V (s) LsI (s) I (s) Ls V (s) Ls Ls Capacit. V (s) I (s) I (s) V (s) Another way of looking at these relationships is to consider them transfer functions. If the output is voltage and the input is current then each component s TF is its impedance as shown above. If current is the output and voltage is the input, then the TF is the component s admittance. Impedance is defined as follows, Z(s) V (s) I (s) which can be viewed as a TF. Note that it has the same form as Ohm s Law, R v(t) i(t) Recall our first example (series LRC circuit). Since the components are in series they must have the same current flowing through them. Therefore, we can treat each component as its own mini-system with current as input and voltage as output. Hence, we can represent each component by its impedance, Admittance is just the reciprocal of impedance. Note that for a resistor, impedance resistance admittance conductance We can again apply KVL, which says that the sum of these mini-outputs (voltages) around the loop must be zero, LsI (s) + RI (s) + V (s) The same algebraic steps as before produce the same final result.

3 KVL and KCL apply in the frequency domain because they are expressions about linear combinations of time-domain signals. The linearity of the LT implies that such expressions remain valid in the frequency domain. Another perspective is to represent components by their impedances (or admittances) and then treat them as if they were pure resistances in a DC circuit only with weird levels of resistance (Ls for inductors, for capacitors). Following this strategy the impedances of the three components can be summed as series resistances: ( Ls + R + ) I (s) V (s) ( Ls + R + ) I (s) V (s) I (s) V (s) Ls + R + But the problem calls for the voltage of the capacitor, I (s) I (s) V (s) V (s) V (s) Ls + R + / V (s) s 2 + R L + Just as KVL and KCL apply in the frequency domain, so do all other techniques for pure resistive circuits: voltage division, mesh analysis, and nodal analysis. Find the transfer function I2(s)/V (s) Voltage Division: The source voltage V (s) is split across the inductor and resistor (treated as a unit) and the capacitor: Hence, / Ls + R + V (s) V (s) / / Ls + R + s 2 + R L s + Mesh : Mesh 2: RI(s) + Ls(I(s) I2(s)) V (s) R2I2(s) + I2(s) + Ls(I2(s) I(s)) 0

4 Express in terms of I(s) and I2(s) on the LHS, (R + Ls)I(s) LsI2(s) V (s) LsI(s) + (R2 + + Ls)I2(s) 0 The solution of this system for I2(s) is, L 2 I2(s) (R + R2)L 2 V (s) + (RR2C + L)s + R The resulting transfer function is shown in the block diagram for this system below, The mesh equations for this example can be re-written in the following form: Mesh : Mesh 2: (R + Ls)I(s) (Ls)I2(s) V (s) (Ls)I(s) + (R2 + + Ls)I2(s) 0 Notice that each mesh equation adheres to the following pattern, ( mesh i impedances)ii(s) ( shared impedances)ij(s) mesh i voltage sources We will see this pattern again in subsequent examples. Inverting Op-Amp Op-amps are important in control systems for amplification, and for the summation and subtraction of analog signals. Our analysis will assume an ideal op-amp which has the following characteristics: High input impedance: Zi Low output impedance: Zo 0 High gain: A An op-amp has two inputs and produces an output voltage that amplifies the difference between them: vo(t) A(v2(t) v(t)) Clearly, real op-amps will not have the same properties as ideal op-amps. However we will utilize the ideal op-amp model because it is simple and produces very reasonable results. Some of the main restrictions in the use of the ideal model are as follows: The output voltage vo is limited by the power supply. If the power inputs are +5V and 5V then the maximum and minimum values of vo would lie in this range (and probably closer to [ 0, 0]). For best performance the impedance at the two inputs should be matched (the book does not adhere to this).

5 Inverting Op-Amp Example Find the transfer function for the following circuit, This shows the wiring for an inverting op-amp. The ideal op-amp model allows us to make the following statements: Input imped. therefore current into both inputs 0 The + input is grounded; The voltage at the input will be amplified by an extremely high gain; Therefore V(s) 0 Hence, we know that I(s) I2(s), I(s) Vi(s) Z(s) I2(s) Vo(s) Z2(s) Vo(s) Vi(s) Z2(s) Z(s) Z(s) Cs R Cs + R Notice that the impedance of two components in parallel is the reciprocal of the sum of their admittances, Z2(s) R2 + Subsituting into Z2(s) Z(s) we obtain, Vo(s) Vi(s) s s2 s This implements a PID controller (studied later this term). This shows the wiring for a noninverting op-amp. The ideal op-amp model allows us to make the following statements: Input imped. therefore current into both inputs 0 The + input is at Vi(s); The voltage difference Vi(s) V(s) will be amplified by an extremely high gain; Therefore V(s) Vi(s) Apply nodal analysis at the input (COVERED ON BOARD). The result is, Vo(s) Z(s) + Z2(s) Vi(s) Z(s)

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

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