Solving a RLC Circuit using Convolution with DERIVE for Windows
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1 Solving a RLC Circuit using Convolution with DERIVE for Windows Michel Beaudin École de technologie supérieure, rue Notre-Dame Ouest Montréal (Québec) Canada, H3C K3 mbeaudin@seg.etsmtl.ca - Introduction In ODE courses, motions of the mass-spring system and RLC-circuits are classical applications. With a Computer Algebra System (CAS), we can focus on these applications, leaving the resolution of the corresponding differential equation to the system. What we propose to do, here, is to connect the solution of a RLC-circuit with convolution. Teaching mathematics to engineers with DERIVE for windows is among the new ways of teaching mathematics. This system easily integrates piecewise continuous functions : we will make and abundant use of this statement in the following paragraphs. We will try to show that DERIVE is powerful, easy to use and appropriate for a large class of students. - The problem If we connect in series an Ohm s resistor of resistance R (ohms), an inductor of inductance L (henrys), a capacitor of capacitance C (farads) to a source of electromotrice force E(t) (volts) where t is the time, the corresponding differential equation is (by Kirchhoff s law) LC d v RC dv C C vc E( t) dt dt
2 where v C is the drop across the capacitor. And if we want the current I(t), we simply calculate I( t) C dv. For simplicity, we will assume zero charge and current when t =. Applying the dtc Laplace transform to the above ODE leads to V ( s ) H ( C s ) ^ E ( s ) z t the transfer function. So, vc( t) h( t) E( t) E( ) h( t ) d where H( s) is LCs RCs, where h(t) is the impulse response (the inverse Laplace transform of H(s)). In particular, we will assume that the we are in a case of underdamping (R C 4 L). Completing the square by hand yields h t LC e Rt ( L) ( ) sin t where LC R 4L Let us note that, because this ODE comes with zero initial conditions and zero output if t <, we are in the situation of a LTI system. So, we can compute the current in some other way, namely F H I ( t ) dh t C E ( t ) ( ) dt I K F H C d dt E ( t ) h ( t ) I K In the following paper, we will focus on the drop across the capacitor only. 3- Preliminary functions and overview of DFW We will need to define convolution (in the Laplace sense or causality) in DERIVE. Instead of defining the two signals, we will use a limit argument to implement the definition x( t) h( t) z t x( ) h( t ) d : The CAS DERIVE has a built-in indicator function, the CHI function : CHI(a, x, b) simplifies to if a < x < b and otherwise. The Heaviside function is STEP in DERIVE and the signum
3 function is SIGN. The system does not possess a Dirac delta function but, with appropriate limit of indicator function, we can «simulate» the Dirac delta function! Let see an example : we will illustrate the fact that y( t) ( t a) y( t a) where y( t) x( t) h( t) comes from the convolution of two rectangular signals x(t) = CHI(, t, ) anb h(t) = CHI(, t, ). We will take a = : Here is the graph of y : Here is the graph of the simplification of the above limit : 3
4 4- Live examples In our RLC-circuit, the numerical values will be : R = satisfy R C 4 L) and the impulse response is, L =.H and C = 5 F (those values So, giving the emf input E(t), the corresponding output (drop across the capacitor) will simply be Example : illustration that an RLC-circuit with zeros I.C. and causal input is a linear, timeinvariant (LTI) system. By definition, a LTI system is a system that is linear and time-invariant : so, if the corresponding output to the input x i (t) is y i (t) (i =, ) and if c i are scalar (i =, ), then the output for c x ( t) c x ( t) is c y ( t) c y ( t) (linearity) and, if a is a scalar and y(t) is the output corresponding to the input x(t), then the input x(t a) produces the output y(t a) (timeinvariance). 4
5 Let u(t) be the unit step function or Heaviside function (STEP(t) in DERIVE). We will find and plot the corresponding outputs to : u(t), u(t), u(t.5). This will not, of course, prove that we have a LTI system! But, for students, this is a way to understand linearity. Example : we can make use of a CAS and continue to teach mathematics! For example, students learn that the solution of RLC-circuit ODE is of the form vc( t) vc tr ( t) vc ss ( t) where the drop across the capacitor tends to the steady-state v t Css ( ) (v t C tr ( ) stands for the transient drop across the capacitor). In particular, if the emf is a constant, say V, a particular solution of our ODE is. That can be seen from example and motivates the input CHI(, t,.5). A graph of the input shows clearly a drop at t =.5. What will be the output if the input is of the form CHI(, t, ) with approaching from the right? Because lim CHI(, t, ) ( t ) where (t) is the Dirac delta function, the output must simply be close to h(t) where h(t) is our impulse response. And the output to the input lim CHI(, t, ) is, in fact, h(t). Recall that The «vector» command in DERIVE helps students to visualize these concepts. Example 3 : a sinusoid input. Let the emf be the periodic input 5sin t of period P = /. A particular solution of the ODE (the steady-state drop across the capacitor) will be of the form Acos( t ) as can be shown using «trigcollect» : 5
6 Example 4 : finite sum of indicator functions, «close» to a square wave. What will be the output if the input is a F H g t I a f? a where g( t) 5 CHI(, t,. 5) CHI(. 5, t,. ) K Example 5 : a periodic square wave as input. What will be the output if the input emf is a real periodic square wave, say E( t) RST 5 t. 5 P 5. 5 t..? We will apply the superposition principle, finding the corresponding Fourier series of E. This will lead us to investigate the maximum amplitude of each term of the steady-state drop across the capacitor and we will show that the steady-state drop across the capacitor is almost a harmonic oscillation whose frequency equals 3 times that of the emf. More precisely, only the first two terms of the output are important. Here are, on the same graph, the input and the steadystate output : 6
7 5- Conclusion We hope that those last examples proved that DFW is powerful enough to be used in engineering. And, with the introduction of the TI-9/9+/89, it is now possible to do similar examples on a «portable» CAS! 7
Using the TI-92+ : examples
Using the TI-9+ : eamples Michel Beaudin École de technologie supérieure 1100, rue Notre-Dame Ouest Montréal (Québec) Canada, H3C 1K3 mbeaudin@seg.etsmtl.ca 1- Introduction We incorporated the use of the
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