Designing Information Devices and Systems II Fall 2017 Note Theorem: Existence and Uniqueness of Solutions to Differential Equations

Transcription

1 EECS 6B Designing Information Devices an Systems II Fall 07 Note 3 Secon Orer Differential Equations Secon orer ifferential equations appear everywhere in the real worl. In this note, we will walk through how to solve them in orer to unerstan secon orer circuits.. Degree of a ifferential equation Consier a ifferential equation of the form, n y n (t) + a n y n n (t) + + a y (t) + a 0y(t) = 0 The egree of the above ifferential equation is n.. Theorem: Existence an Uniqueness of Solutions to Differential Equations Given a n th orer ifferential equation an n initial conitions, y(0) = 0, y (t 0) =,, n y n (t 0 ) = n () there exists a singe unique solution (say, f ). We will not prove this in this class..3 First orer ifferential equations Consier the following simple equation. y (t) = y(t) with y(0) = () This is our starting point. The solution to () will set the first builing block to solving secon orer circuits. We are looking for the "eigenvector" of the ifferentiation operator that correspons to an eigenvalue of. Given the linear nature of the erivative operator, attempting to characterize the eigenvector is a big step towars unerstaning its nature. Getting back to the problem at han, we see that the function f (t) = e t satisfies the equation in () as well as the initial conitions. The Existence an Uniqueness of Solutions to Differential Equations Theorem EECS 6B, Fall 07, Note 3

2 Figure : A secon orer circuit states that there exists only one solution for a given initial conition, so we o not nee to search for more. Furthermore, f (t) = e t is an eigenvector of the ifferentiation operator. What about fining these "eigenfunctions" with some eigenvalue λ? y (t) = λy(t) with y(0) = c The solution is, y(t) = ce λt With this, we have all the builing blocks we nee to solve much more complicate ifferential equations..4 Example of a secon orer circuit Consier a circuit like in Figure. Assume the switch is open up to time 0. Let the voltage of Capacitor at time 0 be 0 an let the current at passing through Capacitor at time 0 be 0. This tells us that, V C (0) = 0 an V C (0) = 0 (3) We want to figure out how the voltage across Capacitor evolves after we close the switch at time 0. To o so, we will apply KCL to Noe A. We get that, i + i + i 3 = 0 (4) This tells us that, V R V +C C V R +C C = 0 (5) We have use the capacitor equation. Note the negative signs. This is to take into account the manner in which we have rawn the voltage signs in the iagram. In this case, because of the zero, it oes not matter but it is goo practice to always keep in min the irection of voltage rops accoring to the iagram. EECS 6B, Fall 07, Note 3

3 Observe that, V R = V C (6) This is because we are comparing the same two points across two ifferent paths. Recall that the voltage ifference between two points shoul always be the same regarless of the ifferent paths between the two points. Using a similar argument, V C = V C + i 3 R where i 3 = C V C (7) Combining (5), (6) an (7), we get the following. ( ) C C R V C R V + C +C +C C R + V C = 0 (8) R Equations like (8) pop up all the time. We re going to now present a straight forwar way of solving such ifferential equations using linear algebra!.5 Solving a general secon orer ifferential equation Consier a general secon orer ifferential equation of the form, y (t) + a y (t) + a 0y(t) = 0 (9) This can be written in matrix-vector form as, [ ] [ ][ ] y (t) 0 y(t) y = y (t) a }{{ 0 a (t) }}{{}}{{} x A x (0) Observe that x is a D vector consisting of x(t) an x (t) as coorinates. (0) is written in short form as, x = A x () We recast (9) to () in orer to exploit the matrix structure of A an the linear properties of the ifferentiation operator ( ). Particular, we are going to iagonalize A to exploit what we know about first orer ifferential equations. Let P be a matrix whose columns consist of eigenvectors of A. Let D be a iagonal matrix consisting of the eigenvalues (λ,λ ) of A an let P be the inverse of matrix P. We know that, Applying this to (), we get, A = PDP x P = DP x () EECS 6B, Fall 07, Note 3 3

4 Recall that multiplying vector x by P fins the coorinates of x with respect to the columns of P. Define z as follows. z = P x Then, we have that, x(t) = A x(t) x(t) = PDP x(t) P x(t) = DP x(t) ( P x ) (t) = DP x(t) ( z)(t) = D z(t) Note that we use the fact that A is inepenent of the time t an that P an ( ) are linear functions on x(t) to conclue that, P x(t) = ( ) P x (t) This transformation allows us to reach the following equation. z (t) = λ z (t) an z (t) = λ z (t) Recall that λ an λ are the eigenvalues of A. We have reuce a secon orer equation into two smaller first orer equations that we know how to solve. The close form values of the eigenvalues are, λ = ( ) a a 4a 0 an λ = ( ) a + a 4a 0 This tells us that, z (t) = c e λ t,z (t) = c e λ t (3) After we solve of z, we can get back x be observing that, x = P z Sie note. We observe a lot of share structure between λ an λ. Particularly, if a 4a 0 < 0, we observe that the eigenvalues are complex an are in fact conjugates of each other. λ = λ EECS 6B, Fall 07, Note 3 4

5 .6 A shortcut to solving secon orer ifferential equations As you can imagine, these steps can get teious sometimes. We present to you a shortcut to solving such equations that hinges on the funamental theorem of ifferential equations to avoi using the matrices P an P.. Cast a given ifferential equation into matrix form ().. Fin the eigenvalues of A. Let this be λ an λ. 3. (a) If λ an λ are complex, they will be complex conjugates of each other. The solution will be, x(t) = c e σt cos(ωt) + c e σt sin(ωt) where, λ = σ + jω,λ = σ jω (b) If λ an λ are real an istinct, the solution will be, x(t) = c e λt + c e λ t This is similar to (3) 4. Use the initial conitions to solve for c an c. You might be wonering what happens when λ = λ = λ. The solution is of the form, x(t) = c e λt + c te λt The reason why is out of scope for this course as it requires the concept of Generalize Eigenvectors. You can verify that the above is inee a solution using the funamental theorem of solutions to ifferential equations..7 Solving the secon orer circuit Now that we have the theory, we can solve (8). We can recast this equation to, V C + C R +C +C R V C V C C C R + = 0 (4) C C R R We get that, a = C +C +C R R C C R an a 0 = C C R R Let us simplify things an plug in values. Let C = C = 00µF,R = R = kω. Then, a = 30,a 0 = 00 EECS 6B, Fall 07, Note 3 5

6 This tells us that, λ 6,λ 4 We know that, V C (t) = c e λt + c e λ t an, V C (t) = c λ e λt + c λ e λ t We can now use the initial conitions in (3) to solve for c an c. EECS 6B, Fall 07, Note 3 6