The Laplace Tranform 8.3, Hayne Miller and Jeremy Orloff Laplace tranform baic: introduction An operator take a function a input and output another function. A tranform doe the ame thing with the added twit that the output function ha a different independent variable. The Laplace tranform take a function f(t) and produce a function F (). We will allow the variable to be complex. A with the tranfer function you learned about earlier the will be thought of a complex frequency You hould think of f(t) and F () a two view of the ame underlying object. If we have a ignal, then f(t) i the familiar view of that ignal in time and F () i the le familiar view in frequency. Everything about the ignal i preent in both view, but ome thing are eaier to ee in one view or the other. Uing them together give u a powerful tool for undertanding ytem and ignal. You have already been uing the Laplace tranform without knowing it. For the ytem P (D)x = Q(D)f, (conidering f to be the input and x the output) the tranfer function Q()/P () i a Laplace tranform of a function w(t). Over time we will give w(t) everal different name. To highlight it importance here we will call it the fundamental olution to the ytem. Alo, jut like the tranfer function, the Laplace tranform of any function ha a pole diagram. In practice the Laplace tranform ha the following benefit: It make explicit the long-term behavior of f(t). The pole diagram of F () give a uccinct ummary of ome of the important propertie of f(t). It allow u to eaily compute the tranfer function for LTI ytem other than P (D)x = Q(D)f. In particular, it will do thi for ytem with delay. Since it i the bridge between the time domain and the frequency domain we can credit it with all the benefit of uing tranfer function: It allow u to analyze LTI ytem, and in particular block diagram, uing imple algebra. For the tranfer function, when we retrict by etting it equal to iω we get the frequency repone of the ytem The pole diagram of a tranfer function can how at a glance the tability and frequency repone of a ytem. It i an important engineering deign tool.
8.3 The Laplace Tranform 2 2 Definition of Laplace tranform The Laplace tranform of a function f(t) of a real variable t i another function depending on a new variable, which i in general complex. We will denote the Laplace tranform of f by Lf. It i defined by the integral (Lf)() = f(t)e t dt, () for all value of for which the integral converge. There are a few thing to note. Lf i only defined for thoe value of for which the improper integral on the righthand ide of () converge. We will allow to be complex. The ue of, in the definition () i neceary to accomodate what we ll call delta function. Until we learn about uch function it will not be important. In thoe cae where in t needed we will allow ourelve to ue the le precie form (Lf)() = f(t)e t dt. ( ) The limit of integration mean that the Laplace tranform i only concerned with function on (, ). What happen before time t = doe not play a role. 3 Notation, F () We will adopt the following convention:. Writing (Lf)() can be cumberome o we will often ue an uppercae letter to indicate the Laplace tranform of the correponding lowercae function: (Lf)() = F (), For example, in the formula (ee the Laplace table) it i undertood that F () mean L(f). 2. Another notation we will ue i L(f(t); ) (Lg)() = G(), etc. L(f ) = F () f( ) 3. If our function doen t have a name we will ue the formula intead. For example, the Laplace tranform of the function t 2 can written L(t 2 ; ) or more imply L(t 2 ). 4. If in ome context we need to modify f(t), e.g. by applying a tranlation by a number a, we can write L(f(t a); ) or L(f(t a)) for the Laplace tranform of f(t a). 5. You ve already een everal different way to ue parenthee. Sometime we will even drop them altogether. So, if f(t) = t 2 then the following all mean the ame thing (Lf; ) = F () = Lf() = L(f(t); ) = L(t 2 ; ); Lf = F = L(t 2 ).
8.3 The Laplace Tranform 3 4 Why i called frequency The Laplace tranform variable i thought of a complex frequency. We alreadt aw thi in the tranfer function: if H() i the tranfer function of an LTI ytem, then when = iω we have H() = H(iω) i the complex gain of the ytem. A couple of other mall point are in order here. The firt i that for the exponential e t to make ene the exponent t mut be dimenionle. Therefore the unit of mut be /time, which are the ame a the unit of frequency. The econd point i really a reiteration of what happen when we ued complex replacement to find the Sinuoidal Repone Formula. Euler formula ay e iωt = co(ωt) + i in(ωt) and we call ω the angular frequency. By analogy we call any exponent in e t a complex frequency. 5 Firt example For the firt few example we will explicitly ue a limit for the improper integral. Soon we will do thi implicitly without comment. Example. Let f(t) =, find F () = Lf(). anwer: Uing the definition ( ) we have L() = F () = e t dt = lim T e t ] T = lim T e T ] T. The limit depend on whether the real part of i poitive or negative. { if Re() > lim T e T = if Re() <. Therefore, L() = F () = { if Re() > diverge if Re(). (We didn t actually compute the cae Re() =, but it i eay to ee it diverge.) Example 2. Compute L(e at ). anwer: Uing the definition ( ) we have L(e at ) = e at e t dt = lim T e (a )t ] T a = lim T e (a )T ] T. a The limit depend on whether the real part of a i poitive or negative. { if Re() > Re(a) lim T e(a )T = if Re() < Re(a). Therefore, L(e at ) = { a diverge if Re() > Re(a) if Re() Re(a).
8.3 The Laplace Tranform 4 (We didn t actually compute the cae Re() = Re(a), but it i eay to ee it diverge.) We now have the firt two entrie in our table of Laplace tranform: f(t) = F () = /, Re() > f(t) = e at F () = /( a), Re() > Re(a). Note that the lat field in each line give the range of where the Laplace integral converge. 6 Linearity You will not be urpried to learn that the Laplace tranform i linear. For function f, g and contant c, c 2 L(c f + c 2 g) = c L(f) + c 2 L(g) Thi i clear from the definition () of L becaue integration i linear. 7 Domain of F (): complex and the region of convergence A we ve een, we allow to be complex and ue, a needed, propertie of the complex exponential. Example 3. Previouly we aw that L() = /, valid for all with Re() >. The region Re() > i called the region of convergence of the tranform. It i a right half-plane in the complex -plane. Imag. axi Re() > Real axi Region of convergence: right half-plane Re() >. Likewie, the region of convergence for f(t) = e at i the right half-plane Re() > Re(a). We will ee that the region of convergence for any function i a right half-plane. 7. The domain of F () For f(t) we have F () = / with region of convergence Re() >. But, the function / i well defined for all. So we can extend the domain of F () beyond the region of convergence to all. The proce of extending the domain of F () beyond the region of convergence i called analytic continuation. Though it ha wider applicability: In thi cla analytic continuation will alway conit of extending F () to the complex plane minu the zero of the denominator.
8.3 The Laplace Tranform 5 8 Function of exponential order and piecewie continuou function Thi i a technical ection which aure u that the Laplace tranform make ene for all the function we care about in 8.3 A we computed Laplace integral we were careful to note for which value of they converged. It can happen the integral doe not converge for any value of. In thi cae we ay that the function fail to have a Laplace tranform. Example 4. It i eay to ee that f(t) = e t2 The problem i the e t2 grow too fat a t get large. ha no Laplace tranform. 8. Function of Exponential Order The cla of function that do have Laplace tranform are thoe of exponential order. We will ee that every function of exponential order ha a Laplace tranform valid in a right half-plane Re() > a for ome value a. A function i aid to be of exponential order if there are number a and M uch that f(t) < Me at. In thi cae, we ay that f ha exponential order a. We will give detail below, but the baic idea i that in the Laplace integral the exponential decay of the e t term will compenate for the growth of f(t) a long a f(t) grow lower than ome exponential. Fortunately for u, all the function we ue in thi cla are of thi type. Example 5. The function, co(ωt), in(ωt), t n all have exponential order. The function e at ha exponential order a. 8.2 Piecewie continuou function A function f(t) i piecewie continuou if it i continuou everywhere except at a finite number of point in any finite interval and if at thee point it ha a jump dicontinuity (i.e. a jump of finite height). Example 6. The quare wave i piecewie continuou. 3π 2π π π 2π 3π t The quare wave i piecewie continuou. The main point of thi ection i the following theorem which aure u that the Laplace tranform converge for all the function we ue in 8.3. Theorem: If f(t) i piecewie continuou and of exponential order a then the Laplace tranform Lf() converge for all with Re() > a. Proof: Suppoe Re() > a, o = (a + α) + ib for ome poitive α. We are given that
8.3 The Laplace Tranform 6 f(t) < Me at. So, f(t)e t = f(t)e (a+α)t e ibt = f(t)e (a+α)t) < Me at e (a+α)t = Me αt, Here, we have ued that e ibt =. Since Me αt dt converge for α >, the Laplace tranform integral alo converge. 9 Partial fraction and invere Laplace tranform In order to ue the Laplace tranform we need to be able to invert it and find f(t) when we re given F (). Often thi can be done by uing the Laplace tranform table. So for example, if F () = /( 5) then the table tell u that f(t) = e 5t. More often we have to do ome algebra to get F () into a form uitable for the direct ue of the table. Our main technique for doing thi i the partial fraction decompoition. You probably ued partial fraction in calculu a a method for computing integral. If you need to relearn thi technique we have poted a note on it. The note include a decription of the Heaviide coverup method. Thi i a imple and extremely ueful computing device. If you do not know it already you hould read the firt ection of the partial fraction note which explain it. 9. Laplace invere by table lookup We ve added a few function to our table of Laplace tranform. Soon we will add ome more. But right now we will learn to ue the table to find the invere Laplace tranform. We will illutrate thi entirely by example. Before we tart you hould open the copy of the table from the cla webite or ele ue the copy at the end of thee note. Notation: The invere Laplace tranform will be denoted L. Example 7. Find L (/( 2)). anwer: Ue the table entry L(e at ) = /( a): L (/( 2)) = e 2t. Example 8. Find L (/( 2 + 9)). anwer: Ue the table entry L(in(ωt)) = ω/( 2 + ω 2 ) and linearity: ( ) L 2 = ( ) 3 + 9 3 L 2 + 3 2 = 3 in(3t). Example 9. Find L (4/ 2 ). anwer: Ue the table entry L(t) = / 2 : L (4/ 2 ) = 4t. Example. Find L (4/( 2) 2 ).
8.3 The Laplace Tranform 7 anwer: We will ue the -hift formula L(e at f(t)) = F ( a). In thi cae we take F () = 4/ 2, which by Example 9 ha f(t) = t. Therefore, L (4/( 2) 2 ) = L (F ( 2)) = e 2t f(t) = e 2t 4t. ( ) Example. Find L 2. + 4 + 3 anwer: We firt need to complete the quare 2 + 4 + 3 = 2 + 4 + 4 + 9 = ( + 2) 2 + 9. We have a hifted function F ( + 2), where F () = /( 2 + 9). Uing Example 8, we know that f(t) = in(3t)/3, o uing the -hift rule we get ( ) L 2 = L 2t in(3t) (F ( + 2)) = e. + 4 + 3 3 (Note that L(ω/(( a) 2 + ω 2 )) = e at in(ωt) i in the Laplace table, o we could have done thi example directly.) ( ) Example 2. Find L ( 2 + ω 2 ) 2. anwer: We haven t een thi formula yet, but there i a table entry, which give: t 2ω in(ωt). Example 3. Find L ( ( 2 +ω 2 ) 2 ). anwer: Thi i alo a table entry, anwer: (in(ωt) ωt co(ωt)). 2ω3 More entrie for the Laplace table In thi ection we will add ome new entrie to our table of Laplace tranform. Note: poted on the cla webite i the complete Laplace table that we will need in thi cla. For convenience it i alo appended at the end of thee note.. Laplace tranform of ine and coine. L(co(ωt)) = 2, with region of convergence Re() >. + ω2 ω 2. L(in(ωt)) = 2, with region of convergence Re() >. + ω2 Proof: We know that co(ωt) = eiωt + e iωt, and in(ωt) = eiωt e iωt 2 2i
8.3 The Laplace Tranform 8 (Thee formula are extremely ueful, if they are not immediately familiar you hould take a moment to undertand them. You can prove them by expanding each of the exponential into co() + i in().) We already know that L(e at ) = /( a). Uing thi and the formula above, we obtain ( e iωt + e iωt ) L(co(ωt) = L = ( 2 2 iω + ) = + iω 2 + ω 2 ( e iωt e iωt ) L(in(ωt) = L = ( 2i 2i iω ) ω = + iω 2 + ω 2 The region of convergence follow from the fact that co(ωt) and in(ωt) both have exponential order. Another approach would have been to ue integration by part to compute the tranform directly from the Laplace integral..2 Laplace tranform of derivative.3 t-derivative rule Thi coure make heavy ue of differential equation, o we hould try to compute L(f ). (We ue the notation f intead of f imply becaue we think the dot doe not it nicely over the tall letter f.) A uual, we write L(f; ) = F (). The t-derivative rule i L(f ) = F () f( ) (2) L(f ) = 2 F () f( ) f ( ) (3) L(f (n) ) = n F () n f( ) n 2 f ( ) +... + f (n ) ( ). (4) Technically we need to aume that f(t) ha exponential order a and Re() > a. With thi aumption e t f(t) i when t =. Proof: Rule (2) i a imple conequence of the definition of Laplace tranform and integration by part. The integration by part formula i uv dt = uv u v dt. We ll ue thi with u = e t and v = f (t), o that u = e t and v = f(t). Now, by definition L(f ) = f (t)e t dt. We tart by writing the definition of L(f ) and then we apply integration by part: L(f ) = f (t)e t dt = f(t)e t ] + f(t)e t dt = f( ) + f(t)e t dt The lat equality follow from our aumption that f(t)e t i at t =. Now notice that the integral in the expreion at the right i none other than F (). Thu we have proved the t-derivative law L(f ) = F () f( ).
8.3 The Laplace Tranform 9 Rule (3) follow by applying rule (2) twice. L(f ) = L(f ) f ( ) Rule (4) Follow by applying rule (2) n time. = (L(f) f( )) f ( ) = 2 F () f( ) f ( ). Note:. Calculation will be eaiet when f( ) =, f ( ) =, etc. We will call thi ret initial condition. 2. A good way to think of the t-derivative rule i L(f) = F () L(f ) = F () + term at. L(f ) = 2 F () + term at. Roughly peaking, Laplace tranform differentiation in t to multiplication by. Example 4. Let f(t) = e at. We can compute L(f ) in two way: directly and by uing rule (2). Let compute both way and check that they give the ame anwer. Directly: f (t) = ae at L(f ) = a/( a). Rule (2): L(f) = F () = /( a) L(f ) = F () f( ) = /( a) = a/( a). Both method give the ame anwer. Example 5. Let f(t) = t 2 +2t+. Compute L(f ) both directly and uing the t-derivative rule. anwer: Directly: Uing rule (4): f (t) = 2 L(f ) = 2/. L(f ) = 2 F () f( ) f ( ) = 2 (2/ 3 + 2/ 2 + /) 2 = 2/. Both method give the ame anwer. -derivative rule There i a certain ymmetry in our formula. If derivative in time lead to multiplication by then multiplication by t hould lead to derivative in. Thi i true, but, a uual, there are mall difference in the detail of the formula. The -derivative rule i L(tf; ) = F () (5) L(t n f; ) = ( ) n F (n) (). (6)
8.3 The Laplace Tranform Proof: Rule (5) i a imple conequence of the definition of Laplace tranform. F () = L(f) = F () = d d Rule (6) i jut rule (5) applied n time. f(t)e t dt f(t)e t dt = tf(t)e t = L(tf(t)).. Power of t and repeated factor in Now that we have the derivative rule we can ue them to avoid having to compute a number of integral. We derive more formula for our Laplace table in a erie of example. Example 6. Ue the -derivative rule to find L(t), L(t 2 ), etc. anwer: Start with f(t) =, then F () = /. The -derivative rule now ay L(t) = F () = / 2 Uing L(t) = / 2 we get L(t 2 ) = L(t t) = dl(t) d = (/ 2 ) = 2/ 3. Continuing we get L(t 3 ) = (2/ 3 ) = 3 2/ 4. In general, we have L(t n ) = n!/ n+. Note:. You hould check that thi rule give the correct anwer when n = 2. The region of convergence for L(t n ) i the ame a for L(), i.e. when Re() >. The eaiet way to ee thi i to notice that multiplying by power of t doe not change the exponential order of a function. Example 7. Ue the -derivative rule to find L(te at ) and L(t n e at ). anwer: Start with f(t) = e at, then F () = /( a). L(te at ) = F () = /( a) 2. Continuing: L(t 2 e at ) = F () = 2/( a) 3 L(t 3 e at ) = F () = 3 2/( a) 4 L(t 4 e at ) = F (4) () = 4 3 2/( a) 5... L(t n e at ) = ( ) n F (n) () = n!/( a) n+. The -derivative rule now ay
8.3 The Laplace Tranform.2 Repeated quadratic factor Look at the table entrie for repeated quadratic factor ( ) L (in(ωt) ωt co(ωt)) = 2ω3 ( ) t L 2ω in(ωt) = ( ) L (in(ωt) + ωt co(ωt)) = 2ω ( 2 + ω 2 ) 2 (7) ( 2 + ω 2 ) 2 (8) 2 ( 2 + ω 2 ) 2 (9) Let prove them uing the -derivative rule. Proof of (8). Let f(t) = in(ωt). We know that F () = implie L(t in ωt) = F () = 2ω ( 2 + ω 2 ) 2. ω 2, o the -derivative rule + ω2 Thi i the ame a formula (8) except the factor of 2ω i moved from one ide to the other. The other two formula can be proved in a imilar fahion. We won t give the proof here..3 -hift formula If a i any complex number and f(t) i any function then the -hift formula i L(e at f(t)) = F ( a). Proof. A uual we write F () = L(f; ). If the region of convergence for L(f) i Re() > a then the region of convergence for L(e at f(t)) i Re() > Re(a) + a. Now we imply calculate directly from the definition of Laplace tranform: L(e at f(t)) = e at f(t)e t dt = f(t)e ( a)t dt = F ( a). Example 8. Find the Laplace tranform of e t co(3t). anwer: Since L(co(3t)) = /( 2 + 9) the -hift formula give L(e t co(3t) = + ( + ) 2 + 9. Note: we could do thi almot a eaily by uing Euler formula to write ( e t co(3t) = (/2) e ( +3i)t + e ( 3i)t). We record here two important cae of the -hift formula: L(e at co(ωt)) = z ( a) 2 + ω 2 L(e at in(ωt)) = ω ( a) 2 + ω 2.
8.3 The Laplace Tranform 2 Conitency. It i a good exercie to check for conitency among our variou formula:. We have L() = /, o the -hift formula give L(e at ) = /( a). Thi matche our formula for L(e at ). 2. We have L(t n ) = n!/ n+. If n = we have L(t ) =!/ = /. Thi matche our formula for L(). 2 Laplace: olving initial value problem 2. Introduction We now have everything we need to olve initial value problem uing the Laplace tranform. We will how how to do thi through a erie of example. To be honet we hould admit that ome initial value problem are more eaily olved by other technique. However, there are cae where the Laplace machinery can help keep thing traight. 2.2 Example of olving initial value problem (IVP) Example 9. Solve ẋ + 3x = e t with ret initial condition (ret IC). anwer: Ret IC mean that x(t) = for t <, o x( ), ẋ( ),... are all. A uual, we let X = L(x). Uing the t-derivative rule we can take the Laplace tranform of (both ide) of the DE. (X() x( )) + 3X() = /( + ). Next we ubtitute the known value x( ) = and olve for X() ( + 3)X() = + X() = ( + )( + 3). () Finally, we find x(t) = L (X) by uing cover-up to do the partial fraction decompoition. X() = ( + )( + 3) = /2 + /2 + 3, thu x(t) = 2 e t 2 e 3t for t >. Note:. The term e t /2 i what the exponential repone formula would give u. The term e 3t /2 i the homogenou part of the olution, needed to match the IC. 2. Thi technique found x(t) for t >. The ret IC tell u x(t) = for t <. 3. The factor of /( + 3) in the expreion for X() in () i none other than the tranfer function of the ytem: ẋ + 3x = f(t). Example 2. Solve ẋ + 3x = e t, x( ) = 4. anwer: Laplace: X(x) x( ) + 3X() = /( + ) ( + 3)X() = 4 + /( + ).
8.3 The Laplace Tranform 3 Solve for X(): X() = 4 + 3 + ( + )( + 3) We can ue the partial fraction work from Example 9. () x(t) = 4e 3t + 2 e t 2 e 3t for t > = 2 e t + 7 2 e 3t for t >. Note: (Same remark a in the previou example.) Example 2. Solve ẋ + 2x = 4t, with initial condition x( ) =. anwer: Taking the Laplace tranform of the equation: X x( ) + 2X = 4/ 2. Therefore, 4 X() = 2 ( + 2) + + 2 = A + B 2 + C + 2 + + 2. Therefore the invere Laplace tranform give u x(t) = A + Bt + Ce 2t + e 2t, for t >. The coverup method get u B = 2, C =. Some algebra (undetermined coefficient) give A =. Finally we record the exact olution: Example 22. Solve ẍ + 4x = co(2t), x(t) = + 2t + 2e 2t, for t >. with ret initial condition anwer: Laplace: ( 2 + 4)X() = /( 2 + 4) X() = /( 2 + 4) 2. Thi i a repeated quadratic factor and it i in our table: x(t) = t in(2t)/4. Note:. Thi i a repone of pure reonance. 2. We could have turned the logic around and ued our previou knowledge of the olution to thi equation to give yet another proof for the table entry L(t in(ωt)/2ω) = /( 2 +ω 2 ) 2. 3 The tranfer function Conider the ytem with input f(t) and repone x: P (D)x = Q(D)f. In cla we defined the tranfer function a Q()/P (). Thi aroe for u a when we tried olving the above ytem with input f(t) = e t. In that cae, the Exponential Repone Formula gave u a particular olution x(t) = (Q()/P ())e t. Now that we have the Laplace tranform we will ee how it give u the tranfer function naturally and for arbitrary f. Problem. Ue the Laplace tranform to olve the equation P (D)x = Q(D)f tarting with ret initial condition.
8.3 The Laplace Tranform 4 anwer: Let X = L(x) and F = L(f). Since we tart from ret there are no term, o we get P ()X() = Q()F (), or equivalently X() = P () F (). (2) Q() There we have it! Viewed in the frequency domain, the output i imply the tranfer function time the input. We could ay that the tranfer function tranfer the input to the output. Another tandard way of aying thi i tranfer function = output input. Be ure to recognize that thi i all taking place on the frequency ide. Alo be ure to marvel at the fact that from thi ide the output/input alway give the ame function. Here a ummary of what we have:. For the ytem P (D)x = Q(D)f the tranfer function i W () = Q()/P (). 2. In the frequency domain, olving for X with ret initial condition i purely algebraic: X() = W (S)F (). 4 Block diagram 4. Introduction We already dicued ome imple block diagram when we introduced the notion of ytem, input, and output. Here, we will look at ytem with more complicated block diagram and how how to ue them to compute the tranfer function. A we do thi, it will be ueful to keep in mind the deciption of the tranfer function a multiplying the input to get the output. 4.2 Simple example Example 23. Suppoe we have the ytem mẍ+bẋ+kx = f(t), with input f(t) and output x(t). The Laplace tranform convert thi to function and equation in the frequency variable : X() = m 2 + b + k F (). The tranfer function for thi ytem i W () = /(m 2 + b + k), and we can write the relation between input and output a input F () output X() = W ()F () A a block diagram we can repreent the ytem by F () W () X() Block diagram for a ytem with tranfer function W ().
8.3 The Laplace Tranform 5 Sometime we write the formula for the tranfer function in the box repreenting the ytem. For the above example thi would look like F () m 2 + b + k X() Block diagram giving the formula for the tranfer function. Example 24. (Cacading ytem) Conider the cacaded ytem P (D)x = Q (D)f, P 2 (D)y = Q 2 (D)x, ret IC. The input to the cacade i f and the output i y. That i, the firt equation take the input f and output x. Then x i the input to the econd equation, which ouput y. Thi i eay to olve on the frequency ide. Let W () = Q ()/P () and W 2 () = Q 2 ()/P 2 () be the tranfer function for the two differential equation. Conidering the two equation eparately we have X() = W () F () and Y () = W 2 () X(). It follow immediately that Y () = W 2 () W () F (). Therefore the tranfer function for the cacade i output/input = Y ()/F () = W 2 () W (). In other word, for cacaded ytem the tranfer function multiply. Repreenting thi a block diagram we have two equivalent diagram F () X() Y () W () W 2 () F () W ()W 2 () Y () Equivalent block diagram for a cacaded ytem. Example 25. (Parallel ytem) Suppoe that we have a ytem coniting of two ytem in parallel a hown in the block diagram. F () W () W 2 () + Y () Sytem in parallel. Find the tranfer function for the entire ytem. anwer: The plu ign in the circle indicate the two ignal coming into the junction hould be added. The plit near the tart indicate the ame input F () i ent into each ytem. The way to figure out the tranfer function i to name the output of each individual ytem.
8.3 The Laplace Tranform 6 F () F F W () W 2 () X X 2 + Y () Sytem with intermediate output labeled. For each ytem we know output = tranfer function input. Thu, X = W F, X 2 = W 2 F, Y = X + X 2. So, we eaily compute Y = X + X 2 = W F + W 2 F = (W + W 2 ) F. Therefore the tranfer function i W + W 2. Example 26. An example of a parallel ytem i everal microphone feeding the ame ound into a mixing dek which in turn feed an amplifier and peaker ytem. 4.3 Feedback loop Many ytem are bet decribed with feedback loop. In a broad ene, in a feedback loop the output i ued to modify the input which in turn produce the output. In feedback control, the output of the ytem i monitored and ued to modify the input in uch a way a to ultimately produce a deired output. It i very hard to control a ytem without a feedback loop. For example, imagine trying to walk without enory feedback about your urrounding. For mot people, jut hutting their eye make it hard to tand on one foot for very long. Block diagram are an excellent way to decribe feedback. Suppoe we tart with a ytem with tranfer function W (). F () W () X() and modify it to have the feedback loop hown in the next figure. F V W Y + gy g The original ytem i known a the open loop ytem and the correponding ytem with feedback i known a the cloed loop ytem. We ve labeled the output from each ytem element. The ymbol for the ytem element indicate what it doe to it input(). The ymbol g mean the input to that element i caled by g, that i apply a gain of g to the input. The ymbol mean the two input are combined; the plu and minu ign indicate whether to add or ubtract the correponding input. The method of finding the tranfer function i the ame a in the previou example. A bit of algebra give V = F gy, Y = W V Y = W (F gy ) Y = W + gw F.
8.3 The Laplace Tranform 7 A uual, the tranfer function i output/input = Y/F = W/( + gw ). Thi formula i one cae of what i often called Black formula. Example 27. Suppoe we have an open loop ytem, ay a circuit, with tranfer function W () = /(a 2 + b + c). If we add a feedback loop with gain g then uing Black formula the cloed loop tranfer function i /(a 2 + b + c) + g/(a 2 + b + c) = a 2 + (b + g) + c. Example 28. Feedback can turn an untable ytem into a table one. Conider the open loop ytem with tranfer function /( ). F () X() Thi ha a pole at =, o it i untable. Find all the value of a contant gain g that make the cloed loop ytem table. F V Y + gy g anwer: Black formula tell u that the cloed loop tranfer function i /( ) + g/( ) = + g. Thu the cloed loop ytem ha one pole and it at g. A long a g > thi pole i negative and the cloed loop ytem i table. Note. The oppoite i alo true: feedback can make a table ytem untable. 5 Stability of a function In thi dicuion of tability we will aume L(f) i a rational function. It i poible to deal with time hifted function, but it would unduly complicate the preentation, o in thi ection we will aume that all function have a Laplace tranform of the form F () = Q()/P (). 5. Definition of tability for a function We ay that a function i exponentially table if it goe to zero fater than ome decaying exponential. Formally: f(t) i exponentially table if it i of negative exponential order, i.e. if f(t) < Me at for ome poitive M and a. Exponential tability i equivalent to all the pole of f have negative real part.
8.3 The Laplace Tranform 8 For a rational function F () = Q()/P () a pole i called imple if it i not a repeated root of P (). If all the pole have negative real part except for at leat one imple pole with zero real part, then we ay that f(t) i marginally table. In thi cae, f(t) may not decay to, but it tay bounded. Example 29. The function e 3t, e 3t co(5t), t 3 e 2t are all exponentially table. Example 3. The function co(t),, e 2t + are all marginally table Example 3. The function t in(ωt) ha a double pole at = iω. It i not table or marginally table and it i not bounded a t get large. Example 32. The function e 2t ha a pole at = 2. It i not bounded and not table. 5.2 The final value theorem We are often intereted in the long term behavior of a ytem repone. If we are working in the frequency domain it help to have a way of determining the long term behavior directly from the Laplace tranform of a function. The next theorem i ueful in thi regard. Theorem. Final value theorem. Let f(t) be a function and F () = L(f). If all the pole of F () have negative real part then lim f(t) = lim F (). t An equivalent way of tating the criterion i that all the pole of F () have negative real part except for poibly a imple pole at =. Proof. Rather than give a formal proof we will indicate an eay way to undertand thi theorem. The criterion that the pole of F () have negative real part mean that the partial fraction decompoition of F () i of the form F () = A + B + B 2 C +... + + a + a 2 ( + a ) 2 +... + D ( + a ) 3 +..., where Re(a j ) <. Thi mean that f(t) ha the form f(t) = A + B e a t + B 2 e a 2t +... + C te a t +... + D t 2 e a t +... Only the firt of thee term doe not go to a t grow large. That i, On the frequency ide we have F () = A + B + a + lim f(t) = A. (3) t B 2 + a 2 +... C ( + a ) 2 +..., D ( + a) 3 +..., Only the firt of thee term doe not go to a goe to. That i, lim F () = A. (4) Taken together Equation 3 and 4 prove the final value theorem.
8.3 The Laplace Tranform 9 Example 33. The function f(t) = 2 + e 3t + te 5t ha Laplace tranform F () = 2/ + /( + 3) + /( + 5) 2. It ha imple pole at = and = 3 and a double pole at = 5. If atifie the criterion of the final value theorem and we can check: and lim f(t) = lim 2 + t t e 3t + te 5t = 2 lim F () = lim 2 + Thu we have verified the theorem in thi cae. + 3 + ( + 5) 2 = 2. Example 34. The function f(t) = 2 + e 3t + e 5t ha Laplace tranform F () = 2/ + /( + 3) + /( 5). It ha imple pole at =, = 3 and = 5. Becaue one of it pole i poitive it doe not atifie the criterion of the final value theorem. Checking the two limit: and lim f(t) = lim 2 + t t e 3t + e 5t = lim F () = lim 2 + + 3 + 5 = 2. We ee the limit are not the ame! Thi how u a valuable leon: Leon. You mut check that the hypothei of the final theorem hold, otherwie you can get a mileading reult. Here another example howing thi leon. Example 35. The function f(t) = in(2t) ha Laplace tranform F () = 2/( 2 + 4). It ha imple pole at = ±2i. Thee pole do not atiy the criterion of the final value theorem and we ee and lim f(t) = lim in(2t) doe not exit t t lim F () = lim 2 + 4 =. Again, the limit are not the ame, but thi doe not violate the final value theorem becaue it doe not apply to thi cae. A good example of the ue of the final value theorem i in the note on PID controller poted on the cla webite.
Propertie and Rule Function Laplace Tranform Table 8.3, January 25 Tranform f(t) F () = f(t)e t dt (Definition) a f(t) + b g(t) a F () + b G() (Linearity) e at f(t) F ( a) (-hift) f (t) F () f( ) f (t) 2 F () f( ) f ( ) f (n) (t) n F () n f( ) f (n ) ( ) tf(t) t n f(t) F () ( ) n F (n) () u(t a)f(t a) e a F () (t-tranlation or t-hift) u(t a)f(t) e a L(f(t + a)) (t-tranlation) (f g)(t) = t + f(τ) dτ t + f(t τ) g(τ) dτ F () G() F () Intereting, but not included in thi coure. f(t) F (σ) dσ t (integration rule) (The function table i on the next page.)
Laplace Table, 8.3 2 Function Table Function Tranform Region of convergence / Re() > e at /( a) Re() > Re(a) t / 2 Re() > t n n!/ n+ Re() > co(ωt) /( 2 + ω 2 ) Re() > in(ωt) ω/( 2 + ω 2 ) Re() > e at co(ωt) ( a)/(( a) 2 + ω 2 ) Re() > Re(a) e at in(ωt) ω/(( a) 2 + ω 2 ) Re() > Re(a) δ(t) all δ(t a) e a all coh(kt) = ekt + e kt 2 /( 2 k 2 ) Re() > k inh(kt) = ekt e kt 2 k/( 2 k 2 ) Re() > k 2ω (in(ωt) ωt co(ωt)) 3 ( 2 + ω 2 ) 2 Re() > t 2ω in(ωt) ( 2 + ω 2 ) 2 Re() > 2ω (in(ωt) + ωt co(ωt)) 2 ( 2 + ω 2 ) 2 Re() > u(t a) e a / Re() > t n e at n!/( a) n+ Re() > Re(a) Intereting, but not included in thi coure. π t Re() > t a Γ(a + ) a+ Re() >