6.0 Laplace Transforms In many applications we have interconnections of LTI systems. We can determine the output of the system by using convolution in the time-domain, but this often proves to be difficult when we have more than just a few interconnecting systems. Sometimes we don’t want to just compute the output, but rather we want to be able to determine properties of the system in a simple way. We will utilize Laplace transforms for this, though in some applications the use of Fourier transforms may be more appropriate. 6.1 Laplace Transform Definitions The two-sided Laplace transform of a signal()xtis defined to be () ()stXsxte∞−−∞=∫dt while the one-sided Laplace transform is defined as 0() ()stXs xte dt−∞−=∫ The only difference between the two definitions is the lower limit, and one and two sided refers to integrating one on side of zero or both sides of zero. In this course we will only use the one-sided transform, and when we refer to the Laplace transform we mean the one sided transform. However, you should be aware the two-sided transform has some use when we are dealing with noncausal systems. Looking at the form of the integral, since the exponent must be dimensionless, we can conclude that the variable has the units of 1/time. s There are a few conventions we need to know about. First of all, the lower limit on the one sided Laplace transform is generally written as 0−, or starting at a time just before zero. This is particularly useful when determining the Laplace transform of an impulse centered at zero. Secondly, a very common convention is to use lower case letters for time-domain functions, and capital letters for the corresponding transform domain. We usually write )(()xtXs↔or {}() ()xtX=L s to show that ()xt and ()Xssare transform pairs. We dill denote the Laplace transform operator as . Finally, the complex variable is sometimes written in terms of its real and imaginary parts as Lsjσω=+ . This is particularly useful in determining if the integral is finite (or can be made to be finite) or is infinite. 6.2 Basic Laplace Transforms Let’s start off by determining the Laplace transform of some basic signals. As you will see sometimes we need to put conditions on σ to be sure the integral converges. This condition defines the region of convergence. Example 6.2.1. For )(()xt tδ= we have 000() ( ) ( ) 1ssXs e d e dλδλ λ δλ λ−∞∞−−==∫∫= ©2009 Robert D. Throne 1since the delta function is contained in the region of integration. Hence we have the transform pair () () () 1xt t X sδ=↔= Example 6.2.2. For 0() ( )xtttδ=−, where , we have 00t ≥000000() () ()stsstXstedetdλδλ λ δλ λ−∞∞−−−−=− = −=∫∫e since again the delta function is located in the region of integration. Hence we have the transform pair 00() ( ) ()stxtttXseδ−=−↔ = Note that if the integral will be zero. 00t < Example 6.2.3. For () ()xtut= we have 000() ( )sssedeXs usedλλλλλλλ λ−=∞∞∞−−−===−=∫∫ At this point we cannot really evaluate this integral unless we put some conditions on σ. Let’s make the substitution sjσω=+ and we have ()00()jjeeeeXssssλλσωλσλωλσλωλλλ0jλλ=∞=∞−+ −+ −=∞======−−− Using Euler’s identity cos( ) sin( )jejωλωλω=+λ We can determine the magnitude of jeωλas 22||cos()sin()jeωλωλ ωλ1=+= Hence the term jeωλdoes not contribute to the convergence of the integral. That means that ω does not contribute to the convergence of the integral. That leaves us withσ. If 0σ>, then the exponent in the exponential is negative, when we evaluate it at the limit of infinity we get zero. Hence we have the transform pair 1() () () , 0xt ut X ssσ=↔= > Note that the condition 0σ>, or the real part of must be positive, defines the region of convergence. Hence the integral will converge if and we can rewrite the transform pair as s{s} > 0ℜ1() () () , {} 0xt ut X s ss=↔ =ℜ> Example 6.2.4. For 0() ( )xtutt=− we have 00000() ( ) , {} 0stssstteeXs u ted ed sssλλλλλλλλ−=∞∞∞−−−−==− = = = ℜ>−∫∫ The region of convergence is the same as for the non-delayed unit step. So we have the transform pair ©2009 Robert D. Throne 200() ( ) () , {} 0stext ut t X s ss−=−↔ = ℜ> Example 6.2.5. For we have (())atxt e ut−=()()000() ( ) d()saas saeXs e u e d esaλλλλ λλλλ λ−=∞∞∞−+−− −+====−+∫∫ Now in order for the integral to converge, we need{}sa 0ℜ+>.We can rewrite this as {} {} 0saℜ+ℜ >or . Hence the region of convergence is defined as and we have the Laplace transform pair {} {}sℜ>−ℜa a{} {}sℜ>−ℜ1() ()() , {} {}atut X s s asxt ea−↔= ℜ>−ℜ+= Example 6.2.6. For 0() cos( ) ()xttutω= we will need to use Euler’s identity in the form 000cos( )2jt jteetωωω−+= Computing the Laplace transform we have 000000000() ()() ()0000001() cos( )( )211 1122 22(jjssjs jsjs jsXs u e d e e deeed e djs jseωλ ωλλλλλωλ ωλωλ ωλλλωλ λ λ λλλωω−∞∞−−−)=∞=∞∞−−−−∞==++⎡⎤==+⎣⎦=++=−−∫∫∫∫ Both integrals will converges if , and then we have {} 0sℜ>0022 2200 011 1)122(12sj sjssj sj sXss0ωωωωωω++−+=−+ +==+ The Laplace transform pair is then 0220() cos( ) () () , {} 0sxt t ut X s ssωω=↔=ℜ+> Example 6.2.7. Let’s assume we want to find the Laplace transform of the derivative of ()xt. Then we have 0() ( )sdx t dxeddt dλλλλ−∞−⎧⎫=⎨⎬⎩⎭∫L ©2009 Robert D. Throne 3In order to evaluate this we will need to use integration by parts. For two functions()uλand ()vλ, we can write . Rearranging this we get the usual form for integration by parts, . For our integral we have ()d uv vdu udv=+vdu=−∫udv uv∫() ()dv dxddλλλλ= so or dv dx=() ()vxλλ=. We also have ( )sueλλ−= , so ()sdusedλλλ−=− or sdu se dλλ−=− Combining these we have 000() ( )() () (0) ()ss sdx t dxed x e sx ed x sXsdt dλλλ λλλλλ λ λλ−−−∞∞=∞−− −−=⎧⎫==+=−+⎨⎬⎩⎭∫∫L So we have the transform pair ()() (0)dx tsX s xdt−⎧⎫=−⎨⎬⎩⎭L Table 6.1 summarizes some common Laplace transforms. 6.3 Laplace Transforms of RLC Circuits In this section we will determine the Laplace transforms of resistors, capacitors, and inductors, and then use these relationships in some examples to