Solutions to Laplace Transform Problems in Chapter 12, Assignments of Microelectronic Circuits

Download Solutions to Laplace Transform Problems in Chapter 12 and more Assignments Microelectronic Circuits in PDF only on Docsity! 97 CHAPTER 12 PROBLEMS 12.1 If f(t) = e-at sin bt, find F(s) using (a) the definition of the Laplace Transform and (b) the fact that L[e-at f(t)] = F(s + a). 12.2 Find f(t) if F(s) is given by the expression ( ) ( ) ( ) ( )6s4s2s s24s +++ =F 12.3 Find f(t) if F(s) is given by the expression ( ) ( ) ( )20s8ss 4s4s 2 ++ + =F 12.4 Find f(t) if F(s) is given by the expression ( ) ( ) ( ) ( )3s1s2s 2s12s 2 +++ + =F 12.5 Given the function ( ) ( ) ( ) ( )4s2ss 10s24s ++ + =F Find the initial and final values of the function by evaluating it in both the s-domain and time domain. 98 CHAPTER 12 SOLUTIONS 12.1 (a) By definition ( ) ( ) dtetfs st 0 − ∞ ∫=F And since f(t) = e-at sin bt ( ) dtebtsines st 0 at − ∞ − ∫=F Using Euler’s identity ( ) ( ) ( ) ( ) ∫ − =       − ∫= ∞ ++−−+− − ∞ +− 0 tjbastjbas jbtjbt 0 tas dt j2 ee dt j2 eeesF Evaluating the integral ( ) 22 bas b jbas 1 jbas 1 j2 1 ++ =       ++ − −+ = (b) In this case f(t) = sin bt. Then ( ) dtbtsines 0 st ∫= ∞ −F Again, using the Euler identity ( ) ( ) ( )( )∫ −=       − ∫= ∞ +−−− − ∞ − 0 tjbstjbs jbtjbt 0 ts dtee j2 1 dt j2 eeesF Evaluating the integral 101 Multiplying the entire equation by s and evaluating it at s = 0 yields ( ) 0 0 0s 2 k 5 4 k 20s8s 4s4 = = ++ + = Using the same procedure for k1, we obtain ( ) ( ) ( ) 1 1 1 1 1 2j4s k56.206 5 1 k 5 j2 k j2 1 k j2 1 k 2j4ss 4s4 =°∠ = +− = − − = +− = ++ + +−= Then, we know that * 1k56.2065 1 =°−∠ Now using the fact that ( )θ+=      ++ θ−∠ + −+ θ∠ − btcosek2 jbas k jbas k at 1 111-L The function f(t) is ( ) ( ) ( )tu56.206t2cose 5 2 5 4tf t4       °++= − 12.4 In order to perform a partial fraction expansion on the function F(s), we need to factor the quadratic term. We can use the quadratic formula or simply note that (s + 1) (s + 1) = s2 + 2s + 1. Therefore, F(s) can be expressed as 102 ( ) ( ) ( ) ( )3s1s 2s12s 2 ++ + =F or in the form ( ) ( ) ( ) ( ) ( ) 3s k 1s k 1s k 3s1s 2s12s 22 1211 2 + + + + + = ++ + =F If we now multiply the entire equation by (s + 1)2, we obtain ( ) ( ) ( ) 3s 1sk k1sk 3s 2s12 22 1211 + + +++= + + Now evaluating this equation at s = -1 yields ( ) 12 12 1s k6 k 3s 2s12 = = + + −= In order to evaluate k11 we differentiate each term of the equation with respect to s and evaluate all terms at s = -1. Note that the derivative of k12 with respect to s is zero, the derivative of the last term in the equation with respect to s will still have an (s + 1) term in the numerator that will vanish when evaluated at s = -1, and the derivative of the first term on the right side of the equation with respect to s simply yields k11. Therefore, ( ) ( ) ( ) ( ) ( ) ( ) 11 11 1s 2 11 1s k3 k 3s 12s12123s k 3s 2s12 ds d = = + +−+ =      + + −= −= Finally, ( ) ( ) 2 2 3s 2 k3 k 1s 2s12 =− = + + −= And therefore, F(s) can be expressed in the form 103 ( ) ( ) 2s 3 1s 6 1s 3s 2 + − + + + =F Using the transform pairs, we find that ( ) [ ] ( )tue3te6e3tf t2tt −−− −+= 12.5 First, let us use the Theorems to evaluate the function in the s-domain. The initial value can be derived from the Theorem lim f(t) = lim sF(s) t → 0 s → ∞ Therefore, ( ) ( ) ( ) ( ) 0 s 8 s 61 s 240 s 24 s lim 8s6s 240s24 s lim 4s2s 10s24 s lim sslim 2 2 2 =             ++ + ∞→ =       ++ + ∞→ =       ++ + ∞→ =F The final value is derived from the expression lim f(t) = lim sF(s) t → ∞ s → 0 Hence, ( ) ( ) ( ) ( ) 30 8 240 4s2s 10s24 0s lim 0s sslim = =       ++ + → = → F The time function can be derived from a partial fraction expansion as