Gargicollege.in



7.Laplace transform of functions obtained with multiplication by tnDefinition 7.1Leibniz's rule states that integration and differentiation are interchangeable i.e.ddxy1y2fx,ydy=y1y2?f(x,y)?x dy with the condition that fx,y and ?f(x,y)?x are continuous within the interval (x1,x2) for the variable x and the corresponding interval (y1, y2) for the variable y. In general,ddxy1xy2xfx,ydy=y1xy2x?fx,y?xdy+ f(x,y2x)dy2dx-f(x,y1x)dy1dx In this problem ,the limits on t are independent of s. ∴ the last 2 terms are 0 . L(tn ft} Fs= L ft= 0∞e-st ftdt taking the derivative with respect to s on both the sides , we obtain dF(s)ds= dds0∞e-st ftdt dF(s)ds=0∞??s e-st ftdt ... Leibniz rule dF(s)ds=- 0∞e-st ft tdt=-L{tft} D.U. If we consider the derivative once more , we obtain d2Fds2= 0∞e-st ft t2dt = L{t2ft}We can consider the higher order derivatives and obtain ,Ltnft= (-1)ndnF(s)dsn (12)Example 7.1:Determine L{ te-4t}From the above theorem , ft= e-4t ;n=1 , Fs=1s+4d F(s)ds= -1(s+4)2 therefore L te-4t= -1(s+4)2 Example 7.2:Determine L{t2sin2t} In this case we need to consider the second derivative of L{sin2t}.d2ds22(s2+4) =dds -2ss2+42=-2s2+42+8s2s2+43 Lt2sin2t= -2s2+42+8s2s2+43 (13)Example 3: Find L{t2cosat}= -12d2ds2ss2+a2 D.U.8s3a2+s23 -6s(a2+s2)2 or L{t2cosat}= 8s3a2+s23 -6s(a2+s2)2 (14)8. Laplace transform of functions which are divided by tL { ftt} = s∞Fudu (15)Proof: F(s) is L {f(t)}Using the definition of Laplace transform, we obtain s∞Fudu=s∞[0∞f(t)e-tudt] du due to the convergence of the function we can change the order of integration to obtain[ 0∞s∞f(t)e-tudu] dt = 0∞-f(t)e-stdt-t = L {F(t)t} since the value of the integral at the upper limit is 0.Example 8.1:Evaluate L0x sinttdtComparing with the above formula , ft=sint and Lsintt= s∞1u2+1 du= π2-tan-1s=tan-11s We next apply the previous theorem L0xftdt=F(s)s ft= sintt and therefore L0x sinttdt= 1s tan-11s (16) 9. Laplace transform of periodic functionsWe consider a periodic function f(t) with a period T. Thus f(t+T) = f(t)substituting the expression for LT we have,0∞fte-stdt= 0Tfte-stdt+T2Tfte-stdt+2T3Tfte-stdt+.. in the second integral, we substitute, t-T = u; t= u+ T, dt= du; and the limits change to ( 0,T). Thus the second integral is 0Tfu+Te-s(u+T)du=e-sT0Tfue-sudu as f(u+T)=f(u) for the next integral we substitute t-2T=u; and the integral reduces to 0Tfu+2Te-su+2Tdu= e-2sT0Tfue-sudu similarly by substituting in the following integrals we obtain an infinite geometric series of the form(1+e-sT+ e-2sT+ e-3sT+…)0Tfue-sudu=[0Tfue-sudu]11-e-sT Thus Laplace transform of a periodic function f(t) with period T isLft=11-e-sT0Tfue-sudu (17)Example 9.1:Determine the Laplace transform of the periodic function ft which has a period of 4. ft=2 for 0≤t≤2 ;ft= -2 for 2≤t≤429184602451100022371055016500We evaluate 04fte-stdt=022e-stdt+24-2e-stdt = 2(1-e-2s)s+2s(e-4s-e-2s)Hence the Laplace transform of the given square wave is 2(1-2e-2s+4e-4s)1-e-4s Example 9.2: Determine the Laplace transform of a saw tooth wave with period 2ft= t for 0≤t≤2 ; Fig 3 saw tooth wave with period 202te-stdt= 2e-2s-s+02e-stsdt=2e-2s-s-[e-2s-1s2] Thus the Laplace transform of a sawtooth wave of period 2 is given by 2e-2s-s-[e-2s-1s2]1-e-2s Example 9.3 :Find the Laplace transform of the half wave rectified waveft=Vsin ωt for 0≤ωt≤πft= 0 for π≤ωt≤2π Fig 4 Output of half wave rectifier0Tfte-stdt=0πωVsin ωte-stdt+0=V[0π/ωe-t(s-iω)-e-t(s+iω)2idt] V2ie-t(s-iω)s-iω-e-t(s+iω)s+iωπ/ω0=V2i1-e-π(s-iω)/ωs-iω-1-e-π(s+iω)/ωs+iω V2i2-eπs/ωs-iω-2-e-πs/ωs+iω =V2i4iω+se-πsω-eπsω-iω(e-πsω+eπsω)s2+ω2 Thus the Laplace transform of the rectified half wave sine wave isV2i4iω+se-πsω-eπsω-iω(e-πsω+eπsω)s2+ω21-e-2sπ/ω 10. More Laplace transformsExample10.1: Determine Lsint We consider the sine series sint=t-t33!+t55!-…Lsint=Lt-t33!+t55!-… L{t}= Γ(32)/s32 ; Lt33!= Γ(52)3!s5/2 ; Lt55!= Γ(72)5!s7/2 Γ12=π where Γ represents the gamma function which is defined by the following integral. Γn+1=0∞xn-1e-xdx and Γn+1=nΓ(n) Substituting these values in eqn (1) , L{sint}=π2s3/2-πs5/2.12.32.13!+πs7/2.12.32.52.15!-… L{sint}= π2s3/21-14s+132s2-…= π2s3/21-14s+1(4s)22!-… L{sint}= π2s3/2e-14s (18) 11. Evaluation of integrals using Laplace transforms Example11.1: Evaluate0∞sintxxdx using Laplace transform. Lsintt= s∞1u2+1 du= π2-tan-1s=tan-11s We can change the variable tx=y and the result is unaltered.s=0 , 0∞sintxxdx = π2- tan-10= π2 L0∞sintxxdx= π2 (19) 12. Laplace Transforms of Integrals of Functions 1. Determine the Laplace transform of t∞e-ppdp Let ft= t∞e-ppdp Taking the derivative on both sides and considering the derivative of an integral of a function is the function, we obtain , f't= e-tt ; tf't= e-t Ltf't= 1s+1 ; using Ltngt=-1ndndsnGs Gs= L{gt} Thus Ltf't=-ddsLf't=-ddssFs-f0= -dds[sFs] as f0 is not a function of s . -dds[sFs] = 1s+1 integrating both sides with respect to s , -sFs=ln1s+1 orFs=1sln(s+1) Lt∞e-ppdp= 1sln(s+1) (20)Exercises:1. Determine the Laplace transform of the following functions.a) sinh24t b) sinh?at/t c) et(t+1) d) cos2(2 t)e) et(t+1)2 2. Determine the Laplace transform of the following functions. a) ft= (t-3)u(t-3) t>0 b) ft=0 for 0<t≤3 ft=1, 3≤t≤5 ft=0, t≥5 3. Determine the Laplace transform of a step function. ft=a for t≥2;ft=0 for 0<t≤a 4. Find Laplace transform of (a) ft= e3t-1t (b) ft= 2te-tsin2t 5. Evaluate the following integrals using Laplace transformsa) 0∞e-2tsin2tt dt b) 0∞te-2tcost dt 6. Draw the output of a full wave rectifier and determine its Laplace transform given that ft= sinωt , a is the period of the output.7. Verify the initial value theorem for the following functions a) t2-cost b) tsin2t 8. Verify the final value theorem for the following functionsa) e-t+4 b) e-tcos2t 9. Evaluate the Laplace transform of cos3t.10. Draw a triangular wave and determine the Laplace transform of the function representing this wave. Treat the time period of the wave as 2a and its amplitude as 1.11. Using L{sint}= π2s3/2e-14s and Lf'(t)=sFs-f(0), determine L{2COSt t}. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download