Laplace Transforms - Coursework Example

work on Laplace Transforms Q1. Laplace Transforms is an integral transforms used to transform a real valued variable t (t?0) into a complex valued phenomenon s. The function of t say f(t) is transformed into a function of s say F(s) such that the functional relationship is written as L[f(t)]=F(s) where f(t) is of exponential order and the limit for f(t) exists. Laplace transforms are used to solve many complicated differential equations which otherwise are difficult to solve. The foremost advantage in using Laplace transforms is it convert difficult integral and differential equations into simple algebraic equation which are easily solvable than the original differential and integral equations which are otherwise very difficult to solve.

The Laplace transforms formula is given below: L[f(t)]= here f(t) is piecewise continuous and the derivative of f(t) should exist in the domain of t (t?0) Q2. L[]==, s>k. Q3. The following are the Laplace transforms for the given S.No. Function y(t) Laplace Transform L{y(t)} Region of Convergence (ROC) 1 y(t) Y(s) n/a 2 a (=a constant) a/s Re(s)>0 3 t 1/s2 Re(s)>0 4 t2 2/s3 Re(s)>0 5 tn[n>0 n?N] [n(n–1)(n–2)….1]/sn+1 n!/sn+1 Re(s)>0 n>–1 6 e–at 1/(s+a) s>a 7 e–attn n!/(s+a)n+1 Re(s)>0 8 H(t–a) e–as/s Re(s)>0 9 ?(t) 1 For all s 10 ?(t–a) 1 For all s 11 cos(?t) s/(s2+?2) Re(s)>0 12 sin(?t) ?/(s2+?2) Re(s)>0 13 e–at cos(?t) (s+a)/[(s+a)2+?2] Re(s)>0 14 e–at sin(?t) ?/[(s+a)2+?2] Re(s)>0 15 dy/dt sY(s)–y0 Y(s)=L[y(t)],y0=y(0) 16 d2y/dt2 s2Y(s)–sy0–y’(0) y’(0)=dy/dt|t=0 Q4. (i) 3/s–5/s2+12/s3 (ii) 10/(s+4)+7/(s–1/2)=10/(s+4)+14/(2s–1) (iii) 2 (iv) 4s/(s2+9)+12/(s2+4) (v) 36/[(s+1)2+9] (vi) sY(s)–f0 where Y(s)=L[y(t)] and y0=y(0) Q5. (i) ?(t)+6+4t (ii) e2t (iii) 2e–5t+sin3t (iv) 1/3(e2t–e–t) (using partial fractions) (v) et/2sin5t (using shifting theorem) (vi) e–3tcos(2t) {using shifting theorem} Q6.

The final solution for (i) y(t)=3e2t–2e–t The procedure is: given that y’–2y=6e–t Taking laplace transform on both sides we get sY(s)–y(0)–2Y(s)=6/(s+1) (s–2)Y(s)=1+6/(s+1) since y(0)=1 (s–2)Y(s)=(s+7)/(s+1) Y(s)=(s+7)/(s–2)(s+1)–––––(*) By splitting into partial fractions, the right hand side becomes [3/(s–2)]–[2/(s+1)] by taking inverse Laplace transform on both sides of (*) we get y(t)=3e2t–2e–t which is the final solution. (ii) The solution is y(x)=3e–3xcos(2x) Given that y’’(x)+6y’(x)+13y=?(1) taking Laplace transform on both sides we get s2Y(s)–sy(0)–y’(0)+6sY(s)–6y(0)+13Y(s)=1 since L[?(1)]=1 for any value of t. s2Y(s)–3s+10+6sY(s)–18+13Y(s)=1 since y(0)=3; y’(0)=–10 by combining like terms we get Y(s)[s2+6s+13]=3s–10+18+1=3s+9 Y(s)=(3s+9)/(s2+6s+13) =3[(s+3)/{(s+3)2+22}] Taking inverse Laplace transform on both sides and applying shifting theorem we get y(x)=3e–3xcos(2x) Q7.

The use of Laplace Transforms in Engineering is very much appreciable and trustworthy. Many Engineering applications which are not solvable using ordinary differential equations can be solved by Laplace transforms provided the Laplace transforms exist. The Laplace transforms seek the exponential order to exist. Many real life Engineering applications can be brought into the purview of exponential order and also it insists the derivative to exist and piecewise continuity must be present. The foremost advantage of Laplace transforms is it converts difficult differential and integral equations to simple algebraic equations which is easily solvable rather than the original differential/integral equations.

It can be used in electrical circuit theory to know the original current passing through a circuit for a given electromotive force, inductance, resistance and capacitance. It is used in vibration theory because it resolves function into moments. It is used to find the bending moment of any RC column in Civil Engineering applications. In circuit theory it is more used. In harmonic oscillators, optical devices also Laplace transforms are used. In spring vibrations relating to Mechanical Engineering problems it is much used.

In chemical reactions involving differential equations, Laplace transforms are used. Because of the nature of converting from time domain to frequency domain it is used in Biological (genetical) and stochastic applications. Because of its usage in solving ordinary and partial differential equations involving complex nature, it is widely used in Engineering since many Engineering problems involve complex differential and integral equations. Q8. List of books referred: 1. “Advanced Engineering Mathematics” by Erwin Kreyszig, 9th Edition, 2005, Wiley Publications.

ISBN: 978–0–470–00820–1. 2. “Advanced Engineering Mathematics” M.D.Greenberg, 2nd Edition, 1998, Prentice Hall. ISBN: 0133214311.