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Laplace transform of a periodic function1 16 November 2001 If f (t) is a periodic function with period a we have f (t + a) = f (t). The Laplace transfrom of f (t) can be found by integrating over only one period: Z a 1 f (t)e−st dt (1) L(f ) = 1 − e−as 0 This equation can be derived as follows: Z ∞ f (t)e−st dt L(f ) = 0 Z Z a −st f (t)e dt + = 0 = ∞ Z X n=0 2a f (t)e−st dt + . . . a (n+1)a f (t)e−st dt (2) na Next, we integrate one term of the sum using the change of variable t0 = t − na: Z (n+1)a f (t)e −st dt = na Z a 0 f (t + na)e −st0 −nas 0 dt = 0 Z a 0 f (t0 )e−st −nas dt0 (3) 0 Hence, dropping the prime on t we get L(f ) = ∞ X e−san n=0 !Z a f (t)e−st dt (4) 0 and, finally, we use ∞ X xn = n=0 to make the replacement ∞ X e−san = n=0 and this proves the formula. 1 1 1−x (5) 1 1 − e−sa (6) Conor Houghton, houghton@maths.tcd.ie please send me any corrections. 1