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Additional resources for Fourier and Laplace Transform (Solutions)

Example text

So g(z) is continuous on G = C \ {1, −i, 2i}. 11 f (w) − f (z) w2 − z 2 = lim = lim (w + z) = 2z. w→z w − z w→z w−z a The derivative is 4(z − 1)3 ; the function is differentiable on C, so it is analytic on C. b The derivative is 1 − 1/z 2 ; since the function is not differentiable at z = 0, it is analytic on C \ {0}. c The derivative is ((z 3 + 1)(2z − 3) − 3z 2 (z 2 − 3z + 2))/(z 3 + 1)2 ; the function is not differentiable when z 3 = −1. √ Solving this equation one obtains that it is analytic on C \ {−1, 12 ± 12 i 3}.

2 a Table 7 gives f (t) = eat . b The convolution theorem implies that g(t) = eav ∗ ebv ; to determine g(t) explicitly, we need toR calculate this convolution. From the definition t it follows that g(t) = ebt 0 eτ (a−b) dτ . If a = b, then g(t) = teat . If a = b at bt then g(t) = (e − e )/(a − b). Next we can verify the convolution theorem. Write G(s) = (Lg)(s). If a = b, then G(s) = 1/(s − a)2 (table 7, no. 10). If a = b then G(s) = (1/(s − a) − 1/(s − b))/(a − b) (table 7, no. 2), which equals 1/(s − a)(s − b).

5) it follows for iω−1 t t −2t −1 1 2 4 that − (−t)e ↔ . Hence, h(t) = δ(t) + (−t)e − (t)e . i(−ω)+1 iω−1 3 3 b The impulse reponse h(t) is not causal, so the system is not causal. 7). 13 a From the differential equation we immediately obtain the frequency response: H(ω) = ω 2 − ω02 √ . ω 2 − i 2ω0 ω − ω02 b Write the cosine as a combination of exponentials, then it follows from eiωt → H(ω)eiωt that y(t) = (H(ω0 )eiω0 t + H(−ω0 )e−iω0 t )/2. However H(±ω0 ) = 0, so y(t) = 0 for all t. c Note that we cannot use the method from part b.