EXAM TFY4280 Signal Processing Sat. 2 June 2012. 09:00 - NTNU

TFE4280 EXAMEN page 2 of 8For x(t) = ε(t) we need to find Laplace transform of h 1 (t):L {h 1 (t)} = L { ( e −t − e −2t) ε(t)} = 11 + s − 1s + 2 = s + 2 − s − 1(s + 1)(s + 2) = 1(s + 1)(s + 2)Y (s) = H(s)X(s) =0.5s − 1(s + 1) + 0.5(s + 2)X(s) = 1 s1s(s + 1)(s + 2) = A s + B(s + 1) + Cs + 2A = 1 2B = −1 C = 1 2check:0.5(s + 2)(s + 1) − s(s + 2) + 0.5s(s + 1)= =s(s + 1)(s + 2)0.5s 2 + 3/2s + 1 − s 2 − 2s + 0.5s 2 + 0.5ss(s + 1)(s + 2)=1s(s + 1)(s + 2) OK!Then:Y (s) = 0.5s − 1(s + 1) + 0.5(s + 2)y(t) = ε(t) ( 0.5 − e −t + 0.5e −2t)B) For random signal, first we consider what happens with the meanh 1 (t) = ( e −t − e −2t) ε(t)y(t) = x(t) ∗ h 1 (t)µ y (t) = E{x(t) ∗ h 1 (t)} = E{x(t)} ∗ h 1 (t) = µ x (t) ∗ h 1 (t)since(time independent)∫∞µ y = µ xh 1 (t)dt = µ x∫∞µ x (t) = µ x(e −t − e −2t) ∫∞ε(t)dt = µ x(e −t − e −2t) dt = 0.5µ x−∞For the ACF:−∞ϕ yy (τ) = ϕ hh (τ) ∗ ϕ xx (τ)ϕ hh (τ) = h 1 (τ) ∗ h 1 (−τ)0Where ϕ hh (τ) can be calculated from analytical expression for h 1 .Q2 (25p) Consider LTI system described by:[ d2dt + 5 d ]2 dt + 4 y(t) =[2 d ]dt + 6 x(t)A) (10p) Find the impulse response h(t)