The transfer function of a low-pass filter is G(s) = Y(5) U (5) K = 1+5 The input to the filter is u(t) = sin 3t, -
Posted: Sat May 21, 2022 1:09 am
K value in G(s) is K = 4. Using the partial fraction expansion approach, derive the filter output y(t) that contains both the transient response and the steady-state response. j+e-16 Hint: You are aware of Euler's identities: = cos e and 916-9-29 = sin e 2j 2
The transfer function of a low-pass filter is G(s) = Y(5) U (5) K = 1+5 The input to the filter is u(t) = sin 3t, -<t<0. =
Posted: Sat May 21, 2022 1:09 am
K value in G(s) is K = 4. Using the partial fraction expansion approach, derive the filter output y(t) that contains both the transient response and the steady-state response. j+e-16 Hint: You are aware of Euler's identities: = cos e and 916-9-29 = sin e 2j 2