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 1 (10.01 KiB) Viewed 12 times
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. =
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
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!