Question

The system given by equation y(t-2a)-3y(t-a)+2y(t)=x(t-a) is?​

Answer 1

Hello friend

Explanation: Given ,y(t)=Cos(t) – u(t) and x(t) = 5Sin(t) – u(t),

Explanation: Given ,y(t)=Cos(t) – u(t) and x(t) = 5Sin(t) – u(t),Hence, transfer function H(s)=[ss2+1−1s][5s2+1−1s]=−1s(s2+1)(5s−s2−1)s(s2+1)=1S2−5S+1

Explanation: Given ,y(t)=Cos(t) – u(t) and x(t) = 5Sin(t) – u(t),Hence, transfer function H(s)=[ss2+1−1s][5s2+1−1s]=−1s(s2+1)(5s−s2−1)s(s2+1)=1S2−5S+1Roots of equation s2 – 5s + 1 = 0 is s = 4.79, 0.208.

Given, \frac{d^2}{dt^2} y(t)-\frac{d}{dt} y(t)+y(t)-\int_0^t x(t)dt=x(t)

Given, \frac{d^2}{dt^2} y(t)-\frac{d}{dt} y(t)+y(t)-\int_0^t x(t)dt=x(t)Now, Taking laplace, we get, (s^2-s+1)Y(s)=(1+\frac{1}{s})X(s)

Given, \frac{d^2}{dt^2} y(t)-\frac{d}{dt} y(t)+y(t)-\int_0^t x(t)dt=x(t)Now, Taking laplace, we get, (s^2-s+1)Y(s)=(1+\frac{1}{s})X(s)H(s)=\frac{s+1}{s(s^2-s+1)}

Given, \frac{d^2}{dt^2} y(t)-\frac{d}{dt} y(t)+y(t)-\int_0^t x(t)dt=x(t)Now, Taking laplace, we get, (s^2-s+1)Y(s)=(1+\frac{1}{s})X(s)H(s)=\frac{s+1}{s(s^2-s+1)}Roots of s3-s2+s=0, are 0,.5±0.866

hope it will help you

THANK YOU