Find y(t)
\(\displaystyle \frac{dy}{dt} + 3y = 6u(t-1) - 3u(t-2)\)
\(\displaystyle y(0) = 0\)
Here is my work:
Note: u(t) is the unit step function
\(\displaystyle Y = \frac{3}{s(s+3)}(2e^{-s}-e^{-2s})\)
\(\displaystyle H(s) = \frac{3}{s(s+3)} = \frac{1}{s} - \frac{1}{s+3}\)
\(\displaystyle h(t) = 1 - e^{-3t}\)
\(\displaystyle Y = H(s)(2e^{-s} - e^{-2s}) = 2H(s)e^{-s} - H(s)e^{-2s}\)
\(\displaystyle y(t) = 2h(t-1)u(t-1) - h(t-2)u(t-2)\)
\(\displaystyle y(t) = 2[1-e^{-3(t-1)}] - [1 - e^{-3(t-2)}]u(t-2)\)
The correct answer given by my professor is:
\(\displaystyle y(t) = 2 - 2e^{-3t}-u(t-1)(1-e^{-3t + 3}\)
\(\displaystyle \frac{dy}{dt} + 3y = 6u(t-1) - 3u(t-2)\)
\(\displaystyle y(0) = 0\)
Here is my work:
Note: u(t) is the unit step function
\(\displaystyle Y = \frac{3}{s(s+3)}(2e^{-s}-e^{-2s})\)
\(\displaystyle H(s) = \frac{3}{s(s+3)} = \frac{1}{s} - \frac{1}{s+3}\)
\(\displaystyle h(t) = 1 - e^{-3t}\)
\(\displaystyle Y = H(s)(2e^{-s} - e^{-2s}) = 2H(s)e^{-s} - H(s)e^{-2s}\)
\(\displaystyle y(t) = 2h(t-1)u(t-1) - h(t-2)u(t-2)\)
\(\displaystyle y(t) = 2[1-e^{-3(t-1)}] - [1 - e^{-3(t-2)}]u(t-2)\)
The correct answer given by my professor is:
\(\displaystyle y(t) = 2 - 2e^{-3t}-u(t-1)(1-e^{-3t + 3}\)
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