Consider a causal LTI system characterised by differential equation dytdt-16yt - 3xt- The response of the system to the input xt - 3e-t-3 ut- where ut denotes the unit step function- is

The differential equation, nbsp; dy(t)dt+16 y(t) = 3 x(t) nbsp; So, sY(s) + 16Y(s) = 3 X(s) Y(s) = 3 X(s) ( s+1/6 ) X(s) = 3 ( s+1/3 ) So, Y(s) = 9( s+ ...

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Consider a causal LTI system characterised by differential equation $\frac{d y (t)}{d t} + \frac{1}{6} y (t) = 3 x (t)$ . The response of the system to the input $x (t) = 3 e^{- t / 3} u (t)$ , where u(t) denotes the unit step function, is

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Solution

The differential equation,
$\frac{d y (t)}{d t} + \frac{1}{6} y (t) = 3 x (t)$

$S o, s Y (s) + \frac{1}{6} Y (s) = 3 X (s)$

$Y (s) = \frac{3 X (s)}{(s + \frac{1}{6})}$

$X (s) = \frac{3}{(s + \frac{1}{3})}$

$S o, Y (s) = \frac{9}{(s + \frac{1}{3}) (s + \frac{1}{6})}$

$= \frac{54}{(s + \frac{1}{6})} - \frac{54}{(s + \frac{1}{3})}$

$S o, y (t) = (54 e^{- 1 / 6 t} - 54 e^{- 1 / 3 t} u (t))$

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Consider a causal LTI system characterised by differential equation dy(t)dt+16y(t)=3x(t). The response of the system to the input x(t)=3e−t/3u(t), where u(t) denotes the unit step function, is

The differential equation, dy(t)dt+16y(t)=3x(t) So, sY(s)+16Y(s)=3X(s) Y(s)=3X(s)(s+16) X(s)=3(s+13) So, Y(s)=9(s+13)(s+16) =54(s+16)−54(s+13) So, y(t)=(54e−1/6t−54e−1/3tu(t))

Consider a causal LTI system characterised by differential equation $\frac{d y (t)}{d t} + \frac{1}{6} y (t) = 3 x (t)$ . The response of the system to the input $x (t) = 3 e^{- t / 3} u (t)$ , where u(t) denotes the unit step function, is