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Standard XII

Physics

Energy of a Capacitor

A stationary ...

Question

A stationary random process X(t) is applied to the circuit shown below:

If the mean value of the input process X(t) is 5, then the mean value of the output process Y(t) will be

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Solution

The correct option is D 4
$E [X (t)] = 5$

$E [Y (t)] = H (0) E [X (t)]$

$H (0) H (s)_{s = 0}$

H(s)= Transfer funciton of the given circuit

From the given circuit,
$H (0) = \frac{4 M Ω}{4 M Ω + 1 M Ω} = \frac{4}{5} = 0.80$

$So, [Y (t)] = 0.80 \times 5 = 4$

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A stationary random process X(t) is applied to the circuit shown below: If the mean value of the input process X(t) is 5, then the mean value of the output process Y(t) will be

The correct option is D 4E[X(t)]=5 E[Y(t)]=H(0)E[X(t)] H(0)H(s)s=0 H(s)= Transfer funciton of the given circuit From the given circuit, H(0)=4MΩ4MΩ+1MΩ=45=0.80 So, [Y(t)]=0.80×5=4

A stationary random process X(t) is applied to the circuit shown below:

If the mean value of the input process X(t) is 5, then the mean value of the output process Y(t) will be

The correct option is D 4
$E [X (t)] = 5$

$E [Y (t)] = H (0) E [X (t)]$

$H (0) H (s)_{s = 0}$

H(s)= Transfer funciton of the given circuit

From the given circuit,
$H (0) = \frac{4 M Ω}{4 M Ω + 1 M Ω} = \frac{4}{5} = 0.80$

$So, [Y (t)] = 0.80 \times 5 = 4$