The measured system shown in the figure uses thee sub-system in cascade whose gains are specified as $G_{1}, G_{2} a n d 1 / G_{3}$ . The relative small errors associated with each respective subsystem $G_{1}, G_{2} a n d G_{3}$ are $ε_{1}, ε_{2} a n d ε_{3}$ . The error associated with the output is:

ε_{1} - ε_{2} + ε_{3}

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\frac{ε_{1} . ε_{2}}{ε_{3}}

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ε_{1} + ε_{2} - ε_{3}

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ε_{1} + ε_{2} + ε_{3}

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Solution

The correct option is D $ε_{1} + ε_{2} + ε_{3}$
$\frac{d G_{1}}{G_{1}} = ϵ_{1}, \frac{d G_{2}}{G_{2}} = ϵ_{2}, a n d \frac{d G_{3}}{G_{3}} = ϵ_{3}$

$O u t p u t (y_{0}) = \frac{G_{1} G_{2}}{G_{3}} x$

where,
x=input

$l n y = l n G_{1} + l n G_{2} - l n G_{3} + l n x$

Differentiating both sides,

$\frac{d y_{0}}{y} = \frac{d G_{1}}{G_{2}} + \frac{d G_{2}}{G_{2}} - \frac{d G_{3}}{G_{3}} + \frac{d x}{x}$

No error is specified in input so,
$\frac{d x}{x} = 0$
$\frac{d y_{0}}{y} = ϵ_{1} + ϵ_{2} - ϵ_{3}$
But when the quantities are either multiplied or divided, their relative errors are always added. Hence, option (d) is correct.

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