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Tolerance Design 579

Example 16.6. RL Circuit A 100-V, f-Hz power supply across a resistance R
in series with an inductance L will result in a current of y amperes:
100
y
R (2 fL ) 2
2
Assuming that f 50 Hz, the nominal value for the resistor is T R 9.5 ,
the current tolerance R 1.0 , the nominal value for the inductor, T L
0.01 H (henry), and the current tolerance L 0.006 H.
If the customer’s tolerance for the circuit is y 10.0 1.0 A, we would
like to ascertain whether the design parameter tolerances are adequate.
From Eqs. (16.2) and (16.3)
100
Max(y) Max
R (2 fL ) 2
2
100
11.64
(9.5 1.0) (2
3.1416
5 0
(0 .01 0.006)) 2

2
100
Min(y) Min
R (2 fL ) 2
2
100
8.59

(9.5 1.0) (2
3.1416
5 0
(0 .01 0.006)) 2

2
100
E(y) E
R (2 fL ) 2
2
100
9.99

(9.5) (2
3.1416
50
(0 .01)) 2

2
Clearly, from a worst-case tolerance perspective, the design parameter tol-
erance is not adequate to ensure customer tolerance.
16.2.2 Nonlinear worst-case
tolerance analysis
If the transfer function equation y f(x 1 ,x 2 ,…,x i ,…,x n ) is nonlinear, the
tolerance analysis is difficult. From the Taylor expansion formula
(Chap. 6) we obtain
∂f ∂f ∂f ∂f
y x 1 x 2 ... x i ... x n (16.4)
∂x 1 ∂x 2 ∂x i ∂x n
According to Chase and Greenwood (1988), the worst-case tolerance
limit in the nonlinear case is
... ...
∂f
∂f
∂f
∂f
0 1 2 i n (16.5)
∂x 1 ∂x 2 ∂x i ∂x n