By William G. Andrew, H.B. Williams

This quantity covers software engineering details, together with time-saving charts, tables, graphs, and calculations for designers, engineers, and operators.

**Read or Download Applied Instrumentation in the Process Industries, Volume 3: Engineering Data and Resource Manual PDF**

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**Additional info for Applied Instrumentation in the Process Industries, Volume 3: Engineering Data and Resource Manual**

**Example text**

7. 1. Harmonic analysis Let us consider a stable LTI whose impulse response h(θ) is canceled after a period of time t R . For the models of physical systems, this period of time t R is in fact rejected infinitely; however, for reasons of clarity, let us suppose t R as finite, corresponding to the response time to 1% of the system. When this system is subject to a harmonic excitation x ( t ) = Ae2π jf0t from t = 0 , we obtain: y (t ) = t ∫0 h (θ ) Ae 2π jf0 ( t −θ ) dθ = Ae2π jf0t t ∫0 h (θ ) Ae −2π jf0θ dθ For t > t R , the impulse response being zero, we have: t ∫0 h(θ ) e −2π jf0θ dθ =H ( f0 ) = +∞ ∫0 h(θ ) e−2π jf0θ dθ = H ( f0 ) e jΦ ( f0 ) and hence for t > t R , we obtain y(t ) = AH ( fo ) e2π jf0 t = A H ( f0 ) e j (2π f0 t +Φ ( f0 )) .

Unit-step response The theorems of initial and final values make it possible to easily comprehend the asymptotic features of the unit-step response: zero initial value, final value equal to K, tangent at the origin with zero slope. Based on the value of ξ with respect to 1, the transfer function poles have a real or complex nature and the unit-step response looks different. ξ > 1 : the transfer function poles are real and the unit-step response has an aperiodic look (without oscillation): ( ⎛ ⎛ − ξ + ξ 2 −1 ⎜ 1 i(t ) = K ⎜ 1 − ⎜⎜ (1 − a) e ⎜ 2⎜ ⎝ ⎝ where a = ) Tt 0 ( + (1 + a) e − ξ − ξ 2 −1 ) Tt ⎞⎟ ⎞⎟ ξ ξ 2 −1 The tangent at the origin is horizontal.

7. 1. Harmonic analysis Let us consider a stable LTI whose impulse response h(θ) is canceled after a period of time t R . For the models of physical systems, this period of time t R is in fact rejected infinitely; however, for reasons of clarity, let us suppose t R as finite, corresponding to the response time to 1% of the system. When this system is subject to a harmonic excitation x ( t ) = Ae2π jf0t from t = 0 , we obtain: y (t ) = t ∫0 h (θ ) Ae 2π jf0 ( t −θ ) dθ = Ae2π jf0t t ∫0 h (θ ) Ae −2π jf0θ dθ For t > t R , the impulse response being zero, we have: t ∫0 h(θ ) e −2π jf0θ dθ =H ( f0 ) = +∞ ∫0 h(θ ) e−2π jf0θ dθ = H ( f0 ) e jΦ ( f0 ) and hence for t > t R , we obtain y(t ) = AH ( fo ) e2π jf0 t = A H ( f0 ) e j (2π f0 t +Φ ( f0 )) .