### Process Transfer Functions AnalysisConsider the following process transfer functions. Determine the characteristic parameters (not the poles and zeros) of each process. Describe the dynamic response that would be expected for a unit step change input, and identify any special features of the response.#### (a) Transfer Function \( G(s) \)\[ G(s) = \frac{8e^{-3s}}{16s+4} \]#### (b) Transfer Function \( G(s) \)\[ G(s) = \frac{15}{3s+1} - \frac{18}{4s+1} \]### Analysis and Expected Responses1. **Transfer Function (a)**: - **Form**: This transfer function includes a delay component \( e^{-3s} \) and a first-order system \( \frac{8}{16s+4} \). - **Characteristic Parameters**: The system has a time delay of 3 units and a time constant derived from \( 16s + 4 \). - **Dynamic Response**: We expect an initial delay in response, followed by a gradual rise to a steady-state value. The time delay indicates that the system will not respond immediately to a unit step input, but will instead start to respond after 3 time units.2. **Transfer Function (b)**: - **Form**: This is a combination of two first-order systems with different time constants. - **Characteristic Parameters**: The first component has a time constant \(\frac{1}{3}\) from \( 3s + 1 \) and a gain of 15. The second component has a time constant \(\frac{1}{4}\) from \( 4s + 1 \) and a gain of -18. - **Dynamic Response**: The system exhibits dynamics characterized by the superposition of the two responses. One effect will rise with a characteristic time of \(\frac{1}{3}\) and then decay, while the opposing effect will rise and decay with a time constant of \(\frac{1}{4}\). The negative sign indicates that this component contributes an inverse response, leading to potential overshoot or undershoot in the overall system response.### Special Features- **Delay in (a)**: The delay feature in (a) significantly impacts the initial response timeframe.- **Interaction in (b)**: The interacting first-order systems in

Consider the process transfer function given We have to determine the characteristics (not the poles and zeroes) of each process The transfer system of any process is G(s)=Y(s)X(s) where Y(s) is the output in s-domain X(s) is the input in s-domain (a) The Given process transfer function is G(s)=8e-3s16s+4 .......(1) =8e-3s4(4s+1)=2e-3s(4s+1) ......(2) As we know the transfer function for the first-order system with dead time G(s)=kpe-tdsτs+1 .....(3) Compare the equation (2) and (3) the characteristics parameter obtained is kp=2, Time delay td=3 Time constant τ=4 Gain =kp=2 The value of Y(s)=8e-35s(16s+4) Use partial fraction 1s(16s+4)=As+B16s+4A=14 and B=-4 1s16s+4=14s+416s+4=14s-14s+1=14s-14s+14 ⇒L-114s-L14s+14⇒14L-11s-14L-11s+14 ⇒814-14e-14t We know that L-1G(s)e-as=g(t-a)u(t-a) ⇒y(t)=841-e-14(t-3)u(t-3) y(t)=2(1-e-14t-3)u(t-3)(b) The given process transfer function G(s)=153s+1-184s+1 ....(1) The above transfer function can be written as G(s)=G1(s)-G2(s) G1s=153s+1 and G2(s)=184s+1 Compare it with the transfer system we get G1s=Y(s)Xs=153s+1 and G2s=Y(s)X(s)=184s+1 We know that the first-order transfer functions a(s)=kpτs+1 In case of G1s, kp1=15, τ1=3 In case of G2s, kp2=18, τ2=4 When u(t)=1, Lu(t)=1s=Xs G(s)=Y(s)X(s)=153s+1-184s+1 Y(s)=1s15(3s+1)-18s(4s+1) Take inverse Laplace L-1y(s)=L-11s.15(3s+1)-18s4s+1 Take L-11s15(3s+1) Use partial fraction 15s(s+1)=As+B3s+1 After solve we get A=15 and B=-45 ⇒L-115s-L-1453s+1⇒15-453e-13⇒15-15e-13t Take 18s(4s+1) Use fraction method 18s(4s+1)=As+B4s+1 A=18 and B=-72 ⇒L-118s-L-1724s+1⇒18.(1)-724e-14t⇒18-18e-14t L-1y(s)=y(t)=15-15e-13t-(18-18e-14t) y(t)=15-15e-13t-18+18e-14ty(t)=18e-14t-15e-13t-3

Engineering

Electrical Engineering

Consider the following process transfer functions. Determine the characteristic parameters (not the poles and zeros) of each process. Describe the dynamic response that would be expected for a unit step change input, and identify any special features of the response. 8e-3s 15 (b) G(s) = 18 (a) G(s) = 16s+4 3s+1 4s+1

Consider the following process transfer functions. Determine the characteristic parameters (not the poles and zeros) of each process. Describe the dynamic response that would be expected for a unit step change input, and identify any special features of the response. 8e-3s 15 (b) G(s) = 18 (a) G(s) = 16s+4 3s+1 4s+1

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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