analogy rlc en mass spring dash pot

- 17 -

CHAPTER (2)

Transfer function and system responses

Objectives:

This chapter will consider the system representation using block diagram, transfer function, and study the system response. After you have read this chapter, you should be able to

• Simplify a block diagram representation • Analyze the transfer function properties • Study the system responses; transient and steady state • Investigate the system stability

Transfer function and system responses ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

- 18 -

2.1 System modeling Modeling of dynamic systems can described mathematically using physical laws. Based on the use of the conservation law of mass and energy, we can write the ordinary differential equations (ODEs) and algebraic equations describing system. Usually linear and time invariant parameters models are required to simplify the analysis and control tasks. Many systems can be treated using the same approach such as mechanical, electrical, hydraulic, pneumatic, thermal, fluid,…etc. Once the ordinary differential equation is obtained, the system can be analyzed and developed to be controlled. Example (Mechanical system) Consider a mechanical system that consists of mass-spring-damper as illustrates in the figure. Where: f(t) is the applied force; v(t) is the resulted velocity m is the inertia c is the damping coefficient k is the stiffness coefficient

Figure 2.1 Mass-damper-spring System

By applying the force balance on the mass m, the ODE model is:

∫ =++ f(t)v(t)dtkcv(t)dt

dv(t)m

(2.1) Example (Electrical system) Consider an electrical system that consists of R-L-C circuit as illustrates in the figure. Where: V(t) is the applied voltage; i(t) is the resulted current R is the resistance in Ohm L is the inductance in Henry C is the capacitance in Farad

Figure 2.2 R-L-C Circuit

c

k

m f(t)

v(t)

R

C

Li(t)

+ V(t) -


- 19 -

By applying Ohm's law on the circuit, the ODE model is:

∫ =++ V(t)i(t)dtC1Ri(t)

dtdi(t)L (2.2)

The following table represents an analogy between the two systems:

Table 2.1 Mechanical/Electrical systems analogy

The above two examples show that, the relationship between input and output is a differential equation, because they involve dynamic elements. Dynamic element has not in phase a cause and effect (input-output) relationship and it can store energy inside itself. While a static element has in phase a cause and effect relationship and cannot store energy. Example: Mechanical system System input is the force f(t), while its input is the translational velocity v(t). Mass (m) is a dynamic element due to its inertia (store energy). Damping (c) is a static element, dissipates heat and cannot store energy. Stiffness (k) is a dynamic element, it can store potential energy. Example: Electrical system System input is the voltage V(t), while its input is the electrical current i(t). The capacitor (C) is a dynamic element, it can store electrostatic energy The resistor (R) is a static element, dissipates heat and cannot store energy The Inductance (L) is a dynamic element, it can store electromagnetic energy The differential equation is however, not a very convenient way of seeing how the output depends on the input. One method of simplifying the form of relationship and making it easier to handle is to use what is called the transfer function. The transfer function is often described in two multiplied parts, the static part and the dynamic part. The static transfer function describes the input/output relationship when the input and output are not changing in time. It can be linear or non linear, in equation or in tabulation form.

statesteady in input Systemstatesteady in output Systemfunction transfer Static = (2.3)

The dynamic transfer function describes the input/output relationship when there is a time variation for the system variables (during the transient response).


- 20 -

2.2 Block diagram The block diagram is a functional block that illustrates the input-output. The functional block is a symbol for the mathematical operation on the input signal to produce the output signal. The advantages of the block diagram representation of a system lie in the fact that it is easy to form the overall block diagram for the entire system by merely connecting the blocks of the components and that it is possible to evaluate the contribution of each component to the overall performance of the system. The block diagram may be consists of:

• Functional blocks either on the feed-forward or in the feedback path • Branch points from which signals can go either to a functional block or

summing point • Summing points to add or subtract signals

The following diagram illustrates the closed loop control system

Figure 2.3 Closed loop system

For the system shown, we have the relations: Y = G * E E = R – H* Y

(2.4) By eliminating E from these equations, it gives Y = G [R – H*Y]

(2.5) That gives the relation,

HG 1

GRY

+=

(2.6)

The transfer function relating Y to R is called the closed loop transfer function. The denominator in this transfer function is called characteristic equation. The characteristic equation reserves the dynamic properties of the system as it will be shown later. The feed-forward function block may be consists of several blocks in cascade (series) such as:

• Controller block • Actuator to amplify the power of the control signal • Process block that is required to be under control

Error E

-

Set point R +

Feed-forward path

G

Feedback path

H

Output Y


- 21 -

While, the feedback functional block consists of the primary senor and the transducer components. Disturbance rejection in closed loop system Consider a block diagram as shown in figure (2.4). Gp represents the transfer function of the process under control and Gc represents the controller transfer function. H is the feedback transfer function of the measurement system.

Figure 2.4 Closed loop system subject to a disturbance

The output is Y, the input is R, and the disturbance signal is D. We may assume that the system is at rest initially with zero error, we may then calculate the response Y1 to the disturbance D only. This response can be found from

H G G 1G

DY1

cp

p

+=

(2.7)

On other hand, in considering the response to the reference input R, we may assume that the disturbance is zero. Then the response can be found from

H G G 1GG

RY2

cp

cp

+=

(2.8)

As the system is linear, we can apply the superposition theorem. That means, the final response Y is sum of the two individual ones Y1 and Y2.

]DRG[H G G 1

G Y c

cp

p ++

=

(2.9)

R +

+

+

-

Y

Disturbance D

Gp

Gc

H


- 22 -

Consider now the case where Gp H >>1 and GpGc H >>1. In this case, the system response due to the disturbance Y1/D is almost zero, and the effect of the disturbance is suppressed. This is an advantage of the closed loop system. On the other hand, the closed loop transfer function Y2/R approaches to 1/H as the loop gain GpGcH increases. This means that if the loop gain is greater than one the closed loop transfer function becomes independent of Gp and Gc and becomes inversely proportional to H so that variations of Gp and Gc do not affect the closed loop transfer function Y2/R. This is another advantage of the closed loop system. It can easily be seen that any closed loop system with unity feedback, H=1, tends to equalize the input and output. Block diagram reduction It is important to note that blocks can be connected in series only if the output of one block is not affected by the next following block. If there are cascaded blocks, we can replace them by their multiply in one block. Any simple feedback loops can be reduced to one block. Complicated block diagram can be treated step by step to reduce it using the following rules: Rule (1): Negative feedback

is equivalent to Rule (2): Cascade blocks

is equivalent to Rule (3): Parallel blocks

is equivalent to

R Y G1 + G2

+

R Y G1

G2

R Y G1G2

R Y G2

G1

-

R + Y G

H

R Y G

1 + G H


- 23 -

Rule (4): Moving signal backward

is equivalent to Rule (5): Moving signal forward

is equivalent to Rule (6): Moving summing point

is equivalent to Rule (7): Moving summing point

is equivalent to Rule (8): Order of summing points

is equivalent to B - C +

A-B+CA +

C + B -

A-B+CA +

B +

A AG +BG G

G

+

B

A AG +BG G

B

+

A AG +B G

1/G B

+

A AG +B G

1/G

R RG G

R R

R RG G

R RG G

G RG

R R G G


- 24 -

2.3 Transfer function The overall transfer function is the product of static part ad dynamic part. ds G G G =

(2.10) Where G is the overall transfer function Gs is the static part of G Gp is the dynamic part of G The dynamic part Gp approaches to a unity value at the end of transient response time. The static transfer function Gs of a process control loop element specifies how the output is related to the input if the input is constant. An element also has a time dependence that specifies how the output changes with in time when the input is changing in time. The dynamic transfer function is often simply called the time response. It is particularly important for sensors because they are the primary element for providing knowledge of the controlled variable value. In this section, we will be interest to discuss the two most common types of sensor time responses (first order and second order only). The following figure shows a sensor that produces an output y(t) as a function of the input u(t).

Figure 2.5 Sensor input/output The static transfer function determines the output when the input is not changing in time. To specify the time response, the nature of the time variation in output y(t), is given when the input exhibits a step change as shown. Note that at time t=0 the input to the sensor is suddenly changed from an initial value ui to a final value uf. If the sensor were perfect, its output would be determined by the static transfer function to be yi before t=0 and yf after t=0. However, all sensors will exhibit some lag between the output and the input in time before settling on the final value. The lag in time is due to the physical construction of the sensor from dynamic elements.

t

Sensor u(t)

y(t)

ui

t

uf yf

yi


- 25 -

2.3.1First order system The first order system is described by first order ordinary differential equation as

u(t)dty(t) d y(t) =+τ

(2.11) Where u(t) is the input y(t) is the output τ is the sensor (system) time constant The general equation for the system response in solving the above ODE is given by:

( ) ( )t/τifi e1 y-y y y(t) −−+= (2.12)

The steady state gain k is the ratio between the steady state output change to the steady state input change and can defined as.

if

ifs u-u

y-yuy G =

∆∆

=

(2.13) Where Gs = steady state gain (static part of the transfer function) yf = final output value yi = Initial input value uf = final input value ui = Initial input value The system response depends on the values of k and τ as it will be shown in the next section. The time constant is the required time to reach 63% of its steady state output. Also, we can define settling time ts as the time to reach 98% of the steady state output value (ts = 4 τ) or as the time to reach 95% of steady state output (ts =3τ). Example A sensor measures temperature linearly with a first order transfer function of static gain = 33 mV/oC and has a 1.5 s constant time. Find the output o.75 s after the input changes from 20oC to 41oC. Find the error in temperature measurement at this instant of time. Solution We first find the initial and final values of the sensor output yf = (33mV/oC)(20oC)= 660 mV yi = (33 mV/oC)(41oC)= 1353 mV Now, y(o.75)= 660 +(1353-660)(1-e-0.75/1.5) = 932.7 m V This value corresponds to a temperature of the following value T T = 923 mV/(33 mV/oC) = 28.3oC The error = 41- 28.3 = 12.7 oC


- 26 -

2.3.2 Second order system In some systems or sensors, a step change in the input will cause the output to oscillate for a short period of time before settling down to a value that corresponds to the new input. Such oscillation (and the decay of the oscillation itself) is a function of the sensor parameters. This is called a second order response because for this type of system, the time behavior is described be a second-order differential equation as

u(t) dt

y(t) ddty(t) d 2 y(t) 2

n2

2

n2n ωωζω =++

(2.14) Where u(t) is the input y(t) is the output ζ is the system damping factor ωn is the natural frequency of the system It is not possible to develop a universal solution, as it is for the first-order system. Instead, we simple describe the general nature of the response. The system response depends on the values of ζ and ωn as it will be shown in the next section.


- 27 -

2.4 System responses We will apply the following procedure to simulate linear systems:

• Solve linear constant-coefficient ordinary differential equations, initial value problem. • Find system response (output vs. time) given input history and initial values. • We will consider 1st and 2nd order. Number of initial values must equal order of

ODE. For example, the second order equation of x(t) is given by ( r(t) is the system input)

r(t) x(t)a (t)x a (t)x a 012 =++ &&& (2.15)

with the initial conditions 00 v (0) x ; xx(0) == & General Solution Method Total solution = Homogeneous (transient response to ICs) + Particular (steady-state response to input) (t) x (t) x (t)x PHT +=

(2.16) Homogeneous Solution: transient response to initial conditions

stH012 eA (t) xAssume 0 x(t)a (t)x a (t)x a ==++ &&&

(2.17) Substitute xH(t) and its successive derivatives in the above equation, then evaluate si - roots of nth order polynomial in s, leave Ai coefficients unknown for now (n of them) Particular Solution: steady-state response to input forcing Assume xP(t) has the same form as input forcing function (see the following table)

Table 2.2 forcing function

Determine the unknown coefficients Bi by substituting xP(t) and its successive derivatives into equation (2.15) Total Solution (equation 2.16): Determine the unknown homogeneous coefficients Ai using initial conditions on xT(t) where number of roots = number of initial conditions = ODE order = n


- 28 -

Example: (1st order Mechanical system) Consider the c-k (spring-damper) translational mechanical system with force input f(t) and displacement output x(t) with the following parameters: f(t) = 5 N, step input of force at t=0 c = 1 Ns/m ; k = 50 N/m x(0) = 0, initial displacement

From force balance equation in equation (2.1) with m=0, the ODE is:

0 x(0)subject to 5 x(t)50 (t)x ==+& Homogeneous solution Assume xH(t)= Aest , substitute in ODE with forcing input=0, then solve and compute s = -50 Particular solution Assume xP(t) = B(constant) and substitute in the ODE, then B=0.1 Total solution xT(t) = 0.1 + A e-50t

Substitute in xT(t) to compute A by using the initial condition, then A= -0.1. Final solution is x(t) = 0.1 (1-e-50t) is obtained and then Plot x(t) vs. time gives

Figure 2.6 Mechanical system response (First order)

Displacement starts at zero, as specified by initial condition. Steady-state current displacement is xSS = 0.10 m (constant since input force is constant); agrees with Hooke’s Law: f = kxSS; xSS = 5/50 = 0.10 (the steady state gain = 0.1). The time constant τ = 0.02 sec = c/k =1/50 it is the same result as it was expected. A system with a smaller time constant rises faster than one with a larger time constant. The settling time = 0.08 sec (98% of final value).


- 29 -

Example: (1st order Electrical system) Consider the electric system R-C circuit in series, voltage v(t) is input and current i(t) is output (see equation 2.2). The model equation ODE will be

(t)qi(t)

q(t)dt i(t)

:q charge substitute ; v(t)i(t)dtC1 i(t) R

&=

=

=+

∫∫

(2.18) The ODE will be

F 0.2 C and 50RGiven ; v(t)q(t)C1 (t)q R µ=Ω==+&

(2.19)

Solve v(t)stepunit a and 0 q(0) subject to 1 q(t) 5000 (t)q 50 ==+& Plot the response of q(t) and i(t)

Figure 2.7 Electrical system response (First order) As expected from the circuit physics, the charge q in a capacitor builds up to a constant given a constant voltage input; also the capacitor current goes to zero at steady-state. In this example one can see the same steady state value (1/5000) and time constant behavior ( 3 τ = 3 RC = 0.03. 95% of final value, dotted lines).


- 30 -

Example: 2nd order system (over and critical damped) Consider the mass-spring-damper translational mechanical system as seen in figure 2.1, where F(t) is an input and x(t) is an output to the system. The ODE is given by:

F(t) k x(t) (t)xc (t)xm =++ &&& (2.20)

Subject to initial conditions 0at t step N 30 F(t)

and m/s 0.05 (0)x ; 0.1mx(0)==

== &

Case (1): over-damped (real distinct roots, sluggish response, no overshoot) System parameters: m =10kg, c = 70 Ns/m, k = 120 N/m Solve: 0.05 (0)x and 0.1 x(0)subject to 3 x(t)12 (t)x 7 (t)x ===++ &&&& Case (2): critical damped (real roots and repeated, fastest response without overshoot) System parameters: m =10kg, c = 60 Ns/m, k = 90 N/m Solve: 0.05 (0)x and 0.1 x(0)subject to 3 x(t)9 (t)x 6 (t)x ===++ &&&&

Homogeneous solution: 0 x(t)9 (t)x 6 (t)x =++ &&& xH(t) = Aest ; (s2 + 6 s + 9) Aest Characteristic polynomial s2 + 6 s + 9 = (s+3)2 = 0 s1,2 = -3, -3 real, repeated roots Homogeneous solution form xH(t) = A1 e-3t + A2 te-3t Particular solution 3 x(t)9 (t)x 6 (t)x =++ &&& Assume xP(t) B ; 0+ 6(0) +9B = 3 Then B=1/3 =0.33 Total solution xT(t) = A1 e-3t + A2 te-3t + 0.33 d/dt (xT(t)) = -3 A1 e-3t + A2 e-3t -3 A2 t e-3t Now apply initial conditions: xT(0) = 0.1 =A1 + A2(0) + 0.33 d/dt(xT(0)) = 0.05= -3 A1 + A2 The solution of the above equations is A1 = -0.233 A2 = -0.65


- 31 -

Final solution is: xT(t) = 0.33 – (0.233 +0.65 t) e-3t Plot of x(t) vs. Time

Figure (2.8) Second order response (over and critical damped) x(t) starts at 0.1 m, while the speed is non-zero, as specified by initial condition. The steady state value of x = 1/3 m. Transient approaches zero after t = 2.5 sec, critically damped -3 root goes to zero slightly faster. Example: Second order system (under-damped case) Consider the same model with parameters: m =10kg c = 20 Ns/m k = 100 N/m Solve: 0.05 (0)x and 0.1 x(0)subject to 3 x(t)10 (t)x 2 (t)x ===++ &&&& By solving the ODE, we obtain under-damped case (complex – conjugate roots with negative real part). It has fast response, but with overshoot – damping envelope. xT(t) = 0.3 – e-t (0.2 cos 3t + 0.05 sin 3t) = 0.3 – 0.2026 e-t sin (3t + 1.326)


- 32 -

Plot of x(t) vs. Time

Figure 2.9 Second order response (under-damped case)

x(t) starts at 0.1 m, the speed is non-zero, as specified by initial condition. Transient approaches zero after t = 5 sec; under-damped; damping envelope ± +- 0 2026 0 30 . . e t ; steady state value of x = 0.3 m. Damping Condition Summary, Stable 2nd-order systems: OVERDAMPED (ζ>1) Real distinct negative roots, sluggish response, no overshoot CRITICALLY DAMPED (ζ=1) Real, repeated negative roots, fastest response without overshoot UNDERDAMPED (0<ζ<1) Complex-conjugate roots, negative real part, fast response, overshoot and oscillation, damps out UNDAMPED (ζ= 0) Complex-conjugate roots, zero real part Simple harmonic motion oscillation, theoretically never damps out Example

Solve

undamped ; 1 y(t) 4 (t)y dunderdampe ; 1 y(t) 4 (t)y 2 (t) y

damped critical ; 14y(t)(t)y 4 (t)yoverdamped ; 1 y(t) 4 (t)y 6 (t)y

=+=++

=++=++

&&

&&&

&&&

&&&

With zero initial conditions and step input.


- 33 -

Results plot:

Figure 2.10 Second order system (step response)

For the same example, replace the step input by an impulse input signal and plot the results.

Figure 2.11 Second order system (impulse response)


- 34 -

Second order system performance:

1) Peak time (time to reach to the maximum peak)

2P

1 t

ζω

π

−=

n

(2.21) 2) Settling time (time to reach to 98% of the steady state output)

n

s ζω4t =

(2.22) Or (time to reach 95 % of the steady state output)

n

s ζω3t =

(2.23)

3) Rise time (time between 10-90% of the output)

)0.8 3.0 ( 0.6 2.16 tn

R ≤≤+

≅ ζωζ for

(2.24) 4) Maximum overshoot (maximum value of the output in under-damped case)

2-1

-

P e M ζ

ζπ

=

(2.25) For the given system,

i 3.15 0.25- roots

0.08 ; 16.310±=

=== ζωn

The complex poles indicate that the system is under-damped with overshoot The system response for a unit step input is given in the following figure.


- 35 -

Figure 2.12 System response (ζ=0.08, ωn =3.16)

We can use equations (2.21 2.25) to measure the system performance. The measured parameters are: tP = 1.0 sec. ts = 16.0 sec. MP= 77.9 % (too high value). tR= 0.25 sec. (bad approximation ζ < 0.3 – from data tR =0.36 sec.). This 2nd-order system is highly under-damped; ζ is close to 0, the percent overshoot is very large, and the settling time is very high. This is typical of real-world systems such as flexible space robots (both joint motion and Cartesian motion, on all degrees-of freedom). Now let us create another example, to specify more desirable behavior for such a system. Let us specify ts = 1.5 sec and MP = 5%; the result will still be under-damped but not severely so. From equation (2.21), calculate ζ to obtain maximum overshot in order of 5%. Then, from equation (2.22) calculate the natural frequency ωn to obtain settling time equal to 1.5 sec. this gives the following parameters: ζ= 0.69 ; ωn= 3.86 For the given system,

i 2.8 2.67 - roots

0.69 ; 86.3±=

== ζωn

Note that, the negative real part is increased due to increase damping and imaginary part is decreased to decrease the overshoot. The obtained response is given in the following figure.


- 36 -

Figure 2.13 System response (ζ=0.69, ωn =3.86) This 2nd-order system has ts = 1.5 sec, tR = 0.54 sec. and MP = 5%. From the study of the above responses, we can summarize controller goals as:

• Ensure stability • Achieve desired transient response • Reduce steady-state error • Decrease system sensitivity to changes, uncertainty (Robust) • Reduce effects of unknown, unwanted disturbances

Desired transient response

• Rise time • Peak time • Percent overshoot • Settling time • Damping, frequency

Generic second order system for any damping factor ζ and any ωn natural frequency, the system has the roots s1,2 as

1- - s 21,2 ζωζω nn ±=

(2.26) The following Fig. 2.14 represents the transient response versus s-plane pole locations (unit step response for the above generic 2nd order system).


- 37 -

Figure (2.14) Generic 2nd order system

Case (a): overdamped, real distinct negative roots, sluggish response ( ζ>1). Case (b): critically damped, real repeated negative roots ( ζ=1). Case (c): underdamped, complex conjugate roots with negative real part ( 0<ζ<1). Case (d): undamped, simple harmonic motion ( ζ=0). Case (d - special): double integrator, parabolic response ( ζ=0). Case (e): (unstable system), complex conjugate roots with positive real part ( -1<ζ<0). Case (f): (unstable system), negative critical-damping, real repeated positive roots ( ζ= -1). Case (g): (unstable system), negative damping, real distinct positive roots ( ζ<-1). It is not shown in figure.


- 38 -

2.5 System stability A system is stable if the output is bounded for all bounded inputs (BIBO stability).

Figure 2.15 BIBO Stability

Stability is property of a system, independent of input signal. Equilibrium states can be a) unstable equilibrium, b) neutral equilibrium, or c) stable equilibrium:

Figure 2.16 Stability configuration

Bottom-line in controls design: must be stable, otherwise no good to improve transient response, error if the response blows up! Simple test for system stability: The real part of all poles must be negative. This dictates how the exponential term behaves (must be negative in the exponent so transient will go to zero - positive exponent causes transient to go to infinity). Poles: sj = aj ± i bj If all aj < 0 stable If any aj = 0 marginally stable If any aj >0 unstable So, given a system, stability analysis is simple: roots(den) - look at sign of all real parts of the poles. Example Consider the system that has i 3 2- s1,2 ±= Complex roots with negative real parts, the system is stable (underdamped).

a b c

Bounded input signal

Stable system

Bounded output signal

Bounded input signal

Unstable system

Unbounded output signal


- 39 -

2.6 Basic concept (MCQ)

Place the letter of statement that best completes the sentence in space provided. 1] The dynamic element can store ___________. A) Heat B) Energy C) Magnetic field 2] The static element has a cause and effect relationship _________. A) Out of phase B) In phase C) With lag 3] One of the following elements is a dynamic element; it is _________. A) Resistor B) Damper C) Capacitor 4] The superposition theorem can be applied to a ___________ system. A) Linear B) Nonlinear C) Transfer function 5] The dynamic part of the transfer function describes the output when the input is _______

in time. A) Changing B) Not changing C) Constant 6] The time constant of the first order system means that the output is reached to ______

from its steady state value. A) 98 % B) 63 % C) 50 % 7] A ______________ can oscillate for a step change of input.

A) First order system B) Second order system C) Dynamic system


- 40 -

8] The settling time for the first order system is equal to _______ of its time constant to

reach 95 % of steady state output.

A) Two B) Three C) Four

9] The 2nd order system with _______ roots has a response without overshoot. A) Real B) Complex C) Real and negative 10] The second order system is highly under-damped when the damping factor is close to

______. A) One B) Zero C) Negative value 12] As the damping factor increases, the overshoot ___________. A) Decreases B) Increases C) Is stable 13] Increasing the negative real part of complex roots, __________ the system damping. A) Decreasing B) Increasing C) Not change 14] As the imaginary part of complex roots increases, the maximum overshoot __________. A) Decreases B) Increases C) Is stable 15] A stable system has all roots with __________ real parts. A) Positive B) Negative C) Zero


- 41 -

2.7 Problems 1] Find the over all transfer function for the given system

Figure 2.16 Problem 2.1 2] A control system is shown in the following figure. The transfer functions G2 and H2 are

fixed. Determine the transfer function G1 and H1 so that the closed loop transfer function Y/R is exactly equal to 1.

Figure 2.17 Problem 2 3] Determine the transfer function for the given system.

Figure 2.18 Problem 3

4] A control system is shown in the following figure. The process Gp is a first order with unity steady state gain and a 10 s time constant. The controller Gc is a gain with value 10.

- Compute the closed loop time constant and comment on the results.

Figure 2.19 Problem 4

Y

-

- R + +

G1

G2

R +

-

Gc Y

Gp

R + + +

-

G1 Y

G2

H2

H1

+

-

G2 R + Y

G1

G3

+


- 42 -

5] A mercury thermometer is a first-order measurement system with a time constant of 5

seconds. Describe how the reading given by the thermometer will change with time when it is suddenly placed in a hot liquid. About how long will it take for a steady reading to be obtained?

6] Explain how the responses of the given systems vary with time when subject to a unit step

input: a) Static gain = 12 b) Static gain =12 and time constant = 4 sec. c) Static gain = 12, ζ= 0.5, ωn= 0.78 7] The open loop transfer function of a system is given with ζ= 0.4, ωn= 0.5. Determine the

approximate values of peak overshoot, time to peak, and settling time. Plot the approximate unit step response of the system.

8] A thermometer requires 1 min to indicate 98% of the response to a step input. Assuming

the thermometer is a first order system, find the time constant. If the thermometer is placed in a bath, the temperature of which is changing linearly at a rate 10oC/min, how much error does the thermometer show?

9] A photocell with a 35 ms time constant is used to measure light flashes. How long after a

sudden dark to light flash before the cell output is 80 % of the final value? 10] A pressure sensor has a resistance that changes with pressure according to R = (0.15kΩ/psi) P + 2.5 kΩ This resistance is then converted to a voltage with the transfer function V = 10 R/(R+10 kΩ) volts The sensor time constant is 350 ms. At t=0, the pressure P changes suddenly from 40 psi

to 150 psi. What is the voltage output at 0.5 seconds? What is the indicated pressure at this time?

analogy rlc en mass spring dash pot

Documents