Class 4 Systems and Solutions. 1 st Order Systems

Class 4

Systems and Solutions

1st Order Systems

1st Order Systems

oxx

tfxdt

dx

0

(input) measured be to Quantity (output) response Instrument

(s) constant time

tf

x

1st Order Systems

oxx

tfxdt

dx

0

tKtf

ttuKtf

tuKtf

h

sr

ss

cos)(

)(

)(

Step Input

Ramp Input

Harmonic Input

1st Order Systems with Step Input

o

s

xx

tuKxdt

dx

0

0,1

0,0

t

ttu -1 0 1 2

0

1

2

Time, t

u(t)

1st Order Systems with Step InputSolution by Integration

o

s

xx

Kxdt

dx

0

tx

x s

s

s

dtxK

dx

dtxK

dx

xKdt

dx

o 0

tsos

t

so

s

os

s

oo

tu

u

s

tx

x s

eKxKtx

eKx

Kx

xK

xK

tu

ut

u

u

dtu

du

dxduxKu

dtxK

dx

o

o

ln,ln

,

,

0

0

Error Ratio

1st Order Systems with Step InputError Ratio and Excitation Ratio

t

so

s

os

serr

err

err

eKx

Kx

xK

xKr

r

r

value input from deviation Startingvalue input from deviation Current

Error StartingError Current

Ratio Error

terr

os

sex

os

o

os

osex

os

o

os

os

os

oex

ex

ex

erxK

xKr

xK

xx

xK

xKr

xK

xx

xK

xK

xK

xxr

r

r

111

1

1

output initial from deviation Inputvalue initial its from deviation Output

Excitation DesiredExcitation Current

Ratio Excitation

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

time, t

Excitation Ratio

Error Ratio

Error:

Output deviation from input

Excitation:

Output deviation from its initial value

1st Order Systems with Step InputSolution by Superposition

o

s

xx

Kxdt

dx

0

th

rtrtrthh

rth

rth

hh

Cetxrr

CerCerCexx

rCetxCetx

xx

1,1

01,0,0

0

ODE shomogeneou into ngSubstituti

Assume

ODE shomogenoeu the For

spp

hh

ph

Kxx

xx

txtxtx

and

where

form the of solution a Assume

0

tsos

soos

tshp

sp

spp

eKxKtx

KxCxCKx

CeKtxtxtx

Ktx

Kxx

,0

condition, initial the satisfy To

thereofore is solution complete The

nobservatio by found is

equation particular the of Solution

1st Order Systems with Step InputSolution by Laplace Transform

1)(

1)(

1)(

)()(

)()(

0

ss

xKsxsX

ss

xsKsX

s

xsKssX

xs

KsXssX

s

KsXxssX

xx

Kxdt

dx

oso

os

os

os

so

o

s

tsos

sos

ososo

osos

s

os

os

s

os

o

oso

eKxKtx

s

Kx

s

KsX

s

xK

s

xKxsX

xKxK

B

ss

xKssB

xKss

xKssA

s

B

s

AxsX

ss

xKsxsX

1)(

1

1)(

11

1

11

10

1)(

1)(

1

0

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Am

plit

ude

1

1)(

1)(

)()(

00

ssF

sX

sFssX

sFsXssX

x

tfxdt

dx

Transfer Function

>> num=1;

>> den=[1 1];

>> sys = tf(num,den);

>> step(sys)

>> grid

1st Order Systems with Unit Step Input and Unit Time Constant

MATLAB Simulation by Transfer Function


MATLAB Simulation by Simulink

xtudt

dx

tuxdt

dx

1

1st Order Systems with Ramp Input

o

ro

xx

ttuKxxdt

dx

0

0,1

0,0

t

ttu

ttu

1st Order Systems with Ramp InputSolution by Superposition

o

ro

xx

tKxxdt

dx

0

th

rtrtrthh

rth

rth

hh

Cetxrr

CerCerCexx

rCetxCetx

xx

1,1

01,0,0

0


Assume


tKxxx

xx

txtxtx

ropp

hh

ph

and

where


0

trro

trro

roro

trohp

orp

roo

o

r

ro

pp

ropp

eKtKxtx

eKtKxtx

KCxCKxx

CetKxtxtxtx

xtKtx

KxAxB

xBA

KA

tKxBAtA

AtxBAttx

tKxxx

1

,0

,

,

condition, initial the satisfy To

thereofore is solution complete The

ODE the into Substitute



1st Order Systems with Ramp InputSteady State Error and Relative Error

rss

tr

tss

tss

tr

ro

Ke

eKe

errore

eKerror

txtKxerror

error

1lim

lim

1

error state Steady

Output-Input

trro eKtKxtx 1

o

ro

xx

ttuKxxdt

dx

0

t

r

tr

r eK

eKErr

11

Error State SteadyError

error Relative

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

rErr

1st Order Systems with Ramp InputRelative Input and Relative Excitation

trro eKtKxtx 1

t

r

trr

r

r

etK

eKtKex

ex

11

Error State SteadyExcitation Current

Excitation Relative

tK

tK

r

r

Input Relative

S.S.EValue Initial-Input

Input Relative

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

3.5

4

Relative Excitation

Relative Input

t

1

o

ro

xx

ttuKxxdt

dx

0

1st Order Systems with Ramp InputSolution by Laplace Transform

1)(

1

1

)(

1)(

1)(

)()(

)()(

0,

2

2

2

2

2

2

2

2

2

s

C

s

B

s

AxsX

ss

xK

ssxsX

ss

KsxsxsX

s

KsxsxssX

xs

K

s

xsXssX

s

K

s

xsXxssX

xxtKxxdt

dx

o

o

r

o

roo

roo

oro

roo

oro

trro

trror

rror

or

s

o

r

o

ro

s

o

r

or

s

o

r

eKtKxtx

eKKxtKtx

s

K

s

Kx

s

KsX

xKss

xK

sssC

x

Kx

ssxK

sss

ds

dB

xKss

xK

sssA

1

1)(

1

1

1

1

1

1

1

2

1

2

2

0

2

2

2

0

2

2

2

sssss

sGsH

ssF

sXsG

sFssX

sFsXssX

x

tfxdt

dx

2

1

)1(

1)()(

1

1)()(

1)(

)()(

00

1st Order Systems with Unit Ramp Input and Unit Time Constant

MATLAB Simulation by Transfer Function

MATLAB does not have a ‘ramp’ command to plot the ramp

response of the system. However, note that the response, Rst

(s)

of a system with transfer function G(s) to unit step input is Rst

(s)

= G(s)/s, and its response to a unit ramp input is Rrmp

(s) =

G(s)/s2 = (G(s)/s)/s . Thus, the response of G(s) to unit ramp is

equal to the response of H(s) = G(s)/s to unit step.

We may use MATLAB’s ‘step’ command to obtain the ramp

response of a system G(s) simply by obtaining the step response

of H(s) = G(s)/s to unit step.

0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

>> num = [0 0 1];

>> den = [1 1 0];

>> t=0:0.1:5;

>> sys = step(num,den,t);

>> plot(t,sys,'o',t,t,'-')

>> grid


MATLAB Simulation by Simulink

xttudt

dx

ttuxdt

dx

1

1st Order Systems with Harmonic Input

00

cos

x

tKxdt

dxh

0 2 4 6 8 10 12

-1

0

1

1st Order Systems with Harmonic InputSolution by Superposition

1

2

222

tan

1

)cos()cos()2cos(

)2cos(

)sin(

)cos(

)cos(

h

h

h

p

p

p

hpp

KA

KAA

tKtAtA

tAtx

tAtx

tAtx

tKxx

plot vector the From

ODE the into Substitute



th

rtrtrthh

rth

rth

hh

ropphh

ph

h

Cetxrr

CerCerCexx

rCetxCetx

xx

tKxxxxx

txtxtx

x

tKxdt

dx

1,1

01,0,0

0

0

00

cos


Assume


and where


sin2sinsin2coscos2cos

cos2sincos2cossin2sin

ωt

ϕ

τAω

A

Kh

10-2

10-1

100

101

102

0

0.5

1

1.5

1st Order Systems with Harmonic InputAmplitude Ratio and Phase

1

2

tan

1

)cos(

00

cos

h

h

KA

tAtx

x

tKxdt

dx

1

2

tan

11

Phase

AmplitudeInput AmplitudeOutput

Ratio Amplitude

ha

a

KAr

r

0 2 4 6 8 10 12

-1

0

1 f(t)

x(t)

ϕ

ra

τω

ra

(τω)ϕ

(τω)

1st Order Systems with Harmonic InputSolution by Complex Exponential

iyxz

tiAtAz

Aez

ie

ti

i

sincos

sincos

ωt

x

A

y

z = Aeiωt

Euler’s Identity

ωt

x

A

y

z = Aeiωt

-ωt

z* = Ae-iωt

*2

1cosRe

*

zztAzx

Aez

Aezti

ti

Complex Conjugate

21

2

1

2121

2

1

2

1

2

1

2

1

212121

22

11

ii

i

iii

i

i

eA

A

eA

eA

z

z

eAAeAeAzz

eAz

eAz

Multiplication & Division Rules

niin

innnin

i

eAAez

eAAez

Aez

nn

111

Power Rules

Second Order Systems In the system shown, the input displacement, x

i, will

cause a deflection in the spring, and some time will be

needed for the output displacement xo

to reach the

input displacement.

m

k

c

xi

xo

Second Order Systems

If m/k << 1 s2 and c/k << 1 s, the system may be approximated as a zero order system with unity gain.

If, on the other hand, m/k << 1 s2 , but c/k is not, the system may be approximated by a first order system. Systems with a

storage and dissipative capability but negligible inertial may be modeled using a first-order differential equation.

m

k

c

xi x

o

iiooo

iiooo

ooioi

o

kxxcxxk

cx

k

m

kxxckxxcxm

xmxxcxxk

xmF

Example – Automobile Accelerometer Consider the accelerometer used in seismic and vibration engineering to

determine the motion of large bodies to which the accelerometer is

attached.

The acceleration of the large body places the piezoelectric crystal into

compression or tension, causing a surface charge to develop on the

crystal. The charge is proportional to the motion. As the large body

moves, the mass of the accelerometer will move with an inertial

response. The stiffness of the spring, k, provides a restoring force to

move the accelerometer mass back to equilibrium while internal

frictional damping, c, opposes any displacement away from equilibrium.

m

k

c

xi x

o

Piezoelectric crystal

Zero-Order systems Can we model the system below as a zero-order system? If the mass, stiffness, and damping coefficient satisfy certain

conditions, we may.

m

k

c

xi

xo

i

ioioioi

ooioi

o

xmmck

xmxxmxxcxxk

xmxxcxxk

xmF

First Order Systems Measurement systems that contain storage elements do not respond

instantaneously to changes in input. The bulb thermometer is a good

example. When the ambient temperature changes, the liquid inside the

bulb will need to store a certain amount of energy in order for it to reach

the temperature of the environment. The temperature of the bulb

sensor changes with time until this equilibrium is reached, which

accounts physically for its lag in response.

In general, systems with a storage or dissipative capability but negligible

inertial forces may be modeled using a first-order differential equation.

𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝐾𝑢(𝑡)

𝑥ሺ0ሻ= 𝑥0 𝑥ሺ𝑡ሻ= 𝐾+ሺ𝑥0 − 𝐾ሻ𝑒−𝑡/𝜏 𝑥ሺ𝑡ሻ− 𝐾𝑥0 − 𝐾 = 𝑒−𝑡/𝜏

𝑥ሺ𝑡ሻ− 𝑥0𝐾− 𝑥0 = 1− 𝑒−𝑡/𝜏

1st Order Systems with Step Input

Error Ratio

Excitation Ratio

Note that the excitation ratio also represents the system response in case of

x0=0 and K=1

error ratio = current errorstarting error = current deviation from final valuestarting deviation from final value

excitation ratio = current excitationdesired (input) excitation= current deviation from initial valueinput deviation from initial value

Excitation ratio may also be called response ratio = current response / desired

response

Example 1

A bulb thermometer with a time constant τ =100 s. is subjected to a step

change in the input temperature. Find the time needed for the response

ratio to reach 90%

Example 1 Solution

A bulb thermometer with a time constant τ =100 s. is subjected to a step change in the input temperature. Find the time

needed for the response ratio to reach 90%

𝑥ሺ𝑡ሻ− 𝑥0𝐾− 𝑥0 = 1− 𝑒−𝑡/𝜏 = 0.9

𝑥ሺ𝑡ሻ− 𝐾𝑥0 − 𝐾 = 𝑒−𝑡/𝜏 = 0.1

𝑡 𝜏= lnሺ10ሻ= 2.3Τ

𝑡 = 2.3× 𝜏= 230 𝑠.≈ 4 minutes

1st Order Systems with Ramp Input excitation ratio = current excitationdesired (input) excitation

excitation ratio = current deviation from initial valueinput deviation from initial value

𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝑥0 + 𝐾𝑟𝑡𝑢(𝑡)

𝑥ሺ0ሻ= 𝑥0 𝑥ሺ𝑡ሻ= 𝑥0 + 𝐾𝑟𝑡− 𝐾𝑟𝜏(1− 𝑒−𝑡/𝜏)

𝑆.𝑆.𝐸= lim𝑡→∞ሺ𝑥(𝑡) − 𝑓(𝑡)ሻ 𝑆.𝑆.𝐸= 𝐾𝑟𝜏

𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 = 𝑥ሺ𝑡ሻ− 𝑥0𝐾𝑟𝑡

𝐸𝑥𝑐𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜 = 1− (1− 𝑒−𝑡/𝜏)𝑡 𝜏Τ

limሺ𝑡 𝜏Τ ሻ→0ቆ1− (1− 𝑒−𝑡/𝜏)𝑡 𝜏Τ ቇ

= limሺ𝑡 𝜏Τ ሻ→0ሺ1ሻ− lim

ሺ𝑡 𝜏Τ ሻ→0൫1− 𝑒−𝑡/𝜏൯𝑡 𝜏Τ

= 1− limሺ𝑡 𝜏Τ ሻ→0𝑒−𝑡/𝜏1 = 1− 1 = 0

Error = 𝑥ሺ𝑡ሻ− 𝑓ሺ𝑡ሻ= −𝐾𝑟𝜏(1− 𝑒−𝑡/𝜏)

Steady State Error

Note that using L’Hospital rule


𝜏𝑑𝑥𝑑𝑡 + 𝑥= 𝐹 cos (ωt) 𝑥ሺ𝑡ሻ= 𝐶𝑒−𝑡/𝜏 + 𝑋cos (ωt − φ)

𝑋= 𝐹ඥ1+ሺ𝜏𝜔ሻ2

φ = tan−1ሺ𝜏𝜔ሻ

C depends on the initial conditions and the exponential term will vanish with time. We are interested in the particular steady solution 𝑋cos (ωt − φ). Solving for 𝑋 and φ, we find


𝐴𝑟 = 𝑋𝐹= 1ඥ1+ሺ𝜏𝜔ሻ2 = 1

ඥ1+ 4𝜋2𝑇𝑟2

φ = tan−1ሺ𝜏𝜔ሻ=tan−1ሺ2𝜋𝑇𝑟ሻ

Define the amplitude ratio 𝐴𝑟 = 𝑋 𝐹Τ and the time ratio 𝑇𝑟 = 𝜏 𝑇Τ where 𝑇= 2𝜋 𝜔Τ is the period of the excitation function,


The amplitude ratio, Ar(ω), and the corresponding phase shift,

ϕ, are plotted vs. ωτ. The effects of τ and ω on frequency

response are shown.

For those values of ωτ for which the system responds with Ar

near unity, the measurement system transfers all or nearly all

of the input signal amplitude to the output and with very little

time delay; that is, X will be nearly equal to F in magnitude

and ϕ will be near zero degrees.


At large values of ωτ the measurement system filters out any frequency information of the input signal by responding with very small

amplitudes, which is seen by the small Ar(ω) , and by large time delays, as evidenced by increasingly nonzero ϕ.


Any equal product of ω and τ produces

the same results. If we wanted to measure

signals with high-frequency content, then

we would need a system having a small τ.

On the other hand, systems of large τ may

be adequate to measure signals of low-

frequency content. Often the trade-offs

compete available technology against

cost.

dB = 20 log Ar(ω)


The dynamic error,δ(ω), of a system is defined as

δ(ω) = (X(ω) – F)/F

δ(ω) = Ar(ω) –1

It is a measure of the inability of a system to adequately reconstruct the amplitude of the input signal for a particular input frequency.

We normally want measurement systems to have an amplitude ratio at or near unity over the anticipated frequency band of the input

signal to minimize δ(ω) .

As perfect reproduction of the input signal is not possible, some dynamic error is inevitable. We need some way to quantify this. For a

first-order system, we define a frequency bandwidth as the frequency band over which Ar(ω) > 0.707; in terms of the decibel defined as

dB = 20 log Ar(ω)

This is the band of frequencies within which Ar(ω) remains above 3 dB

Example 2

A temperature sensor is to be selected to measure temperature within a reaction vessel. It is suspected that the temperature will

behave as a simple periodic waveform with a frequency somewhere between 1 and 5 Hz. Sensors of several sizes are available, each

with a known time constant. Based on time constant, select a suitable sensor, assuming that a dynamic error of 2% is acceptable.

Example 2. Solution

A temperature sensor is to be selected to measure

temperature within a reaction vessel. It is suspected that the

temperature will behave as a simple periodic waveform with a

frequency somewhere between 1 and 5 Hz. Sensors of several

sizes are available, each with a known time constant. Based

on time constant, select a suitable sensor, assuming that an

absolute value for the dynamic error of 2% is acceptable.

Accordingly, a sensor having a time constant of 6.4 ms or less

will work.

aA𝛿ሺωሻaA≤ 0.02 −0.02 ≤ 𝛿ሺωሻ≤ 0.02 −0.02 ≤ 𝐴𝑟 − 1 ≤ 0.02 0.98 ≤ 𝐴𝑟 ≤ 1.02

0.98 ≤ 1ඥ1+ሺ𝜏𝜔ሻ2 ≤ 1.0

0 ≤ 𝜏𝜔≤ 0.2

0 ≤ 2𝜏𝜋(5) ≤ 0.2

The smallest value of 𝐴𝑟 will occur at the largest frequency

𝜔= 2𝜋𝑓= 2𝜋(5)

𝜏≤ 6.4× 10−3s.

Example 2. Solution

A temperature sensor is to be selected to measure

temperature within a reaction vessel. It is suspected that the

temperature will behave as a simple periodic waveform with a

frequency somewhere between 1 and 5 Hz. Sensors of several

sizes are available, each with a known time constant. Based

on time constant, select a suitable sensor, assuming that an

absolute value for the dynamic error of 2% is acceptable.

Accordingly, a sensor having a time constant of 6.4 ms or less

will work.

aA𝛿ሺωሻaA≤ 0.02 −0.02 ≤ 𝛿ሺωሻ≤ 0.02 −0.02 ≤ 𝐴𝑟 − 1 ≤ 0.02 0.98 ≤ 𝐴𝑟 ≤ 1.02

0.98 ≤ 1ඥ1+ሺ𝜏𝜔ሻ2 ≤ 1.0

0 ≤ 𝜏𝜔≤ 0.2

0 ≤ 2𝜏𝜋(5) ≤ 0.2

The smallest value of 𝐴𝑟 will occur at the largest frequency

𝜔= 2𝜋𝑓= 2𝜋(5)

𝜏≤ 6.4× 10−3s.

2nd Order Systems Example:

Spring – mass damper

RLC Circuits

Accelerometers

Mathematical Model:

𝑑2𝑥𝑑𝑡2 + 2𝜁𝜔𝑛 𝑑𝑥𝑑𝑡 + 𝜔𝑛2𝑥= 𝑓ሺ𝑡ሻ 𝜁 Damping ratio (dimensionless) 𝜔𝑛 Natural frequency (1/s) 𝑓ሺ𝑡ሻ: Input (quantity to be measured) 𝑥: Output (instrument response)

2nd Order Systems with step input

𝑓ሺ𝑡ሻ= 𝐾𝑢(𝑡)

𝑢ሺ𝑡ሻ= ቄ0 𝑡 < 01 𝑡 ≥ 0

ds

𝑑2𝑥𝑑𝑡2 + 2𝜁𝜔𝑛 𝑑𝑥𝑑𝑡 + 𝜔𝑛2𝑥= 𝐴𝑓ሺ𝑡ሻ 𝜁 Damping ratio (dimensionless) 𝜔𝑛 Natural frequency (1/s) 𝑓ሺ𝑡ሻ: Input (quantity to be measured) 𝑥: Output (instrument response) 𝐴: Arbitrary constant



2nd Order Systems with periodic input


Documents

Class 4 Systems and Solutions. 1 st Order Systems