H83PDCH83PDCH83PDCH83PDC ProcessProcessProcessProcess Dynamic
& Control Dynamic & Control Dynamic & Control Dynamic
& ControlLecture 4 Dynamic Behaviour of ProcessesH83PDC Process
Dynamic & Control Lecture 4 - 2Lecture Outline Response of
First-Order Processes Response of Second-Order Processes Response
of Processes with Time DelaysH83PDC Process Dynamic & Control
Lecture 4 - 3Transfer Function Use transfer function to investigate
how processes respond to changes in their inputs Time-domain
solution can be found once the nature of the input change is
specified = + 1 +1 + 1 = + Output = controlled/ process variable
Inputs = manipulated & disturbance variablesG1(s) Q(s)G2(s) Ti
(s)T(s)++H83PDC Process Dynamic & Control Lecture 4 - 4Types of
Process Inputs Changes Sudden & sustained input changes e.g.
reactor feedstock change, heater input changeStep InputsMs Xs= )
(Gradually change with a roughly constant slope e.g. ambient
temperature changeRamp Input2) (sas XR=Sudden step change then
returns to its original value e.g. feed to a reactor is shut off
for an hourRectangular Pulse( )s tRPweshs X = 1 ) (H83PDC Process
Dynamic & Control Lecture 4 - 5Types of Process Inputs Changes
Complex manner, not possible to be described as deterministic
functions of timeRandom InputCyclic process changes e.g. 24 h
variations in cooling water temperatureSinusoidal Input2 2sin)
(+=sAs XA short transient disturbance e.g. injection of a tracer
dye into the processImpulse Input( ) 1 = s XimpulseH83PDC Process
Dynamic & Control Lecture 4 - 6Response of First-Order
ProcessesGeneral first-order transfer functionwhere K = process
gain; = time constant1 ) () (+=sKs Xs Y Investigate some particular
forms of input X(s) Deriving expressions for Y(s) Determine the
resulting response y(t)H83PDC Process Dynamic & Control Lecture
4 - 71. Step ResponseFor a step input of magnitude M,Thus,The
corresponding time-domain response: The process response is still
only 63.2% complete at t = The process output does approximate the
new steady-state value when t 5Step response of a first-order
processsMs X = ) (sMsKs Y +=1) (( ) te KM t y = 1 ) (H83PDC Process
Dynamic & Control Lecture 4 - 8 After an initial transient
period, the ramp input yields a ramp output with slope equal to Ka,
but shifted in time by the process time constant 2. Ramp
ResponseFor a ramp input,Thus,The corresponding time-domain
response:For large value of time (t >> )) ( ) ( t Ka t y2)
(sas X =2221 ) 1 () (sKasKasKas sKas Y + +=+=( ) Kat e Ka t yt+ =1
) (/Ramp response of a first-order process (comparison of input and
output)x(t)x(t)H83PDC Process Dynamic & Control Lecture 4 - 93.
Sinusoidal ResponseFor a sinusoidal input xsin(t) = A sin
t,Thus,The corresponding time-domain response:By using
trigonometric identities,where 2 2sin) (+=sAs X((
++++ +=+ +=2 2 2 222 2 2 21 1 ) )( 1 () ( s sssKAs sKAs Y( ) t t
eKAt yt sin cos1) (2 2+ +=) sin(11) (2 22 2 ++++=tKAeKAt yt) ( tan1
=Trigonometric Identitieswhere) sin( cos sin2 2 + + = + b a b a) (
tan1a b= H83PDC Process Dynamic & Control Lecture 4 - 10 The
exponential term goes to zero as t , leaving a pure sinusoidal
response Frequency Response ! (will be discussed later) For large
t, the y(t) is also sinusoidal, therefore the output sine is
attenuated byAmplitude attenuation !) sin(11) (2 22 2 ++++=tKAeKAt
yt112 2+ Phase lag A12 2+ KASinusoidal responseProcess112233) (t x)
(t ytt) ( AH83PDC Process Dynamic & Control Lecture 4 -
11Response of Second-Order ProcessesGeneral second-order transfer
functionwhere K = process gain; = time constant (determine response
time); = damping coefficient (dimensionless)A second-order transfer
function arises physically whenever: A second-order differential
equation process model is transformed Two first-order processes are
connected in series2 12 1 2 12 1) 1 )( 1 ( ) 1 )( 1 ( ) () () ( K K
Ks sKs sK Ks Xs Ys G =+ +=+ += = where 1 2) (2 2+ +=s sKs G X(s)
X(s)Y1(s)X(s) X(s)H83PDC Process Dynamic & Control Lecture 4 -
12 Equating the denominators, Factor left side of equation, Roots,
Three important classes of second-order systems( )( ) 1 1 1 22 12
2+ + = + + s s s s ;121 = ) 1 ( ;122 += |||
\|+ +|||
\|+ = + + 11111 22 22 2 s ss sH83PDC Process Dynamic &
Control Lecture 4 - 131. Step ResponseFor a step input of magnitude
M,The corresponding time-domain response:Case 1: Overdamped ( >
1)Case 2: Critically Damped ( = 1)Case 3: Underdamped (0 < 1)s s
sKMs Y) 1 2 () (2 2+ += )]}1sin(1)1)[cos( exp( 1 { ) (222t ttKM t y
+ =)]}1sinh(1)1)[cosh( exp( 1 { ) (222t ttKM t y + =)] exp( ) 1 ( 1
[ ) ( t tKM t y + =H83PDC Process Dynamic & Control Lecture 4 -
14 Large values of yield a sluggish (slow) response The faster
response without overshoot is obtained for critically damped case (
= 1)Step response of critically damped andoverdamped second-order
processesH83PDC Process Dynamic & Control Lecture 4 - 15
Responses exhibit a higher degree of oscillation and overshoot
(y/KM> 1) as approaching zeroStep response of underdamped
second-order processesH83PDC Process Dynamic & Control Lecture
4 - 16Performance characteristics for the step response of an
underdamped processDynamics of underdamped processes1. Rise time
(tr) = time that process output takes to first reach the new
steady-state value2. Time to first peak (tp) = time that required
for the output to reach its first maximum value3. Settling time
(ts) = time that required for the process output to reach and
remain inside a band whose width is equal to 5% of the total change
in y for 95% response time4. Overshoot, OS = a/b (% overshoot = 100
a/b)5. Decay ratio, DR = c/a (c = height of 2ndpeak)6. Period of
oscillation (P) = time between two successive peaks or two
successive valleys of the responsesH83PDC Process Dynamic &
Control Lecture 4 - 17Analytical expressions:a. Time to first
peakb. Overshootc. Decay ratiod. Period of oscillation) 1 2 exp( )
(2 2 = = OS DR212= P21 =pt|||
\|=21expOSH83PDC Process Dynamic & Control Lecture 4 -
18Relation between some performance characteristics of an
underdamped second-order process and the process damping
coefficientH83PDC Process Dynamic & Control Lecture 4 -
192.Sinusoidal ResponseFor a sinusoidal input xsin(t) = A sin t,
the output for large values of time:WhereOutput amplitude:Amplitude
ratio:Normalized amplitude ratio:) sin() 2 ( ] ) ( 1 [) (2 2 2 ++ =
tKAt y((
=21) ( 12tan2 2 2) 2 ( ] ) ( 1 [ + = =KAAAR)2 2 2) 2 ( ] ) ( 1
[1 + = =KAAARN)2 2 2) 2 ( ] ) ( 1 [ + =KAA)H83PDC Process Dynamic
& Control Lecture 4 - 20 The maximum value of ARNcan be found
by:707 . 0 0 ;2 1;1 212max2maxmax
