Control System – Transient Response Specification

As we already seen in my previous post, in transient response analysis , the underdamped response is the most important. We defined the transient response specification for a unit step input as:

c(t)  =  1 -  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta)

Where,   \theta = { \tan ^{ -1 }{{ \frac { { \sqrt { (1-{ \delta }^{ 2 }) } }}{\delta} } } }  and  { { \omega }_{ n } }\sqrt { (1-{ \delta }^{ 2 }) } ={ {  \omega }_{ d } }

,

 

c(t)  =  1 –  \frac { { e }^{-\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_d

t  + { \tan ^{ -1 }{{ \frac { { \sqrt { (1-{ \delta }^{ 2 }) } }}{\delta} } } }  )

 

A typical transient response for a step input is shown below:

Transient Response

Important specifications of the transient response are listed below :

  • Delay Time (T_d )
  • Rise Time (T_r)
  • Peak Time (T_p)
  • Peak Overshoot (M_p)
  • Settling Time (T_s)

Now let us define each of these term individually:

Delay Time (T_d) : It can be defined as the quantity of time required for any response in order to attain 50% of that of the final value in the very first try . It can be represented as,

                                           T_d    =  \frac { { 1+0.7\delta } }{ \omega _{ n } }  sec

Rise Time (T_r)  :  It is the amount of time required by the response to increase from 10% to 90% of the ultimate value of an overdamped system. For an underdamped system the rise time is the time taken by the response in order to increase from 0% to 100% of that of the  final value in the first attempt. The formula for this is,

                                            T_r     =  \frac { { \Pi -\theta } }{ \omega _{ d } } sec

Peak Time (T_p) : It is the amount of time needed by any response in order to reach its initial (first) peak. The first peak is eventually the highest peak achieved.

                                             T_p    =   \frac{\Pi}{\omega_d} sec

Peak Overshoot (M_p) :  It is the highest peak value of any response that is calculated from the initial input signal .Further it is the highest possible error in between the output and input . It is normally  represented in the form of percentage as shown below:

   percentage of { \quad M }_{ p }\quad   =    { e }^{ { { \frac { -\Pi \delta }{ \sqrt { (1-{ \delta }^{ 2 }) } } } } }\quad \quad \times \quad 100   

Settling time (T_s) : It can be defined as time needed for the transient damped oscillations to reach and then stay for around a given designated tolerance band ( ranging between 2 to 5 % of the input value).

                                             T_s     =     \frac{4}{\delta\omega_n} sec

Note : Rise time, peak time, and settling time gives us information about the speed of the transient response.

We shall now derive each of these response specifications (except T_d  for better clarity).

1.Derivation of Rise Time (T_r) :

For an underdamped system, it is the time taken for the output to reach 100% of the input value in the 1st attempt.

Hence at t = T_r   , c(t) = 1

We know, c(t)  =  1 –  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta) , 

At t = T_r  , c(t) = 1  ( for all the ‘t’ , consider it as ‘T_r  in the below equations)

1 –  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta) =  1  ,

Then ,     \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta ) =  0,

But    \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }  can not be zero,    \therefore sin ( \omega_dt  + \theta ) = 0 ,

Hence,   \omega_dt  +  \theta =  0 ,

We know, sin n\Pi  = 0     where n = 1, 2 …..

Then   sin (\omega_dt  +  \theta) = 0    only if      \omega_dt  +  \theta  =  n\Pi

\omega_dt =  n\Pi\theta    ,   then   t  =  \frac{n\Pi-\theta}{\omega_d}  ,   

T_r  is calculated when the output reaches 100% of the input in the first attempt, i.e at n = 1 ,

T_r = \frac { { \Pi -\theta } }{ \omega _{ d } }    sec

2.Derivation of Peak Time (T_p) :

At  T_p  ,the value of c(t) is at its maximum value, we know,

c(t)  =  1 –  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta)  ,  

At t = T_p ,  c(t)   becomes maximum . We thus should use the Maxima theorem ( from derivatives property in maths), which states that the derivative of the maxima is zero.

\frac{d}{dt}[c(t)] = 0  ( at time t = T_p) ,

Thus,  \frac{d}{dt}[ 1 –  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta)] = 0  ( at time t =

T_p) ,

We differentiate c(t) wrt zero,

\frac{d}{dt}[ 1 –  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta)] = 0  ,  

Then,  

\therefore \quad\frac { -{ e }^{ -{ \omega }_{ n }\delta t }.(-\delta { \omega }_{ n }).\sin { { (\omega }_{ d }t } +\theta ) }{ \sqrt { (1-{ \delta }^{ 2 } )} } \quad -\quad \quad \frac { -{ e }^{ -{ \omega }_{ n }\delta t } }{ \sqrt { (1-{ \delta }^{ 2 } )} } .{ \omega }_{ d }\cos { ({ \omega }_{ d }t+\theta ) } \quad =\quad \quad 0\quad ,

Also we know, 

{ { \omega }_{ n } }\sqrt { (1-{ \delta }^{ 2 }) } ={ {  \omega }_{ d } } ,

On squaring both sides and putting the value of \omega_d in the equation,

\therefore\frac { -{ e }^{ -\delta { \omega }_{ n }t } }{ \sqrt { (1-{ \delta }^{ 2 }) } } { \quad \omega }_{ n }[\quad \delta \sin { ({ \omega }_{ d }t+\theta ) } -\quad \sqrt { (1-{ \delta }^{ 2 }) } \cos { ({ \omega }_{ d }t+\theta ) } ]\quad =\quad 0  ,

 

But    \frac { -{ e }^{ -\delta { \omega }_{ n }t } }{ \sqrt { (1-{ \delta }^{ 2 }) } }  can not be zero ,

Then ,   \quad \delta \sin { ({ \omega }_{ d }t+\theta ) } -\quad \sqrt { (1-{ \delta }^{ 2 }) } \cos { ({ \omega }_{ d }t+\theta ) } \quad =\quad 0

\quad \delta \sin { ({ \omega }_{ d }t+\theta ) } = \quad \sqrt { (1-{ \delta }^{ 2 }) } \cos { ({ \omega }_{ d }t+\theta ) } \quad   ,

\frac { { \sin { ({ \omega }_{ d }t+\theta ) } } }{ \cos { ({ \omega }_{ d }t+\theta ) } }    =   \frac { \sqrt { 1-{ \delta }^{ 2 } } }{ \delta }   ,

\therefore {{\tan { ({ \omega }_{ d }t+\theta ) }}  =  \frac { \sqrt { 1-{ \delta }^{ 2 } } }{ \delta } ,

From properties of trigonometric relations we know,

n\Pi = \omega_dt   ,

t =    \frac { { n\Pi } }{ \omega _{ d } }  or      T_p =  \frac { { \Pi } }{ \omega _{ d } }  (for first overshoot i.e n = 1),

Also,   T_p =  \frac { { \Pi } }{ { { \omega }_{ n } }\sqrt { (1-{ \delta }^{ 2 }) } } sec

When n = 2 , we get time taken for 1st    overshoot

When n = 3 , we get time taken for 2nd   overshoot

3.Derivation of Peak Overshoot (M_p) : 

We already know that M_p is the maximum value of c(t) measured over and above the input value. So we can easily say that M_p  is to be calculated at time t =

T_p , then we have,

M_p =  c(t) – 1 (at time t = T_p  ) ,

c(t)  =  1 –  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta)  –  1  ,  ( since t = T_p  =  \frac{\Pi}{\omega_d} )

sin(\Pi + \theta)  =  – sin\theta  ,   

M_p\frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin\theta

Also, from the standard pole diagram (we have seen in my earlier post),

sinθ =  \frac { { \omega _{ n } }{ \sqrt { (1-{ \delta }^{ 2 }) } } }{ \omega _{ n } }  ,

sinθ =  { \sqrt { (1-{ \delta }^{ 2 }) } } }  ,

M_p\frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }{ \sqrt { (1-{ \delta }^{ 2 }) } } }       ,

M_p  =   { e }^{ -\delta { \omega _{ n } }t }   = { e }^{ -\delta { \omega _{ n } }\frac{\Pi}{\omega_d} }   =  { e }^{ -\frac { \delta \Pi } {\sqrt { (1-{ \delta }^{ 2 }) }} }   ( since    { { \omega }_{ n } }\sqrt { (1-{ \delta }^{ 2 }) } ={ {  \omega }_{ d } }   )   ,

% M_p = { e }^{ -\frac { \delta \Pi } {\sqrt { (1-{ \delta }^{ 2 }) }} }   x  100

4.Derivation of Settling Time ( T_s) :

A  system having underdamped transient response consists of poles which are complex conjugate , given by:

s_{ 1,2 }=\quad -\delta { \omega _{ n } }\quad \pm \quad j{ \omega _{ d } }

Also,  c(t)  =  1 -  \frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } }sin ( \omega_dt  + \theta)

We already seen that exponential part of above equation (\frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } })  governs  the speed of the response. Also the sinusoidal part merely provides the oscillations . Hence for calculation of Settling time, our prime focus should be on the exponential part of the above equation.

If we consider 2% tolerance , then  

(\frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } })=  0.02 ,

Also since we are dealing with underdamped system for which   ‘\delta‘<<< 1 ,

We can ignore the denominator,

Then ,   (\frac { { e }^{ -\delta { \omega _{ n } }t } }{ { \sqrt { (1-{ \delta }^{ 2 }) } } })   = 0.02,

Taking log both the sides,

{{ -\delta { \omega _{ n } }t } }  =   -3.912 ( at time t = ‘T_s‘  )

{{ \delta { \omega _{ n } }t } }  =   4  ( on approxiation) ,

Thus     T_s = \frac{4}{{\omega_n\delta}}  sec

Note: above formula is valid only for 2% tolerance band.

Finally, we reach the end of this wonderful post( really ??).I know the derivations part was a little boring and difficult to interpret at a first glance. I must say that before you dig deep inside derivations, please learn by heart all the formulas first( in the examination you can expect direct questions from the formulas). See you soon in my next post.

Spread the Wisdom !!

Techie Aric

Aric is a tech enthusiast , who love to write about the tech related products and 'How To' blogs . IT Engineer by profession , right now working in the Automation field in a Software product company . The other hobbies includes singing , trekking and writing blogs .

Leave a Reply

Your email address will not be published. Required fields are marked *