The settling time in matlab differs from the calculated value

Hi,

Why does Matlab calculate the settling time which measures the time for the error to fall below 2% of the peak value of the error for the closed-loop function T given below differently from the calculated value?

Cheers,

T= 4/( s^2 + 1.6 s+4)

Calculated settling time=4/(wn*zeta)=5 s;

where wn=2, zeta=0.4.

s = stepinfo(T)

s =

struct with fields:

RiseTime: 0.7326

TransientTime: 4.2047

SettlingTime: 4.2047

SettlingMin: 0.9065

SettlingMax: 1.2537

Overshoot: 25.3741

Undershoot: 0

Peak: 1.2537

PeakTime: 1.7269

[Mathieu NOE]

just to recall the exact definition :

• Settling time (ts) is the time required for a response to become steady. It is defined as the time required by the response to reach and steady within specified range of 2 % to 5 % of its final value.
• Calculated settling time=4/(wn*zeta)=5 s; is only an approximation of the true value

Matlab gives the correct value

Time Response of Second Order Control System (Worked Example) | Electrical4U

[fatih]

Since the formula Ts=4/(wn*zeta) gives a value that is very different from Matlab's answer, what is the general formula in Matlab that gives a value close to the settling time?

Answer 1

[Sam Chak]

Hi @fatih

From the following computations and the step response plot, you may gain insight into how the actual settling time is measured.

s       = tf('s');

%% plant
T       = 4/(s^2 + 1.6*s + 4)

T =
 
         4
  ---------------
  s^2 + 1.6 s + 4
 
Continuous-time transfer function.

[num, den] = tfdata(T, 'v');

%% properties
wn      = sqrt(den(end))

wn = 2

zeta    = den(2)/(2*wn)

zeta = 0.4000

%% estimated settling time
tsEst1  = 4/(wn*zeta)           % textbook formula (lazy version)

tsEst1 = 5

tsEst2  = -log(0.02)/(wn*zeta)  % true formula, yet a conservative approximation

tsEst2 = 4.8900

%% measured settling time
ST      = stepinfo(T);
tsMeas  = ST.SettlingTime

tsMeas = 4.2047

%% step response
step(T), grid on, hold on
plot(tsMeas, 0.98, 'ro')
yline(1.02,   '--', sprintf('1.02 threshold'),                  'color', '#7F7F7F', 'LabelVerticalAlignment', 'top',    'LabelHorizontalAlignment', 'left')
yline(0.98,   '--', sprintf('0.98 threshold'),                  'color', '#7F7F7F', 'LabelVerticalAlignment', 'bottom', 'LabelHorizontalAlignment', 'left')
xline(tsEst1, '--', sprintf('Estim1 Settling Time: %.3f s', tsEst1), 'color', '#7F7F7F', 'LabelVerticalAlignment', 'bottom')
xline(tsEst2, '--', sprintf('Estim2 Settling Time: %.3f s', tsEst2), 'color', '#7F7F7F', 'LabelVerticalAlignment', 'bottom', 'LabelHorizontalAlignment', 'left')
xline(tsMeas, '--', sprintf('Actual Settling Time: %.3f s', tsMeas), 'color', '#7F7F7F', 'LabelVerticalAlignment', 'bottom', 'LabelHorizontalAlignment', 'left')
hold off

[fatih]

Since the formula Ts=4/(wn*zeta) gives a value that is very different from Matlab's answer, what is the general formula in Matlab that gives a value close to the settling time?

[Sam Chak]

Hi @fatih

There is no closed-form expression to determine the exact settling time. To find it, one must solve the transcendental equation using numerical methods.

s       = tf('s');

%% Plant
T       = 4/(s^2 + 1.6*s + 4)

T =
 
         4
  ---------------
  s^2 + 1.6 s + 4
 
Continuous-time transfer function.

[num, den] = tfdata(T, 'v');

%% Extract data
a       = den(1);
b       = den(2);
c       = den(3);

sigma   = - b/2;                % real part of the pole
omega   = sqrt(4*a*c - b^2)/2;  % imaginary part of the pole
coeff   = b/sqrt(4*a*c - b^2);  % coefficient of the sine component

%% Analytical Solution of the Transfer Function
t   = linspace(0, 8, 8001);
sol = @(t) 1 - exp(sigma*t).*cos(omega*t) - coeff*exp(sigma*t).*sin(omega*t);

For underdamped 2nd-order linear system:

%% Step response
plot(t, sol(t)), grid on, 
xlabel('t'), ylabel('y(t)'), title('Step response')

%% Solve for Settling Time, ts
t0  = 4;    % initial guess
ts1 = fsolve(@(t) sol(t) - 1.02, t0)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

ts1 = 5.1417

ts2 = fsolve(@(t) sol(t) - 0.98, t0)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

ts2 = 4.2047

[fatih]

Thanks.