## Time Delay in Dynamic Systems

Time delay is a shift in the effect of an input on an output dynamic response. A first-order linear system with time delay is:

$$\tau_p \frac{dy(t)}{dt} = -y(t) + K_p u\left(t-\theta_p\right)$$

has variables y(t) and u(t) and three unknown parameters with

$$K_p \quad \mathrm{= Process \; gain}$$

$$\tau_p \quad \mathrm{= Process \; time \; constant}$$

$$\theta_p \quad \mathrm{= Process \; dead \; time}$$

The time delay \theta_p is expressed as a time shift in the input variable u(t).

$$u\left(t-\theta_p\right)$$

$$y(t \lt \theta_p) = y(0)$$

$$y(t \ge \theta_p) = \left( e^{-\left(t - \theta_p \right) / \tau_p}\right) y(0) + \left( 1 - e^{-\left(t - \theta_p \right) / \tau_p} \right) K_p \Delta u$$

For a step change \Delta u, the analytical solution for a first-order linear system without time delay (x(t) = y(t) with theta_p=0)

$$\tau_p \frac{dx(t)}{dt} = -y(t) + K_p u\left(t\right)$$

is

$$x(t) = K_p \left(1-\exp\left(\frac{-t}{\tau_p}\right)\right) \Delta u$$

$$y(t) = x \left( t-\theta_p \right) S \left( t-\theta_p \right)$$

$$\quad = K_p \left( 1- \exp\left(-\frac{ t-\theta_p } { \tau_p } \right) \right) \, \Delta u \, S \left( t-\theta_p \right)$$

where

$$S\left(t-\theta_p\right)$$

is a step function that changes from zero to one at t=theta_p.

Analytic Solution Derivation with Laplace Transforms

$$\tau_p \frac{dy(t)}{dt} = -y(t) + K_p u\left(t-\theta_p\right)$$

Perform a Laplace transform from the tables on each part of the equation:

$$\mathcal{L}\left(\tau_p \frac{dy(t)}{dt}\right) = \tau_p \left(s Y(s) - y(0)\right)$$

$$\mathcal{L}\left(-y(t)\right) = -Y(s)$$

$$\mathcal{L}\left(K_p u\left(t-\theta_p\right)\right) = K_p U(s) e^{-\theta_p s}$$

If the input U(s) is a step function of size \Delta u then:

$$U(s) = \frac{\Delta u}{s}$$

Combining all of the individual Laplace transforms, the equation in Laplace domain with zero initial condition y(0)=0 is then:

$$\tau_p \, s \, Y(s) = -Y(s) + K_p \frac{\Delta u}{s} e^{-\theta_p s}$$

At this point, the equation is algebraic and can be solved for Y(s) by combining terms on each side:

$$\tau_p \, s \, Y(s) + Y(s) = K_p \frac{\Delta u}{s} e^{-\theta_p s}$$

and factoring out the Y(s) term:

$$Y(s) \left(\tau_p \, s + 1\right) = K_p \frac{\Delta u}{s} e^{-\theta_p s}$$

A final steps are to isolate Y(s) on the left side of the equation and perform an inverse Laplace transform to return to the time domain:

$$Y(s) = K_p \frac{\Delta u}{s\,\left(\tau_p \, s + 1\right)} e^{-\theta_p s}$$

$$\mathcal{L}^{-1}\left(Y(s)\right) = y(t) = K_p \left( 1-\exp \left( -\frac{ \left( t-\theta_p \right) } { \tau_p } \right) \right) \, \Delta u \, S \left( t-\theta_p \right)$$

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

# specify number of steps
ns = 100
# define time points
t = np.linspace(0,ns/10.0,ns+1)

class model(object):
# process model
Kp = 2.0
taup = 200.0
thetap = 0.0

def process(y,t,u,Kp,taup):
# Kp = process gain
# taup = process time constant
dydt = -y/taup + Kp/taup * u
return dydt

def calc_response(t,m):
# t = time points
# m = process model
Kp = m.Kp
taup = m.taup
thetap = m.thetap
# specify number of steps
ns = len(t)-1

delta_t = t[1]-t[0]

# storage for recording values
op = np.zeros(ns+1)  # controller output
pv = np.zeros(ns+1)  # process variable

# step input
op[10:]=2.0

# Simulate time delay
ndelay = int(np.ceil(thetap / delta_t))

# loop through time steps
for i in range(0,ns):
# implement time delay
iop = max(0,i-ndelay)
y = odeint(process,pv[i],[0,delta_t],args=(op[iop],Kp,taup))
pv[i+1] = y[-1]
return (pv,op)

# calculate step response
model.Kp = 2.5
model.taup = 2.0
model.thetap = 3.0
(pv,op) = calc_response(t,model)

pv2 = np.zeros(len(t))
for i in range(len(t)):
pv2[i] = model.Kp * (1.0 - np.exp(-(t[i]-model.thetap-1.0)/model.taup))*2.0

pv3 = np.zeros(len(t))
for i in range(len(t)):
pv3[i] = model.Kp * (1.0 - np.exp(-(t[i]-1.0)/model.taup))*2.0

plt.figure(1)

plt.subplot(2,1,1)
frame1 = plt.gca()
frame1.axes.get_xaxis().set_visible(False)
plt.plot(t,pv3,'r--',linewidth=1,label=r'$y(t)$')
plt.plot(t,pv2,'k.-',linewidth=2,label=r'$y(t-\theta_p)$')
plt.plot([0,4,4.0001,10],[0,0,1,1],'g:',linewidth=3,label=r'$S(t- \theta _p)$')
plt.plot(t,pv,'b-',linewidth=4,label=r'$x(t)$')
plt.legend(loc='best')
plt.ylabel('Process Output')
plt.ylim([-4,5])

plt.subplot(2,1,2)
frame1 = plt.gca()
frame1.axes.get_xaxis().set_visible(False)
plt.plot(t,op,'r-',linewidth=3,label=r'$u(t)$')
plt.plot(t+3.0,op,'k--',linewidth=3,label=r'$u(t-\theta_p)$')
plt.ylim([-0.1,2.1])
plt.xlim([0,10])
plt.legend(loc='best')
plt.ylabel('Process Input')

plt.xlabel('Time')

plt.savefig('output.png')
plt.show()