Example: A process has a transfer function with the following, With k=2,τ_1=10,τ_2, case 1= τ_a=20

A process has a transfer function with the following equation
G_s=(K (τ_a s+1))/((τ_1 s+1)(τ_2 s+1))

With k=2,τ_1=10,τ_2=2. If τ_a has the following cases
case 1= τ_a=20
case 2= τ_a=4
case 3= τ_a=1
case 4= τ_a=-2

For each case obtain process response curve in the time domain for a step input of magnitude of 0.5 and plot the response curve

where:
K is the process gain,
τ determines the speed of response (or,equivalently,the response time)of the system / the process time constant

Function:
G_s=(K (τ_a s+1))/((τ_1 s+1)(τ_2 s+1))

Step input (1/s)
G_s=(K (τ_a s+1))/(s(τ_1 s+1)(τ_2 s+1))

Convert to the time domain
y(t)=KM[ 1+((τ_a-t_1 ))/((τ_1-t_2 ) ) e^(-t/τ_(1 ) ) - ((τ_a-t_2 ))/((τ_2-t_1 ) ) e^(-t/τ_(2 ) )]

Given: k=2,τ_1=1,τ_2=2,M=0.5

y(t)=2(0.5)[ 1+((τ_a-10))/((10-2) ) e^(-t/10) - ((τ_a-2))/((2-10) ) e^(-t/2)]

y(t)=2(0.5)[ 1+((τ_a-10))/((8) ) e^(-t)/10 - ((τ_a-2))/((-8) ) e^(-t/2)]

case 1= τ_a=20
y(t)=1[ 1+((20-10))/((8) ) e^(-t/10) - ((20-2))/((-8) ) e^(-t/2) ]= 1+1.25e^(-t/10) -2.25e^(-t/2)

case 2= τ_a=4
y(t)=1[ 1+((4-10))/((8) ) e^(-t/10) - ((4-2))/((-8) ) e^(-t/2) ]= 1-0.75e^(-t/10) -0.25e^(-t/2)

case 3= τ_a=1
y(t)=1[ 1+((1-10))/((8) ) e^(-t/10) - ((1-2))/((-8) ) e^(-t/2) ]= 1-1.125e^(-t/10) -0.125e^(-t/2)

case 4= τ_a=-2
y(t)=1[ 1+((-2-10))/((8) ) e^(-t/10) - ((-2-2))/((-8) ) e^(-t/2) ]= 1-1.5e^(-t/10) -0.5e^(-t/2)

[vuyiswampanza]

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