1. A process can be represented by the first order equation
##4\frac{dy(t)}{dt}+y(t)=3u(t)##
Assume the initial steady state is steady (y=0 at t=0)
(a) Determine the transfer function of this process in the s domain
(b) If the input is a ramp change in u(t)=4t, determine the value of y(t) when t=10s
2. For part (a) i had to use laplace transform on ##4\frac{dy(t)}{dt}+y(t)=3u(t)## term by term
For part (b) I used ##y(t) = 4[\tau+t+\tau e^\frac{t}{\tau}##]
3. First of all for part a
##4\frac{dy(t)}{dt}+y(t)=3u(t)##
##4[sy(s)y(s)]+y(s)=3u(s)##
So ##4[sy(s)+y(s)=3u(s)##
or ##y(s)[4s+1]=3u(s)##
So the transfer function (output / input) or gain for this process in the `s` domain would be:
##G(s)=\frac{y(s)}{3u(s)}=\frac{1}{4s+1}##
The attachment (part A example) is from my notes which I have basically tried to adapt to this solution and follow
I am still working on part b and will post my solution in a couple of days but I was wondering if someone could look at this so far and advise.
##4\frac{dy(t)}{dt}+y(t)=3u(t)##
Assume the initial steady state is steady (y=0 at t=0)
(a) Determine the transfer function of this process in the s domain
(b) If the input is a ramp change in u(t)=4t, determine the value of y(t) when t=10s
2. For part (a) i had to use laplace transform on ##4\frac{dy(t)}{dt}+y(t)=3u(t)## term by term
For part (b) I used ##y(t) = 4[\tau+t+\tau e^\frac{t}{\tau}##]
3. First of all for part a
##4\frac{dy(t)}{dt}+y(t)=3u(t)##
##4[sy(s)y(s)]+y(s)=3u(s)##
So ##4[sy(s)+y(s)=3u(s)##
or ##y(s)[4s+1]=3u(s)##
So the transfer function (output / input) or gain for this process in the `s` domain would be:
##G(s)=\frac{y(s)}{3u(s)}=\frac{1}{4s+1}##
The attachment (part A example) is from my notes which I have basically tried to adapt to this solution and follow
I am still working on part b and will post my solution in a couple of days but I was wondering if someone could look at this so far and advise.
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