Modeling a second order system in time and frequency domain

Question

For school I was asked to model the transfer function $H(s)=\frac{3}{(s+4)(s+5)}$ in both the time and frequency domains using initial conditions $y(0)=2,y'(0)=3$, a step input, and Simulink. I would appreciate any help in better understanding this problem and any real world example would also be appreciated as the coursework is not typically related to real applications.

TL;DR:

1. How do I pick which factor to apply each initial condition to in the frequency domain ($s+4$ vs. $s+5$)?
2. What changes should I make in my Simulink models to reach the same system response in both the time and frequency domain?
3. Do you have any words of wisdom to help me understand this topic more in depth?

The differential equation of this system is $3(e^{-4t}-e^{-5t})$ which I used to set up the frequency domain simulation, summing $\frac{3}{s^2-9s-20}$ with $y(0)*3e^{-4t}$ and $y'(0)*3e^{-5t}$.

My professor claimed the following response was accurate, but I have not been able to get the time domain representation using integrators to match.

Here is my attempt at the time domain model:

I passed $y'(0)=3$ into Integrator and $y(0)=2$ into Integrator1 as initial conditions, which yields the response:

It is my understanding the two responses should be the same. I can get a very similar response in the frequency domain by switching which expression to multiply with the initial conditions (that is swapping the constant blocks used as product inputs) but it is not exact.

Answer 1

For the Simulink model, you can incorporate the initial conditions in the Simulink model. If $y(t)$ is the output and $r(t)$ is the input, then the differential equation for the transfer function is: $$\frac{\mathbb{d}^2 y(t)}{\mathbb{d}t} + 9\frac{\mathbb{d}y(t)}{\mathbb{d}t} +20y(t)=3 r(t)$$

This means that: $$Y(s)=\frac{(s+9)y(0) + y'(0)}{(s+4)(s+5)}+\frac{3}{(s+4)(s+5)}R(s)$$ Of course, the transfer function $H(s)$ is obtained when all initial conditions are zero: $$H(s)=\frac{Y(s)}{R(s)}=\frac{3}{(s+4)(s+5)}$$ Therefore you can simply model the following in simulink and use the values for $y(0)=2$ and $y'(0)=3$: $$Y(s)=\frac{2(s+9) + 3}{(s+4)(s+5)}+\frac{3}{(s+4)(s+5)}R(s)$$

Now, the time domain solution is: $$y(t)=\frac{49}{4} \mathrm{e}^{-4t}-\frac{52}{5}\mathrm{e}^{-5t}+\frac{3}{20}$$

The Simulink diagram is shown below:

The resulting plots are given below:

You can see that both solutions are identical.

Note: In modelling the initial conditions, we want the system response when the input $r(t)$ is zero. This is the zero input response which can be achieved by using an impulse response to the homogenous solution. And an impulse response can be generated by differentiating a step response. Hence the $s$ in the transfer functions that represent the initial conditions.

Answer 2

Only a partial answer.

... model the transfer function $H(s)=\frac{3}{(s+4)(s+5)}$ in both the time and frequency domains using initial conditions $y(0)=2,y'(0)=3 ...$

Transfer functions are derived under the assumption of zero initial conditions AFAIK.

The differential equation of this system is $3(e^{-4t}-e^{-5t})$ which I used to set up the frequency domain simulation, summing $\frac{3}{s^2-9s-20}$ with $y(0)*3e^{-4t}$ and $y'(0)*3e^{-5t}$.

It is not an equation and it is not a differential equation.

Differential equation is something like $y\prime\prime (t) + 9 y\prime (t) + 20 y(t) = 3 x(t)$

In the "frequency domain approach", what you appear to be doing is computing the natural (?) and forced response (?) separately (I'm not sure).

... with $y(0)*3e^{-4t}$ and $y'(0)*3e^{-5t}$

I do not know how we can apportion each initial condition to a pole of the system. AFAIK, each pole would contribute to $y(0)$ as well as $y^n\prime (0)$

Do you have text book references or lecture notes where the initial conditions are paired to a pole in the system ?