# Write the differential equation that's equivalent to this transfer function

• Engineering
• s3a
In summary: It seems as if the solution is only consistent with the G(s) = C(s) / R(s) part, but not the other parts.
s3a
Homework Statement
Write the differential equation that is mathematically equivalent to the transfer function below: G(s) = Y(s) / R(s).

G(s) = C(s) / R(s) = (s^4 + 2s^3 + 5s^2 + s + 1) / (s^5 + 3s^4 + 2s^3 + 4s^2 + 5s + 2)
Relevant Equations
• ##G(s) = Y(s) / R(s)##
• ##r(t) = t^3 u(t)##
I have the solution to the problem, and I mechanically, but not theoretically (basically, why do the C(s) and R(s) disappear?), understand how we go from

##(s^5 + 3s^4 + 2s^3 + 4s^2 + 5s + 2) C(s) = (s^4 + 2s^3 + 5s^2 + s + 1) R(s)##

to

##c^{(5)}(t) + 3c^{(4)}(t) + 2c^{(3)}(t) + 4c^{(2)}(t) + 5c^{(1)}(t) + 2c^{(0)}(t) = r^{(4)}(t) + 2r^{(3)}(t) + 5r^{(2)}(t) + r^{(1)}(t) + r^{(0)}(t)##.

And, then I understand that the ##r(t)## needs to be replaced by ##t^3 u(t)##, but the final answer in the solution is

##c^{(5)}(t) + 3c^{(4)}(t) + 2c^{(3)}(t) + 4c^{(2)}(t) + 5c^{(1)}(t) + 2c^{(0)}(t) = 18δ(t) + (36 + 90t + 9t^2 + 3t^3) u(t)##, and I don't understand the step that leads the previous step to this final answer.

Any input would be GREATLY appreciated!

Hi, are you sure that ##r(t)=t^3u(t)## and not ##r(t)=3t^3u(t)## ?
Ssnow

I'm not sure of anything ( ;P ), but that ( ##r(t) = t^3 · u(t)## ) is technically what the solution says.

Maybe the solution is wrong?

Since no one else answered, @Ssnow, could you please explain to me what you've done just in case you are in fact correct?

I simply used ##r(t)=3t^3u(t)## instead yours... , so using the product rules: ## r^{(0)}(t)=3t^3u(t)##, ##r^{(1)}(t)=9t^2u(t) + 3t^3u^{(1)}(t)##, ##r^{(2)}(t)=18tu(t) + ...## higher derivatives for ##u(t)## ... so putting inside your second member that contains the derivatives of ##r(t)##, and collecting the ##u(t)##, we arrive to your solution ...
Ssnow

Actually, is the going from (s^5 + 3s^4 + 2s^3 + 4s^2 + 5s + 2) C(s) = (s^4 + 2s^3 + 5s^2 + s + 1) R(s) to ##c^(5) + 3c^(4) + 2c^(3) + 4c^(2) + 5 c^(1) + 2c^(0) = r^(4) + 2r^(3) + 5r^(2) + r^(1) + r^(0)## from the L{f^(n) (t)} = ##s^n F(s) – s^{n-1} f(0) – s^{n-2} f'(0) - . . . – s f^(n-2)(0) – f^(n-1)(0)## property, where all derivatives of f(t), including the 0th derivative, f(t) itself are 0 when t = 0 because the problem statement says to assume that "all initial conditions are equal to zero"?

As for going from the before-last step to the last step, thanks for your response; I now see how plugging in 3t^2 into the derivatives of r(t) and then multiplying by the appropriate coefficients yields the (36 + 90t + 9t^2 + 3t^3) part, but I'm not 100% sure why it's just a u(t) and not a bunch of derivatives of u(t) as well. Am I expected to to compute a monstrosity such as this one ( https://www.wolframalpha.com/input/?i=d^3/dt^3+(3t^3+*+u(t)) ) and then plug in t = 0? If it's the case that t = 0, wouldn't u(t) be u(0) instead? Also, what's with the 18 δ(t)? Does the δ(t) have something to do with the initial condition stuff having it be the case that t = 0?

Edit:
P.S.
I recently realized that the problem statement is inconsistent with the solution for the G(s) = C(s) / R(s) part, but it is consistent in stating that r(t) = u(t) t^3 (whether it's correct or not).

## 1. What is a transfer function?

A transfer function is a mathematical representation of the relationship between the input and output of a system. It describes how the output of a system changes in response to changes in the input.

## 2. How is a transfer function different from a differential equation?

A transfer function is a simplified version of a differential equation that only considers the input and output variables. It does not take into account the internal dynamics of the system, making it easier to analyze and manipulate. A differential equation, on the other hand, includes all variables and their relationships within the system.

## 3. How do you write a differential equation from a transfer function?

To write a differential equation from a transfer function, you need to perform a Laplace transform on both sides of the transfer function. This will convert the transfer function into a differential equation that can be solved using standard methods.

## 4. Are there any limitations to using transfer functions?

Yes, there are some limitations to using transfer functions. They are only applicable to linear systems, meaning that the output is directly proportional to the input. They also do not take into account any initial conditions or non-linearities within the system.

## 5. Can a transfer function be used to model any type of system?

No, transfer functions are specifically used to model linear systems. They are not suitable for modeling non-linear systems, such as biological or chemical systems, which may have complex and non-linear relationships between their input and output variables.

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