I've been working with control systems and transfer functions recently, but I've come across a format that I haven't seen before using coefficients of theta, and I can't find any information about it. Does anyone recognize the format below, used to describe a control system. As I was writing this question I realised what it is probably saying, but I'm unable to fully understand it. My thought process is written below so could somebody help me fill in the gaps?

$$\ddot{\Theta_0} + 4\dot{\Theta_0} + 68\Theta_0 = 34\Theta_i$$

I'm assuming that the dots above theta are the powers. 2 dots means squared etc. $\Theta_0$ is representing the output while $\Theta_i$ is representing the input. (The 'O' is actually a zero not a letter but I'm assuming thats a typo?). Since a transfer function is: $$T(s) = \frac{Y(s)}{U(s)}= \frac{output}{input}$$ we could divide by $\Theta_i$ to get the transfer function. But this is where I get lost. I think this has something to do with the characteristic equation - the denominator of the transfer function when equal to $0$ - but I dont see how?

  • 3
    $\begingroup$ $\ddot{Θ}$ means second derivative etc. And just look for the way to go from time domain to frequency domain. You will find plenty of resources. $\endgroup$ Commented May 16, 2020 at 14:26
  • $\begingroup$ @TeoProtoulis so is this the same as writing the differential equation $\frac{d^2y}{dy^2} + 4\frac{dy}{dx}+68y = 34f(x)$? $\endgroup$
    – MendelumS
    Commented May 17, 2020 at 8:37
  • $\begingroup$ Check the source where you came across this format and follow any references they make. $\endgroup$
    – Solar Mike
    Commented May 17, 2020 at 8:52
  • $\begingroup$ @SolarMike thats kind of the point...the source has no references, its just a few equations written in this form and it asks for a sketch of the step response for each system. (I'm not asking for help with the questions, I'm not doing them, just need help as I haven't seen the format before) $\endgroup$
    – MendelumS
    Commented May 17, 2020 at 9:04
  • $\begingroup$ So this "source" is a homework question? Or an assignment where you are meant to show understanding? $\endgroup$
    – Solar Mike
    Commented May 17, 2020 at 9:11

1 Answer 1


In mathematics,which are very closely related and broadly applied in control engineering, the derivatives with respect to time are often notated using the so-called: dot notation. You can check this link for more information: Derivative. This means that the particular equation (the one at the question) can be rewritten in the following form:

$$ \ddot{Θ}_ο+4\dot{Θ}_ο+68Θ_ο=34Θ_i \Rightarrow \ddot{y}+4\dot{y}=68y = 34u \Rightarrow $$

$$ \frac{d^2y}{dt^2}+4\frac{dy}{dt}+68y = 34u $$

This notation is used for up to the third derivative usually. For higher order derivatives, due to appearance reasons, the notation used changes and becomes: $y^{(n)}$ for the n-th order derivative with respect to time.

Now, what's left to do is to convert the equation from time-domain to $s$-domain. In order to do so the following formula is used:

$$ \mathscr{L}\{\frac{d^nf}{dt^n}\}=s^nF(s)-s^{n-1}f(0)-s^{n-2}f^{(1)}(0)- \ ... \ -sf^{(n-2)}(0)-f^{(n-1)}(0) $$


  • $f^{(n-a)} \ \rightarrow \ \text{(n-a)-th derivative order of the function f}$
  • $f^{(n-a)}(0) \ \rightarrow \ \text{initial value of the (n-a)-th derivative order of the function f}$

For this particular ordinary differential equation (ODE) this formula yields:

  • $\ddot{y}=s^2Y(s)-sy(0)-\dot{y}(0)$
  • $\dot{y}=sY(s)-y(0)$

By considering the usual fact of $0$ initial conditions it all comes down to:

$$ \frac{d^2y}{dt^2}+4\frac{dy}{dt}+68y = 34u \Rightarrow s^2Y(s)+4sY(s)+68Y(s)=34U(s) $$

$$ Y(s)\cdot (s^2+4s+68) = 34U(S) \Rightarrow \frac{Y(s)}{U(s)} = \frac{34}{s^2+4s+68} $$

The overall transfer function of this particular system is:

$$ \frac{Y(s)}{U(s)} = T(s) = \frac{34}{s^2+4s+68} $$


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