Finding the output time response of a transfer function with a step response

Question

The question is finding the output time response of the transfer function below. I solved that it is an underdamped response. The step response is 1/s.

$$T(s)=\frac{20}{s^{2}+6s+144}$$

According to the answer key, the answer is $$A+Be^{-3t}cos(3\sqrt{15}t+\Phi )$$

My goal is to find A and B.

And I do understand that the poles of an underdamped response is two complex numbers: $$-\sigma_{d}\pm j\omega_{d}$$

So I already know what happened in the second term. Now, I tried solving for A and B via partial fractions. I found out that A = 5/36, B = -5/6, and C = -5/36. Is this the right approach to find what are the proper values of A and B in the answer, even more so that doing the partial fraction brought out three variables: $$\frac{20}{s(s^{2}+6s+144)}=\frac{A}{s}+\frac{B+Cs}{s^{2}+6s+144}$$

Answer 1

You can do this in several ways but I like to use the Heaviside expansion formula more than the partial fraction decomposition. If you re-arrange your transfer function with a unitless denominator then multiplied by the step function \$\frac{1}{s}\$, you have \$D(s)=s(\frac{s^2}{144}+\frac{6}{144}s+1)\$. The forced value is the dc gain multiplied by the 1-V step and it is 20/144 or 139 mV.

The roots for this expression are \$s_1\$, \$s_2\$ and \$s_3\$. Now, find the derivative of \$D(s)\$ with respect to \$s\$ (call it \$D'(s)\$ and determine the value of \$D'(s_1)\$, \$D'(s_2)\$ and \$D'(s_3)\$. This is it, you have your time-domain expression if you apply the formula as shown below:

The next step is to re-arrange the complex exponential to unveil the sine and cosine terms but this is the extra mile I leave to the student legs. I like this method as it yields insight on how the time-domain response is directly affected by the natural time constants in the denominator. You also clearly see how the zeroes (the numerator roots) if any, affect the coefficients only while the decays are solely fixed by the poles.

Answer 2

Verbal Kint's answer is one way to do it (+1), another way is to avoid the need for a 3rd order transfer function and, instead, find the impulse response, directly, which might be simpler to be integrated in time domain to give the step response (or not):

$$\begin{align} H(s)&=\dfrac{20}{s^2+6s+144} \tag{1} \\ \omega_{1,2}&=-3\pm j3\sqrt{15} \tag{2} \\ \mathcal{L}^{-1}(H(s))&=\dfrac{20}{3\sqrt{15}}\mathrm{e}^{-3t}\sin(3\sqrt{15}t) \tag{3} \\ \mathcal{L}^{-1}\left(\dfrac{H(s)}{s}\right)&=-\dfrac{\sqrt{15}}{108}\mathrm{e}^{-3t}(\sin(3\sqrt{15}t)+\sqrt{15}\cos(3\sqrt{15}t))+\dfrac{15}{108} \tag{4} \\ \int{\dfrac{20}{2\sqrt{15}}\mathrm{e}^{-3t}\sin(3\sqrt{15}t)\,\mathrm{d}t}&=-\dfrac{\sqrt{15}}{108}\mathrm{e}^{-3t}(\sin(3\sqrt{15}t)+\sqrt{15}\cos(3\sqrt{15}t)) \tag{5} \end{align}$$

If you plot (4) and (5) side by side you'll see that they are identical except the DC level. That's because integrating adds a constant, C. That constant is just (5) evaluated at zero, which is exactly the missing term:

Answer 3

Well, in general, we are trying to find:

$$\text{y}_\eta\left(\alpha,\beta,t\right):=\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\cdot\frac{\beta\cdot\eta}{\left(\text{s}+\alpha\right)^2+\eta^2}\right]_{\left(t\right)}\tag1$$

Using the convolution theorem of the Laplace transform, we can write:

$$\text{y}_\eta\left(\alpha,\beta,t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\right]_{\left(t-\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\beta\cdot\eta}{\left(\text{s}+\alpha\right)^2+\eta^2}\right]_{\left(\tau\right)}\space\text{d}\tau\tag2$$

Using the table of selected Laplace transforms, we can see that:

$$\text{y}_\eta\left(\alpha,\beta,t\right)=\beta\cdot\int_0^t1\cdot\exp\left(-\alpha\tau\right)\sin\left(\eta\tau\right)\space\text{d}\tau=\beta\int_0^t\exp\left(-\alpha\tau\right)\sin\left(\eta\tau\right)\space\text{d}\tau=$$ $$\frac{\beta\left(\eta-\exp\left(-\alpha t\right)\left(\eta\cos\left(\eta t\right)-\alpha\sin\left(\eta t\right)\right)\right)}{\alpha^2+\eta^2}\tag3$$

And in your circuit, we can find \$\eta,\alpha\$ and \$\beta\$ by solving the following system of equations:

$$ \begin{cases} 2\alpha=6\\ \\ \alpha^2+\eta^2=144\\ \\ \beta\cdot\eta=20 \end{cases}\tag4 $$