How to find the system response given the transfer function without using Laplace transforms?

The given transfer function is:

$$\frac{6p+100}{p^2+12p+100}$$

and inputted into the system is a unit step response:

$$r(t) = \left\{\begin{array} f0 & t < 0\\1 & t > 0 \end{array}\right.$$

However I'm not supposed to use Laplace transforms to solve this. I believe it has something to do with finding the poles but I'm not exactly sure how to accomplish this.

• One way would be expanding that as a function of v(in) and v(out); then p represents the derivative of each quantity. From then on it's about changing the differential equation into a more solvable one. May 19 '18 at 18:37
• I'm not sure I understand. Yes, you can calculate the step response by looking at the poles and their residues, but that is basically performing an inverse Laplace transform which is what you're not supposed to use... May 19 '18 at 19:04
• To examine the Bode Plot take the log of the function and compute the value for p=0. THen what is the breakpoint at -3dB = 20logf(x) then what is the dB/octave May 19 '18 at 19:09
• Use differential equation analysis. A step is quite a friendly input for this approach.
– Chu
May 20 '18 at 9:36

I assume you are permitted to perform partial fractions, even if you aren't supposed to use $\mathscr{L}^{-1}$. The roots of your denominator are $p_1=-6+j\:8$ and $p_2=-6-j\:8$ and the root of the unit step function, $\frac{1}{p}$ is $p_3=0$. You have:

\begin{align*}\frac{1}{p}\cdot \frac{6p+100}{p^2+12p+100}&=\frac{6p+100}{p\cdot\left(p-p_1\right)\cdot\left(p-p_2\right)}\\\\&=\frac{A}{p-p_1}+\frac{B}{p-p_2}+\frac{C}{p-p_3}\\\\&=\frac{-0.5}{p-p_1}+\frac{-0.5}{p-p_2}+\frac{1}{p-p_3}\end{align*}

We know that the proposed solutions take the form of $V_{\left(t\right)}=A\: e^{\:p\: t}$. So it follows that:

\begin{align*} V_{\left(t\right)}&= A\: e^{\:p_1\: t}+B\: e^{\:p_2\: t}+C\: e^{\:p_3\: t}\\\\&=-\frac{1}{2}\: e^{\:p_1\: t}-\frac{1}{2}\: e^{\:p_2\: t}+ e^{\:p_3\: t}\\\\&=-\frac{1}{2}\: e^{-6\: t}\: e^{8j\: t}-\frac{1}{2}\: e^{-6\: t}\: e^{-8j\: t}+ 1\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\left[\operatorname{cos}\left(8t\right)+i\operatorname{sin}\left(8t\right)\right]+\left[\operatorname{cos}\left(-8t\right)+i\operatorname{sin}\left(-8t\right)\right]\right)\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\operatorname{cos}\left(8t\right)+i\operatorname{sin}\left(8t\right)+\operatorname{cos}\left(8t\right)-i\operatorname{sin}\left(8t\right)\right)\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\operatorname{cos}\left(8t\right)+\operatorname{cos}\left(8t\right)\right)\\\\&=1- e^{-6\: t}\cdot\operatorname{cos}\left(8t\right) \end{align*}

I believe it has something to do with finding the poles but I'm not exactly sure how to accomplish this.

I can only tell you how to find the poles because, if I were to properly analyze this I would use a table of Laplace transforms. However, I can furnish a few more details by examining the equation. Hopefully, my observations are not sub-conciously based on my slight knowledge of laplace transforms but I can't rule it out.

You find the poles by looking at the denominator and equating it to zero. I'm going to use s instead of p: -

$s^2+ 12s+100 = 0$

The solution to that quadratic equation yields: -

s = -6 + j8 and s = -6 - j8 Picture source

So you have two poles at -6 on the horizontal axis and +/- 8 on the jw axis. Going back to your original equation, the "100" in the denominator represents the square of the natural frequency ($\omega_n$ in the above diagram). This means that the radius of the circle in that diagram is 10.

So, your damping factor ($\zeta$) is 0.6 and the damped frequency (the "ringing" frequency that the step will induce) is 0.8 x $\omega_n$ = 8 rad/sec.

Sanity check using a numerical inverse Laplace solver (with the OP's equation multiplied by the step function 1/s): - Great numerical Laplace solver

• I think the unit step is $\frac{1}{p}$ and would add a pole at (0,0).
– jonk
May 19 '18 at 19:54
• @jonk I agree but I'm trying to give information that bypasses use of 1/s and just looking at what the denominator would tend to predict. Have I done this without prior knowledge of Laplace or has that somehow unconciously affected my answer I wonder? May 19 '18 at 19:58
• Wasn't the object to not utilize the Laplace Transform (or maybe its inverse)? Maybe I could be misinterpreting what the OP says.
– user103380
May 19 '18 at 22:21
• The latter part of my answer was a sanity check to make sure that my prediction of natural resonant frequency, damping ratio and ringing frequency were correct and, of course, the only way to demonstrate this is by using an inverse Laplace numerical solver. Are there any other hoops you need me to jump through lol @KingDuken May 19 '18 at 22:55

If you do not want to go through the calculation of an inverse Laplace transform, you can first rearrange your expression in a low-entropy form as suggested by the fast analytical circuit techniques or FACTs: factor 100 in the numerator and the denominator. You obtain $H(s)=\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}$. You can see that you have a zero located at 2.6 Hz, a quality factor $Q$ of 0.833 and a resonant frequency of 1.592 Hz. The frequency response of such a network is below:

Now that you have these elements on hand, you realize that it is the description of a 2nd-order low-pass filter featuring a zero. A $LC$ network featuring ohmic losses and associated with a zero in the form of the capacitor equivalent series resistance (ESR) is a good model to obtain the transient response with SPICE. The circuit is below with values computed from the Mathcad sheet: The transfer function can quickly be derived with the help of FACTs and is determined as: $H(s)=\frac{1+sr_CC_1}{1+s(r_C+r_L)C_1+s^2L_1C_1}$

If the input source is a 1-V step, you have the following time-domain response: The time-domain response derived by Mathcad and matching Monsieur jonk's solution is given below and confirms the SPICE simulation: The key here was to rewrite the transfer function in the right way and obtain the necessary insight on where poles and zeros are located. Once it is in the correct low-entropy format, it is easier to infer what the time-domain response should look like.

One way to solve this without using the laplace transform is by taking this back to the differential equation which produced this transfer function.

The transfer function:

$$\dfrac{Y(s)}{r(s)}=\dfrac{6s+100}{s^2+12s+100}$$

This could be re-written as:

$$(s^2+12s+100)Y(s)=(6s+100)r(s)$$

Back to the time domain:

$$y''+12y'+100y=6r'+100r$$

Here, $r(t)$ is the unit step function. The derivative of the step function is the dirac delta function, so the differential equation becomes:

$$y''+12y'+100y=6\delta(t)+100, \text{for } t>0$$

Now you have a differential equation, which you can readily solve. The tricky part here is the dirac delta function, but remember it's zero everywhere but at $t=0$. Then for $t>0$:

$$y''+12y'+100y=100, \text{for } t>0$$

After solving the differential equation (homogeneous and particular solutions added together),

$$y(t)=1+K_1e^{-6t}\cos(8t)+K_2e^{-6t}\sin(8t) \tag1$$

And the constants are found by using the initial conditions.

Here is the trick, you have to somehow account for the impulse (dirac delta function), right at t=0 (we have disregarded it so far). Laplace transform are usually defined from $t=0^-$ in order to include the impulses (such as the dirac delta function). So we need to include this impulse if we want to get the same result one would with the Laplace transform.

This is a math problem from here on (don't wanna have to explain more extensively) but you can read this and get familiar with the details. In general though, when you have a differential equation, with a forcing function which includes a dirac delta, you need to find the solution to the case when $t>0$ (just what we did before in (1)), with the initial conditions $y(0)=0$, $y'(0)=a/m$ (for this case, $a=6$ and $m=1$). This is what brings in the fact the you have a dirac delta as part of the forcing function.

$a$ is just the coefficient multiplying the dirac delta function—6 in this problem. $m$ is the coefficient for the highest order derivative ($y''$ is highest order and its coefficient is 1).

With those initial conditions ($y(0)=0$, $y'(0)=6$), you can find the values of the constants, $K_1$ and $K_2$. These turn out to be $K_1=-1$ and $K_2=0$. Therefore:

$$y(t)=1-e^{-6t}\cos(8t)$$

A lot simpler to solve this problem with Laplace transform, but still possible to find a solution using time domain analysis. Also, the nature of the dirac delta function is what makes this problem a bit trickier.