I was reading this PDF which tries to explain the concept of poles and zeros of Laplace transform. I started with second paragraph on page #590 and read it up to page #592 before the new section "Analysis of Electric Circuits" begins. Please have a look on FIGURE 32-5 on page #591. The following is an excerpt from the same PDF.

Our goal is to find combinations of σ and ω that exactly cancel the impulse response being investigated. This cancellation can occur in two forms: the area under the curve can be either zero, or just barely infinite. All other results are uninteresting and can be ignored. Locations in the s-plane that produce a zero cancellation are called zeros of the system. Likewise, locations that produce the "just barely infinite" type of cancellation are called poles. Poles and zeros are analogous to the mountains and valleys in our train story, representing the terrain "around" the frequency response.

It does make sense but I'm facing a problem to interpret a zero in the context of spring-mass system, as I explain below. My question is not directly related to electrical engineering but I'd say an engineer would be in a better position to answer it.

The following differential equation describes a mass-spring system with input force set to zero.


\$y(t)\$ represents displacement from equilibrium position, its first derivative would be velocity function, \$v(t)\$. Two initial conditions are as follows: \$y(0^-)=4\$, \$y'(0^-)=v(0^-)=0\$.

Laplace transform: \$Y(s)= \frac{4s+8}{s^2+2s+10}\$

Impulse response: \$y(t)=4\exp(-t)\cos(3t)+1.333\exp(-t)\sin(3t)\$

There are two poles at \$-1\pm 3i\$, and one zero at \$-2\$.

The zero could be written as \$-2+j(0)\$

It doesn't make sense to multiply the impulse response with \$e^{2t}\$. As you can see that area under the curve cannot be zero as the function is diverging.

$$e^{-\sigma t}(4e^{-t}\cos(3t)+1.333e^{-t}\sin(3t))$$



Edit #1:

This edit was made after after the comment of @jDAQ.

I also tried to do the convolution in Matlab.

clear all; close all; clc;



output=conv(impulse_response, input);

Please see the plot here: https://imagizer.imageshack.com/img923/231/Dr6MR4.jpg

You can see that the output is not zero as expected.

Edit #2:

The picture shows the spring-mass system with a damper: https://imagizer.imageshack.com/img924/8415/gK84LJ.jpg. At the beginning the spring is stretched four units toward the right of screen, and then released. The function y(t) shows the displacement of mass around the equilibrium position.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Jun 13 at 15:20

On the integration mentioned in the pdf

To check for the zero supposedly sitting at \$-2+0i\$ , the integration required is

\$\int_0^\infty k e^{-1t}sin(3t+\phi) e^{-(-2+0 i) t} dt\$.

But that value of \$s = -2+0 i\$ is outside the Region of convergence (ROC) for the given signal. You already saw that it is not converging. The Laplace transform exists only to the right side of the right most pole which in this case is \$-1\pm 3i\$.

See slide #10 of this presentation and Wikipedia

In the pdf that you have linked, the zero happens to be in the ROC. It is to the right of the right most pole. Figure is shown below. Pole and zero for the notch in the pdf linked by O.P.

You can adjust the spring mass damper example in your case so that the zero is to the right of the right most pole and try again to see if you are getting the result you expect (convergence of the integral).

Also, I still think there may be some confusion regarding the zeros of a specific signal (solved using initial conditions) and that of a system. The pdf does not mention initial conditions; they would have assumed IC as zero. However, non-zero initial conditions appear in your spring mass example.

On the concept of Zeroes for a spring mass system

A spring mass system with one spring and mass can be written in many ways (Your spring mass system input has been defined as zero, making the determination of zeroes of the system difficult). Each way changes the zeroes of the system. Some examples are given below.

4 examples of spring mass system

system (A)

Input : force f(t)

Output : displacement of mass y(t)

Net force on the mass : f(t) + spring force

Zeros of system : none (or two zeroes at \$s = \infty i\$)

Intuition : When force is varying with infinite frequency, the mass doesn't move due to its inertia.


\$ \frac{d^2 y(t)}{dt^2} = (f(t) - Ky(t))/M \$

\$ \frac{Y(s)}{F(s)} = \frac{1/M}{s^2 + K/M} \$


Input : force f(t)

Output : a force measurement sensor output m(t) (which happens to depend on the displacement and velocity of the moving mass)

Net force on the mass : f(t) + spring force (\$K_1\$ can be combined with \$K\$) + dash pot force

Zeros of system : one zero at \$s = -K_1/C_1 + 0i\$

Intuition : When \$y(t) > 0\$ spring \$K_1\$ excerts a right-ward force on the sensor. At the same time if \$dy(t)/dt < 0\$, then the dash pot excerts a left-ward force on the sensor. If \$K_1y(t) = -C_1 dy(t)/dt\$, the sensor senses zero force; i.e output of system is zero even when the internal states are not zero.

i.e. a zero of the system.

The condition \$K_1y(t) = -C_1 dy(t)/dt\$ means that the displacement is expressed as \$y(t) = e^{-t K_1/C_1}\$; Thus connecting the whole thing to an exponential signal.


\$ \frac{d^2 y(t)}{dt^2} = (f(t) - (K+K_1) y(t))/M - C_1 \frac{d y(t)}{dt} \$

\$ \frac{Y(s)}{F(s)} = \frac{1/M}{s^2 + C_1 s + (K+K_1)/M} \$

\$ m(t) = K_1 y(t) + C_1 \frac{d y(t)}{dt} \$

\$ \frac{M(s)}{F(s)} = \frac{M(s)}{Y(s)} \frac{Y(s)}{F(s)} = \frac{(s C_1 + K_1) 1/M}{s^2 + C_1 s + (K+K_1)/M} \$

system (C)

Input : displacement x(t)

Output : displacement of mass y(t)

Net force on the mass : spring force (spring extension is \$x(t)-y(t)\$)

Zeros of system : none (or two zeroes at \$s = \infty i\$)


\$ \frac{d^2 y(t)}{dt^2} = (x(t)-y(t))K/M \$

\$ \frac{Y(s)}{X(s)} = \frac{K/M}{s^2 +K/M} \$

system (D)

Input : displacement x(t)

Output : displacement of mass y(t)

Net force on the mass : spring force + dash pot force

Zeros of system : one zero at \$s = -K/C + 0i\$

Intuition : When \$y(t) = 0, Kx(t)= -Cdx(t)/dt\$, the net force on the mass is zero. Hence the mass remains stationary;

i.e. output of system is zero, i.e. a zero of the system.


\$ \frac{d^2 y(t)}{dt^2} = (x(t)-y(t))K/M + \frac{d(x(t)-y(t))}{dt}C/M \$

\$ \frac{Y(s)}{X(s)} = \frac{s C/M + K/M}{s^2 + s C/M + K/M} \$

It can be noted that in all cases, the initial condition of the system is not mentioned / assumed zero. The zeroes of the system depends strongly on the way in which input and output of the system are defined as well as the dynamics of the system.

| improve this answer | |
  • 1
    \$\begingroup\$ A1: They cannot be disregarded just because you cant evaluate the integral at that complex value for two reasons; a) The zero has a physical meaning as illustrated by the conditions \$K x(t) = -C dx/dt\$ in my example systems B&D. b) You can still evaluate the integral on the imaginary axis (i.e. sinusoid inputs) and for most applications, this is sufficient. A2: In my derivation of the transfer functions for the 4 example systems, I needed only a) differential equations representing the dynamics, b)definition of input and output c) assumption of zero initial conditions. contd \$\endgroup\$ – AJN Jun 14 at 13:54
  • 1
    \$\begingroup\$ As @jDAQ said in their answer, if we are using Laplace txform to solve differential equations, we use initial conditions (but then we don't talk about zeros). If we are using it for finding transfer functions then we assume zero initial conditions (and talk about concept of zeroes and poles). \$\endgroup\$ – AJN Jun 14 at 13:56
  • 1
    \$\begingroup\$ Since the input signal in your system is already fixed (as \$\delta (t)\$), Laplace transform is used as a tool for solving ODEs; the concept of zero need not be brought in. Q3 It might seem that the input for all spring mass systems should be forces and not displacements. However in many practical applications the force is not known, but the movement of the fixture is known; e.g. A skyscraper represented as a spring mass system and the earth represented as a fixture and earth moves during an earthquake. contd \$\endgroup\$ – AJN Jun 14 at 14:04
  • 1
    \$\begingroup\$ We don't know the actual values of the forces transferred from the earth to the building; but we do know the displacement (or acceleration) of the earth. In such an example \$x(t)\$ or \$d^2 x(t) / dt\$ of earth is known from sensors such as seismographs. An we can still solve for the response of the building if we setup our problem as in example system C or D. That is why didn't draw a 'fixed' fixture; the fixture itself is moving with respect to some real or virtual stationary point. \$\endgroup\$ – AJN Jun 14 at 14:08
  • 1
    \$\begingroup\$ Some useful links google brought up. lpsa.swarthmore.edu/Systems/MechTranslating/… me.utexas.edu/~dsclab/leks/DSC_Vibration_Modeling.pdf both links show modelling of the spring mass systems with force or displacement as input. \$\endgroup\$ – AJN Jun 14 at 14:12

@AJN is correct regarding the region of convergence, but with

$$ F(s) = \frac{4 s + 2}{s^{2} + 2 s + 10} $$

the zero will be \$ s_0 = -1/2\$ and will be in the region of convergence. Resulting in the plot of \$h(t)=f(t)e^{-(-0.5)}\$, which is the output of the system with some initial conditions (and no input) multiplied by a function (related to the zero of your system), enter image description here

with \$ \int_0^\infty f(t) e^{-st}dt \Big\rvert_{s=-0.5} \left(= \mathcal{L}\{f(t)\} = F(s)\Big\rvert_{s=-0.5} \right) = 0.\$

If you had \$ \int_0^k f(t) e^{-(-0.5)t}dt = g(k)\$ it would have a plot as the one bellow, and \$ g(k) \xrightarrow{ k \xrightarrow{} \infty} 0 \$

enter image description here

One more thing, you used initial conditions, so you are not dealing with transfer-functions just using Laplace transforms to solve ODEs. In the case of an actual transfer function the zero would be a result of the system dynamics, how it is built and its physics, not due to an initial condition (which would die out in a stable system).

In the book you are following they use those examples in figure 32-5 to exemplify the Laplace transform, which would be the "area under the curve" of that product \$p(s,t)h(t)\$, $$ F(s)=\int_0^\infty f(t) e^{-st}dt = \int_0^\infty f(t) p(s,t)dt ,$$ although they write that

In some mathematical techniques it is important to know what portions of the s-plane are within the region-of-convergence. However, this information is not needed for the applications in this book.

It seems that your problem was exactly that.

| improve this answer | |
  • \$\begingroup\$ Thanks a lot for the help. I'll get back to you once I have been through your answer more carefully. \$\endgroup\$ – PG1995 Jun 13 at 7:47
  • 1
    \$\begingroup\$ For figure 2b, the cancellation they mention is (from the pdf) "This occurs because the area above the x-axis (from the delta function) is exactly equal to the area below (from the rectified sinusoid)". The area under the up-arrow representing the impulse also has to be accounted. In that example, numerical integration might not have turned up a correct answer even in the region of convergence. \$\endgroup\$ – AJN Jun 13 at 11:55
  • \$\begingroup\$ @AJN that does explain it, you are right. I'll remove that part from my answer. \$\endgroup\$ – jDAQ Jun 13 at 15:42
  • \$\begingroup\$ @jDAQ Thanks a lot for your help. Since AJN posted the answer first and his is more comprehensive, I accepted his answer. I do have some follow-up questions which I'd ask later. Best wishes! \$\endgroup\$ – PG1995 Jun 14 at 2:41
  • 1
    \$\begingroup\$ @PG1995, h(t) is not an impulse response, just a name that I used for that function. Also since you are solving an ODE using Laplace transforms, saying something is the something response is kind of wrong. But in this case f(t) would be the time-domain version of F(s) (so f(t) is the 'impulse response'). The expression on top of the second plot is g(k), if you evaluate that integral you should get to that result (I used sympy to do it). \$\endgroup\$ – jDAQ Jun 15 at 2:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.