While studying the Laplace transform using Steven W. Smith Book I found something uncomprehending. In the 32th chapter - The Laplace Transform, page 590, last paragraph describes the cancelling phenomena when an impulse response is cancelled using an exponentially weighted sinusoid (see picture below). When cancelling occurs then we are dealing with zero or pole at the s-plane. What is not clear for are the products of the probing waveform and impulse response examples (3rd column in the figure below):

a) Decreasing with time: how it can be said that $p(t) \times h(t)$ is finite?
b) Exact cancellation (zero): how it can be said that $p(t) \times h(t)$ is zero?
c) Too slow of increase: how it can be said that $p(t) \times h(t)$ is finite?
d) Exact cancellation (pole): how it can be said that $p(t) \times h(t)$ is infinite?
e) Too fast of increase: how it can be said that $p(t) \times h(t)$ is undefinied?

I would be glad if someone could explain me what is the connection between $p(t) \times h(t)$ shape and if it is pole or zero.

enter image description here

  • $\begingroup$ It says Area is finite. I think that means it’s integrating the product. $\endgroup$
    – Peter K.
    Commented Dec 16, 2021 at 16:52

1 Answer 1


Peter's comment is correct, it's about the integral of the product $p(t)h(t)$:


The impulse response $h(t)$ has the following form:

$$h(t)=\delta(t)+c_1\, e^{\sigma t}\cos(\omega_0t),\qquad t\ge 0\tag{2}$$

with some constant $c_1$ and some $\sigma<0$.

From what I understand, the function $p(t)$ must look like


with some constant $c_2$. I can't be sure but it seems likely that $c_1=c_2$.

Now we consider $5$ cases:

  1. $\sigma_p<0$
  2. $\sigma_p=0$
  3. $\sigma_p>0$ and $\sigma+\sigma_p<0$
  4. $\sigma_p=-\sigma$
  5. $\sigma_p>0$ and $\sigma+\sigma_p>0$

In the first three cases we obtain for the product $p(t)h(t)$ a decaying exponential times $\cos^2(\omega_0t)$, plus a Dirac impulse, the integral of which is finite in all cases. In the second case, the exponent equals $\sigma$, and it appears that the constants can be chosen in such a way that the value of the integral can be made zero. I don't see any clear explanation of this in the book chapter, but I might be missing something.

In cases $4$ and $5$, the exponential is constant ($4$) or growing indefinitely ($5$), hence the integral diverges in both cases.

  • $\begingroup$ Thank you for the answer. My understanding of the integral was always the area of the region enclosed by the graph and the horizontal x-axis. In the first example where it is said that "area is finite", the integral is indefinite and in converges to zero but it will never be achieve zero. Doesn't it mean that then integral would be infinite? Please, correct me if I am wrong $\endgroup$ Commented Dec 17, 2021 at 9:06
  • 1
    $\begingroup$ @MarcinPuchlik: Your understanding of the integral as the area under the curve is correct. In the first example the curve will never be zero but the area under the curve converges to a finite number because the envelope decays sufficiently fast due to the exponential with a negative exponent. As a simple example, note that also the function $f(t)=e^{-t}$ will never become zero, but its integral from $0$ to $\infty$ is finite. $\endgroup$
    – Matt L.
    Commented Dec 17, 2021 at 10:07
  • $\begingroup$ Sure, you are right - the integration result in the first example will be finite (I needed to refresh the common integrals). Now, lets take a look at the second example where it is said that "area is exactly zero". It can be seen that this situation is quite the same as in the first example - exponent is decaying. How then the $p(t)$ can cancel $h(t)$ - I mean, how integration of the $p(t) \times h(t)$ produces zero in the second example? $\endgroup$ Commented Dec 17, 2021 at 11:33
  • $\begingroup$ @MarcinPuchlik: The second case is the tricky one, and I think it's not explained properly in the text. The fact that the value of the integral is finite is clear, just like in cases 1 and 3. The fact that it's exactly zero is not clear, but it has to do with the constants. They turn out to be such that the positive contribution from the integration of the Dirac impulse cancels the negative contribution from the function for $t>0$. The exact values of those constant must be determined from the RLC circuit described elsewhere in the book. Again, it's not obvious at all. $\endgroup$
    – Matt L.
    Commented Dec 17, 2021 at 12:25
  • $\begingroup$ I forgot about the Dirac impulse, I was just looking at the decaying sinusoid - thanks for reminding. Are you suggesting that we just have to assume that integration in second example would give zero result? $\endgroup$ Commented Dec 17, 2021 at 13:43

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