If Laplace transform is expressed as :

$$\int_{-\infty}^{+\infty} h(t)e^{-st}dt $$

with :

$$s = \sigma + j\omega$$

and $h(t)$ an impulse response expressed as :

$$h(t) = Ae^{-\sigma_0t}\cos(\omega_0t+\phi) = e^{-\sigma_0t}\cos(\omega_0t)$$ ($A=1$ and $\phi = 0$ for simplification, $h(t)=0$ if $t<0$)

Then, each vertical line (parallel to the imaginary axis) in the $s$ plane corresponds to the Fourier transform of $f(t) = h(t)e^{-\sigma t}$ for a fixed $\sigma$.

For $\sigma = -\sigma_0$, the decaying exponential of $h(t)$ is canceled and we get the Fourier transform* of $h(t) = \cos(\omega_0t)$, that is : diracs at $\omega_0$ and $-\omega_0$ (not accurate, see (*) just below), hence two poles : $-\sigma_0 + j\omega_0$ and $-\sigma_0 - j\omega_0$ as in the following picture (illustration only, poles not located correctly) :


Indeed, we can understand that :

(*)Please, note that the following is not accurate : since $h(t) = 0$ if $t<0$, we should use the unilateral Laplace transform, not bilateral ! So here we would get the unilateral Fourier transform of a sinusoid, not the bilateral (with diracs only) one ! To see what this would be, please see the link given at the end of the accepted answer

$$\int_{-\infty}^{+\infty} h(t)e^{-j\omega t}dt $$ $$= \int_{-\infty}^{+\infty} \cos(\omega_0t)e^{-j\omega t}dt$$ $$= \int_{-\infty}^{+\infty} \frac{e^{j\omega_0t}-e^{-j\omega_0t}}{2}e^{-j\omega t}dt$$ $$= \frac{1}{2}\int_{-\infty}^{+\infty} e^{j(\omega_0-\omega)t}-e^{-j(\omega_0+\omega)t}dt$$

If $\omega = \omega_0$ or $-\omega_0$, then the integral would blow up due to the $$\int_{-\infty}^{+\infty} e^0dt $$ member, hence the poles in the s plane.

So as shown in ch.32, p.24 of The Scientist and Engineer's Guide to DSP (see figures below), with Laplace transform we multiply $h(t)$ with $e^{-st}$ = $e^{-\sigma}e^{-j\omega}$, that is we multiply $h(t)$ with sinusoids that are either :

  • (a) Exponentially decaying ($\sigma$ > 0)
  • (b) Stable ($\sigma = 0$)
  • (c) Exponentially growing slower than our impulse response decay ($ -\sigma_0 < \sigma < 0$)
  • (d) Exponentially growing, compensating our impulse response decay ($\sigma = -\sigma_0$) : OK, as studied above.
  • (e) Exponentially growing quicker ($\sigma < - \sigma_0$ and $\sigma < 0$)

(letters correspond to pairs of points in the s plane shown in figures below, always at a fixed $\omega$ or $-\omega$ value)

Different values of s... Lead to different values of the integral

I understand case d : since we cancel the exponential part, we get only the (unilateral !!) Fourier transform of a sinusoid. That is : infinite at $\omega_0$ and $-\omega_0$ hence the poles (though I don't know why we have a continuous function of omega with infinite values at $\omega_0$ and $-\omega_0$ instead of diracs as in the original Fourier transform of a sinusoid -> Because we use unilateral Laplace hence Fourier, see end of accepted answer !).

Case a, c and e are intuitive. In case a, we multiply $h(t)$ with a decaying exponential. The integral will be some finite complex value (for all values of $\sigma > 0$. In case c, we multiply by an exponential growing slower than the decaying exponential of $h(t)$, hence some finite complex value for the integral (for all values of $-\sigma_0 < \sigma < 0$). In case e, we multiply the $h(t)$ by an exponential that grows quicker than exponential of $h(t)$ decays : hence integral does not converge (for all values of $\sigma < -\sigma_0$).

But for case b, I can't get the intuition of why the integral would be zero as shown with the area under the curve (red in the above figures) ? In other words, I understand the vertical line in the s plane at $\sigma = -\sigma_0$, it is the Fourier transform of $h(t)e^{-\sigma_0 t}$ so it is Fourier transform of $h(t)$ once its exponential component is removed, hence 2 poles due to sinusoid. We get poles whenever $e^{-st}$ is identical (compensates) to the impulse response. But what would cause Fourier transform of $h(t)e^{-\sigma t}$ to be 0 at some $\omega$ ? For which $h(t)$ and how it would impact the area under the curve (integral) ?


The definition of the Laplace transform you're using is called the bilateral Laplace transform, which is less common than the unilateral Laplace transform. The difference between the two is that the first has a lower integration limit of $-\infty$ whereas the second has a lower limit of $0$. This difference becomes irrelevant if the signal under consideration is zero for $t<0$. This is the case in the example of the book. Note, however, that the bilateral Laplace transform of the impulse response $h(t)$ that you defined in your question doesn't exist for any value of $s$. It does exist if you set $h(t)$ to zero for $t<0$ (i.e., multiply it with a unit step $u(t)$).

The figures in the book refer to a causal notch filter. Note that the region of convergence (ROC) of the Laplace transform of that impulse response is to the right of the poles. Consequently, the Laplace transform evaluated at any fixed value of $s$ to the right of the poles will be finite, simply because we're inside the ROC, i.e., the integral converges. If we choose $s$ exactly at the zero of the filter then we evaluate the filter's response at the notch frequency, which must be zero, simply by the definition of "notch frequency". The filter's response to a signal at that frequency must be zero. Finally, if we choose the value of $s$ exactly at or to the left of the filter's poles, we are outside the ROC, in which case the integral doesn't converge.

  • $\begingroup$ Thanks for your answer. I understand that a zero in the transfer function means the filter output will be zero. But I don't understand how this would happen if we look at the integral $\int_{-\infty}^{+\infty} e^{-\sigma_0t}\cos(\omega_0t)e^{-st}dt $ : how can it be 0 ? If our impulse response is in the form of an exponentially decaying sinusoid, I can't think of any values of $s$ so that the integral is zero (for example at points B and B' of fig. 32-5 in the original post, we have zeros at $s=j\omega_{zero}$ and $s=-j\omega_{zero}$ : what could possibly be $\omega_{zero}$ ? $\endgroup$ – Elaws Nov 23 '20 at 13:29
  • $\begingroup$ In other words (to complete previous comment), which values of $\omega_{zero}$ makes $\int_{-\infty}^{+\infty} e^{-\sigma_0t}\cos(\omega_0t)e^{-j\omega_{zero}t}dt $ be zero ? $\endgroup$ – Elaws Nov 23 '20 at 13:37
  • $\begingroup$ @Elaws: You need to take the impulse response of a notch filter, that 's what the example is about. And in case b) you simply get $\int_{-\infty}^{\infty}h(t)e^{-j\omega_0 t}dt$ where $\omega_0$ is the notch frequency. That integral is just the Laplace transform of $h(t)$ evaluated at the filter's zero, hence its value is zero. $\endgroup$ – Matt L. Nov 23 '20 at 13:38
  • $\begingroup$ Thanks, but $h(t)$ can't be in the form of an exponentially decaying sinusoid, as shown in fig. 32-5 (second row) ? The integral can't be zero for any $\omega$ ? $\endgroup$ – Elaws Nov 23 '20 at 13:41
  • $\begingroup$ @Elaws: The impulse response looks similar but it's not just equal to $e^{-\sigma_0t}\cos(\omega_0t)u(t)$ (note the step function $u(t)$, otherwise the integral doesn't exist anyway). That simple exponentially decaying sinusoid has no zeros in its Laplace transform, just two poles, so the Laplace integral can't be zero. $\endgroup$ – Matt L. Nov 23 '20 at 13:44

The original post has been updated to add information of why the integral diverges or has some finite complex value.

Figure 32.5 (original question) can't be understood (especially "b. Exact cancellation") if we consider :

$$ h(t) = e^{-\sigma_0t}\cos{\omega_0t} $$

($h(t) = 0$ for $t<0$)

$h(t)$ in fig. 32-5 is not a simple exponentially decaying sinusoid: if it was, the integral could indeed not equal 0 for any value of s, as raised by the original question.

Instead, as pointed out by Matt L., $h(t)$ is the impulse response of a Notch filter. How does this help in understanding why the integral would go to 0 for some $s$ ? Well, this impulse response has the peculiarity of having a dirac in it (and also some combination of exponentially decaying sinusoids) ! And if you pay attention to fig.32-5, this dirac is indeed shown in the impulse response (missed this thinking it was the ordinate axis...), see figure below :

Dirac, not ordinate axis !

And it is the area under this dirac that will compensate for the area under the exponentially decaying sinusoids components of $h(t)$ for the appropriate values of $s$, hence the zeros !

For a more detailed explanation of both the calculation involved in this, as well as the physical meaning of a dirac in an impulse response, please see the answers given to this question.

Another question was the following :

(though I don't know why we have a continuous function of omega with infinite values at ω0 and −ω0 instead of diracs as in the original Fourier transform of a sinusoid).

I think this is due to having an unilateral Laplace transform instead of bilateral. Indeed, see in this example the unilateral Fourier transform of sine waves. It's as if we multiplied the sine wave with a unit step function. So the unilateral Fourier transform of a sine wave is the Fourier transform of a sine wave convoluted by the Fourier transform of a unit step function (see details in given link). This is why in a given vertical slice (for a fixed $\sigma$) of $s$ plane, we won't get the usual Fourier transform, but the unilateral one, which is a bit different.


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