Poles and the impulse response

If our impulse response is in the form :

$$h(t) = e^{-\sigma_0 t}\cos(\omega_0 t) \, u(t)$$

(where $u(t)$ is the unit step function)

And its Laplace transform is :

$$H(s) = \frac{N(s)}{D(s)} = \int_{0}^{+\infty} h(t)e^{-st}dt$$ $$s = \sigma + j\omega$$

Poles are values of $s$ so that $$D(s) = 0 \rightarrow H(s) = +\infty $$ But to understand this, I prefer to look at the integral : it will go to infinity (poles) when $s$ reflects components of $h(t)$. In a way, $e^{-st}$ "probes" $h(t)$. Indeed :

  • A single real pole ($s = -\sigma_0$) means $h(t) = e^{-\sigma_0t}u(t)$ because : $$\int_{0}^{+\infty} e^{-\sigma_0t}e^{-(-\sigma_0)t}dt = \int_{0}^{+\infty} 1dt = +\infty $$.

  • Complex conjugate poles ($s = -\sigma_0 \pm j\omega_0$) mean $h(t)$ is an exponentially decaying sinusoid (say $h(t) = e^{-\sigma_0t}\cos(\omega_0t)$) because : $$\int_{0}^{+\infty} e^{-\sigma_0t}\cos(\omega_0t)e^{-(-\sigma_0)t}e^{-j\omega t}dt = \int_{0}^{+\infty}\cos(\omega_0t)e^{-j\omega t}dt $$ which is infinite at $\omega = \pm\omega_0$ (Fourier transform of $h(t)$ without its exponential component, which is a sinusoid).

  • Complex conjugate poles with $\sigma = 0$ ($s = \pm j\omega_0$) mean $h(t)$ has no decaying component (say $h(t) = \cos(\omega_0t) u(t)$) because : $$\int_{0}^{+\infty} \cos(\omega_0t)e^{-j\omega t}dt$$ which is infinite at $\omega = \pm\omega_0$ (Fourier transform of $h(t)$ which is a sinusoid).

Zeros : a dirac in the impulse response ?

Now, let's look at $H(s)$ for a Notch filter, as shown in ch.32,p.17 of "The Scientist and Engineer's Guide to DSP" and see if similar reasoning on the integrals can be done.

Notch filter

Let's use the following filter (figure above for illustration only, I use different poles and zeros here) :

$$H(s) = \frac{s^2+1}{(s-(-1+i))(s-(-1-i))}$$

This filter has 2 poles and 2 zeros :

  • Zeros : $z_1,z_2 =\pm i$
  • Poles : $p_1,p_2 =-1 \pm i$

Let's find $h(t)$ and see why the integral would indeed go to 0 or $+\infty$ for these values of zeros and poles, respectively.

If it makes sense, this tool gives the following inverse Laplace transform for $H(s)$ :

$$h(t) = \delta(t) - 2e^{-t}\cos(t) u(t) + e^{-t}\sin(t) u(t)$$

  • Poles : for $s=p_1$ or $p_2$ in the Laplace transform, the exponentials of h(t) get canceled and remain the Fourier transform of some sinusoid which is indeed infinite at $\omega = \pm 1$ (I'm not discussing the $\delta(t)$ but I suppose it won't change this result).

  • Zeros : for $s=z_1$ or $z_2$ in the Laplace transform, the result is 0 if real part and imaginary of the Laplace transform are 0. Real part is :

$$\int_{0}^{+\infty} (\delta(t) - 2e^{-t}\cos(t)+e^{-t}\sin(t))\cos(t)dt$$

$$=\int_{0}^{+\infty} \delta(t)\cos(t)dt + \int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\cos(t)dt$$


$$\int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\cos(t)dt = -1$$

Imaginary part is :

$$\int_{0}^{+\infty} \delta(t)\sin(t)dt + \int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\sin(t)dt$$


$$\int_{0}^{+\infty} (- 2e^{-t}\cos(t)+e^{-t}\sin(t))\sin(t)dt = 0$$


  1. If the inverse Laplace transform is correct, how to handle $\int_{0}^{+\infty} \delta(t)\cos(t)dt$ and $\int_{0}^{+\infty} \delta(t)\sin(t)dt$ to show that $H(s)$ is indeed 0 at $z_1$ and $z_2$ ?
  2. If all of this is correct, what does it (physically) mean for an impulse response to have a dirac in its expression ? I thought impulse response of most physical systems was only a combination of decaying exponentials and sinusoids ?

2 Answers 2


For you first question you can use the following

$$ \int_{-\infty}^{\infty} \delta (t-a)\,f(t)\,dt = f(a), $$

with $f(t)$ any function. In your case those integrals would thus yield the values one and zero respectively.

For your second question I will only considering linear time invariant systems. In that case the impulse response of such a system can only contains a Dirac delta function if the transfer function of that system has a numerator of the same order as the denominator. Namely, any transfer function of the form

$$ G(s) = \frac{b_n\,s^n + b_{n-1}\,s^{n-1} + \cdots + b_1\,s + b_0}{s^n + a_{n-1}\,s^{n-1} + \cdots + a_1\,s + a_0}, $$

with $b_n \neq 0$ can also be written as

$$ G(s) = b_n + \frac{b'_{n-1}\,s^{n-1} + \cdots + b'_1\,s + b'_0}{s^n + a_{n-1}\,s^{n-1} + \cdots + a_1\,s + a_0}, $$

with $b'_k = b_k - b_n\,a_k$. The inverse Laplace transform of the constant $b_n$ would contribute a Dirac delta term. For the remaining part of the transfer function one could use partial fraction expansion to show that it can't contribute a Dirac delta term.

If a physical system would have a numerator of the same order as the denominator then it would require that the output of the system is directly affected by the input. An example of such physical system might be some electrical motor where you input a voltage and measure the angular position with some voltage leakage from the input signal to the output. However, most physical systems have numerator of a lower order as the denominator. It is more likely that you might encounter equal order numerators and denominators in digital filters (though, those would be z-domain and not s-domain, but roughly the same argument holds) such as notch filters. Those filters are however often used in series with physical systems, so their combined transfer function would also have a lower order numerator.

  • $\begingroup$ Thanks for you answer ! For your first equation, shouldn't the result of integral from $0$ to $+\infty$ be half the one from $-\infty$ to $+\infty$ though ? Also, could you please add detail about how the electrical system you describe will output a dirac (with its decaying exponentials) if it is excited by a dirac ? $\endgroup$
    – Elaws
    Nov 23, 2020 at 23:10
  • $\begingroup$ @Elaws the integral from $-\infty$ to $\infty$ can be cut up into two parts, namely from $-\infty$ to $0$ and from $0$ to $\infty$. $\endgroup$
    – fibonatic
    Nov 23, 2020 at 23:16
  • $\begingroup$ Ok, I meant I suppose 0 is excluded from one of these 2 integrals otherwise we would integrate the dirac at 0 twice. $\endgroup$
    – Elaws
    Nov 23, 2020 at 23:25
  • $\begingroup$ @Elaws the Laplace transform is essentially a Fourier transform with the assumption that $h(t)=0$ for all $t<0$ so from that one could already guess that what I stated could be correct. But I agree it is still a little unambiguous of how one would split the integral. However, if you also look at Laplace transform tables you would see that the Laplace transform of a Dirac delta is equal to one, which matches the expectation when using $f(t)=1$. $\endgroup$
    – fibonatic
    Nov 23, 2020 at 23:45

If the function to be transformed has an impulse at $t=0$, the unilateral Laplace transform is commonly defined as


(note the lower integration limit $0^-$). The bilateral Laplace transform doesn't have that problem anyway.

The consequence of this definition is that the integrals in your derivation become




which gives the expected result.

Impulse responses containing a Dirac impulse are nothing special. A simple (ideal) amplifier or attenuator with input-output relation $y(t)=\alpha x(t)$ has a (scaled) Dirac impulse as its impulse response. Note that you only get a Dirac impulse at the output if you input a Dirac impulse, which doesn't happen in practice. A Dirac impulse in the impulse response just means that part of the output is a (possibly scaled and delayed) copy of the input. Any system with a frequency response that has a finite non-zero limit $\lim_{\omega\to\infty}H(j\omega)$ has a Dirac impulse in its impulse response. Some examples of such systems for which that limit exists and is finite are high-pass filters, band-stop filters and all-pass filters. Your notch filter is a special case of a band-stop filter.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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