Zeros of a system and the region of convergence

Question

Say, I have the Laplace transform of an impulse response:

$$\frac{s+1}{\frac{1}{2}s^2+s+1}$$ I've learned that the zero of this system will be such $s$ for which the above will equal zero. So $s=-1$.

But the above is laplace transform of:

$$2e^{-t}\cos{t}$$ with the assumption that $s > -1$.

And it makes sense since for $s = -1$, $\lim_{t \rightarrow \infty} 2e^{-t}\cos{t} \cdot e^{-(-1)t}$ does not converge. So it seams for me that $s=-1$ is outside of the region of convergence.

If it is so, what is the zero of this system then?

Answer 1

Typically, there is a constraint on the real part of $s = \sigma + i\omega$.

Indeed, we can write the Laplace transform the given time domain function as

$$2\int_0^\infty\mathrm{d}t\,e^{-t}\cos(t)e^{-st} = 2\int_0^\infty\mathrm{d}t\,\cos(t)e^{-(\sigma+1)t}\,e^{-i\omega t}$$

Clearly, for the integral to exist, we need $\sigma + 1 > 0 \rightarrow \sigma > -1$

However, for the case that $\sigma = -1$, the Laplace transform is

$$2\int_0^\infty\mathrm{d}t\,\cos(t)\,e^{-i\omega t} = \int_{-\infty}^\infty\mathrm{d}t\,\cos(t)\,e^{-i\omega t}$$

which is the Fourier transform of $\cos(t)$. As you may know, this integral exists in the distribution (generalized function) sense:

$$\int_{-\infty}^\infty\mathrm{d}t\,\cos(t)\,e^{-i\omega t} = \pi\left[\delta(\omega - 1) + \delta(\omega + 1)\right]$$

which is zero for all $\omega$ with the exception of $\omega = \pm 1$. This is consistent with the transfer function given since the denominator is zero for $s = -1 \pm i$, i.e., the poles of the transfer function are $s = -1 \pm i$.

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So the zero of this system is actually s=−1+iω for all ω with exception of ω=±1 not just s=−1?

Well, let's go back to the system itself. We can read the differential equation more or less directly from the given transfer function:

$$\frac{1}{2}\ddot{y} + \dot{y} + y = \dot{x} + x$$

Let $x(t) = x_s(t) = e^{st}$ and then try $y(t) = y_s(t) = H(s)e^{st}$ directly in the differential equation to find

$$\left(\frac{1}{2}s^2 + s + 1\right)H(s) = s + 1$$

and so, assuming zero initial conditions,

$$y_s(t) = \frac{s+1}{\frac{s^2}{2} + s + 1}e^{st}$$

Now, when $s = -1 \pm i\omega$, we get

$$y_s = \frac{\pm i\omega}{1 - \omega^2}e^{\pm i\omega t}e^{-t}$$

and so

• the output is zero only if $\omega = 0$
• the output 'blows up' as $\omega \rightarrow \pm 1$