## Pole-zero cancellation

Consider a causal continuous-time LTI system $\mathcal{H}$ with transfer function

$H (s) = \displaystyle\frac{s+b}{s+a}$

where $a, b \in \mathbb{R}$. There is a zero at $s = - b$, as $H(s)\mid_{s = -b} = 0$. There is also a pole at $s = -a$, as $H(s)$ blows up at $s = -a$.

If $b = a$, then we have $H(s) = (s + a) / (s + a) = 1$, and $\mathcal{H}$ becomes the identity system. In this case, the zero is “on top” of the pole, and we say that pole-zero cancellation has occurred. Usually, this phenomenon is discussed in the $s$-domain, but what happens in the time-domain?

We can view $\mathcal{H}$ as the cascade connection of two causal LTI systems

Let $H_1 (s) = 1 / (s + a)$, and $H_2 (s) = s + b$ be the transfer functions of $\mathcal{H}_1$ and $\mathcal{H}_2$, respectively. Then, we have $H (s) = H_1 (s) H_2 (s)$. Suppose we apply a Dirac delta impulse, $\delta (t)$, to the system. The output of the first filter in the cascade will then be the natural mode of $\mathcal{H}_1$

$h_1 (t) = e^{- a t} u(t)$

where $u (t)$ is the Heaviside step function. If $a > 0$, then the natural mode decays exponentially. Recall that the inverse Laplace transform of $(s + b) H_1 (s)$ is $\dot{h}_1 (t) + b h_1 (t)$. Then, we have that the impulse response of the overall system $\mathcal{H}$ is

$h (t) = \dot{h}_1 (t) + b h_1 (t) = -a e^{- a t} u(t) + e^{- a t} \delta (t) + b e^{- a t} u(t)$

and, since $e^{- a t} \delta (t) = \delta (t)$ for all $t \in \mathbb{R}$, we obtain

$h (t) = \delta (t) + (b - a) e^{- a t} u(t)$.

If $b = a$, then the zero cancels the pole and we get $h (t) = \delta (t)$ at the output of $\mathcal{H}$. In other words, the zero at $s = - a$ annihilates the natural mode created by the pole at $s = -a$. In theory, we could use pole-zero cancellation to kill an unstable natural mode. In theory…

What if the pole-zero cancellation is not “exact”? Let $b = a \pm \varepsilon$, where $\varepsilon > 0$ is “small”. Then the impulse response of $\mathcal{H}$ is

$h(t) = \delta (t) \pm \varepsilon e^{- a t} u(t)$.

If $a > 0$ we have a stable natural mode and the “weak” exponential will eventually vanish. The closer the zero is to the pole, the “weaker” the natural mode will be, until it completely disappears when there is exact pole-zero cancellation. However, if $a < 0$ then we have an unstable natural mode that diverges regardless of how small $\varepsilon$ is.

Alternatively, as $H (s) = 1 + (b-a) / (s+a)$, we can view $\mathcal{H}$ as the parallel connection of the identity system and a causal LTI system

If $b = a$, then $\mathcal{H}$ becomes the identity system. In my humble opinion, viewing $\mathcal{H}$ as a parallel connection of two systems is not very illuminating, as the annihilation of the natural mode in the time-domain (due to differentiation) is not evident.

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### 2 Responses to “Pole-zero cancellation”

1. Rod Carvalho Says:

We can generalize, of course. For example, consider a causal continuous-time LTI system $\mathcal{H}$ with transfer function

$H (s) = \displaystyle\frac{s+b}{(s+a)^n}$

where $n \geq 2$. We can write this transfer function as $H(s) = H_1 (s) H_2 (s)$, where $H_1 (s) = 1 / (s+a)^n$, and $H_2 (s) = s + b$. If the input of the cascade is the Dirac delta, then the output of $\mathcal{H}_1$ will be its impulse response

$h_1 (t) = \displaystyle\frac{t^{n-1}}{(n-1)!} e^{-a t} u(t)$.

Therefore, the output of the cascade will be $h (t) = \dot{h}_1 (t) + b h_1 (t)$. Note that the derivative of $h_1 (t)$ is as follows

$\dot{h}_1 (t) = \displaystyle\frac{t^{n-2}}{(n-2)!} e^{-a t} u(t) - a \displaystyle\frac{t^{n-1}}{(n-1)!} e^{-a t} u(t) + \displaystyle\frac{t^{n-1}}{(n-1)!} e^{-a t} \delta(t)$

where $t^{n-1} \delta(t) = 0$ for all $t \in \mathbb{R}$, because $n \geq 2$. Finally, we obtain

$h (t) = \displaystyle\frac{t^{n-2}}{(n-2)!} e^{-a t} u(t) + (b - a) \displaystyle\frac{t^{n-1}}{(n-1)!} e^{-a t} u(t)$.

If $b = a$, then pole-zero cancellation occurs, and the impulse response of the cascade becomes

$h (t) = \displaystyle\frac{t^{n-2}}{(n-2)!} e^{-a t} u(t)$

whose Laplace transform is $H (s) = 1 / (s+a)^{n-1}$, as expected. After pole-zero cancellation, we have no finite zeros, and the pole at $s = -a$ has degree $n-1$.

2. Dave Says:

You’re a legend, this helps a ton for my signals and systems exam!