Archive for October, 2010

Cascading (almost) identical LTI systems

October 23, 2010

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

H (s) = \displaystyle\frac{1}{(s+a) (s+b)}

where a, b \in \mathbb{R}. Note that we have two poles (at s = -a and s = -b), but no finite zeros. We can view \mathcal{H}, a 2nd order system, as the cascade connection of two causal 1st order LTI systems

If b = a, then the transfer function becomes H (s) = 1 / (s+a)^2 and we have a double pole at s = -a. Taking the inverse Laplace transform of H (s) we obtain the impulse response of the overall system \mathcal{H}

h (t) = t e^{- a t} u (t)

where u (t) is the Heaviside step function. If a > 0, then the impulse response will eventually decay to zero. However, note that the exponential is multiplied by t, which means that there is a transient “peak”. If a is positive but “close” to zero, this transient could be quite “wild”! This phenomenon is somewhat similar to resonance, although we are dealing with decaying exponential excitations, not sinusoidal ones. If we apply a Dirac delta impulse to the system depicted above, the output of the first system in the cascade will be the impulse response of \mathcal{H}_1

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

which happens to be also the impulse response of \mathcal{H}_2, the second system in the cascade. The lesson is the following: cascading identical LTI systems leads to resonant-like behavior.

If b \neq a, then we have a simple pole at s = -a and another simple pole at s = -b. Via partial fraction expansion of H (s), we obtain

H (s) = \displaystyle\frac{1}{b - a} \left(\displaystyle\frac{1}{s+a} - \displaystyle\frac{1}{s+b}\right)

and, taking the inverse Laplace transform, we get the impulse response

h (t) = \displaystyle\frac{1}{b - a} \left(e^{-a t} - e^{-b t}\right) u (t).

Note that these are valid only if b \neq a. If we make b = a, we get pesky indeterminate stuff: H (s) = 0 / 0 and h (t) = 0 / 0.

Let us see what happens if \mathcal{H}_1 and \mathcal{H}_2 are “almost identical”. To be precise, let us investigate the impulse response of the cascade when the (simple) poles at s = -a and s = -b are almost “on top of each other”. Let b = a \pm \varepsilon, where \varepsilon > 0 is “small”. Then, the impulse response of \mathcal{H} can be written as follows

h (t) = \pm \displaystyle\frac{1}{\varepsilon} \left(e^{-a t} - e^{-(a \pm \varepsilon) t}\right) u (t) = \pm \displaystyle\frac{1}{\varepsilon} \left(1 - e^{- (\pm \varepsilon) t} \right) e^{-a t} u (t).

For convenience, let us define

\gamma_{\varepsilon} (t) := \pm \displaystyle\frac{1}{\varepsilon} \left(1 - e^{- (\pm \varepsilon) t} \right)

so that h (t) = \gamma_{\varepsilon} (t) e^{-a t} u (t). Taking the Taylor expansion of \gamma_{\varepsilon} (t) about t = 0, we get

\gamma_{\varepsilon} (t) = \pm \displaystyle\frac{1}{\varepsilon} \left(\pm \varepsilon t - \displaystyle\frac{\varepsilon^2}{2} t^2 \pm \displaystyle\frac{\varepsilon^3}{3!} t^3 - \dots\right)

and, taking the limit as \varepsilon approaches zero, we obtain

\displaystyle\lim_{\varepsilon \to 0} \gamma_{\varepsilon} (t) = \displaystyle\lim_{\varepsilon \to 0} \pm \displaystyle\frac{1}{\varepsilon} \left(\pm \varepsilon t - \displaystyle\frac{\varepsilon^2}{2} t^2 \pm \displaystyle\frac{\varepsilon^3}{3!} t^3 - \dots\right) = t

and, therefore, we finally get

\displaystyle\lim_{\varepsilon \to 0} h (t) = \displaystyle\lim_{\varepsilon \to 0} \gamma_{\varepsilon} (t) e^{-a t} u (t) = t e^{-a t} u (t).

This shows, albeit in a sinfully non-rigorous manner, that the impulse response of H (s) = 1 / ((s+a) (s+b)) approaches the impulse response of H (s) = 1 / (s+a)^2 as the pole at s = -b gets “closer” and “closer” to the pole at s = -a. I concede that this result is not terribly exciting…

Let us now perform a numerical experiment! Let a = 1, and let b = a + \varepsilon, so that b approaches a as \varepsilon \to 0. The plot below illustrates the impulse responses for various values of \varepsilon

where \varepsilon = 0 corresponds to the cascade of two identical systems (which creates a double pole at s = -1). Note that, as expected, as \varepsilon approaches zero, the impulse response of the cascade approaches the resonant one, h (t) = t e^{- a t} u (t).

Last but not least, here’s the MATLAB script that generates the plot:

% build transfer function of H1
a = 1; H1 = tf(1,[1 a]);

% create time grid
t = [0:0.01:5];

% compute impulse responses of the cascade
h = [];
for epsilon = [1 0.5 0.1 0];

 % build transfer function of H2
 b = a + epsilon; H2 = tf(1,[1 b]);

 % compute and store impulse response
 h = [h, impulse(H2 * H1, t)];

end

% plot impulse responses
figure;
plot(t, h(:,1), 'r-', t, h(:,2), 'm-', t, h(:,3), 'b-', t, h(:,4), 'k--');
legend('\epsilon = 1', '\epsilon = 0.5', '\epsilon = 0.1', '\epsilon = 0');
xlabel('t (seconds)'); title('Impulse responses of the cascade');

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Pole-zero cancellation

October 22, 2010

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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