13.9: Delay and Feedback

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Let \(f(t) = 0\) for \(t < 0\). Fix \(a > 0\) and let \(h(t) = f(t - a)\). So, \(h(t)\) is a delayed version of the signal \(f(t)\). The Laplace property Equation 13.5.8 says

\[H(s) = e^{-as} F(s), \nonumber \]

where \(H\) and \(F\) are the Laplace transforms of \(h\) and \(f\) respectively.

Now, suppose we have a system with system function \(G(s)\). (Again, called the open loop system.) As before, can feed the output back through the system. But, instead of just multiplying the output by a scalar we can delay it also. This is captured by the feedback factor \(ke^{-as}\).

The system function for the closed loop system is

\[G_{CL} (s) = \dfrac{G}{1 + ke^{-as} G} \nonumber \]

Note even if you start with a rational function the system function of the closed loop with delay is not rational. Usually it has an infinite number of poles.

Example \(\PageIndex{1}\)

Suppose \(G(s) = 1\), \(a = 1\) and \(k = 1\) find the poles of \(G_{CL} (s)\).

Solution

\[G_{CL} (s) = \dfrac{1}{1 + e^{-s}}. \nonumber \]

So the poles occur where \(e^{-s} = -1\), i.e. at \(in \pi\), where \(n\) is an odd integer. There are an infinite number of poles on the imaginary axis.

Example \(\PageIndex{2}\)

Suppose \(G(s) = 1\), \(a = 1\) and \(k = 1/2\) find the poles of \(G_{CL} (s)\). Is the closed loop system stable?

Solution

\[G_{CL} (s) = \dfrac{1}{1 + e^{-s}/2}. \nonumber \]

So the poles occur where \(e^{-s} = -2\), i.e. at \(-\log (2) + in\pi\), where \(n\) is an odd integer. Since \(-\log (2) < 0\), there are an infinite number of poles in the left half-plane. With all poles in the left half-plane, the system is stable.

Example \(\PageIndex{3}\)

Suppose \(G(s) = 1\), \(a = 1\) and \(k = 2\) find the poles of \(G_{CL} (s)\). Is the closed loop system stable?

Solution

\[G_{CL} (s) = \dfrac{1}{1 + 2e^{-s}}. \nonumber \]

So the poles occur where \(e^{-s} = -1/2\), i.e. at \(\log (2) + in\pi\), where \(n\) is an odd integer. Since \(\log (2) > 0\), there are an infinite number of poles in the right half-plane. With poles in the right half-plane, the system is not stable.

Remark

If \(\text{Re} (s)\) is large enough we can express the system function

\[G(s) = \dfrac{1}{1 + k e^{-as}} \nonumber \]

as a geometric series

\[\dfrac{1}{1+ke^{-as}} = 1 - ke^{-as} + k^2 e^{-2as} - k^3 e^{-3as} + \ ... \nonumber \]

So, for input \(F(s)\), we have output

\[X(s) = G(s) F(s) = F(s) - ke^{-as} F(s) + k^2 e^{-2as} F(s) - k^3 e^{-3as} F(s) + \ ... \nonumber \]

Using the shift formula Equation Equation 13.5.8, we have

\[x(t) = f(t) - kf(t - a) + k^2 f(t - 2a) - k^3 f(t - 3a) + \ ... \nonumber \]

(This is not really an infinite series because \(f(t) = 0\) for \(t < 0\).) If the input is bounded and \(k < 1\) then even for large \(t\) the series is bounded. So bounded input produces bounded output –this is also what is meant by stability. On the other hand if \(k > 1\), then bounded input can lead to unbounded output –this is instability.