Transfer function $h(t)$ of a positive feedback system

Question

I want to find the transfer function h(t) of the below positive feedback system. I came out till this.

How can i get the inverse laplace of this function? say β = 1 and γ = 1

Answer 1

We seek $h(t)$, with the Laplace Transform of $h(t)$ as

$$H(s) = \int_0^\infty h(t)e^{-st}dt \label{1}\tag{1}$$

Given:

$$H(s) = \frac{1}{1-\beta e^{-s\tau}}\label{2}\tag{2}$$

The solution to this is:

$$h(t) = \sum_{n=0}^\infty \beta^{n\tau}\delta(t-n\tau)$$

Derivation:

Given the transcendental equation, solving this directly as an inverse Laplace Transform would be challenging; one approach is to approximate it to any precision by expanding $\beta e^{-s\tau}$ as a Taylor Series and then look for a convergence in the limit. Avoiding taking that approach however is a primary motivation for the z-transform. We could apply the z-transform in this case if it would be sufficient to determine samples of the solution. I will show the solution for samples of the system every $\tau$ seconds, as that would be trivial to solve. Solving for multiple samples every $\tau$ seconds would become increasingly more complicated, but the continuous time impulse response (which is the inverse Laplace Transform of the Transfer Function) can be easily determined from the implementation block diagram directly.

Rewriting equation \ref{1} for the case of $h(t)$ sampled every $\tau$ seconds as $h(n\tau)$ results in:

$$H(s) = \sum_{n=0}^\infty h(n\tau)e^{-s n\tau} \tag{3} \label{3}$$

Substitute:

$$z = e^{s\tau}$$

And we get what is known as the z-transform:

$$H(z) = \sum_{n=0}^\infty h(n\tau) z^{-n}\tag{4} \label{4}$$

Similarly, if we make the same substitution $z=e^{-s\tau}$ in equation \ref{2} we get:

$$H(s) = \frac{1}{1-\beta e^{s\tau}}$$

$$H(z) = \frac{1}{1-\beta z^{-1}}= \frac{z}{z-\beta} \tag{5} \label{5}$$

From the resulting geometric series in \ref{4}, the inverse z-transform for \ref{5} is as follows representing the samples of $h(t)$ sampled at every $\tau$ seconds:

$$h(n \tau) = \beta^{n\tau} \tag{6}\label{6}$$

The inverse z-transform only exists (converges) for $|\beta/z| < 1$, but the resulting equation \ref{6} is true for all $\beta$; when $\beta<1$ equation \ref{6} will decay to zero as we increase through time with index $n$, when $\beta>1$ then equation \ref{6} will grow without bound.

When $\beta =1$ and $\tau=1$ this reduces to:

$$h(n \tau) = \beta^{n\tau} = 1^{n} = 1$$

But this does not mean $h(t) = 1$ in this case, but only the samples every $\tau$ seconds will be one. As we increase the sampling rate and follow the process above, we will find that the solution of the impulse response for the case when $\beta =1$ and $\tau=1$ converges to an impulse every $\tau$ seconds:

$$h(t) = \sum_{n=0}^\infty \delta(t-n), \text{ for $\beta=1, \tau=1$}$$

This also matches our intuition; consider passing an impulse into the input of the system in the diagram given by the OP: we would immediately see this impulse appear at the output, and then due to the gain of $\beta=1$ and delay of $\tau = 1$, the impulse will again appear at the output every one seconds. With that we also see that for all other cases of $\beta$ and $\tau$ the solution will be a weighted series of impulses either growing or decaying depending on if $\beta$ is greater than or less than one:

$$h(t) = \sum_{n=0}^\infty \beta^{n\tau}\delta(t-n\tau)$$