In a hybrid system, the sampler and ZOH perform the operations: a continuous signal is sampled (analogue to digital conversion, ADC) to produce a discrete signal, and the discrete signal is returned to continuous form (digital-to-analogue conversion, DAC). Usually there is some digital processing between these two operations, but the combined DAC/ADC operation is normally represented as a single Laplace TF in the block diagram.
For simplicity, consider the ZOH in isolation, i.e. no data processing between the sample and hold. Let the input signal = x(t), the output signal = y(t), and the sampling increment = T sec. At the k'th sampling instant (ie at t=kT), the input to the ADC may be denoted x(k) and the output from the DAC is held constant at x(k) until the next sample arrives.
Thus, the DAC output between x(k) and x(k+1) is a rectangular pulse of height, x(k), and duration, T. This can be modelled, for Laplace transform representation, as a step of height x(k) at t=kT and a step of height -x(k) at time t=(k+1)T, and may be realised via the LT delay operator, e^-skT, which delays any signal by kT sec
Hence the LT of the isolated pulse is {e^(-skT)} x(k)/s - {e^[-s(k+1)T} x(k)/s, or:
e^(-skT) {1-e^-sT} x(k)/s
For the entire signal, from t=0 to t=kT this function reduces to Y(s) = X(s) {1-e^-sT}/s where X(s) and Y(s) are the Laplace transformed ZOH input and output. Hence Y(s)/X(s) = (1-e^-sT)/s
If, now, we wish to proceed by connecting a G(s) block to the ZOH output, the overall TF becomes (1 - e^-sT) G(s)/s. But there's a problem: if we are required to close the loop around this TF, the resultant CLTF will not be analytic due to e^-sT appearing in the CLTF denominator. So we must revert to z-transforms to render it analytic.
The z-transform of the exponential bit is easy; it's (1 - z^-1), because z^-1 is the delay operator in the z-domain (and is therefore equivalent to e^-sT) and we must then find the z-transform of G(s)/s to complete the picture.
For control purposes you need to linearize the system around a single operating point. Considering the synchronous DQ frame, the grid voltage is DC and therefore can be neglected.
Do not replace the grid voltage with a resistor. You are mixing the power delivery circuit with the filter circuit.
The resistor will cause huge damping where you don't want the damping to be. And your inductances are way too small.
First, you need to select the maximum current ripple. This will set your inductance. Then you need to select the shunt capacitor value based on maximum current harmonic distortion that is allowed into the utility.
For stability purposes you need to add passive (resistor) or active (measurement + controls) damping into the system.
Inductance L2 could be partly physical and partly the inherent grid inductance. So you need to make sure the design and controls are stable for any grid output inductance (which varies based on the grid loading and time of day).
So, to answer your question:
You need to control the current so calculate the inverter voltage to grid current (L2) transfer function. No reason to calculate voltage to voltage transfer function since the grid voltage is established by the utility.
\$ \frac{I_{L2}}{V_{inverter}} = ... \$
Btw. this is a simple sophomore level calculation. But the implications, the control and other dependencies make it a grad-level problem. Good luck!
Best Answer
You have used the correct expression for the zero-order hold, and the transfer function is indeed: \begin{equation} H(s) = \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)} \end{equation} However, it's unclear how you proceeded from here. It looks like you've simply replaced s by z (z is actually equal to esT) and then split the function into partial fractions.
Alternatively, you can first find h(t) and then find H(z):
\begin{eqnarray} H(s) &=& \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)} \\ &=& (1-e^{-sT}) \left \{ \dfrac{3}{s} - \dfrac{4}{s+1} + \dfrac{1}{s+4} \right \} \end{eqnarray} Taking the Inverse Laplace Transform: \begin{eqnarray} h(t) &=& f(t)u(t) - f(t-T)u(t-T) \end{eqnarray} where \begin{equation} f(t) = 3 - 4e^{-t} + e^{-4t} \end{equation} H(z) can now be found from h(t). You can refer to this link for a table of formulae: http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html