Question

The block-diagram representation of a closed-loop automatic regulating system, in which generator voltage control is accomplished by controlling the exciter voltage, is shown in Figure 12.14. ${\mathit{T}}_{\mathbf{a}}\mathbf{,}{\mathit{T}}_{\mathbf{e}}$ and ${\mathit{T}}_{\mathbf{f}}$are the time constants associated with the amplifier, exciter, and generator field circuit, respectively. (a) Find the open-loop transfer function $\mathbf{G}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$. (b) Evaluate the minimum open-loop gain such that the steady-state error${\mathbf{\Delta e}}_{\mathbf{ss}}$ does not exceed $\mathbf{1}\mathit{\%}$. (c) Discuss the nature of the dynamic response of the system to a step change in the reference input voltage.

Short answer

(a) The open loop transfer function$\mathrm{G}\left(\mathrm{s}\right)$ is$\frac{{\mathrm{K}}_{\mathrm{f}}{\mathrm{K}}_{\mathrm{E}}{\mathrm{K}}_{\mathrm{A}}}{(1+{\mathrm{T}}_{\mathrm{f}}\mathrm{s})(1+{\mathrm{T}}_{\mathrm{e}}\mathrm{s})(1+{\mathrm{T}}_{\mathrm{a}}\mathrm{s})}$ .

(b) The minimum open loop gain for the steady state error does not exceed$1\%$ is $\mathrm{K}\ge 99$.

(c) If the characteristic equation has complex conjugate roots, then response will have dynamic response as ${\mathrm{Ae}}^{\mathrm{at}}\sin (\mathrm{\omega t}+\mathrm{\varphi })$.

Step-by-step solution

Write the given data from the question.

The time constant and gain of amplifier is $T_a$and $K_A$respectively.

The time constant and gain of exciter is $T_e$and $K_E$respectively.

The time constant and gain of generator is $T_f$and $K_F$respectively.

Determine the open loop transfer function .

(a)

The output of generator $\Delta V_t$is given by,

${\mathbf{\Delta V}}_t\mathbf{=}\left(\frac{K_F}{1+T_{f}s}\right){\mathbf{\Delta V}}_f$

The output of exciter$\Delta V_f$is given by,

${\mathbf{\Delta V}}_f\mathbf{=}\left(\frac{K_E}{1+T_{e}s}\right){\mathbf{\Delta V}}_R$

The output of amplifier $\Delta V_R$is given by,

${\mathbf{\Delta V}}_R\mathbf{=}\left(\frac{K_A}{1+T_{a}s}\right)\mathbf{\Delta e}$

Calculate the open loop transfer function,

$G\left(s\right)=\frac{\Delta V_t}{\Delta e}$

Substitute $\left(\frac{K_f}{1+T_{f}s}\right)\Delta V_f$ for $\Delta V_t$into above equation.

$G\left(s\right)=\frac{\left(\frac{K_f}{1+T_{f}s}\right)\Delta V_f}{\Delta e}$

Substitute $\left(\frac{K_E}{1+T_{e}s}\right)\Delta V_R$ for $\Delta V_f$ into above equation.

$G\left(s\right)=\frac{\left(\frac{K_f}{1+T_{f}s}\right)\left(\frac{K_E}{1+T_{e}s}\right)\Delta V_R}{\Delta e}$

Substitute $\left(\frac{K_A}{1+T_{a}s}\right)\Delta e$for $\Delta V_R$into above equation.

$\begin{array}{c}\mathrm{G}\left(\mathrm{s}\right)=\frac{\left(\frac{{\mathrm{K}}_{\mathrm{f}}}{1+{\mathrm{T}}_{\mathrm{f}}\mathrm{s}}\right)\left(\frac{{\mathrm{K}}_{\mathrm{E}}}{1+{\mathrm{T}}_{\mathrm{e}}\mathrm{s}}\right)\left(\frac{{\mathrm{K}}_{\mathrm{A}}}{1+{\mathrm{T}}_{\mathrm{a}}\mathrm{s}}\right))\mathrm{\Delta e}}{\mathrm{\Delta e}}\\ \mathrm{G}\left(\mathrm{s}\right)=\left(\frac{{\mathrm{K}}_{\mathrm{f}}}{1+{\mathrm{T}}_{\mathrm{f}}\mathrm{s}}\right)\left(\frac{{\mathrm{K}}_{\mathrm{E}}}{1+{\mathrm{T}}_{\mathrm{e}}\mathrm{s}}\right)\left(\frac{{\mathrm{K}}_{\mathrm{A}}}{1+{\mathrm{T}}_{\mathrm{a}}\mathrm{s}}\right)\\ \mathrm{G}\left(\mathrm{s}\right)=\frac{{\mathrm{K}}_{\mathrm{f}}{\mathrm{K}}_{\mathrm{E}}{\mathrm{K}}_{\mathrm{A}}}{(1+{\mathrm{T}}_{\mathrm{f}}\mathrm{s})(1+{\mathrm{T}}_{\mathrm{e}}\mathrm{s})(1+{\mathrm{T}}_{\mathrm{a}}\mathrm{s})}\end{array}$

Hence the open loop transfer function $G\left(s\right)$ is$\frac{K_f{K}_E{K}_A}{(1+T_{f}s)(1+T_{e}s)(1+T_{a}s)}$ .

Calculate the minimum open-loop gain such that the steady-state error Δess  does not exceed 1%.

(b)

From the block diagram,

$\frac{\Delta e}{\Delta V_{ref}}=\frac{1}{1+G\left(s\right)}$

Substitute $\frac{K_f{K}_E{K}_A}{(1+T_{f}s)(1+T_{e}s)(1+T_{a}s)}$for $G\left(s\right)$into above equation.

At the steady state, the $s$ is zero.

$\begin{array}{l}\frac{\Delta e}{\Delta V_{ref}}=\frac{(1+T_f\left(0\right))(1+T_e\left(0\right))(1+T_a\left(0\right))}{(1+T_f\left(0\right))(1+T_e\left(0\right))(1+T_a\left(0\right))+K_f{K}_E{K}_A}\\ \frac{\Delta e}{\Delta V_{ref}}=\frac{1}{1+K_f{K}_E{K}_A}\end{array}$

Let assume$K_f{K}_E{K}_A=K$,

$\frac{\Delta e}{\Delta V_{ref}}=\frac{1}{1+K}$

The steady state error does not exceed $1\%$.

$\begin{array}{c}\frac{1}{1+K}\le 1\%\\ \frac{1}{1+K}\le \frac{1}{100}\\ K+1\ge 100\\ K\ge 99\end{array}$

Hence the minimum open loop gain for the steady state error does not exceed$1\%$ is $K\ge 99$.

The nature of the dynamic response of the system.

(c)

The output voltage is given as follows.

$\Delta V_t\left(t\right)=L^{-1}\left[\frac{G\left(s\right)}{1+G\left(s\right)}\Delta V_{ref}\left(s\right)\right]$

The characteristics equation from the above is $1+G\left(s\right)$and response depends on the roots of the characteristic equation.

The equation will have the three roots,$s_1,s_2$ and$s_3$ .

If the roots $s_1,s_2$and $s_3$are real then the response will have the transient component $A{e}^{st}$.

If the roots are complex conjugate, then response will have dynamic response as $A{e}^{at}\sin (\omega t+\varphi )$.