Question

The Automatic Voltage Regulator (AVR) system of a generator is represented by the simplified block diagram shown in Figure 12.15, in which the sensor is modelled by a simple first-order transfer function. The voltage is sensed through a voltage transformer and then rectified through a bridge rectifier. Parameters of the AVR system are given as follows.

(a) Determine the open-loop transfer function of the block diagram and the closed-loop transfer function relating the generator terminal voltage ${\mathit{V}}_{\mathbf{t}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ to the reference voltage ${\mathit{V}}_{\mathbf{r}\mathbf{e}\mathbf{f}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$. (b) For the range of${\mathit{K}}_{\mathbf{A}}$ from$\mathbf{0}$ to$\mathbf{12}\mathbf{.}\mathbf{16}$, comment on the stability of the system. (c) For ${\mathit{K}}_{\mathbf{A}}\mathbf{=}\mathbf{10}$, evaluate the steady-state step response and steady-state error.

Short answer

(a) The open loop and closed loop transfer function of AVR system is $\frac{K_A}{s^4+33.5s^3+307.5s^2+775s+500}$and $\frac{V_t\left(s\right)}{V_{ref}\left(s\right)}=\frac{25(s+20)K_A}{s^4+33.4s^3+307.5s^2+775s+500+500K_A}$ respectively.

(b) System is marginally stable for the value of $K_A\le 12.16$.

(c) The steady state response and steady error for the system are$0.909\text{\hspace{0.17em}V}$ and$0.091\text{\hspace{0.17em}V}$ respectively.

Step-by-step solution

Write the given data from the question.

The time constant and gain of amplifier is$T_A=0.1$ and$K_A$ respectively.

The time constant for exciter,$T_E=0.4$

The gain for exciter,$K_E=1$

The time constant for generator,$T_G=1$

The gain for generator,$K_G=1$

The time constant for sensor,$T_R=0.05$

The gain for sensor,$K_R=1$

Determine the equation to calculate the open loop, closed loop, stability, steady state response and steady state error.

The equation to calculate the open loop transfer function is given as,

$\mathit{O}\mathit{P}\mathit{T}\mathit{F}\mathbf{=}\mathbf{G}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{H}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ …… (1)

The equation to calculate the closed loop transfer function is given as,

$\frac{{\mathbf{V}}_{\mathbf{t}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}{{\mathbf{V}}_{\mathbf{r}\mathbf{e}\mathbf{f}}\mathbf{s}}\mathbf{=}\frac{\mathbf{G}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}{\mathbf{1}\mathbf{+}\mathbf{G}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{H}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}$ …… (2)

The equation to calculate the steady state response of the system is given as follows.

${\mathbf{\left(}{\mathbf{V}}_{\mathbf{t}}\mathbf{\right)}}_{\mathbf{ss}}\mathbf{=}\underset{\mathbf{s}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\left(}{\mathbf{V}}_{\mathbf{t}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\right)}$ …… (3)

The equation to calculate the steady state error of the system is given as follows

$\mathbf{\Delta e}\mathbf{=}\mathbf{1}\mathbf{-}{\mathbf{\left(}{\mathbf{V}}_{\mathbf{t}}\mathbf{\right)}}_{\mathbf{ss}}$…… (4)

Determine the open-loop transfer function of the block diagram and the closed-loop transfer function.

(a)

The output of generator ${\mathrm{V}}_{\mathrm{t}}\left(\mathrm{s}\right)$is given by,

$V_t\left(s\right)=\left(\frac{K_G}{1+T_{G}s}\right)V_F\left(s\right)$

The output of exciter $V_F\left(s\right)$is given by,

$V_F\left(s\right)=\left(\frac{K_E}{1+T_{E}s}\right)V_R\left(s\right)$

The output of amplifier$V_R\left(s\right)$is given by,

$V_R\left(s\right)=\left(\frac{K_A}{1+T_{A}s}\right)V_e\left(s\right)$

Calculate the open loop transfer function,

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

Substitute $\left(\frac{K_G}{1+T_{G}s}\right)V_F\left(s\right)$for $V_t\left(s\right)$into above equation.

$G\left(s\right)=\frac{\left(\frac{K_G}{1+T_{G}s}\right)V_F\left(s\right)}{V_e\left(s\right)}$

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

$G\left(s\right)=\frac{\left(\frac{K_G}{1+T_{G}s}\right){\displaystyle }\left(\frac{K_E}{1+T_{E}s}\right)V_R\left(s\right)}{V_e\left(s\right)}$

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

$\begin{array}{c}G\left(s\right)=\frac{\left(\frac{K_G}{1+T_{G}s}\right)\left(\frac{K_E}{1+T_{E}s}\right)\left(\frac{K_A}{1+T_{A}s}\right)V_e\left(s\right)}{V_e\left(s\right)}\\ G\left(s\right)=\left(\frac{K_G}{1+T_{G}s}\right)\left(\frac{K_E}{1+T_{E}s}\right)\left(\frac{K_A}{1+T_{A}s}\right)\\ G\left(s\right)=\frac{K_G{K}_E{K}_A}{(1+T_{G}s)(1+T_{E}s)(1+T_{A}s)}\end{array}$

Substitute$1$for$K_G$,$1$for$K_E$,$0.1$for$T_A$, $0.4$for $T_E$,$1$for$T_G$and$0.05$for$T_R$into above equation.

$\begin{array}{l}G\left(s\right)=\frac{\left(1\right)\left(1\right)K_A}{(1+1s)(1+0.4s)(1+0.1s)}\\ G\left(s\right)=\frac{K_A}{(1+1s)(1+0.4s)(1+0.1s)}\end{array}$

Calculate the transfer function of the feedback loop.

$H\left(s\right)=\frac{V_s\left(s\right)}{V_t\left(s\right)}$

Substitute$\frac{K_R}{1+T_{R}s}{V}_t\left(s\right)$for$V_s\left(s\right)$into above equation.

$\begin{array}{l}H\left(s\right)=\frac{\frac{K_R}{1+T_{R}s}{V}_t\left(s\right)}{V_t\left(s\right)}\\ H\left(s\right)=\frac{K_R}{1+T_{R}s}\end{array}$

Substitute $1$for $K_R$and $0.05$for $T_R$into above equation.

$H\left(s\right)=\frac{1}{1+0.05s}$

Calculate the open loop transfer function of AVR system.

Calculate the closed loop transfer function of the AVR system.

$\begin{array}{l}\frac{V_t\left(s\right)}{V_{ref}\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)H\left(s\right)}\\ \frac{V_t\left(s\right)}{V_{ref}\left(s\right)}=\frac{\frac{K_A}{(1+s)(1+0.4s)(1+0.1s)}}{1+\frac{K_A}{(1+s)(1+0.4s)(1+0.1s)(1+0.05s)}}\\ \frac{V_t\left(s\right)}{V_{ref}\left(s\right)}=\frac{(1+0.05s)K_A}{(1+s)(1+0.4s)(1+0.1s)(1+0.05s)+K_A}\\ \frac{V_t\left(s\right)}{V_{ref}\left(s\right)}=\frac{25(s+20)K_A}{s^4+33.4s^3+307.5s^2+775s+500+500K_A}\end{array}$

Hence the open loop and closed loop transfer function of AVR system is $\frac{K_A}{s^4+33.5s^3+307.5s^2+775s+500}$ and

$\frac{V_t\left(s\right)}{V_{ref}\left(s\right)}=\frac{25(s+20)K_A}{s^4+33.4s^3+307.5s^2+775s+500+500K_A}$ respectively.

Determine the stability of the system for the gain  KA, from 0 to 12.16 .

(b)

The characteristic equation to the system is given by,

$\begin{array}{l}1+G\left(s\right)H\left(s\right)=0\\ s^4+33.4s^3+307.s^2+775s+500+500K_A=0\end{array}$

By using the Routh Hurwitz array to find the value of gain $K_A$.

$\begin{array}{l}{s}^{4}1307.5500+500K_A\\ s^{3}33.57750\\ s^{2}284.365500+500K_{A}0\\ s^{1}716.1-58.9K_{A}00\\ s500+500K_{A}00\end{array}$

For the system to be stable, the sign should not be changes in the second column.

Therefore,

$\begin{array}{c}716.1-58.9K_A>0\\ 716.1>58.9K_A\\ K_A<12.16\end{array}$

Again,

$\begin{array}{c}500+500K_A>0\\ 500K_A>-500\\ K_A>-1\end{array}$

Consider the positive value for the gain $K_A$.

$K_A<12.16$

Now derive the equation for $K_A=12.16$.

$\begin{array}{c}248.365s^2+500+500K_A=0\\ 248.365s^2+500+500\times 12.16=0\\ 248.365s^2+6580=0\\ s=\pm j5.147\end{array}$

Hence the root of the equation lies on the imaginary axis and system is marginally stable for the value of $K_A=12.16$.

Determine the steady state response and steady state error for KA=10.

From the closed loop transfer function.

$V_t\left(s\right)=\left(\frac{25(S+20)K_A}{s^4+33.4s^3+307.s^2+775s+500+500K_A}\right)V_{ref}\left(s\right)$

Substitute $\frac{1}{s}$for $V_{ref}\left(s\right)$into above equation.

$V_t\left(s\right)=\left(\frac{25(S+20)K_A}{s^4+33.4s^3+307.s^2+775s+500+500K_A}\right)\times \frac{1}{s}$

Calculate the steady state response of the system,

${\left(V_t\right)}_{ss}=\underset{s\to 0}{\lim }\left(s\right)\left(V_t\left(s\right)\right)$

Substitute $\left(\frac{25(s+20)K_A}{s^4+33.4s^3+307.5s^2+775s+500+500K_A}\right)\times \frac{1}{s}$for $V_t\left(s\right)$into above equation

Apply the limits,

$\begin{array}{l}{\left(V_t\right)}_{ss}=\left(\frac{25(0+20)K_A}{{\left(0\right)}^4+33.4{\left(0\right)}^3+307.5{\left(0\right)}^2+775\left(0\right)+500+500K_A}\right)\\ {\left(V_t\right)}_{ss}=\frac{500K_A}{500(1+K_A)}\\ {\left(V_t\right)}_{ss}=\frac{K_A}{(1+K_A)}\end{array}$

Substitute $10$for $K_A$into above equation.

$\begin{array}{l}{\left(V_t\right)}_{ss}=\frac{10}{1+10}\\ {\left(V_t\right)}_{ss}=\frac{10}{11}\\ {\left(V_t\right)}_{ss}=0.909\text{\hspace{0.17em}V}\end{array}$

Calculate the steady state error of the system,

$\Delta e=1-{\left(V_t\right)}_{ss}$

Substitute$0.909V$ for${\left(V_t\right)}_{ss}$ into above equation,

$\begin{array}{l}\Delta e=1-0.909\\ \Delta e=0.091\text{\hspace{0.17em}V}\end{array}$

Hence, the steady state response and steady error for the system are$0.909V$ and$0.091V$ respectively.