Lag Compensators MCQ Quiz in मल्याळम - Objective Question with Answer for Lag Compensators - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept

The slope in dB/decade of the asymptotic Bode magnitude plot is decided by open loop poles and open loop zeroes.

One pole provides a slope of -20 dB/decade.

One zero provides a slope of +20 dB/decade.

Calculation

Given, $\mathrm{P}(\mathrm{s})=\frac{0.001}{\mathrm{s}(2\mathrm{s}+1)(0.01\mathrm{s}+1)}$

and $\mathrm{C}(\mathrm{s})=\frac{\mathrm{s}+10}{\mathrm{s}+0.1}$

When they are connected in cascade, the modified open loop transfer function is:

$G(s)=P(s)C(s)$

$G(s)=\frac{0.001(s+10)}{s(2s+1)(0.01s+1)(s+0.1)}$

$G(s)=\frac{0.1(1+\frac{s}{10})}{s(1+\frac{s}{0.1})(1+\frac{s}{0.5})(1+\frac{s}{100})}$

Before the cut-off frequency ω = 3, poles at s = 0, 0.1, and 0.5 will exist only.

Total no.of open-loop poles = 3

Total slope provided by poles = -20 × 3 = -60 dB/decade

Given the transfer function of a lag compensator as:

$G_C\left(s\right)=\frac{1+\alpha sT}{1+sT}$

$G_C\left(j\omega \right)=\frac{1+\alpha j\omega T}{1+j\omega T}$

The phase angle of the function is written as:

ϕ = tan^-1 (αωT) – tan^-1 (ωT)

The phase will be maximum if:

$\frac{d\varphi }{d\omega }=0$

$\frac{\alpha T}{1+{\alpha }^2{\omega }^2{T}^2}-\frac{T}{1+{\omega }^2{T}^2}=0$

$\frac{\alpha }{1+{\alpha }^2{\omega }^2{T}^2}=\frac{1}{1+{\omega }^2{T}^2}$

α(1 + ω²T²) = (1 + α²ω² T²)

ω²(α²T² – αT²) = α – 1

${\omega }^2=\frac{\left(\alpha -1\right)}{\alpha T^2\left(\alpha -1\right)}=\frac{1}{\alpha T^2}$

$\omega =\frac{1}{T\sqrt{\alpha }}$

Hence, the phase angle will be maximum for the frequency,

$\omega =\frac{1}{T\sqrt{\alpha }}$ rad/sec       

$G\left(s\right)=\frac{0.5+s}{0.25+s}$

$=\frac{0.5\left(1+2s\right)}{0.25\left(1+4s\right)}=\frac{2\left(1+2s\right)}{\left(1+4s\right)}$

Now, it is in the form of  $\left(\frac{1+aTs}{1+Ts}\right)$

aT = 2 and T = 4

⇒ a = ½

Maximum phase lag  ${\varphi }_m={\sin }^{-1}\left(\frac{a-1}{a+1}\right)$

$={\sin }^{-1}\left(\frac{\frac{1}{2}-1}{\frac{1}{2}+1}\right)$

$={\sin }^{-1}\left(-\frac{1}{3}\right)$

⇒ ϕ_m = -19.47°

Here -ve sign indicates lagging angle.

Lag Compensator :

In Lag Compensator initially, Pole is added to make the system stable.

Advantages of Lag Compensator:

• A phase lag network offers high gain at low frequency. Thus, it performs the function of a low pass filter.
• The introduction of this network increases the steady-state performance of the system.
• The lag network offers a reduction in bandwidth and this provides longer rise time and settling time and so the transient response.
• The angular contribution of the pole is more than that of the compensator zero because the pole dominates the zero in the lag compensator.

Lead Compensator:

In Lead Compensator initially, zero is added and hence the system becomes stable.

Advantages of Lead Compensator:

• It improves the damping of the overall system.
• The enhanced damping of the system supports less overshoot along with less rise time and settling time. Therefore, the transient response gets improved.
• The addition of a lead network improves the phase margin.
• A system with a lead network provides a quick response as it increases bandwidth thereby providing a faster response.
• Lead networks do not disturb the steady-state error of the system.
• It maximizes the velocity constant of the system.

$\frac{V_0\left(s\right)}{V_{in}\left(s\right)}=\frac{1+4s}{1+8s}$

It is in the form of $\frac{1+asT}{1+sT}$

By comparing both, aT = 4 and T = 8

⇒ a = 0.5

As, a < 1, it is a lag compensator.

The frequency at which maximum lag occurs is

${\omega }_m=\frac{1}{T\sqrt{a}}$

The standard form of a compensator is $\frac{1+sT}{1+\alpha sT}$

$$\begin{gathered} \therefore G\left(s\right)=\frac{5\left(\frac{9}{5}s+1\right)}{0.5\left(\frac{3}{0.5}s+1\right)} \\ =\frac{10\left(1.8s+1\right)}{\left(6s+1\right)} \end{gathered}$$

αT = 6, T = 1.8

$\alpha =\frac{6}{1.8}=\frac{10}{3}$

∴ α > 1, lag compensator

Maximum phase shift provided by compensator is,

$={\sin }^{-1}\left(\frac{1-\alpha }{1+\alpha }\right)$

$={\sin }^{-1}\left[\frac{1-\frac{10}{3}}{1+\frac{10}{3}}\right]$

$={\sin }^{-1}(-\frac{7}{13})=-{32.58}^{\circ }$

Since we are asked to find the phase shift of (lag or lead), it will be 32.58° 

Lag compensator:

Transfer function:

If it is in the form of $\frac{1+aTs}{1+Ts}$, then a < 1

If it is in the form of $\frac{s+a}{s+b}$, then a > b

Maximum phase lag frequency: ${\omega }_m=\frac{1}{T\sqrt{a}}$

Maximum phase lag: ${\varphi }_m={\sin }^{-1}\left(\frac{a-1}{a+1}\right)$

ϕm is negative

Pole zero plot:

The pole is nearer to the origin.

Filter: It is a low pass filter (LPF)

Effect on the system:

• Rise time and settling time increases and Bandwidth decreases
• The transient response becomes slower
• The steady-state response is improved
• Stability decreases

Analysis:

The transfer function of a lag compensator is given by $G\left(s\right)=\frac{\tau s+1}{\beta \tau s+1}$ 

Where, $\beta =\frac{Z_c}{P_c}>1$

Given, $\mathrm{G}\left(\mathrm{s}\right)=\frac{\left(1+0.05s\right)}{\left(1+0.5s\right)}=\frac{\left(1+\mathrm{s}\mathrm{T}\right)}{\left(1+2\mathrm{s}\mathrm{T}\right)}$

Where $\mathrm{T}=0.05$

$\alpha \mathrm{T}=0.5$

$\therefore \alpha =\frac{0.5}{0.05}=10$

Here $\alpha =10>1$. So it is lag compensator

Maximum phase lag occurs at ${\omega }_{\mathrm{m}}=\frac{1}{\mathrm{T}\sqrt{\alpha }}=\frac{1}{0.05\sqrt{10}}=6.325$

Maximum phase lag ${\varphi }_{\mathrm{m}}={\sin }^{-1}\left(\frac{1-\alpha }{1+\alpha }\right)$

$={\sin }^{-1}\left(\frac{1-10}{1+10}\right)={\sin }^{-1}\left(-\frac{9}{11}\right)$

$=-54.9°$

It is a lag compensator

$\begin{array}{l}{V}_{o\left(s\right)}=V_i\left(s\right).\frac{\left(R_2+\frac{1}{sc}\right)}{R_1+R_2+\frac{1}{sc}}\\ \frac{V_o\left(s\right)}{V_i\left(s\right)}=\frac{1+s\left(3\right)}{1+s\left(5\right)}=\frac{1+3s}{1+5s}\end{array}$

standard form $=\frac{1+s\tau }{1+\beta s\tau }$

$\begin{array}{l}\therefore \tau =3,\beta \tau =5\to \beta =\frac{5}{3}=1.67\\ {\varphi }_m={\sin }^{-1}\left(\frac{1-\beta }{1+\beta }\right)={\sin }^{-1}\left(\frac{1-\frac{5}{3}}{1+\frac{5}{3}}\right)=-{\sin }^{-1}\left(\frac{1}{4}\right)=-{14.47}^{\circ }\\ {\omega }_{\mathrm{m}}=\frac{1}{\tau \sqrt{\beta }}=\frac{1}{3\sqrt{\frac{5}{3}}}=0.26\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\mathrm{e}\mathrm{c}\end{array}$