Option 1 : lag network

**Concept:**

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: \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

ϕm is negative

Pole zero plot:

The pole is nearer to the origin.

**Given:**

\(T\left( s \right) = \left( {\frac{{1 + 0.5s}}{{1 + s}}} \right) = \frac{{0.5\left( {s + 2} \right)}}{{\left( {s + 1} \right)}}\)

Zero = -2

Pole = -1

**Analysis:**

The pole-zero plot of T(s) is as shown:

Since the pole is closer to the origin than zero.

It is a lag compensator.

__Lead 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 lead frequency: \({\omega _m} = \frac{1}{{T\sqrt a }}\)

Maximum phase lead: \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

ϕm is positive

Pole zero plot:

The zero is nearer to the origin.

Option 3 : 0 dB

\(\left| {G\left( s \right)} \right| = \frac{{\sqrt {1 + {{\left( {0.1\omega \tau } \right)}^2}} }}{{\sqrt {1 + {{\left( {\omega \tau } \right)}^2}} }}\)

For all ω Numerator is less than Denominator maximum value of |G(s)| occurs at ω = 0

\({\left| {G\left( s \right)} \right|_{max}} = 1\)

In dB, 20 log (1) = 0 dB

For the following network to work as lag compensator, the value of R_{2} would be

Option 4 : Any value of R_{2}

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: \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

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

Application:

The equivalent Laplace transform network for the given network is,

By applying voltage division,

\({V_2}\left( s \right) = {V_1}\left( s \right)\left( {\frac{{{R_2} + \frac{1}{{Cs}}}}{{{R_1} + {R_2} + \frac{1}{{Cs}}}}} \right)\)

\( ⇒ \frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{{R_2}Cs + 1}}{{\left( {{R_1} + {R_2}} \right)Cs + 1}}\)

\( = \frac{{{R_2}Cs + 1}}{{\left( {\frac{{{R_1} + {R_2}}}{{{R_2}}}} \right){R_2}Cs + 1}}\)

\( ⇒ \frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{\tau s + 1}}{{\beta \tau s + 1}},\;\tau = {R_2}C,\;\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}}\)

The above system to be lag compensator, β > 1

\(\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}} > 1\)

⇒ R_{1} > 0

Therefore, at any value of R_{2} the given system acts as lag compensator.

Which of the following can be the pole-zero configuration of a phase-lag controller (lag compensator)?

Option 1 :

Let H(f) be TF of phase-lag controller

\(H\left( f \right) = \frac{{s + a}}{{s + b}}\)

\(\angle H\left( b \right) = {\tan^{-1}}\left( {\frac{{\rm{\omega }}}{a}} \right){ - \;{tan^{ - 1}}}\left( {\frac{{\rm{\omega }}}{b}} \right)\)

For lag Network ∠H(b) = -ve

\({\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{a}} \right) < {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{b}} \right)\)

\(\left( {\frac{{\rm{\omega }}}{a}} \right) < \left( {\frac{{\rm{\omega }}}{b}} \right)\)

b < a

Pole is near to j ω axis than zero

Option 1 : for a given relative stability, the velocity constant is increased

**Lag compensator:**

The transfer function of a lag compensator is given by

\(G\left( s \right) = \frac{{\tau s + 1}}{{\beta \tau s + 1}}\)

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

- Both pole and zero lie in LHS of real plane but PC < ZC i.e. zero is farther away from origin
- Lag Compensation adds a pole at origin (or for low frequencies).
- It helps to reduce the steady-state error of the system.
- An unstable system is a system which has at least one pole at the right side of the s-plane,
- Even if we add a lag compensator to an unstable system, it will remain unstable.

**Effects of phase-lag compensator:**

- It improves the steady-state response
**It makes the system dynamic response slower. Reduces steady-state error**- Increases the rise time
- Decreases the bandwidth
- Reduces stability margin i.e. the system becomes less stable

**Explanation:**

Phase lag compensation is an integrator, as it reduces the steady-state error.

We know that

**Velocity constant =1/steady state error**

**The effect of phase-lag compensation on the servo system is that for given relative stability, the velocity constant is increased.**

**Direction:** It consists of two statements, one labelled as Statement (I) and the other as Statement (II). Examine these two statements carefully and select the answers to these items using the codes given below:

Statement (I): Inductor is not used to realize a lag network.

Statement (II): Inductor produces time delay and hysteresis loss.Option 1 : Both statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)

Inductors are not used as part of the lag compensator circuit component in the control circuit design or implementation as it is not convenient to use it.

Inductor produces time delay and hysteresis loss and hence these are not suitable for designing lag networks.

Therefore, both statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)Option 4 : Lag compensator always stabilizes an unstable system.

__Concept:__

- Lag Compensation
**adds a pole at origin**(or for low frequencies). - Lag Compensation to
**reduce the steady-state error**of the system. - An unstable system is a system that has at least one pole at the right side of the s-plane.
- So, even if we add a lag compensator to an unstable system, it will remain unstable.

**Explanation:**

If an already unstable system has a CLTP =\(\frac{K}{{\left( {s - 3} \right)\left( {s + 2} \right)}}\)

After using a lag compensator the CLTP = \(\frac{K}{{s\left( {s - 3} \right)\left( {s + 2} \right)}}\)

Option 1 : \(\frac{1}{ \propto }\frac{{\left( {s + \frac{1}{\tau }} \right)}}{{\left( {s + \frac{1}{{ \propto \tau }}} \right)}}\)

**Concept:**

The transfer function of the phase lag compensator is given by

\(G\left( s \right) = \frac{{1 + \tau s}}{{1 + a\tau s}}\)

∝ > 1 and τ > 0

\(G\left( s \right) = \frac{\tau ({s + \frac{1}{\tau}})}{\alpha \tau ({s + \frac{1}{\alpha \tau}})}\)

\(G\left( s \right) = \frac{ ({s + \frac{1}{\tau}})}{\alpha ({s + \frac{1}{\alpha \tau}})}\)

For a lag compensator poles are more closer to the origin than zeros.

__Important Points__

The transfer function of phase lag compensator is given by

\(G\left( s \right) = \frac{{1 + Ts}}{{1 + aTs}}\)

\(G\left( {j\omega } \right) = \frac{{1 + j\omega T}}{{1 + j\omega aT}}\)

Phase angle, \(\angle G\;\left( {j\omega } \right) = {\tan ^{ - 1}}\omega T - {\tan ^{ - 1}}a\omega T\)

ϕ = tan-1 ωT – tan-1 aωT

The maximum phase lag occurs at ωm such that \(\frac{{d\phi }}{{d\omega }} = 0\)

\(\Rightarrow {\omega _m} = \frac{1}{{T\sqrt a }}\)

It is a Geometric mean of its two corner frequencies

\({\omega _m} = \sqrt {\frac{1}{T} \times \frac{1}{{aT}}} = \frac{1}{{T\sqrt a }}\)

The maximum phase lag,

\({\phi _m} = {\tan ^{ - 1}}\omega T - {\tan ^{ - 1}}a\omega T\)

\(= {\tan ^{ - 1}}\left( {\frac{1}{{T\sqrt \alpha }}} \right)T - {\tan ^{ - 1}}\left( {\frac{1}{{T\sqrt \alpha }}T} \right)\)

\(= {\tan ^{ - 1}}\frac{1}{{\sqrt a }} - {\tan ^{ - 1}}\sqrt a\)

\({\phi _m} = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{\sqrt a }} - \sqrt a }}{{1 + \sqrt a \cdot \frac{1}{{\sqrt a }}}}} \right)\)

\({\phi _m} = {\tan ^{ - 1}}\left( {\frac{{1 - a}}{{2\sqrt a }}} \right)\)

\(= {\sin ^{ - 1}}\left( {\frac{{1 - a}}{{a + 1}}} \right) = {\cos ^{ - 1}}\left( {\frac{{2\sqrt a }}{{a + 1}}} \right)\)An electrical network shown in figure What type of compensator is this?

Option 2 : Phase lag compensator

**Concept:**

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

**Analysis:**

The equivalent Laplace transform network for the given network is,

By applying voltage division,

\({V_2}\left( s \right) = {V_1}\left( s \right)\left( {\frac{{{R_2} + \frac{1}{{Cs}}}}{{{R_1} + {R_2} + \frac{1}{{Cs}}}}} \right)\)

\( ⇒ \frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{{R_2}Cs + 1}}{{\left( {{R_1} + {R_2}} \right)Cs + 1}}\)

\( = \frac{{{R_2}Cs + 1}}{{\left( {\frac{{{R_1} + {R_2}}}{{{R_2}}}} \right){R_2}Cs + 1}}\)

\( ⇒ \frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{\tau s + 1}}{{\beta \tau s + 1}},\;\tau = {R_2}C,\;\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}}\)

The above system to be lag compensator, β > 1

\(\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}} > 1\)

⇒ R1 > 0 always

Therefore, at any value of R2 the given system acts as lag compensator.

Option 4 : None of the above

Let take an example for lag network transfer function.

\(\begin{array}{l} G\left( S \right) = \frac{{S + 3}}{{S + 2}}\\ \angle G\left( {j\omega } \right) = {\tan ^{ - 1}}\left( {\frac{\omega }{3}} \right) - {\tan ^{ - 1}}\left( {\frac{\omega }{2}} \right)\\ \left| {G\left( \omega \right)} \right| = \sqrt {\frac{{{\omega ^2} + 9}}{{{\omega ^2} + 4}}} \end{array}\)

At ω = 0, |M| = 1.5, ϕ = 0°

At ω = ∞, |M| = 1, ϕ = 0°

Always phase is negative and less than 90°, It’s magnitude is more at zero frequency (ω = 0) at ω = ∞