GATE EE

GATE EE Control Systems: Time & Frequency Response, Stability Analysis Notes

Lecture Notes

SEC 01

Introduction

1Analysis of LTI Systems

1Key Topics

Time response analysis (transient and steady-state)
Standard test signals and system responses
Steady-state error analysis
Routh-Hurwitz stability criterion
Root locus technique
Impulse and frequency response
Pole-zero analysis and system behavior

1Objective

Analyze the behavior of Linear Time-Invariant (LTI) systems for GATE EE preparation.

SEC 02

System Representation

1Transfer Function and System Representation

1Transfer Function

\[G(s) = \dfrac{Y(s)}{X(s)} = \dfrac{b_m s^m + b_{m-1} s^{m-1} + \dots + b_0}{a_n s^n + a_{n-1} s^{n-1} + \dots + a_0}\]

1Poles and Zeros

Poles: Values of \(s\) that make denominator zero
Zeros: Values of \(s\) that make numerator zero
Order: Highest power of \(s\) in denominator

1Example

\[G(s) = \dfrac{s+2}{(s+1)(s+3)}\]

Zero at \(s = -2\)
Poles at \(s = -1, -3\)
Second-order system

1Stability Condition

System is stable if and only if all poles lie in the left half of s-plane (negative real parts).

SEC 03

Impulse Response

1Impulse Response Analysis

1Definition

Impulse response \(h(t)\) is the output when input is unit impulse \(\delta(t)\).

\[h(t) = \mathcal{L}^{-1}\{G(s)\}\]

1Properties

Convolution: \(y(t) = x(t) * h(t)\)
Laplace: \(Y(s) = X(s) \cdot G(s)\)
System characterization

1Example

For \(G(s) = \dfrac{1}{s+a}\):

\[h(t) = e^{-at}u(t)\]

where \(u(t)\) is unit step function.

1Stability from Impulse Response

System is BIBO stable if \(\int_0^{\infty} |h(t)| dt < \infty\)

SEC 04

Time Response Analysis

1Time Response Analysis

1Components

Transient response: Initial dynamic behavior
Steady-state response: Long-term behavior

1Standard Test Signals

Impulse: \(\delta(t)\) \(\leftrightarrow\) \(1\)
Step: \(u(t)\) \(\leftrightarrow\) \(\dfrac{1}{s}\)
Ramp: \(tu(t)\) \(\leftrightarrow\) \(\dfrac{1}{s^2}\)
Parabolic: \(\dfrac{t^2}{2}u(t)\) \(\leftrightarrow\) \(\dfrac{1}{s^3}\)

1Example

First-order system:

\[\dfrac{C(s)}{R(s)} = \dfrac{1}{Ts + 1}\]

Step response:

\[c(t) = 1 - e^{-t/T}, \quad t \geq 0\]

where \(T\) is the time constant.

1First Order System Parameters

1System: \(G(s) = \dfrac{K}{Ts + 1}\)

DC Gain: \(K\)
Time constant: \(T\) (time to reach 63.2% of final value)
Settling time: \(t_s = 4T\) (for 2% criterion)

1Step Response

\[c(t) = K(1 - e^{-t/T})u(t)\]

1Ramp Response

\[c(t) = K(t - T + Te^{-t/T})u(t)\]

1Key Point

First-order systems never exhibit overshoot for step input.

SEC 05

Second Order Systems

1Second Order System Analysis

1Standard Form

\[\dfrac{C(s)}{R(s)} = \dfrac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\]

where:

\(\zeta\): Damping ratio
\(\omega_n\): Natural frequency (rad/s)
\(\omega_d = \omega_n\sqrt{1-\zeta^2}\): Damped frequency

Response Characteristics
Parameter	Formula
Peak overshoot (\(M_p\))	\(e^{-\pi\zeta/\sqrt{1-\zeta^2}} \times 100\%\)
Peak time (\(t_p\))	\(\dfrac{\pi}{\omega_n\sqrt{1-\zeta^2}}\)
Rise time (\(t_r\))	\(\dfrac{\pi - \cos^{-1}\zeta}{\omega_n\sqrt{1-\zeta^2}}\)
Settling time (\(t_s\))	\(\dfrac{4}{\zeta\omega_n}\) (for 2% criterion)

1Damping Cases for Second Order Systems

1Response Types Based on Damping Ratio

\(\zeta > 1\): Overdamped - No oscillations, slow response
\(\zeta = 1\): Critically damped - Fastest response without overshoot
\(0 < \zeta < 1\): Underdamped - Oscillatory response with overshoot
\(\zeta = 0\): Undamped - Sustained oscillations

1Step Response Forms

Underdamped (\(0 < \zeta < 1\)):

\[c(t) = 1 - \dfrac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}} \sin(\omega_d t + \phi)\]

where \(\phi = \cos^{-1}\zeta\)

1Pole Locations

\[s_{1,2} = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2}\]

Real part: \(-\zeta\omega_n\)
Imaginary part: \(\pm\omega_d\)

1Important Relations

\(\zeta = \cos\theta\) where \(\theta\) is angle from negative real axis
\(M_p\) decreases as \(\zeta\) increases
\(t_s\) is independent of \(\omega_d\)

1Higher Order Systems

1Dominant Pole Concept

For systems with multiple poles:

Dominant poles: Closest to imaginary axis
Determine the main characteristics of transient response
Other poles should be at least 5-10 times farther from imaginary axis

1Example

\[G(s) = \dfrac{100}{(s+1)(s+2)(s+10)}\]

Dominant poles: \(s = -1, -2\)

Approximate as second-order system for initial analysis.

1Approximation Guidelines

Neglect poles that are 5+ times farther than dominant poles
Use dominant pole approximation for quick analysis
Verify with complete system analysis if needed

SEC 06

Steady-State Error

1Steady-State Error Analysis

1Error Signal and Final Value Theorem

For unity feedback system: \(E(s) = \dfrac{R(s)}{1 + G(s)}\)

Steady-state error: \(e_{ss} = \lim_{t \to \infty} e(t) = \lim_{s \to 0} sE(s)\)

1System Type and Error Constants

System type is determined by the number of integrators (\(1/s\)) in \(G(s)\):

Position constant: \(K_p = \lim_{s\to 0} G(s)\)
Velocity constant: \(K_v = \lim_{s\to 0} sG(s)\)
Acceleration constant: \(K_a = \lim_{s\to 0} s^2G(s)\)

Steady-State Error for Unity Feedback Systems
Input	Type 0	Type 1	Type 2
Step (\(\dfrac{1}{s}\))	\(\dfrac{1}{1+K_p}\)	0	0
Ramp (\(\dfrac{1}{s^2}\))	\(\infty\)	\(\dfrac{1}{K_v}\)	0
Parabolic (\(\dfrac{1}{s^3}\))	\(\infty\)	\(\infty\)	\(\dfrac{1}{K_a}\)

1Generalized Error Analysis

1Non-Unity Feedback Systems

For system with feedback \(H(s)\):

\[E(s) = \dfrac{R(s)}{1 + G(s)H(s)}\]

Error constants are calculated using \(G(s)H(s)\).

1Disturbance Rejection

For disturbance \(D(s)\) at plant input:

\[E(s) = \dfrac{-G(s)D(s)}{1 + G(s)H(s)}\]

Higher loop gain reduces disturbance effects.

1Position Error Constant Formula

For \(G(s) = \dfrac{K \prod (s + z_i)}{s^N \prod (s + p_j)}\):

\[K_p = \lim_{s \to 0} G(s) = \begin{cases} \dfrac{K \prod z_i}{\prod p_j} & \text{if } N = 0 \\ \infty & \text{if } N \geq 1 \end{cases}\]

1Key Insight

Increasing system type reduces steady-state error
Higher gain improves accuracy but may affect stability
Trade-off between accuracy and stability

SEC 07

Stability Analysis

1Routh-Hurwitz Criterion

1Procedure

For characteristic equation \(a_n s^n + a_{n-1} s^{n-1} + \dots + a_0 = 0\):

All coefficients must be positive (necessary condition)
Construct Routh array
Count sign changes in the first column
Number of sign changes = Number of RHP poles

1Example

For \(s^3 + 4s^2 + 6s + 4 = 0\):

\[\begin{array}{c|cc} s^3 & 1 & 6 \\ s^2 & 4 & 4 \\ s^1 & 5 & 0 \\ s^0 & 4 & \\ \end{array}\]

No sign changes \(\Rightarrow\) stable.

1Special Cases

Zero in first column: Replace with small \(\epsilon\)
Row of zeros: Use auxiliary equation
Marginal stability: Poles on imaginary axis

1Advanced Stability Concepts

1Marginal Stability

System has poles on imaginary axis. From row of zeros:

Form auxiliary equation from previous row
Differentiate to get next row
Marginal frequency from auxiliary equation

1Example

For \(s^4 + 2s^3 + 3s^2 + 2s + 2 = 0\):

\[\begin{array}{c|ccc} s^4 & 1 & 3 & 2 \\ s^3 & 2 & 2 & 0 \\ s^2 & 2 & 2 & 0 \\ s^1 & 0 & 0 & 0 \\ \end{array}\]

Auxiliary equation: \(2s^2 + 2 = 0 \Rightarrow s = \pm j1\)

System is marginally stable with oscillation at \(\omega = 1\) rad/s.

1Note

When row of zeros occurs, replace that row with derivative of auxiliary equation.

SEC 08

Root Locus

1Root Locus Technique

1Basic Concept

Root locus shows the path of closed-loop poles as gain \(K\) varies from 0 to \(\infty\).

For unity feedback: \(1 + KG(s)H(s) = 0\)

1Construction Rules

Locus starts at open-loop poles (\(K=0\))
Locus ends at zeros or infinity (\(K=\infty\))
Real-axis segments: Exist where total #poles + #zeros to the right is odd
Asymptotes: \(n-m\) branches to infinity
Asymptote angles: \(\dfrac{(2q+1)\pi}{n-m}\), \(q = 0,1,2,\ldots,(n-m-1)\)
Centroid: \(\sigma_a = \dfrac{\sum \text{poles} - \sum \text{zeros}}{n-m}\)

1Root Locus - Advanced Rules

1Breakaway/Break-in Points

Solve \(\dfrac{dK}{ds} = 0\) where \(K = -\dfrac{1}{G(s)H(s)}\)

Alternative: \(\sum_{i=1}^n \dfrac{1}{s-p_i} = \sum_{j=1}^m \dfrac{1}{s-z_j}\)

1Angle and Magnitude Conditions

For any point \(s\) on root locus:

Angle condition: \(\sum \angle(s-z_i) - \sum \angle(s-p_j) = (2k+1)\pi\)
Magnitude condition: \(K = \dfrac{\prod |s-p_j|}{\prod |s-z_i|}\)

1Example

For \(G(s) = \dfrac{K}{s(s+4)}\), breakaway point:

\[\dfrac{1}{s} + \dfrac{1}{s+4} = 0 \Rightarrow s = -2\]

At \(s = -2\): \(K = \dfrac{|-2| \cdot |2|}{1} = 4\)

1Root Locus - Additional Rules

1Departure/Arrival Angles

For complex poles/zeros:

Departure angle from complex pole:

\[\theta_d = 180^{\circ} - \sum \angle(\text{pole to other poles}) + \sum \angle(\text{pole to zeros})\]
Arrival angle at complex zero:

\[\theta_a = 180^{\circ} + \sum \angle(\text{zero to poles}) - \sum \angle(\text{zero to other zeros})\]

1Intersection with Imaginary Axis

Use Routh-Hurwitz criterion
Find value of \(K\) that makes system marginally stable
Solve auxiliary equation for intersection points

SEC 09

Pole-Zero Analysis

1Effect of Poles and Zeros on Response

1Pole Effects

Real poles: Exponential terms \(e^{-\sigma t}\)
Complex poles: Oscillatory terms \(e^{-\sigma t}\cos(\omega t + \phi)\)
Dominant poles: Closest to imaginary axis (slowest modes)
Distance from origin: Related to natural frequency

1Zero Effects

LHP zeros: Reduce overshoot, speed up response
RHP zeros: Cause undershoot, non-minimum phase behavior
Near pole-zero cancellation: Reduces effect of both
Zeros at origin: Derivative action

1Design Guidelines

Place poles in LHP for stability
Use zeros to shape transient response
Avoid pole-zero cancellation in RHP
Consider sensitivity to parameter variations

SEC 10

Performance Specifications

1Time Domain Performance Specifications

1Standard Specifications

Rise time (\(t_r\)): Time to go from 10% to 90% of final value
Peak time (\(t_p\)): Time to reach maximum overshoot
Settling time (\(t_s\)): Time to stay within 2% or 5% of final value
Maximum overshoot (\(M_p\)): Maximum deviation from final value

1Relationships for Second-Order Systems

Faster rise time \(\Rightarrow\) Higher \(\omega_n\)
Lower overshoot \(\Rightarrow\) Higher \(\zeta\)
Faster settling \(\Rightarrow\) Higher \(\zeta\omega_n\)
Trade-off between speed and overshoot

1Typical Design Requirements

\(M_p < 20\%\) \(\Rightarrow\) \(\zeta > 0.45\)
\(t_s < 4\) sec \(\Rightarrow\) \(\zeta\omega_n > 1\)

SEC 11

GATE Questions

1GATE Practice Questions

1Example 1

A unity feedback system has \(G(s) = \dfrac{K}{s(s+3)}\). The value of \(K\) for damping ratio \(\zeta = 0.5\) is:

1Solution

Characteristic equation: \(s^2 + 3s + K = 0\)
Standard form: \(s^2 + 2\zeta\omega_n s + \omega_n^2 = 0\)
Compare: \(\omega_n^2 = K\), \(2\zeta\omega_n = 3\)
For \(\zeta = 0.5\): \(\omega_n = \dfrac{3}{2 \times 0.5} = 3\)
Thus, \(K = \omega_n^2 = 9\). Answer: C

1GATE Practice Questions (Contd.)

1Example 2

The steady-state error of a unity feedback system with \(G(s) = \dfrac{10}{s(s+2)}\) to a unit ramp input is:

0
0.2
0.5
\(\infty\)

1Solution

System type = 1 (one integrator)
Velocity constant: \(K_v = \lim_{s\to 0} s \cdot \dfrac{10}{s(s+2)} = \dfrac{10}{2} = 5\)
Steady-state error for ramp: \(e_{ss} = \dfrac{1}{K_v} = \dfrac{1}{5} = 0.2\)
Answer: B

1GATE Practice Questions (Contd.)

1Example 3

For the characteristic equation \(s^3 + 4s^2 + Ks + 4 = 0\), the range of \(K\) for stability is:

\(K > 0\)
\(0 < K < 4\)
\(0 < K < 16\)
\(K > 4\)

1Solution

Routh array:

\[\begin{array}{c|cc} s^3 & 1 & K \\ s^2 & 4 & 4 \\ s^1 & \dfrac{4K-4}{4} = K-1 & 0 \\ s^0 & 4 & \\ \end{array}\]

For stability: All first column elements must be positive
\(K-1 > 0 \Rightarrow K > 1\) and \(4 > 0\) (always satisfied)
Also need \(K > 0\). Therefore: \(K > 1\). Answer: None exactly matches, but closest is A

1GATE Practice Questions (Contd.)

1Example 4

The peak overshoot of a second-order system with \(\zeta = 0.6\) is approximately:

9.5%
15.2%
25.4%
30.8%

1Solution

Peak overshoot: \(M_p = e^{-\pi\zeta/\sqrt{1 -\zeta^2}} \times 100\%\)

For \(\zeta = 0.6\): \(M_p = e^{-\pi \times 0.6/\sqrt{1-0.6^2}} \times 100\%\) \(M_p = e^{-\pi \times 0.6/\sqrt{0.64}} \times 100\%\) \(M_p = e^{-\pi \times 0.6/0.8} \times 100\%\) \(M_p = e^{-2.356} \times 100\% = 0.095 \times 100\% = 9.5\%\)

Answer: A

SEC 12

Summary

1Summary

1Key Takeaways

Transfer function analysis provides complete system characterization
Pole locations determine stability and transient response nature
Time domain specifications help in controller design
Steady-state error analysis ensures accuracy requirements
Routh-Hurwitz criterion is essential for stability analysis
Root locus technique shows parameter sensitivity

1GATE Preparation Tips

Practice standard forms and their responses
Master Routh-Hurwitz for all special cases
Understand pole-zero effects on system behavior
Focus on steady-state error calculations
Learn root locus construction rules

1Important Formulas to Remember

Second-order parameters: \(M_p\), \(t_p\), \(t_s\), \(t_r\)
Error constants: \(K_p\), \(K_v\), \(K_a\)
Stability conditions and Routh array