GATE EE

Detailed Notes on System Modeling & Feedback Control Systems for GATE EE Exam

Lecture Notes

SEC 01

Introduction

1System Modeling and Feedback
1Key Topics Covered

  • Mathematical modeling of physical systems

  • Transfer function representation

  • Block diagram reduction techniques

  • Signal flow graphs (Mason’s Gain Formula)

  • Feedback concepts and properties

  • System types and error constants

1GATE EE Objective

Master the fundamentals of modeling and analyzing control systems with emphasis on problem-solving techniques for GATE EE examination.

SEC 02

Mathematical Modeling

1Mathematical Modeling of Physical Systems
1Basic Elements

Mechanical Systems:

  • Mass (\(m\)): \(F = ma\)

  • Spring (\(k\)): \(F = kx\)

  • Damper (\(b\)): \(F = b\dot{x}\)

Electrical Systems:

  • Resistor (\(R\)): \(V = IR\)

  • Inductor (\(L\)): \(V = L\frac{dI}{dt}\)

  • Capacitor (\(C\)): \(I = C\frac{dV}{dt}\)

1Example

Mass-Spring-Damper System:

\[m\ddot{x} + b\dot{x} + kx = F(t)\]
Taking Laplace transform:
\[ms^2X(s) + bsX(s) + kX(s) = F(s)\]
\[G(s) = \frac{X(s)}{F(s)} = \frac{1}{ms^2 + bs + k}\]
1Mechanical System Diagram
Mechanical system representation
Mechanical system representation
1Analogous Systems
1Force-Voltage Analogy
Mechanical Electrical Relationship
Force \(F\) Voltage \(V\) Through variable
Velocity \(\dot{x}\) Current \(I\) Across variable
Mass \(m\) Inductance \(L\) \(F = m\ddot{x} \leftrightarrow V = L\frac{dI}{dt}\)
Damping \(b\) Resistance \(R\) \(F = b\dot{x} \leftrightarrow V = RI\)
Spring \(k\) \(\frac{1}{C}\) \(F = k\int \dot{x}dt \leftrightarrow V = \frac{1}{C}\int I dt\)
1Important Note

Force-Current analogy is dual to Force-Voltage analogy. Choose the appropriate analogy based on the system configuration.

SEC 03

Transfer Function

1Transfer Function Definition

The transfer function \(G(s)\) of a Linear Time-Invariant (LTI) system is defined as:

\[G(s) = \frac{Y(s)}{X(s)} = \frac{\mathcal{L}[\text{output}]}{\mathcal{L}[\text{input}]}\]
with zero initial conditions.
1Key Properties
1Standard Transfer Functions
1First-Order System
\[G(s) = \frac{K}{\tau s + 1}\]
where \(K\) is DC gain and \(\tau\) is time constant.
1Second-Order System
\[G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\]
where \(\omega_n\) is natural frequency and \(\zeta\) is damping ratio.
1GATE Tip

For GATE problems, quickly identify system type by denominator degree:

SEC 04

Block Diagrams

1Block Diagram Reduction Rules
1Series Connection
control_series_connection
1Parallel Connection
control_parallel_connection
1Feedback Connection
control_feedback_connection
1Sign Convention
1Complex Block Diagram Example
complex_example
1Equivalent Transfer Function
\[\frac{C(s)}{R(s)} = \frac{G_1 G_2 G_3}{1 + G_1 G_2 H_2 + G_2 G_3 H_1 + G_1 G_2 G_3 H_1 H_2}\]
SEC 05

Signal Flow Graphs

1Signal Flow Graph Terminology
1Basic Terms

  • Node: Junction point representing a variable

  • Branch: Directed line segment with gain

  • Path: Sequence of connected branches

  • Loop: Closed path

  • Forward Path: Path from input to output

  • Non-touching loops: Loops with no common nodes

1Example
signal_flow
1Mason’s Gain Formula
1Formula

The overall transfer function is:

\[T = \frac{1}{\Delta} \sum_{k=1}^{N} P_k \Delta_k\]
where:
1Graph Determinant
\[\Delta = 1 - \sum L_i + \sum L_i L_j - \sum L_i L_j L_k + \cdots\]
where \(L_i\) are individual loop gains, \(L_i L_j\) are products of non-touching loop pairs, etc.
1GATE Strategy

For complex systems, Mason’s formula is often faster than block diagram reduction.

1Mason’s Formula Example

Consider the signal flow graph:

signal_flow1

Solution:

SEC 06

Feedback Concepts

1Feedback System Analysis
1Closed-Loop System
closed_system
1Key Equations
1Important Relations
1Advantages and Disadvantages of Feedback
1Advantages

  • Reduced sensitivity to parameter variations

  • Improved stability and transient response

  • Reduced effect of noise and disturbances

  • Improved accuracy in steady-state

  • Bandwidth modification possible

1Disadvantages

  • Reduced overall gain

  • Potential instability if poorly designed

  • Increased complexity and cost

  • May introduce noise through feedback path

1Sensitivity Analysis

Sensitivity of closed-loop system to open-loop gain variations:

\[S_G^T = \frac{1}{1 + GH} \text{ vs. } S_G^G = 1 \text{ (open-loop)}\]
Feedback reduces sensitivity by factor \((1 + GH)\).
SEC 07

System Types

1System Types and Error Constants
1System Type Classification

Based on number of poles at origin in \(G(s)H(s)\):

\[G(s)H(s) = \frac{K(s + z_1)(s + z_2)\cdots}{s^N(s + p_1)(s + p_2)\cdots}\]
System is Type \(N\) (N = 0, 1, 2, ...).
1Error Constants
1Steady-State Errors
1Error Constants Table
Steady-State Error Constants
System Type \(K_p\) \(K_v\) \(K_a\)
Type 0 \(K\) \(0\) \(0\)
Type 1 \(\infty\) \(K\) \(0\)
Type 2 \(\infty\) \(\infty\) \(K\)
Steady-State Errors
Input Type 0 Type 1 Type 2
Step \(\frac{1}{1+K_p}\) \(0\) \(0\)
Ramp \(\infty\) \(\frac{1}{K_v}\) \(0\)
Parabolic \(\infty\) \(\infty\) \(\frac{1}{K_a}\)
SEC 08

GATE Practice

1GATE Practice Questions - 1
1Example

The transfer function of a system is \(G(s) = \frac{10}{s^2 + 3s + 2}\). The DC gain is:

  1. 0

  2. 5

  3. 10

  4. \(\frac{10}{2} = 5\)

1Solution

DC gain is found by substituting \(s = 0\):

\[G(0) = \frac{10}{0^2 + 0 + 2} = \frac{10}{2} = 5\]
Answer: B
1GATE Practice Questions - 2
1Example

For a unity feedback system with \(G(s) = \frac{4}{s(s+2)}\), the closed-loop transfer function is:

  1. \(\frac{4}{s^2 + 2s + 4}\)

  2. \(\frac{4}{s^2 + 2s}\)

  3. \(\frac{2}{s^2 + 2s + 4}\)

  4. \(\frac{4}{s + 2}\)

1Solution

For unity feedback: \(H(s) = 1\)

\[T(s) = \frac{G(s)}{1 + G(s)} = \frac{\frac{4}{s(s+2)}}{1 + \frac{4}{s(s+2)}} = \frac{4}{s(s+2) + 4} = \frac{4}{s^2 + 2s + 4}\]
Answer: A
1GATE Practice Questions - 3
1Example

A Type 1 system has \(G(s)H(s) = \frac{20}{s(s+4)}\). The velocity error constant \(K_v\) is:

  1. 4

  2. 5

  3. 20

  4. \(\infty\)

1Solution

For Type 1 system:

\[K_v = \lim_{s \to 0} s \cdot G(s)H(s) = \lim_{s \to 0} s \cdot \frac{20}{s(s+4)} = \lim_{s \to 0} \frac{20}{s+4} = \frac{20}{4} = 5\]
Answer: B
1GATE Practice Questions - 4

The Mason’s gain formula for the system shown requires calculation of:

  1. Forward paths and loops only

  2. Forward paths, loops, and their cofactors

  3. Only the characteristic polynomial

  4. Transfer function directly

1Solution

Mason’s gain formula requires:

Answer: B

1GATE Practice Questions - 5
1Example

For the block diagram shown, if \(G_1 = 2\), \(G_2 = 3\), and \(H = 0.5\), the overall transfer function is:

block_example

  1. \(\frac{6}{1+3}\)

  2. \(\frac{6}{1+6}\)

  3. \(\frac{6}{4}\)

  4. \(\frac{2}{3}\)

1Solution
\[T(s) = \frac{G_1 G_2}{1 + G_1 G_2 H} = \frac{2 \times 3}{1 + 2 \times 3 \times 0.5} = \frac{6}{1 + 3} = \frac{6}{4} = 1.5\]
Answer: C
1GATE Practice Questions - 6
1Example

A second-order system has the transfer function:

\[G(s) = \frac{25}{s^2 + 4s + 25}\]
The natural frequency \(\omega_n\)and damping ratio \(\zeta\) are:

  1. \(\omega_n = 5\) rad/s, \(\zeta = 0.4\)

  2. \(\omega_n = 25\) rad/s, \(\zeta = 0.2\)

  3. \(\omega_n = 5\) rad/s, \(\zeta = 0.2\)

  4. \(\omega_n = 4\) rad/s, \(\zeta = 0.4\)

1Solution

Standard form: \(G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\)

Comparing: \(\omega_n^2 = 25 \Rightarrow \omega_n = 5\) rad/s

\(2\zeta\omega_n = 4 \Rightarrow \zeta = \frac{4}{2 \times 5} = 0.4\)

Answer: A

SEC 09

Summary

1Summary - Key Formulas
1Transfer Function
\[G(s) = \frac{Y(s)}{X(s)} \text{ (zero initial conditions)}\]
1Closed-Loop Transfer Function
\[T(s) = \frac{G(s)}{1 + G(s)H(s)} \text{ (negative feedback)}\]
1Mason’s Gain Formula
\[T = \frac{1}{\Delta} \sum_{k=1}^{N} P_k \Delta_k\]
1Error Constants
1Key Points for GATE EE
1Important Concepts
1Problem-Solving Strategy

  1. Identify system type and structure

  2. Choose appropriate analysis method (block diagram vs. Mason’s formula)

  3. Apply reduction rules systematically

  4. Verify results using alternative methods when possible

  5. Check units and limiting cases