Time-domain Analysis of Control System

Time-domain Analysis of Control System

Contents

Describe the Time-Response of Continuous Data Control System 1

Recall the Transient-State Response and the Steady-State Response 2

Recall the following Signals: Impulse Signal, Step Signal, Ramp signal, and Parabolic Signal 4

Compare Impulse, Step, Ramp, and Parabolic Signals 5

Describe the First Order Control System 6

Determine: i. Unit-Impulse Response of First-Order Control System ii. Unit-Step Response of First-Order Control System iii. Unit-Ramp Response of First-Order Control System 7

Describe the Second Order Control System 8

Determine: i. Unit Impulse Response of Second Order Control System ii. Unit-Step Response of Second Order Control System iii. Unit-Ramp Response of Second Order Control System 10

Describe the Steady-state Error Analysis 12

Determine for a Unit-Step Input Signal: Steady-state Error and Error Constant 13

Find the Steady-state Error for different Types of Transfer Function with Unit-Step Input Signal 16

Determine the following for a Unit-Ramp Input Signal: i. Steady-state Error and Error Constant ii. Steady-state Errors for different Types of Transfer Function 19

Determine the following for a Unit-Parabolic Input Signal: i. Steady-state Error and Error Constant ii. Steady-state Errors for different Types of Transfer Function 21

Describe the Time-Response of Continuous Data Control System

The time-response of a continuous data control system is the behavior of the system as it responds to a change in the input signal over time. The response of the system is dependent on the characteristics of the system’s components, such as the input and output signals, transfer function, and time constants. Understanding the time-response of a control system is important in analyzing its stability, accuracy, and efficiency.

Example:

Consider a continuous data control system that regulates the temperature of a room using a heating and cooling system. The system has a temperature sensor that measures the room temperature and a controller that adjusts the heating and cooling system based on the temperature measurement.

The time-response of the system can be analyzed by applying a step function to the input signal, such as increasing the desired temperature setpoint from 20°C to 25°C. The response of the system can then be observed and analyzed.

The response of the system can be broken down into four phases:

  1. Delay phase: This is the time it takes for the system to begin responding to the input signal. In this phase, the temperature of the room remains constant as the system takes time to detect and respond to the change in the input signal.
  2. Rise phase: This is the time it takes for the temperature of the room to increase from its initial value to its final value. In this phase, the heating system is activated to raise the temperature of the room to the new setpoint.
  3. Overshoot phase: This is the time it takes for the temperature of the room to overshoot the setpoint and then return to the setpoint. In this phase, the heating system continues to operate even after the temperature reaches the setpoint, causing the temperature to overshoot the setpoint before returning to the setpoint.
  4. Settling phase: This is the time it takes for the temperature of the room to settle at the setpoint. In this phase, the heating and cooling system operates to maintain the temperature of the room at the setpoint.

The time-response of the system can be further analyzed by measuring its performance characteristics, such as the rise time, settling time, and overshoot. The rise time is the time it takes for the system to reach 90% of the final value, while the settling time is the time it takes for the system to settle within a specified range of the setpoint. The overshoot is the amount by which the system’s response exceeds the setpoint before settling at the setpoint.

By analyzing the time-response of the system, it is possible to identify areas where the system can be improved, such as by adjusting the controller’s parameters or modifying the system’s components to reduce overshoot or settling time. Additionally, understanding the time-response of the system is important in ensuring the system’s stability and reliability in regulating the room temperature.

Recall the Transient-State Response and the Steady-State Response

The transient-state response is the behavior of a system in response to an input during the time it takes for the system to reach a steady-state. During this time, the system may experience oscillations, overshoot, or undershoot before finally settling down to a stable state. The transient-state response is characterized by the following parameters:

  1. Rise Time: The time taken for the system output to rise from 10% to 90% of its steady-state value.
  2. Overshoot: The maximum deviation of the system output from its steady-state value.
  3. Settling Time: The time taken for the system output to reach and stay within a specified error bound of its steady-state value.
  4. Peak Time: The time taken for the system output to reach its peak value.

Example: Consider a simple RC circuit consisting of a resistor R and a capacitor C. When a step input voltage is applied to the circuit, the voltage across the capacitor initially starts to increase exponentially with a time constant of RC. As time progresses, the voltage across the capacitor approaches the steady-state value, but it never quite reaches it. The time taken for the capacitor voltage to settle within a specified error bound of its steady-state value is the settling time of the circuit. The rise time of the circuit is the time taken for the capacitor voltage to rise from 10% to 90% of its steady-state value.

The steady-state response is the behavior of a system after it has reached a stable state in response to a given input. In steady-state, the system output remains constant and does not change with time, assuming the input remains constant. The steady-state response is characterized by the following parameters:

  1. Steady-state Error: The difference between the actual output and the desired output at steady-state.
  2. Gain Margin: The amount by which the gain of the system can be increased before it becomes unstable.
  3. Phase Margin: The amount by which the phase of the system can be increased before it becomes unstable.

Example: Consider a feedback control system that uses a proportional controller to regulate the temperature of a furnace. The input to the system is the desired temperature setpoint, and the output is the actual temperature of the furnace. In steady-state, the actual temperature of the furnace will approach the desired setpoint, but it may not necessarily be equal to it due to factors such as measurement noise and controller error. The steady-state error is the difference between the desired setpoint and the actual temperature of the furnace at steady-state. The gain margin and phase margin of the system are measures of its stability, indicating the amount of additional gain or phase lag that can be added to the system before it becomes unstable. A system with a high gain margin and phase margin is more stable than a system with a low gain margin and phase margin.

Recall the following Signals: Impulse Signal, Step Signal, Ramp signal, and Parabolic Signal

  1. Impulse Signal: An impulse signal is a signal that has an instantaneous magnitude at a specific time and zero magnitude elsewhere. It is represented by the Dirac delta function, which is zero for all time except at time t=0, where it is infinite. The impulse signal is useful in modelling systems that respond to a sudden change or impact.

Example: When a ball hits a wall, it experiences an impulse force that is applied for a very short time. The impulse force causes the ball’s momentum to change, resulting in a change in its velocity.

  1. Step Signal: A step signal is a signal that has a constant magnitude for all time after a specific time, and zero magnitude before that time. It is represented by the Heaviside step function, which is zero for all time before a specified time t0, and one for all time after t0. The step signal is useful in modelling systems that respond to a sudden change in input or initial conditions.

Example: When a light switch is turned on, the voltage across the light bulb suddenly increases from zero to a constant value, causing the bulb to emit light.

  1. Ramp Signal: A ramp signal is a signal that increases linearly with time. It has a non-zero slope and a zero intercept. The ramp signal is useful in modelling systems that have a gradual increase or decrease in their response.

Example: When a car accelerates, its speed increases gradually over time, creating a ramp signal that represents the increase in speed.

  1. Parabolic Signal: A parabolic signal is a signal that increases or decreases quadratically with time. It has a non-zero second derivative and a zero first derivative at t=0. The parabolic signal is useful in modelling systems that have a smooth change in their response.

Example: The displacement of a particle undergoing constant acceleration is given by a parabolic signal, as it increases or decreases quadratically with time.

Compare Impulse, Step, Ramp, and Parabolic Signals

Impulse, Step, Ramp, and Parabolic signals are all commonly used signals in signal processing and control systems. Here are some comparisons between them:

  1. Shape and Amplitude: The impulse signal has an instantaneous magnitude at a specific time and zero magnitude elsewhere. The step signal has a constant magnitude after a specific time and zero magnitude before that time. The ramp signal increases linearly with time, while the parabolic signal increases or decreases quadratically with time.
  2. Derivatives: The impulse signal has an infinite derivative at t=0, while the step signal has a derivative that is zero for all time except at t=0, where it is infinite. The ramp signal has a non-zero first derivative, while the parabolic signal has a non-zero second derivative.
  3. Usefulness: The impulse signal is useful in modelling systems that respond to a sudden change or impact, such as a ball hitting a wall. The step signal is useful in modelling systems that respond to a sudden change in input or initial conditions, such as turning on a light switch. The ramp signal is useful in modelling systems that have a gradual increase or decrease in their response, such as the acceleration of a car. The parabolic signal is useful in modelling systems that have a smooth change in their response, such as the displacement of a particle undergoing constant acceleration.
  4. Mathematical Representations: The impulse signal is represented by the Dirac delta function, the step signal is represented by the Heaviside step function, the ramp signal is represented by a linear function, and the parabolic signal is represented by a quadratic function.

In summary, impulse, step, ramp, and parabolic signals are all useful in different contexts and have different mathematical representations. Understanding their differences can help in choosing the appropriate signal for a given application.

Describe the First Order Control System

A first-order control system is a type of control system that has a single integrator or differentiation element in its transfer function. It is characterized by a first-order differential equation in its time-domain representation. First-order control systems are commonly used in a wide range of applications, including industrial control, process control, and motion control.

The transfer function of a first-order control system can be represented as:

G(s) = K / (T*s + 1)

Where K is the gain of the system and T is the time constant. The time constant T is the time taken for the system to reach 63.2% of its steady-state value when subjected to a step input.

The response of a first-order control system can be divided into two parts: the transient response and the steady-state response. The transient response is the response of the system to a sudden change in the input, while the steady-state response is the response of the system after it has reached a stable state.

The behavior of a first-order control system can be understood by analyzing its step response. When a step input is applied to the system, the output initially starts to rise rapidly, but then settles down to a steady-state value over time. The time taken for the system to settle down to the steady-state value depends on the time constant T. A larger time constant indicates slower response time, while a smaller time constant indicates faster response time.

Example: An example of a first-order control system is a simple temperature control system, where the temperature of a room is controlled by turning on and off a heater. The temperature sensor provides feedback to the controller, which adjusts the heater output to maintain the desired temperature. The transfer function of the control system can be represented as a first-order system with a time constant that depends on the thermal properties of the room and the heater. If the time constant is too large, the temperature control system may take too long to reach the desired temperature, while a time constant that is too small may result in overshoot and instability.

Determine: i. Unit-Impulse Response of First-Order Control System ii. Unit-Step Response of First-Order Control System iii. Unit-Ramp Response of First-Order Control System

i. Unit-Impulse Response of First-Order Control System:

The unit impulse response of a first-order control system can be found by taking the inverse Laplace transform of its transfer function. The Laplace transform of the unit impulse function is equal to 1, so the inverse Laplace transform of the transfer function gives the impulse response of the system.

For a first-order control system with transfer function G(s) = K / (T*s + 1), the unit impulse response can be found as:

g(t) = (K/T) * e(-t/T) * u(t)

Where u(t) is the unit step function.

The unit impulse response describes how the system responds to an impulse input. Since the unit impulse function has infinite amplitude and zero duration, the impulse response of the system is a decaying exponential function with a time constant of T.

ii. Unit-Step Response of First-Order Control System:

The unit step response of a first-order control system can also be found by taking the inverse Laplace transform of its transfer function. The Laplace transform of the unit step function is equal to 1/s, so the inverse Laplace transform of the transfer function divided by s gives the step response of the system.

For a first-order control system with transfer function G(s) = K / (T*s + 1), the unit step response can be found as:

g(t) = (K/T) * (1 – e(-t/T)) * u(t)

The unit step response describes how the system responds to a step input. At t=0, the output of the system jumps up by a value of K/T, and then increases exponentially to its final value with a time constant of T.

iii. Unit-Ramp Response of First-Order Control System:

The unit ramp response of a first-order control system can be found by differentiating the unit step response of the system. Since the ramp function is the integral of the step function, the derivative of the step response gives the ramp response.

For a first-order control system with transfer function G(s) = K / (T*s + 1), the unit ramp response can be found as:

g(t) = K * (1 – e(-t/T)) * t * u(t)

The unit ramp response describes how the system responds to a ramp input. At t=0, the output of the system starts at zero and then increases linearly with time. The slope of the ramp response is proportional to the gain K of the system, and the time constant T determines how quickly the response increases.

Describe the Second Order Control System

A second-order control system is a system that has two poles in its transfer function. The transfer function of a second-order system can be written in the general form:

G(s) = (K / (s2 + 2ζωn s + ωn2))

where K is the gain of the system, ωn is the natural frequency of the system, and ζ is the damping ratio. The natural frequency is a measure of how fast the system oscillates when it is excited by a sinusoidal input, while the damping ratio is a measure of how quickly the system’s oscillations decay over time.

The poles of a second-order system can be found by solving the characteristic equation:

s2 + 2ζωn s + ωn2 = 0

The roots of this equation are complex conjugate pairs, given by:

s = -ζωn ± ωn√(ζ2 – 1)

The behavior of a second-order system depends on the values of the natural frequency and damping ratio. There are three different types of responses that a second-order system can exhibit, which are:

  1. Underdamped response: If 0 < ζ < 1, the system is said to be underdamped. In this case, the system will oscillate at the natural frequency with a decaying amplitude. The response of an underdamped system is characterized by a set of oscillations, which are also known as damped sinusoids. The number of oscillations and the decay rate of the amplitude depend on the damping ratio. A typical example of an underdamped system is a mass-spring-damper system.
  2. Overdamped response: If ζ > 1, the system is said to be overdamped. In this case, the system will not oscillate, but will instead return to its steady-state value in a smooth and monotonic fashion. The response of an overdamped system is characterized by a slower rise time and a slower settling time than an underdamped system.
  3. Critically damped response: If ζ = 1, the system is said to be critically damped. In this case, the system will return to its steady-state value as quickly as possible without oscillating. The response of a critically damped system is characterized by a fast rise time and a fast settling time.

Overall, second-order control systems are used in many engineering applications, such as control systems for robots, aircraft, and automobiles. The natural frequency and damping ratio can be adjusted to control the response of the system and ensure that it behaves as desired.

Determine: i. Unit Impulse Response of Second Order Control System ii. Unit-Step Response of Second Order Control System iii. Unit-Ramp Response of Second Order Control System

i. Unit Impulse Response of Second Order Control System:

The unit impulse response of a second-order control system can be found by taking the inverse Laplace transform of the transfer function of the system, which is given by:

G(s) = (K / (s2 + 2ζωn s + ωn2))

Taking the inverse Laplace transform of G(s), we get:

g(t) = K/ωn e(-ζωn t) sin(ωn√(1 – ζ2) t) u(t)

where u(t) is the unit step function.

The impulse response of a second-order system is a decaying sinusoid with a frequency of ωn√(1 – ζ2) and a decay rate of ζωn.

ii. Unit-Step Response of Second Order Control System:

The unit-step response of a second-order control system can be found by taking the inverse Laplace transform of the transfer function of the system, which is given by:

G(s) = (K / (s2 + 2 n s + ωn2))

The unit-step response is defined as the output of the system when the input is a unit step function, i.e., U(s) = 1/s. Substituting U(s) into the transfer function, we get:

Y(s) = G(s) U(s)

Y(s) = K / s (s2 + 2ζωn s + ωn2)

Using partial fraction expansion and inverse Laplace transform, we can find the unit-step response of the system. The unit-step response of a second-order system can have three different forms depending on the value of the damping ratio:

  1. Underdamped response: If 0 < ζ < 1, the unit-step response of the system is given by:

y(t) = K(1 – e(-ζωn t) (cos(ωn√(1 – ζ2) t) + (ζ/√(1 – ζ2))sin(ωn√(1 – ζ2) t)))

The underdamped response exhibits oscillations, and the amplitude of the oscillations decays over time. The settling time of the system is the time it takes for the output to reach and stay within a specified range around the final value.

  1. Overdamped response: If ζ > 1, the unit-step response of the system is given by:

y(t) = K(1 – (1/√(ζ2 – 1)) e(-ωnζt) sin(ωn√(ζ2 – 1) t + φ))

The overdamped response does not exhibit oscillations, and the output reaches the final value in a smooth and monotonic fashion. The settling time of the system is slower than that of an underdamped system.

  1. Critically damped response: If ζ = 1, the unit-step response of the system is given by:

y(t) = K(1 – e(-ωn t)(1 + ωn t))

The critically damped response has the fastest settling time, and the output reaches the final value as quickly as possible without oscillating.

Describe the Steady-state Error Analysis

Steady-state error analysis is a method used to analyze the response of a control system to a step input or a ramp input. It is the error that exists in the output of a control system after the transient response has subsided and the output has settled down to a constant value. The steady-state error analysis is an important concept in control systems as it helps to analyze the performance of a control system under different input signals.

There are three types of steady-state errors that can occur in a control system: zero steady-state error, finite steady-state error, and infinite steady-state error.

  1. Zero Steady-State Error:

A zero steady-state error occurs when the output of the control system reaches the desired value without any error. This means that the system has perfect tracking of the input signal. This type of error is possible when the system has a type 0 transfer function.

For example, consider a system that is required to track the position of a robot arm. If the system has a type 0 transfer function, it will have zero steady-state error, and the arm will be able to track the desired position without any error.

  1. Finite Steady-State Error:

A finite steady-state error occurs when the output of the control system does not reach the desired value but instead settles at a constant value that is different from the desired value. This type of error is possible when the system has a type 1 transfer function.

For example, consider a system that is required to maintain the temperature of a room. If the system has a type 1 transfer function, it may have a finite steady-state error. The temperature of the room may settle at a value that is different from the desired temperature.

  1. Infinite Steady-State Error:

An infinite steady-state error occurs when the output of the control system does not reach the desired value and continues to increase or decrease indefinitely. This type of error is possible when the system has a type 2 transfer function.

For example, consider a system that is required to control the speed of a motor. If the system has a type 2 transfer function, it may have an infinite steady-state error. The speed of the motor may continue to increase or decrease indefinitely without reaching the desired value.

In conclusion, the steady-state error analysis is an important concept in control systems that helps to analyze the performance of a control system under different input signals. It is important to identify the type of steady-state error in a system to determine the performance and stability of the control system.

Determine for a Unit-Step Input Signal: Steady-state Error and Error Constant

In control systems, a unit-step input signal is a common type of input signal used to analyze the performance of a control system. It is a signal that starts from zero and increases to a value of one at time t=0. By analyzing the response of a control system to a unit-step input signal, we can determine its steady-state error and error constant.

Steady-state Error:

Steady-state error is the difference between the desired output and the actual output of a control system when it has reached a steady-state condition. In other words, it is the error that exists in the output of a control system after the transient response has subsided and the output has settled down to a constant value.

For a unit-step input signal, the steady-state error can be determined by taking the limit of the error as time approaches infinity. The steady-state error can be calculated using the following equation:

Ess = lim (s->0) sE(s)

where Ess is the steady-state error, s is the Laplace variable, and E(s) is the Laplace transform of the error signal.

Error Constant:

The error constant is a measure of the ability of a control system to track a unit-step input signal. It is defined as the ratio of the steady-state error to the magnitude of the input signal.

For a unit-step input signal, the error constant can be calculated using the following equation:

Kp = lim (s->0) G(s)

where Kp is the error constant and G(s) is the transfer function of the control system.

The error constant is important in control systems as it determines the ability of the system to track different input signals. A higher error constant indicates a better tracking performance of the control system.

Example:

Consider a control system with the following transfer function:

G(s) = 2/(s+1)

To determine the steady-state error and error constant of the system for a unit-step input signal, we can follow the steps below:

  1. Find the Laplace transform of the unit-step input signal:

U(s) = 1/s

  1. Find the output of the control system:

Y(s) = G(s)U(s) = 2/(s+1) * 1/s = 2/(s(s+1))

  1. Find the Laplace transform of the error signal:

E(s) = 1 – Y(s) = 1 – 2/(s(s+1)) = s/(s(s+1)) = 1/(s+1)

  1. Calculate the steady-state error:

Ess = lim (s->0) sE(s) = lim (s->0) s/(s+1) = 0

Therefore, the system has zero steady-state error.

  1. Calculate the error constant:

Kp = lim (s->0) G(s) = lim (s->0) 2/(s+1) = 2

Therefore, the system has an error constant of 2.

In conclusion, by analyzing the response of a control system to a unit-step input signal, we can determine its steady-state error and error constant. The steady-state error is the difference between the desired output and the actual output of a control system when it has reached a steady-state condition. The error constant is a measure of the ability of a control system to track a unit-step input signal.

Find the Steady-state Error for different Types of Transfer Function with Unit-Step Input Signal

In control systems, the steady-state error is an important measure of the accuracy of a control system’s response to an input signal. It is the difference between the desired output and the actual output of the system when it has reached a steady-state condition. The steady-state error depends on the transfer function of the system, the input signal, and the type of control system. In this learning outcome, we will focus on how to find the steady-state error for different types of transfer functions with a unit-step input signal.

  1. Type 0 System:

A Type 0 system is a control system where the transfer function has no poles at the origin. It means that the system has no inherent integrator and cannot eliminate the steady-state error caused by a step input signal. The steady-state error for a Type 0 system with a unit-step input signal can be found using the following equation:

Ess = 1/Kp

where Kp is the error constant, which can be found by taking the limit of the transfer function as s approaches zero.

Example:

Consider a Type 0 system with the following transfer function:

G(s) = 5/(s+3)

To find the steady-state error for a unit-step input signal, we can follow the steps below:

  1. Calculate the error constant:

Kp = lim (s->0) G(s) = lim (s->0) 5/(s+3) = 5/3

  1. Calculate the steady-state error:

Ess = 1/Kp = ⅗

Therefore, the steady-state error for this Type 0 system with a unit-step input signal is 3/5.

  1. Type 1 System:

A Type 1 system is a control system where the transfer function has one pole at the origin. It means that the system has an inherent integrator and can eliminate the steady-state error caused by a step input signal. The steady-state error for a Type 1 system with a unit-step input signal can be found using the following equation:

Ess = 1/Kv

where Kv is the velocity error constant, which can be found by taking the limit of the transfer function multiplied by s as s approaches zero.

Example:

Consider a Type 1 system with the following transfer function:

G(s) = 4/s

To find the steady-state error for a unit-step input signal, we can follow the steps below:

  1. Calculate the velocity error constant:

Kv = lim (s->0) sG(s) = lim (s->0) 4 = 4

  1. Calculate the steady-state error:

Ess = 1/Kv = 1/4

Therefore, the steady-state error for this Type 1 system with a unit-step input signal is 1/4.

  1. Type 2 System:

A Type 2 system is a control system where the transfer function has two poles at the origin. It means that the system has two inherent integrators and can eliminate the steady-state error caused by a step input signal, and can achieve zero steady-state error for a ramp input signal. The steady-state error for a Type 2 system with a unit-step input signal can be found using the following equation:

Ess = 1/Ka

where Ka is the acceleration error constant, which can be found by taking the limit of the transfer function multiplied by s2 as s approaches zero.

Determine the following for a Unit-Ramp Input Signal: i. Steady-state Error and Error Constant ii. Steady-state Errors for different Types of Transfer Function

i. Steady-state Error and Error Constant for Unit-Ramp Input Signal:

The steady-state error is the difference between the actual output of a system and the desired output of the system when the input signal has been applied for a long time, and the system has reached a stable condition. The error constant is the ratio of the steady-state error to the magnitude of the input signal. The steady-state error and error constant are important parameters in the analysis of control systems.

For a unit-ramp input signal, the steady-state error is given by the following equation:

Errorss = 1 / (1 + Kp), where Kp is the position error constant.

The position error constant is a measure of how well the system can track a constant position input signal. The error constant is a measure of how well the system can track a unit ramp input signal.

For example, consider a closed-loop control system with a transfer function given by:

G(s) = K / (s + 1)2

The steady-state error for a unit-ramp input signal is given by:

Errorss = 1 / (1 + Kp) = 1 / (1 + lim(s->0) G(s)) = 1 / (1 + K)

Where K is the gain of the system. The error constant for a unit-ramp input signal is therefore Kp = lim(s->0) G(s) = K.

ii. Steady-state Errors for different Types of Transfer Function:

The steady-state error for a unit-ramp input signal depends on the type of transfer function of the control system. There are three types of transfer functions: Type 0, Type 1, and Type 2.

Type 0 systems have a transfer function that does not have any poles at the origin (i.e., s=0). The steady-state error for a unit-ramp input signal is finite for Type 0 systems. For example, consider a Type 0 system with the transfer function given by:

G(s) = K / s(s + 1)

The steady-state error for a unit-ramp input signal is given by:

Errorss = 1 / (1 + Kp) = 1 / (1 + lim(s->0) G(s)) = 1 / K

Type 1 systems have a transfer function that has one pole at the origin. The steady-state error for a unit-ramp input signal is zero for Type 1 systems. For example, consider a Type 1 system with the transfer function given by:

G(s) = K / s

The steady-state error for a unit-ramp input signal is given by:

Errorss = 1 / (1 + Kp) = 1 / (1 + lim(s->0) G(s)) = 0

Type 2 systems have a transfer function that has two poles at the origin. The steady-state error for a unit-ramp input signal is infinite for Type 2 systems. For example, consider a Type 2 system with the transfer function given by:

G(s) = K / s2

The steady-state error for a unit-ramp input signal is given by:

Errorss = 1 / (1 + Kp) = 1 / (1 + lim(s->0) G(s)) = infinity

In summary, the steady-state error and error constant for a unit-ramp input signal depend on the type of transfer function of the control system.

Determine the following for a Unit-Parabolic Input Signal: i. Steady-state Error and Error Constant ii. Steady-state Errors for different Types of Transfer Function

i. Steady-state Error and Error Constant for Unit-Parabolic Input Signal:

The steady-state error is the difference between the actual output of a system and the desired output of the system when the input signal has been applied for a long time, and the system has reached a stable condition. The error constant is the ratio of the steady-state error to the magnitude of the input signal. The steady-state error and error constant are important parameters in the analysis of control systems.

For a unit-parabolic input signal, the steady-state error is given by the following equation:

Errorss = 1 / (1 + Kv), where Kv is the velocity error constant.

The velocity error constant is a measure of how well the system can track a constant velocity input signal. The error constant is a measure of how well the system can track a unit parabolic input signal.

For example, consider a closed-loop control system with a transfer function given by:

G(s) = K / (s + 1)3

The steady-state error for a unit-parabolic input signal is given by:

Errorss = 1 / (1 + Kv) = 1 / (1 + lim(s->0) sG(s)) = 1 / K

Where K is the gain of the system. The error constant for a unit-parabolic input signal is therefore Kv = lim(s->0) sG(s) = 0.

ii. Steady-state Errors for different Types of Transfer Function:

The steady-state error for a unit-parabolic input signal depends on the type of transfer function of the control system. There are three types of transfer functions: Type 0, Type 1, and Type 2.

Type 0 systems have a transfer function that does not have any poles at the origin (i.e., s=0). The steady-state error for a unit-parabolic input signal is finite for Type 0 systems. For example, consider a Type 0 system with the transfer function given by:

G(s) = K / s(s + 1)2

The steady-state error for a unit-parabolic input signal is given by:

Errorss = 1 / (1 + Kv) = 1 / (1 + lim(s->0) sG(s)) = 1 / K

Type 1 systems have a transfer function that has one pole at the origin. The steady-state error for a unit-parabolic input signal is also finite for Type 1 systems. For example, consider a Type 1 system with the transfer function given by:

G(s) = K / s2

The steady-state error for a unit-parabolic input signal is given by:

Errorss = 1 / (1 + Kv) = 1 / (1 + lim(s->0) sG(s)) = 1 / K

Type 2 systems have a transfer function that has two poles at the origin. The steady-state error for a unit-parabolic input signal is zero for Type 2 systems. For example, consider a Type 2 system with the transfer function given by:

G(s) = K / s3

The steady-state error for a unit-parabolic input signal is given by:

Errorss = 1 / (1 + Kv) = 1 / (1 + lim(s->0) sG(s)) = 0

In summary, the steady-state error and error constant for a unit-parabolic input signal depend on the type of transfer function of the control system.