Find Variable Range For Steady-State Response: Control Systems

by Kenji Nakamura 63 views

Hey everyone! Today, we're diving into a fascinating control systems problem involving steady-state response and how to figure out the range of variables that make our system behave the way we want. It's like being a detective, but instead of solving a mystery, we're solving equations! So, grab your thinking caps, and let's get started.

Understanding the Problem

So, here’s the deal. We have a transfer function, which is basically a mathematical way of describing how a system responds to different inputs. In our case, the transfer function G(s) looks like this:

G(s)=s(s+2)s2+(b+2)s+ab G(s) = \dfrac{s(s+2)}{s^2+(b+2)s+ab}

This equation might look a bit intimidating at first, but don't worry, we'll break it down. The s here represents the Laplace variable, which is a fancy way of dealing with time-domain functions in the frequency domain (more on that later!). The a and b are our mystery variables – the ones we need to find the range for.

Now, we're feeding this system an input u(t), which is a simple sine wave:

u(t)=sin⁑t u(t) = \sin{t}

Think of it like pushing a swing. The sine wave is the push, and the system's response is how the swing moves. What we care about is the steady-state response – what happens to the swing after we've been pushing it for a while, and it settles into a rhythm. We want to make sure that the maximum swing height, or the maximum amplitude of the output y(t), stays less than 1.

In control systems, the steady-state response is a critical concept. It describes the behavior of a system as time approaches infinity. Essentially, it tells us what the output of the system will be after all the initial transients have died down. For stable systems, the steady-state response is predictable and often periodic when subjected to periodic inputs like sine waves. For our problem, we're particularly interested in the amplitude of the steady-state response because we want to ensure it remains within a specified limit.

The transfer function G(s) plays a central role in determining the steady-state response. It mathematically represents the relationship between the input and output of a system in the Laplace domain. The Laplace transform is a powerful tool that allows us to convert differential equations (which often describe dynamic systems) into algebraic equations, making them easier to analyze. By analyzing the transfer function, we can understand how the system will respond to different types of inputs, including sinusoidal inputs. The denominator of the transfer function, in particular, is crucial because its roots (the poles of the system) determine the stability of the system. If the poles have negative real parts, the system is stable, meaning its response will eventually settle down to a steady-state. Otherwise, the system may be unstable, leading to unbounded responses.

Sinusoidal inputs, like our u(t) = sin(t), are commonly used in control systems analysis because they can reveal a lot about a system's frequency response. The frequency response describes how the system's output amplitude and phase change as the frequency of the input signal varies. When we input a sine wave, the system's steady-state output will also be a sine wave with the same frequency, but potentially different amplitude and phase. The ratio of the output amplitude to the input amplitude is known as the magnitude of the frequency response, and the difference in phase between the output and input is known as the phase shift. These characteristics are directly related to the transfer function evaluated at s = jω, where j is the imaginary unit and ω is the frequency of the sine wave. The magnitude and phase information is crucial for designing control systems that meet specific performance requirements, such as limiting the maximum amplitude of the steady-state response.

So, the challenge is this: how do we find the ranges of a and b that guarantee the maximum steady-state response of y(t) remains less than 1? This involves a blend of Laplace transforms, frequency response analysis, and a dash of algebraic manipulation. Let's dive deeper into the solution!

Laplace Transform and Frequency Response

First things first, let's talk about the Laplace transform. It's a mathematical tool that transforms functions of time into functions of a complex variable s. It's super useful in control systems because it turns differential equations into algebraic equations, which are much easier to handle. Think of it as translating a sentence from a complicated language into a simpler one.

In our case, we're interested in the frequency response of the system. This tells us how the system responds to different frequencies of sinusoidal inputs. To find the frequency response, we replace s in the transfer function G(s) with jΟ‰, where j is the imaginary unit (√-1) and Ο‰ is the frequency of our input signal. Our input is sin(t), which has a frequency of 1 rad/s (Ο‰ = 1). So, we're going to plug in s = j into G(s).

G(j)=j(j+2)j2+(b+2)j+ab G(j) = \dfrac{j(j+2)}{j^2+(b+2)j+ab}

Now, let's simplify this expression. Remember that jΒ² = -1. This gives us:

G(j)=j2+2jβˆ’1+(b+2)j+ab=βˆ’1+2j(abβˆ’1)+(b+2)j G(j) = \dfrac{j^2+2j}{-1+(b+2)j+ab} = \dfrac{-1+2j}{(ab-1)+(b+2)j}

The frequency response, G(j), is a complex number, and it tells us two important things: the magnitude and the phase shift of the output signal relative to the input signal. The magnitude, denoted as |G(j)|, represents the ratio of the output amplitude to the input amplitude. The phase shift represents the difference in phase between the output and input signals.

To find the magnitude, we calculate the absolute value of the complex number G(j). For a complex number z = x + jy, the magnitude is given by |z| = √(x² + y²). So, we need to find the magnitude of both the numerator and the denominator of G(j) and then divide them. The magnitude of the numerator is:

βˆ£βˆ’1+2j∣=(βˆ’1)2+22=5 |-1+2j| = \sqrt{(-1)^2 + 2^2} = \sqrt{5}

The magnitude of the denominator is:

∣(abβˆ’1)+(b+2)j∣=(abβˆ’1)2+(b+2)2 |(ab-1)+(b+2)j| = \sqrt{(ab-1)^2 + (b+2)^2}

Therefore, the magnitude of the frequency response is:

∣G(j)∣=5(abβˆ’1)2+(b+2)2 |G(j)| = \dfrac{\sqrt{5}}{\sqrt{(ab-1)^2 + (b+2)^2}}

Remember, we want the maximum steady-state response, which is the magnitude of G(j), to be less than 1. This gives us the inequality:

∣G(j)∣=5(abβˆ’1)2+(b+2)2<1 |G(j)| = \dfrac{\sqrt{5}}{\sqrt{(ab-1)^2 + (b+2)^2}} < 1

Now, it’s time to roll up our sleeves and solve this inequality to find the ranges of a and b.

Solving the Inequality

Alright, guys, let's tackle this inequality! We've got:

5(abβˆ’1)2+(b+2)2<1 \dfrac{\sqrt{5}}{\sqrt{(ab-1)^2 + (b+2)^2}} < 1

To make things easier, let's get rid of the square roots by squaring both sides. This gives us:

5(abβˆ’1)2+(b+2)2<1 \dfrac{5}{(ab-1)^2 + (b+2)^2} < 1

Now, we can multiply both sides by the denominator to get rid of the fraction:

5<(abβˆ’1)2+(b+2)2 5 < (ab-1)^2 + (b+2)^2

Let's expand the squares:

5<(a2b2βˆ’2ab+1)+(b2+4b+4) 5 < (a^2b^2 - 2ab + 1) + (b^2 + 4b + 4)

Now, let's rearrange the inequality to get everything on one side:

0<a2b2βˆ’2ab+1+b2+4b+4βˆ’5 0 < a^2b^2 - 2ab + 1 + b^2 + 4b + 4 - 5

Simplifying, we get:

0<a2b2βˆ’2ab+b2+4b 0 < a^2b^2 - 2ab + b^2 + 4b

This inequality looks a bit complex, but we can rewrite it to make it more manageable. Notice that the first three terms look like a squared term. We can rewrite the inequality as:

0<(abβˆ’1)2+b2+4bβˆ’4 0 < (ab - 1)^2 + b^2 + 4b - 4

Or,

(ab)2βˆ’2ab+b2+4b>0 (ab)^2 - 2ab + b^2 + 4b > 0

Now, let’s think about how to find the ranges of a and b that satisfy this inequality. This is where things get interesting. We need to consider different cases and potentially use graphical methods to visualize the solution space. One way to approach this is to try to complete the square for the terms involving b. Let’s try that:

We have:

(abβˆ’1)2+(b+2)2>5 (ab-1)^2 + (b+2)^2 > 5

Expanding and rearranging, we get:

a2b2βˆ’2ab+1+b2+4b+4>5 a^2b^2 - 2ab + 1 + b^2 + 4b + 4 > 5

a2b2βˆ’2ab+b2+4b>0 a^2b^2 - 2ab + b^2 + 4b > 0

Let's rewrite this as:

b(abβˆ’2a+b+4)>0 b(ab - 2a + b + 4) > 0

This form helps us see that the inequality holds when both b and the expression in the parentheses have the same sign. Now, we can analyze different cases.

Analyzing Cases and Finding Ranges

Okay, team, we've reached the point where we need to dissect this inequality and figure out the possible ranges for a and b. We've got:

b(abβˆ’2a+b+4)>0 b(ab - 2a + b + 4) > 0

This inequality holds if either both factors are positive or both factors are negative. So, let's consider these two cases.

Case 1: Both factors are positive

This means we have two inequalities:

  1. b > 0
  2. ab - 2a + b + 4 > 0

Let's rearrange the second inequality:

ab+b>2aβˆ’4 ab + b > 2a - 4

b(a+1)>2aβˆ’4 b(a + 1) > 2a - 4

Now, we need to consider sub-cases based on the value of (a + 1). If (a + 1) > 0 (i.e., a > -1), we can divide both sides by (a + 1) without changing the inequality sign:

b>2aβˆ’4a+1 b > \dfrac{2a - 4}{a + 1}

So, in this sub-case, we have a > -1 and b > max(0, (2a - 4) / (a + 1)). We need to analyze the expression (2a - 4) / (a + 1) further. If (a > 2), then (2a - 4) > 0, and the inequality is simply b > (2a - 4) / (a + 1). If (-1 < a < 2), then (2a - 4) < 0, and since b > 0, the inequality holds.

If (a + 1) < 0 (i.e., a < -1), we need to flip the inequality sign when dividing:

b<2aβˆ’4a+1 b < \dfrac{2a - 4}{a + 1}

However, since we also require b > 0, we need to find the range of a for which (2a - 4) / (a + 1) > 0. This occurs when both the numerator and denominator have the same sign. The numerator (2a - 4) is negative when a < 2, and the denominator (a + 1) is negative when a < -1. So, the fraction is positive when a < -1. Thus, in this sub-case, we have a < -1 and 0 < b < (2a - 4) / (a + 1).

Case 2: Both factors are negative

This means we have two inequalities:

  1. b < 0
  2. ab - 2a + b + 4 < 0

Let's rearrange the second inequality again:

b(a+1)<2aβˆ’4 b(a + 1) < 2a - 4

If (a + 1) > 0 (i.e., a > -1), we can divide both sides by (a + 1) without changing the inequality sign:

b<2aβˆ’4a+1 b < \dfrac{2a - 4}{a + 1}

Since we require b < 0, we need to find the range of a for which (2a - 4) / (a + 1) < 0. This occurs when the numerator and denominator have opposite signs. The numerator (2a - 4) is negative when a < 2, and the denominator (a + 1) is positive when a > -1. So, the fraction is negative when -1 < a < 2. Thus, in this sub-case, we have -1 < a < 2 and b < min(0, (2a - 4) / (a + 1)). Since (2a - 4) / (a + 1) is negative in this range, the inequality simplifies to b < (2a - 4) / (a + 1).

If (a + 1) < 0 (i.e., a < -1), we need to flip the inequality sign when dividing:

b>2aβˆ’4a+1 b > \dfrac{2a - 4}{a + 1}

However, since we require b < 0, there is no solution in this sub-case because we would need a value of b that is simultaneously greater than and less than zero.

Summarizing the Results

Alright, guys, we've done some serious algebraic maneuvering and case analysis! Let's summarize our findings. The ranges of a and b that satisfy the condition |G(j)| < 1 are:

  • Case 1 (Both positive):
    • a > -1 and b > max(0, (2a - 4) / (a + 1))
    • Specifically:
      • If a > 2, then b > (2a - 4) / (a + 1)
      • If -1 < a < 2, then b > 0
    • a < -1 and 0 < b < (2a - 4) / (a + 1)
  • Case 2 (Both negative):
    • -1 < a < 2 and b < (2a - 4) / (a + 1)

These ranges define the regions in the a-b plane where the maximum steady-state response of the system will be less than 1. To get a clearer picture, it's often helpful to plot these regions graphically. This would involve sketching the curves b = (2a - 4) / (a + 1) and analyzing the areas that satisfy the inequalities. This graphical representation would provide a visual guide for selecting appropriate values of a and b in a real-world control system design.

Practical Implications and Conclusion

So, what does all this mean in the real world? Well, imagine you're designing a control system for a robot arm. The transfer function G(s) might represent how the arm responds to motor commands. The variables a and b could be related to physical parameters like damping and stiffness in the arm's joints. You want to ensure the arm moves smoothly and doesn't overshoot its target (which would be like having a steady-state response greater than 1). By finding the ranges of a and b that satisfy our inequality, you're essentially designing a system that's stable and meets your performance requirements.

This type of analysis is crucial in many engineering applications, from designing aircraft autopilot systems to controlling chemical processes in a factory. Understanding the steady-state response and how it's affected by system parameters is a fundamental skill for any control systems engineer.

In conclusion, we've taken a deep dive into finding the range of variables based on the steady-state response of a system. We used Laplace transforms to analyze the frequency response, solved inequalities, and considered different cases. This problem highlights the power of mathematical tools in control systems engineering and their importance in designing stable and reliable systems. Keep practicing, keep exploring, and you'll become a control systems whiz in no time! You got this, guys!