# Electrical Engineering - Control Systems - Design Via Frequency Respon

• GreenPrint
In summary, the conversation is about designing a lead compensator for a given transfer function to achieve closed-loop stability. The requirements for the compensator are a transient response of 16% overshoot and a settling time of 2 mS for a step input. The conversation includes various equations and attempts at solving the problem, such as finding the damping ratio, bandwidth frequency, and phase margin. The final solution involves using the bandwidth frequency to solve for the break frequencies and ultimately determining the value of β.
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## Homework Statement

I'm given the following transfer function that represents a physical apparatus

$G(s) = \frac{X(s)}{F(s)} = \frac{3.333*10^{4}}{s^{2}}$

I'm asked to design a lead compensator to achieve closed-loop stability. The requirements are that it must have a transient response of $16%$ overshoot, a settling time of $2 mS$ for a step input.

## Homework Equations

$ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}}$
$ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}}$
$Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}})$

## The Attempt at a Solution

I've spent some time trying to solve this problem. I've looked through my notes, my textbook, the internet and still am lost. I've looked at the solutions manual and it doesn't seem to help either. So I was hoping that someone could point me in the right direction.

I start off by finding the damping ratio since

$ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}} = -\frac{ln(\frac{16}{100})}{\sqrt{∏^{2}+ln^{2}(\frac{16}{100})}} ≈ 0.503$

I know find the bandwidth frequency $ω_{BW}$

$ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}} = \frac{4(10^{3} mS)}{(2 mS)(0.503) S}\sqrt{(1-2(0.503)^{2}) + \sqrt{4(0.503)^{4} - 4(0.503)^{2} +2}} ≈ 5,044.122 \frac{1}{S}$

Next I find the gain evaluated at $jω_{BW}$ $G(jω_{BW})$

$G(jω_{BW}) = \frac{3.333*10^{4}}{(j5,044.122 \frac{1}{S}} ≈ -0.131*10^{-2}$

Next I find the phase margin $Θ_{m}$

$Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}}) = arctan(\frac{2(0.503)}{\sqrt{-2(0.503)^{2} + \sqrt{1 + 4(0.503)^{4}}}}) ≈ 52.068°$

I'm not really sure were to go from here but all of the above work seems right and are all formulas that I can find in my book. The solutions manual does this next

$0.013K = \frac{1}{\sqrt{2}}$

and solves for $K$. I'm not sure where this formula comes from. The solutions manual says it's the compensators gain. But that doesn't really help me. The equation is clearly

$G(jω_{BW})K = ?$

I'm not really sure what the right side of the equation is.

Thanks for any help.

Last edited:
Whats the representation for lead compensator? Closed loop stability implies you're using feedback, and the stability condition is:

1 + G(s)*C(s) = 0, where C(s) is your compensator.

Edit: You'll also want to add a gain K in front of the compensator...so K doesn't just appear out of nowhere. Also for stability, the solution for K usually has a given range of possible values (like K between 0 and 5), so you solutions manual exact solution for K is odd to me.

Edit2: Should've been more clear, for the characteristic equation I gave you above, the roots of that equation are equivalent to the poles of the system and therefore must be in the left half plane for stability.

Last edited:
For a lead network $G_{c}(s) = \frac{1}{β}\frac{s + \frac{1}{T}}{s + \frac{1}{βT}}$ I'm still unsure where $\frac{1}{\sqrt{2}}$ came from though. There's a standard representation for the compensator $C(s)$?

Edit 1:

From the characteristic equation

$1 + kG(s)C(s) = 0$
$kG(s)C(s) = -1$
$kG(s) = - \frac{1}{C(s)}$

This is apparently what the solutions manual is doing? And apparently $C(s) = -\sqrt{2}$ But I'm not sure why it's $-\sqrt{2}$ had not had the solutions manual and not sure were that came from

Last edited:
its something like what you wrote: C(s) = (s + a)/(s + b), where conditions on a,b determine if its lead or lag. Your lead network form is more useful because there exists equations for T and beta.

Have you tried to solve for T,beta? Then its likely the 1/sqrt(2) falls out of the stability condition 1 + KGC = 0

My text says that we can find the frequency, $ω_{max}$ at which the maximum phase angle, $Θ_{max}$, occurs

$ω_{max} = \frac{1}{T\sqrt{β}}$

But this doesn't really seem to help since I don't know the frequency $ω_{max}$, unless it's the bandwidth frequency $ω_{BW}$, which I don't think it is.

My text also states that the maximum phase shift of the compensator $θ_{max}$ is

$θ_{max} = arctan(\frac{1 - β}{2\sqrt{β}}) = arcsin(\frac{1 - β}{1 + β})$

Which doesn't really help since I don't know what $θ_{max}$.

My text also states that the compensator's magnitude at $ω_{max}$ is

$|G_{c}(jω_{max})| = \frac{1}{\sqrt{β}}$

This equation looks strongly similar to where I got stuck. If this correct than $β = 2$. Which I'm not sure how was obtained exactly.

My gut says use the bandwidth frequency in this equation

$ω_{max} = \frac{1}{T\sqrt{β}}$
$Tω_{BW} = \frac{1}{\sqrt{B}} = T(5,044.122 \frac{1}{s}) = \frac{1}{\sqrt{β}}$

But I'm unsure what I'm supposed to use for $T$. Apparently if I'm not mistaken $\frac{1}{T}$ and $\frac{1}{βT}$ are the break frequencies which I don't know how to obtain without the Bode Plot

## 1. What is frequency response in electrical engineering?

Frequency response in electrical engineering refers to the measure of a system's output as a function of the input frequency. It is a way to analyze and understand how a system responds to different frequencies of signals.

## 2. How is frequency response used in control systems?

In control systems, frequency response is used to design and analyze the performance of a system. It helps engineers determine the stability, sensitivity, and bandwidth of a system, which are crucial factors in ensuring its proper functioning.

## 3. What is the difference between open-loop and closed-loop control systems?

The main difference between open-loop and closed-loop control systems is that open-loop systems do not have a feedback mechanism, while closed-loop systems do. In open-loop systems, the output is not compared to the desired input, while in closed-loop systems, the output is continuously monitored and adjusted based on the input.

## 4. What is the role of frequency domain analysis in control system design?

Frequency domain analysis is an essential tool in control system design as it allows engineers to analyze the behavior of a system at different frequencies. It helps identify potential issues and design solutions to improve the performance of the system.

## 5. What are some common applications of frequency response in electrical engineering?

Frequency response is used in a wide range of applications in electrical engineering, such as designing filters, amplifiers, and feedback control systems. It is also used in communication systems, audio equipment, and power systems to ensure efficient and reliable performance.

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