Bode Plot Slope based on frequencies

In summary, the Bode plot slope is a graphical representation that illustrates how the gain and phase of a system vary with frequency. The slope of the magnitude plot in a Bode diagram is typically expressed in decibels per decade, where each order of magnitude change in frequency corresponds to a specific slope based on the system's poles and zeros. For example, a first-order low-pass filter exhibits a slope of -20 dB/decade, while a first-order high-pass filter shows a slope of +20 dB/decade. The overall slope is determined by the net effect of the system's frequency response characteristics, making Bode plots a valuable tool for analyzing and designing control systems.
  • #1
guyvsdcsniper
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Homework Statement
Give numerical values of ω that define the ranges where ω can be written as a power law and give the Bode-plot slope in each range, using decibels per dec or oct
Relevant Equations
db=log(v2/v1)
From 0 to ##10^3## ##\omega## there is a dB gain, from ##10^3## to ##10^5## there is another. Finally from ##10^5## to infinity the slope is constant (0).

I know the formula
$$dbV= 20log_{10}\frac{V_2}{V_1}$$

can give me the slope but that is in terms of Volts, but I have frequency and the magnitude of the transfer function. I cant find a formula in my book or online to calculate the dB gain with these two.

Is there an approach to determining the slope of each with the information given in the image?

Screenshot 2023-09-25 at 12.16.55 PM.png
 
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  • #2
Have you learned the concepts of "poles" and "zeros" yet, and how they affect the transfer function?
 
  • #3
berkeman said:
Have you learned the concepts of "poles" and "zeros" yet, and how they affect the transfer function?
I have, each pole corresponds a 6dB/oct gain.

So i'm guessing I would need to take the magnitude of the transfer function and evaluate in the limit of low mid and high frequencies?

I calculated the transfer function to be:
$$H = \frac{-\omega^2 L C}{j\omega RC-\omega^2 LC+1}$$
and then found its magnitude:
$$\lvert{H}\rvert = \frac{\omega^2 LC}{\sqrt{(\omega^4 LC+(\omega LC)^2-2\omega^2 LC+1}}$$

Am I going to have to take the quadratic equation of the denominator to find the poles?
 
  • #4
guyvsdcsniper said:
Am I going to have to take the quadratic equation of the denominator to find the poles?
Basically yes. But since they haven't given you the component values, the real task here is to recognize the various shapes a quadratic frequency response can take and identify where the poles are from the given plot. Look in your lecture notes about constructing a quadratic bode plot using asymptotes.

Hint: Quadratics always have two poles, identifying whether they are real or complex is an important first step in analysis. If they are complex, then you want to get an idea about the damping (or quality factor Q) to proceed. If they are real, you can treat it as two separate 1st order responses combined together.

This article may be helpful, although it doesn't show the plots.
https://www.physicsforums.com/threads/an-engineers-approach-to-the-quadratic-formula.1053797/
 
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FAQ: Bode Plot Slope based on frequencies

What is a Bode plot?

A Bode plot is a graphical representation of a linear, time-invariant system transfer function, typically used in control system and signal processing. It consists of two plots: one showing the magnitude (in dB) versus frequency (on a logarithmic scale) and the other showing the phase angle versus frequency.

How is the slope of the magnitude plot in a Bode plot determined?

The slope of the magnitude plot in a Bode plot is determined by the poles and zeros of the transfer function. Each pole contributes a slope of -20 dB/decade, and each zero contributes a slope of +20 dB/decade. The overall slope is the sum of the contributions from all poles and zeros.

Why does the slope of the Bode plot change at certain frequencies?

The slope of the Bode plot changes at certain frequencies because these frequencies correspond to the locations of poles and zeros in the transfer function. When the frequency approaches a pole, the slope decreases by 20 dB/decade, and when it approaches a zero, the slope increases by 20 dB/decade.

How do you calculate the frequency at which the Bode plot slope changes?

The frequency at which the Bode plot slope changes is determined by the break frequencies, which are the frequencies at which the poles and zeros occur. These break frequencies can be calculated from the transfer function by finding the values of s (complex frequency) that make the denominator (for poles) or numerator (for zeros) equal to zero.

What is the significance of the slope in a Bode plot?

The slope in a Bode plot is significant because it provides insight into the system's behavior and stability. A steeper slope indicates a more rapid change in the system's response, which can affect the stability and performance of the system. Understanding the slope helps in designing controllers and filters to achieve desired system characteristics.

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