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Bode plot Take a typical 2nd order low pass filter amplitude response bode plot: - 3-D picture introducing pole-zero diagram Here's the bigger picture of that response when combined with a pole zero diagram: - Traditional pole zero diagram Looking down from above onto the 3-D picture shows the traditional pole zero diagram: - Pole zero geometry and |H(...


30

Gain and phase margin are usually applied to systems that are amplifiers of some sort with negative feedback around them. The more negative feedback, the tighter the system is controlled. However, you don't want to provide feedback in such a way that the system will oscillate. The gain and phase margin are two metrics to tell you how close the system is ...


13

Bode plots show gain and phase. Gain is measured in V/V, A/A, or W/W, and is thus dimensionless. Decibels are ratios, like percentages or parts per million. The question, then, is why we sometimes use logarithmic scales instead of linear ones for gain. (Phase is always linear!) One reason is that in graphs, linear scales really only work over about one to ...


11

A first order filter has a roll-off of 20dB per decade. That is, every time the frequency increases/decreases (for high/low pass filter) by a factor of 10, the signal power drops by a factor of 100. The corner frequency of a filter (the 'nominal' filter frequency) is where the output power drops by half, or -3dB. Here's a plot I made you in Matlab. See which ...


11

Frequency is not a measure of signal amplitude or power hence, talking about "this or that" frequency in decibels makes no sense. If somebody says "increase the frequency by 3 dB"; it is meaningless because it isn't a signal or power quantity. If someone else says "increase the signal by 3 dB", we know that means the power is ...


11

We could start using dBHz and see how it feels. My guess is it won't make life any simpler or help understanding. NOTE however : logarithmic frequency scale is already in use, and has been for a while ... a few millennia of precedence for the octave (base 2 logarithm) will make a decade-based system difficult to sell. Octaves are in reasonably common use in ...


11

How do I get the slope and intersection point? Start with a very simple example: - $$H(s) = \dfrac{1}{s}$$ When \$s = 1\$, the amplitude of the transfer function is clearly 1. When \$s = 0.1\$, the amplitude is 10. When \$s = 10\$, the amplitude is 0.1. If you convert those amplitudes to dB you would have: - When \$s = 0.1\$, gain = 20 dB When \$s = 1\$, ...


10

The bode plot is a representation of the bigger picture. That bigger picture is the pole zero diagram: - The top three images (all bode plots) give you different examples of a 2nd order low pass filter. The bottom left picture shows you the bigger picture - it combines the bode plot with the pole zero diagram i.e it's 3D. Bottom right is the view of the 3D ...


10

Is -3dB bandwidth defined for any type of transfer function? Maybe you do not realize what -3dB actually means? -3 dB means that the amplitude of the voltage (current as well) has dropped by a factor \$\sqrt2\$. That means that the power (= voltage * current) has dropped a factor \$\sqrt2\$ * \$\sqrt2\$ = \$2\$. And that is the point: the signal power at ...


9

It adds or subtracts multiples of \$2\pi\$ (360 degrees) to each point in the phase plot so that the phase is plotted continuously rather than having jumps in it. We know that physically the phase is going to be continuous from frequency to frequency. However as the output of the analysis is a vector gain at each frequency, all angles are reduced to the ...


8

My answer applies to higher-than-1st-order systems. There will always be a resonant point even if you can't see it. You need to understand how "poles" work. Take a look at this: - Even if there doesn't appear to be a resonance in the bode plot there will be a "pole" that is present and this pole represents the resonant frequency even though the "dampening" ...


8

One of the main innovations Bode proposed with Bode Stability plots was how the plot asymptotes behave for stable systems. A knowledge of these rules allows compensation just by manipulating the asymptotes. Much simpler than mathematical techniques like pole placement. Some main ones spring to mind (but it's not an exhaustive list): When the magnitude ...


7

It's just a conversion from radians to degrees. The expression for a capacitor's impedance uses units of radians/sec. for frequency. The simulator is reporting frequencies in Hertz. 10 rad/sec is 1.6Hz


7

Background Starting with your result and proceeding: $$\begin{align*} H\left(s\right)&=\frac{R}{R\,L\,C\,s^2+L\,s+R}\\\\ &=\frac{1}{L\,C\,s^2+\frac LR\,s+1}\\\\ &=\frac{\frac 1 {L\,C}}{s^2+\frac 1{R\,C}s +\frac 1{L\,C}}\tag{1} \end{align*}$$ The denominator you see in equation (1) is sometimes called the characteristic equation. There are two ...


7

The AC analysis is a linearized analysis. That means it simulates the small-signal behavior of each element at its operating point. In the case of the switch, that means it is held in either the on or off state, and behaves as a resistor with either the ROFF or RON value. The -96 dB transfer gain is due to the leakage through the switch's 1 megohm off ...


7

Perhaps this is where the confusion starts: 's' is complex (re + j*im, or sigma + jw), not just imaginary. Those two terms are often accidentally used interchangeably, and they shouldn't be. G is complex too, it has a phase and a magnitude for each w. It can be written as phase & magnitude or real & imaginary. Conversion between the two is only a ...


6

From your Bode plot (or 'frequency response' is probably a more descriptive term), just by cursory inspection it can be seen that: the system is 2nd order (since the high frequency roll-off is 40dB/decade); underdamped (since it has a resonance peak); probably has natural frequency of 1rad/sec (since the resonance peak is a little lower than 1 rad/sec); Has ...


6

There are two key frequencies you can quickly identify from the bode plot of the closed-loop system. The unity gain frequency is where the gain is 0dB i.e. neither any amplification nor any attenuation. The phase inversion frequency is where the phase is 180 degrees. If, at the "phase inversion" frequency where the phase shift is 180 degrees, there's more ...


6

Do you really think that a "terribly meaningless unit" would be used with enormous success over many decades? Are you not a bit curious if this "meaningless" unit has perhaps some advantages? here are some of them: 1.) Because of the log-log display (resp. dB-log) we have the chance to see the relation gain versus frequency over a much larger range if ...


6

The "origin pole" is indeed the \$1/s\$ term in the transfer function \$H(s)\$. In the bode plot it results in a first order transfer that does NOT flatten out for low frequencies. Your Bode plot is that of a low pass filter $$H(s) = \frac{1}{1 + s}$$ Note how this \$H(s)\$ would result in \$H(0) = 1 = 0\text{ dB}\$ like in your Bode plot. \$H(s) = 1/s\$ ...


6

There are many types of low pass filter. The one that gives your 'weird waveform' is the simplest 1st order RC filter. This passes the fundamental signal. However, it not only passes all the higher harmonics, but also gives them a phase shift. The frequency response of the perfect lowpass filter shown in your image is actually not realisable as a physical ...


6

1) Why does the magnitude Bode Plot of the response of a filter NOT approach Infinity at a pole? Try looking at this picture and recognize that poles may exist as infinities in the bode plot but more usually they are "behind" it: - 2) In the attached image, why is Wp (Omega subscript:P) called the pole frequency when the denominator clearly does not ...


6

A quick look at the fast analytical techniques or FACTs gives you the transfer function of this guy in the blink of an eye. First, turn the stimulus off or reduce \$V_{in}\$ to 0 V: replace the source by a short circuit. Then, "look" through the capacitor terminals to determine resistance \$R\$ between the terminals. It's immediate: \$R=R+R=2R\$. ...


6

Because a typical op-amp is dominant pole compensated, it has a single pole in the open-loop transfer function at low frequency, and a -20 dB/decade roll off. This leads to the concept of "gain-bandwidth". The open loop gain times the open loop pole frequency will be the "gain-bandwidth product". From that you can calculate the new pole ...


5

Your expression for \$H(s)\$ is correct: $$H(s)=\frac{RCs}{1+RCs}$$ where \$\tau=RC\$ is the time constant of the high pass filter. The most important features of this transfer function are the location of the pole (\$s_{\infty}=-1/RC\$) and the location of the zero (\$s_0=0\$). Note that the location of the pole and the zero is of course the same for the ...


5

May I add a 4th answer in short? 1.) A circuit with feedback is unstable in case the loop gain has a phase shift of 360deg at a frequency where the loop gain magnitude is still larger the 0 dB. Note that this phase shift includes the inverting properties of the inverting terminal. Taking this phase inversion NOT into account (as this is done, normally, in ...


5

People tend to make this way too complicated and difficult to understand. Stability margins are only defined for an ideal, linear transfer function model - a model expressed in terms of rational function of polynomials in the complex variable, s. In a feedback loop with a forward transfer function G(s) and feedback transfer function H(s), the input/output ...


5

Calculated values for the poles and zeros look about right, except for the missing integrator pole at the origin. Maybe it would be best to break the loop up into pieces. Write the transfer function of the error amp, from \$V_{\text{out}}\$ to error amp output to start. You should get something like: \$\frac{\text{EA}_{\text{out}}}{V_{\text{out}}}\$ = \$\...


5

Complex Number Diversion Complex numbers have both a Cartesian notation as well as two equivalent polar notations. The Cartesian notation, \$a+b\:i\$, is very hard to use in electronics but it is very easy to plot. A common polar notation used is \$r\left[\operatorname{cos}\left(\theta\right) + i\:\operatorname{sin}\left(\theta\right)\right]\$. But for ...


5

Different applications are sensitive to phase variation with frequency for different reasons. It can impact stability margin if the filter is in a closed-loop system. The total error between a source signal and a filtered signal is a function of phase (as an extreme example, if h(x) defines a filter with a gain of 1 and a phase shift of 180 degrees at some ...


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