Relationship between q factor and bandwidth place
Answer from the author : You probably mean an electrical transmission line with open or closed ends, which can act as a resonator. If the transmission losses and any losses at the ends are low, the Q factor is necessarily large: the initially stored energy will decay only slowly. But once oscillation begins, the Q is not raised by the positive feedback and we do not see a bandwidth reduction, for instance.
Answer from the author : Indeed, you can consider the effective round-trip losses, which are reduced by the gain, and calculate a Q factor from that. If you crank up the amplifier gain, the Q factor becomes larger and larger, and isolation starts when Q becomes infinite. I'm just not sure what you mean with Q during oscillation. Just redefining Q based on the emission linewidth is a problematic concept. The linewidth can be increased by all sorts of effects, not all related to the resonator.
Although the relationship seems obvious in a mathematical point of view, it does not make sense to me in an intuitive way. Is there a way to describe it in a more sensible way? Answer from the author : It may help to consider that the Q factor refers to the loss of energy within one optical oscillation cycle, while the finesse refers to the losses per round trip.
That explains the proportionality factor mentioned in the text. Here you can submit questions and comments. The author will decide on acceptance based on certain criteria. Essentially, the issue must be of sufficiently broad interest. Please do not enter personal data here; we would otherwise delete it soon. See also our privacy declaration.
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Bibliography [1] S. McCall et al. Gorodetsky, A. Savchenkov and V. Armani et al. Locke et al. Express 17 24 , , doi One reason is "this is the convention," but that's not very satisfying. Another reason is that most of the systems for which we will compute "Q" are under-damped, so the two angular frequencies are very, very similar This gives us four time-related quantities that we could compare.
Try to write it in terms of the mass m and resistive force coefficient b. You should end up with a quantity known as Q, the quality factor. Note that Q is a pure number, with no units. Why is it called the "quality factor"? Well, it clearly describes something important about an oscillating system: something related to the number of cycles it takes for the motion to die down.
How many oscillations to lose amplitude or energy? So, one thing that Q tells us is something about the decay of an oscillating system. But how much does the amplitude decrease in just one cycle? In other words, how much does the amplitude decrease during a time P? Well, if you start with the basic exponential decay and if you remember that one way to write Q is then you should be able to write an expression for the factor by which the amplitude decreases during a single period.
Q: By what factor does the amplitude decrease in one cycle? If we know the factor is f for just one cycle, then we know that after TWO cycles, the decrease will be f2, and after THREE cycles, the decrease will be f3, and after N cycles, the decrease will be fN. Consider a standard spring of force constant k, from which we hang a mass m. We let the system come to rest at an equilibrium, marked by the dashed line in the figure below. Q: How large is the resulting displacement of the mass?
How big is the displacement of the mass now? Q: Suppose that the driving frequency is exactly equal to the natural frequency of the system. What will the amplitude of motion be? Q: How can you write the amplitude at resonance in a way that includes Q, and is easy to compare to the static displacement?
And so we see that the amplitude of motion for the forced, oscillating system at resonance is exactly Q times the displacement by a static force of the same size. Bandwidth As you know, the amplitude of a forced harmonic oscillator depends on a number of factors. A general result is that the amplitude is large when the driving frequency is close to the natural frequency of the undamped system. It's clear that the value of Q is related to the shape of this amplitude graph.
It will take a while, but the result in the end will be worth it. First, we write the amplitude as a function of driving frequency in a slightly different manner, so that "Q" appears in it. Now, we are going to switch from dealing with the amplitude of motion to the power dissipated in a driving oscillating system. What does that mean? The system has a resistive force, remember?
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What is the relationship between Q factor and bandwidth? What is significance of Q factor? The Q factor or quality factor gives the sharpness of resonance obtained in the LCR circuit. If Q value is more, then the resonance will be very sharp and if it is less then resonance will be weak giving rise to weaker circuit.
What is the significance of bandwidth in RLC circuit? The bandwidth of any system is the range of frequencies for which the current or output voltage is equal to What is the significance of Q factor in AC circuit? In an AC system, the Q factor represents the ratio of energy stored in the capacitor to the energy dissipated as thermal losses in the equivalent series resistance.
What is the difference between Q and bandwidth? A low resistance, high Q circuit has a narrow bandwidth, as compared to a high resistance, low Q circuit. Bandwidth is measured between the 0. What is the relationship between Q and the bandwidth of a tuned circuit?
The bandwidth of a tuned amplifier depends upon the Q of the LC circuit i. The value of Q and the bandwidth are inversely proportional. What is the significance of Q factor Class 12? The sharpness of resonance obtained in the LCR circuit is determined by the Q factor, or consistency factor. If the Q value is higher, the resonance will be very sharp; if it is lower, the resonance will be weak, resulting in a weaker circuit.
What is Q factor of series resonant circuit also explain features of series resonance? A high-quality bell rings with a single pure tone for a very long time after being struck. A purely oscillatory system, such as a bell that rings forever, has an infinite quality factor.
More generally, the output of a second-order low-pass filter with a very high quality factor responds to a step input by quickly rising above, oscillating around, and eventually converging to a steady-state value. Like an overdamped system, the output does not oscillate, and does not overshoot its steady-state output i. Like an underdamped response, the output of such a system responds quickly to a unit step input.
Critical damping results in the fastest response approach to the final value possible without overshoot. Real system specifications usually allow some overshoot for a faster initial response or require a slower initial response to provide a safety margin against overshoot.
In negative feedback systems, the dominant closed-loop response is often well-modeled by a second-order system. The phase margin of the open-loop system sets the quality factor Q of the closed-loop system; as the phase margin decreases, the approximate second-order closed-loop system is made more oscillatory i.
Some examples[ edit ] A unity-gain Sallen—Key lowpass filter topology with equal capacitors and equal resistors is critically damped i. A second-order Bessel filter i.
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