Can you become an expert with 200 points a week in the Historic District? 5

Yes. This week's third place is a 200-point increase.

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What's the matter with the exchange of q coins?。。。 I came here just to study.

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and level hooks.

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2.Nothing can be changed. Q coins are exchanged for money.

It's not that complicated, and the waveform given later is wrong, which obviously violates the principle that the capacitor voltage cannot be abruptly changed.

Because the passing square wave frequency is fixed at 1k, the square wave period is 1ms, when the integration time constant is 2ms, when the input is a fixed value, uo=-1 rc *ui(t2-t1)+uo(t1).

Here, T2-T1 is half a cycle, i.e., so UO=UO(T1) *UI (assuming that the RC unit is MS), when RC is 2MS, the pre-time, UI=-3V, UO(T1)=0, and then the UO peak is, the post-time, UI=3V, UO(T1)=, and then the integral end value of the first cycle is obtained, which is 0V.

The rest of the RC values can also be used to infer peak and integral endpoint values. When RC becomes larger, the peak value decreases, and the RC value increases the peak, but because the power supply voltage is limited, when the RC is reduced to a certain extent, the integration to the power supply voltage will no longer rise, which is distortion...

This kind of integration circuit will only be saturated distorted when RC is not large enough.

P.S. The voltage on both sides of the capacitor cannot be abruptly changed because it takes time for the charge to be transferred, but here, the voltage at one end is fixed at 0V, so the output voltage cannot be abruptly changed. As for the waveform asymmetry, it can be concluded that although the input voltage is 6V, it is not symmetrical to 0V, resulting in different charge and discharge currents, so the voltage rise and fall rate is different.

For example, the two voltages input are -2V, 4V...

In order to integrate the ideal waveform for a square wave of a particular frequency, the capacitance and resistance must be changed to a certain value. It is recommended to connect the capacitors in parallel with adjustable resistors. Adjust the parameters slowly and carefully, and you are sure to achieve an ideal waveform.

Personally, I think that the voltage at both ends of the capacitor cannot be abruptly changed, so the waveform should be continuous. I think the landlord also knows this principle, so I will have doubts when I see the sudden change in voltage in the second picture. All I can say is that the second picture is wrong.

If the voltage change diagram is feasible, then it will have to wait for a higher master to explain, but I also think that there is a problem with the explanation on the first floor, and his explanation cannot explain the voltage abrupt change. The landlord respects himself.

Suppose the integration time of the first graph is t1, the integration time of the second graph is t2, the integration time is actually increased in the latter graph, and the integration time is increased by t0 in one cycle, because the function of the integral is certain, in the second graph, in one cycle, the time period from the beginning of the integration to the arrival of t1 2 and the time period from t2 + t0 2 to (t2 2) + t1 2 are the same as in Figure 1, and the time period from t1 2 to t2 2 is the same as from (t2 2) + t1 The time period from 2 to t2 is to continue to integrate the function to be integrated, so the deformation of the triangular wave occurs. In this period, the time reaches t2, the cycle ends, and the next integration cycle begins, and the integration amount of the next integration cycle starts from 0, instead of starting from the value of the previous integration, so it is not connected to the curve of the previous period.

Your problem is mostly on p1.

It is indeed equal to the energy p2 radiating from this hemisphere per unit of time – the wavelength is included in its expression.

The problem is that in your calculation p1=s0*, you say that s0 is the average energy flow density of the plane wave, and you think of it as a constant, and this is the source of the error.

Yes, in a plane wave, the average energy flow density vector, that is, the Boyinting vector, does not obviously contain frequency or wavelength, but in fact, due to the influence of the small holes - reflection and absorption at the periphery, diffraction at the edges, etc., the energy flow density around it is very different from the energy flow density at the far left, that is, it can no longer be expressed by the expression of the plane wave - the real plane wave can only be in an infinitely wide uniform medium.

And now, the degree to which it has been altered, as you can imagine, will be related to the size of the aperture and the length of the wavelength, i.e., the electric and magnetic fields will not be the original values, resulting in what you call the "average energy flow density of the plane wave" will be wavelength-related. To use an inappropriate analogy: take a large screen with holes to block the wind, and the wind speed flowing through the hole is not the same as the original wind speed.

It is very complicated to calculate the energy flow density near the left side of the hole, and I have spent a lot of time and still have not got the result - the owner of the watchtower can continue to calculate. But there is no doubt that the energy flow density will be wavelength-dependent.

Check out this article.

I think the landlord is ignoring the scope of application of the formula. The Fresnel-Huygens principle is suitable for cases where diffraction is obvious, and the Poyinting vector will not be suitable for calculating the energy on the left.

I think so myself.

The Fresnel integral gives the amplitude, which contains the frequency.

However, the above Poyintine vector is averaged over a period, so it does not contain frequencies.

According to the amplitude of the Fresnel integral, the corresponding energy can be found, and there should be no difference.

There are deficiencies to continue the discussion.

P2 is more than P1 (2) because the electromagnetic wave near the left side of the hole is no longer a plane wave but has begun to bend slightly outward before passing through the hole.

This phenomenon is especially obvious when the wavelength is much larger than the aperture, so p1 is not applicable at this time, and p2 ->0 at (2) -0, that is, the wavelength is too large for the electromagnetic wave to pass through the aperture.

When the wavelength is much smaller than the hole, p1 is suitable, but p2 is not, because at this time, 2 cannot be directly replaced with d, but the integral of e (2 ir ) r d d d 3r should be calculated honestly.

Let's put it simply:

Time. When calculating the de = (-i )*e0*( 1+cos ) 2 )*e ikr ) r )*d integral.

e ikr ) is a complex exponent of e that cannot be directly regarded as a constant, but is actually a trigonometric function.

As a trigonometric function, even if the independent variable changes by a small range, the final value may change by a lot, although r>>r

But KR will change a lot of arcs in the integral.

2.>r.

In this case, kr can be regarded as a constant, but the diffraction integral formula is no longer correct, and the Fresnel diffraction integral formula and Kirchhoff boundary conditions only apply to

If it is calculated using Kirchhoff's integral formula, it is inherently inconsistent, and the theory contradicts the boundary conditions assumed at the edge of the aperture. That's why there's the Rayleigh-Somerfi formula.

I didn't know if this was the reason for it.