What are the three necessary conditions for a sine wave oscillator?

What are the three necessary conditions for a sine wave oscillator?

1、 The loop gain should be greater than 1
2、 Equilibrium condition t (J ω) = 1
3、 Stability condition at / AUI

What is the phase balance condition of three-point oscillator?

The three-point oscillator satisfies the condition of phase balance
In the AC path, the two reactance elements connected with the emitter of a transistor must be identical, but the reactance properties of the reactance elements not connected with the emitter are opposite to the former

In RT triangle ABC, ∠ ACB = 90 °, CD ⊥ AB at D, bisector AE of ⊥ BAC intersects CD at h, EF ⊥ AB at F, connects FH, and quadrilateral CEFH is a diamond

What's the matter? It seems to be true

Let a be a constant, and the remainder of the polynomial X3 + AX2 + 1 divided by x2-1 is x + 3, then a=______ ．

∵ the remainder of the polynomial X3 + AX2 + 1 divided by x2-1 is x + 3, ∵ let X3 + AX2 + 1 - (x + 3) = (x2-1) (x + b), and we can get: X3 + ax2-x-2 = X3 + bx2-x-b, ∵ a = BB = 2, ∵ a = 2

In the triangle ABC, the vector AB is perpendicular to the vector AC, the vector AP * vector AC = 2, the vector AP * vector AB = 2, and the module length of the vector AP is 2,
Min for finding the module length of (vector AB * vector ac * vector AP)
Min for finding the module length of (vector AB + vector AC + vector AP)

Draw the vector AP, then | AP | = 2. According to the meaning of the topic, point C is on the middle perpendicular of AP, and point B is on the middle perpendicular of the middle perpendicular,
The module length of (vector AB + vector AC + vector AP) in the component parallel to AP is 2 + 1 + 1 / 2 = 3.5
The component perpendicular to AP satisfies xy = 0.5 (according to triangle similarity) x + Y > = radical 2
Min = root (3.5 ^ 2 + 2) = root 57 / 2

In ABC, we know that De is a point on the edge of BC, ad = AE, BD = EC, and prove AB = AC

Because ad = AE, so ∠ ade = ∠ AED, so ∠ ADB = ∠ AEC (the two corners are equal, and their complements are equal)
Because ∠ ADB = ∠ AEC, ad = AE, BD = EC, so the triangle abd ≌ triangle ACE (both sides and angles equal triangle congruent), so AB = AC

(2013 ^ 2 + 2013) / / 2014 factoring problem solving process

(2013^2+2013)÷2014
=2013x（2013+1）x1/2014
=2013

Given that the distance from the point to the right focal point of the ellipse x2 / 25 + Y2 / 9 = 1 is equal to the distance to the straight line x = 6, the trajectory equation of the point is obtained

Right focus coordinate (4,0)
Lie equation: I X-6 I = √ (x-4) * 2 + y * 2
The solution is y * 2 = 20-4x
Note: y * 2 is the second power of Y

It is known that in ⊿ ABC, ab = AC, ad is the middle line, P is the upper point of AD, CF ∥ AB is made through C, extend BP, cross AC to e, cross CF to F, prove: bp2 = PE
Verification: bp2 = PE × PF

Certification:
∵AB=AC
∴∠ABC=∠ACB
∵ ad is the midline, ⊿ ABC is the isosceles triangle
The ad is the vertical bisector of BC [three lines in one]
Connect PC, then Pb = PC [the distance from the point on the vertical bisector to both ends of the line segment is equal]
∴∠PBC=∠PCB
← APB = ∠ ACP [equal minus equal]
∵CF//AB
∴∠CFP=∠APB
∴∠CFP=∠ACP
And ∵ ∠ FPC = ∠ CPE [common angle]
∴⊿FPC∽⊿CPF（AA‘）
∴PF/PC=PC/PE
Convert to PC & # 178; = PE × PF
∴PB²=PE×PF

Can the set of real and imaginary numbers represent all the numbers in mathematics?
RT
What I want to ask is whether there are any other numbers in mathematics except real numbers and imaginary numbers. If so, please tell me what number it is. If not, please make sure not

Real number and imaginary number are called complex number, which is the largest number field for general mathematics
But number is defined by people. There are more people in the world who don't have the function of number
Hamilton once defined quaternion in history. The set of all quaternions constitutes the first division ring in history
So as long as there is no problem with the number defined and there is no contradiction with the known axiom system, it is OK