In order to amplify the slowly changing weak signal, the amplification circuit should adopt? Coupling mode

In order to amplify the slowly changing weak signal, the amplification circuit should adopt? Coupling mode

Because the frequency of slow current is very low, it can't be amplified by AC, and the signal is very weak. If direct coupling amplification is used, the offset voltage drift will affect the amplification accuracy and noise. Usually, a weak low-frequency signal is superimposed on the oscillation signal by an oscillation signal edge, which is also called AC modulation. After modulation, AC amplification is carried out, and then demodulation is carried out. It is widely used in radio, instrument, etc

The reason of zero drift in direct coupled amplifier is ()
A. Resistance error
B. Dispersion of transistor parameters
C. Transistor parameters are affected by temperature
D. Unstable power supply voltage

A. B, C, D can cause zero drift, of which C is generally the main reason. The zero drift caused by C has a special term called temperature drift

It is known that the bottom of a straight quadrangular prism is a parallelogram with 5cm and 6cm sides and 8cm diagonal. The longest diagonal of the quadrangular prism is 10cm
Find the side area of the prism

The key to solve this problem is to find the [height] of a straight quadrangular prism as shown in the figure. Let its height be XBD & # 39; & amp; sup2; = 8 & amp; sup2; - X & amp; sup2; AC & # 39; & amp; sup2; = 10 & amp; sup2; - X & amp; sup2; according to: the sum of squares of two diagonals of a parallelogram is equal to the sum of squares of four sides of a parallelogram

X-x / 7 = 12 (solving equation)

x-x/7=12
7x-x=84
6x=84
x=14

Sin (XY) = ln (x + y) / y + 1, the process of finding y '(0)

The two sides are derived from X
cos(xy)*(x+y')=[y(1+y')/(x+y)-y'ln(x+y)]/y^2
Substituting x = 0 into sin (XY) = ln (x + y) / y + 1 leads to
lny/y=-1
This y can't be solved
If we get the solution, we can directly get y 'by substituting x = 0 and y

How to solve 7x = 5 [12 + x]

7x=60+5x
2x=60
x=30

When k is an integer value, the equation (K + 1) sin ^ 2x-4cosx + 3K = 5 = 0 has a real number solution, and the solution at this time is obtained

K = - 1, cosx = - 2 do not meet the requirements
k≠0
(k+1)sin^2x-4cosx+3k-5=0
(k+1)cos^2x-4cosx+4-4k=0
[(k+1)cosx-2(1-k)](cosx-2)=0
cosx-2≠0,
(k+1)cosx-2(1-k)=0
cosx=2(1-k)/(k+1),-1

Less than the reciprocal of 6 by 7
-Is M & # n a polynomial or a monomial

The result is - 41 / 6
The reciprocal of 6 is 1 / 6, less than 7 is minus 7
1/6-7=-41/6

If x-2y = - 5, then 6-2x + 4Y =?

6-2x+4y
=5-2(x-2y)
=6-2×(-5)
=16

It is known that the function f (x) satisfies f (x) * f (x + 2) = 13 in the domain R. if f (1) = 2, the value of F (99) can be obtained

f(x)*f(x+2）=13
f(x+2)=13/f(x)
f(x+4)=13/f(x+2)=f(x)
So f (x) is a function of period 4
f(1)=2
f(3)=13/f(1)=13/2
f(99)=f(3+24*4)=f(3)=13/2