Answer:
The Answer and Explanation is in the folloing attachment
Explanation:
A direct-coupled amplifier has a low-frequency gain of 40 dB, poles at 2 MHz and 20 MHz, a zero on the negative real axis at 200
MHz, and another zero at infinite frequency. Express the amplifier gain function in the form of Equations and sketch a Bode plot for the gain magnitude. What do you estimate the 3-dB frequency fH to be?
1 answer:
Answer:
Explanation:
Low frequency gain is= 40db= 20logK=>100 poles at 2MHz,20MHz
Zero at -200MHz, zero at infinity.
A) A(s) = 100FH(s)
B) Poles (1): 2 pi × 2 × 10^6= 4pi × 10^6MHz
(2): 2pi × 20 × 10^6= 4pi × 10^6 MHz
Zeroed: 2pi × 10^6 × 200= 400pi × 10^6, at infinity.
T/(S) = (1 + S/400π × 10^6)/S(1 + S/4π × 10^6)(1 + S/4π × 10^6)
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Answer:
The percentage ductility is 35.5%.
Explanation:
Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.
Step1
Given:
Diameter of shaft is 10.2 mm.
Final area of the shaft is 52.7 mm².
Calculation:
Step2
Initial area is calculated as follows:
A = 81.713 mm².
Step3
Percentage ductility is calculated as follows:
D = 35.5%.
Thus, the percentage ductility is 35.5%.
Answer:
The head difference across the soil specimen is 39.29 cm and the discharge velocity is 0.02 cm/s
Explanation:
The head difference across the soil specimen is:
Where
k = hydraulic conductivity = 0.014 cm/s
Q = volume of water collected = 150 cm³/min = 2.5cm³/s
L = length of the soil specimen = 275 mm = 27.5 cm
A = area = 125 cm²
Replacing:
The hydraulic gradient is:
The discharge velocity is:
v = k*i = 0.014 * 1.43 = 0.02 cm/s