Question #284578

The gain of an amplifier is found to be šŗ = 20 ln(š‘‰š‘œš‘¢š‘”). The tasks are to find equations for: a) Draw a graph of Gain against Vout between š‘‰š‘œš‘¢š‘” = 1 and š‘‰š‘œš‘¢š‘” = 10 b) Determine the gradient of the graph at š‘‰š‘œš‘¢š‘” = 2 and š‘‰š‘œš‘¢š‘” = 5 c) Find the derivative š‘‘šŗ/ š‘‘š‘‰š‘‚š‘¢š‘” and calculate its value at š‘‰š‘œš‘¢š‘” = 2 and š‘‰š‘œš‘¢š‘” = 5 d) Compare your answers for part b) and part c) e) Find the second derivative š‘‘2šŗ/ š‘‘š‘‰š‘‚š‘¢š‘” 2 


Expert's answer

Solution:

Given šŗ = 20 ln(š‘‰š‘œš‘¢š‘”)

From š‘‰š‘œš‘¢š‘” = 1 and š‘‰š‘œš‘¢š‘” = 10.

Treat š‘‰š‘œš‘¢š‘”=x and G=y

(a):

then, the graph is:



(b):

Gradient At š‘‰š‘œš‘¢š‘” =2= slope = yx=20ln⁔(2)2=6.931\dfrac yx= \dfrac{20 \ln(2)}{2}=6.931

Gradient At š‘‰š‘œš‘¢š‘” =5= slope = yx=20ln⁔(5)5=6.437\dfrac yx= \dfrac{20 \ln(5)}{5}=6.437

(c):

dšŗdš‘‰š‘œš‘¢š‘”=20Ɨ1š‘‰š‘œš‘¢š‘”=20š‘‰š‘œš‘¢š‘”\dfrac{dšŗ}{d š‘‰_{š‘œš‘¢š‘”}} = 20\times \dfrac{1}{š‘‰_{š‘œš‘¢š‘”}} =\dfrac{20}{š‘‰_{š‘œš‘¢š‘”}}

At š‘‰š‘œš‘¢š‘” =2, dšŗdš‘‰š‘œš‘¢š‘”=202=10\dfrac{dšŗ}{d š‘‰_{š‘œš‘¢š‘”}} =\dfrac{20}{2}=10

At š‘‰š‘œš‘¢š‘” =5, dšŗdš‘‰š‘œš‘¢š‘”=205=4\dfrac{dšŗ}{d š‘‰_{š‘œš‘¢š‘”}} =\dfrac{20}{5}=4

(d):

There is huge difference in answers in part b and c.

(e):

dšŗdš‘‰š‘œš‘¢š‘”=20š‘‰š‘œš‘¢š‘”ā‡’d2Gdš‘‰š‘œš‘¢š‘”2=āˆ’20š‘‰š‘œš‘¢š‘”2\dfrac{dšŗ}{d š‘‰_{š‘œš‘¢š‘”}} =\dfrac{20}{š‘‰_{š‘œš‘¢š‘”}} \\ \Rightarrow \dfrac{d^2G}{d š‘‰_{š‘œš‘¢š‘”}^2} =\dfrac{-20}{š‘‰_{š‘œš‘¢š‘”}^2}


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