Answer to Question #284578 in Calculus for poochie

Question

The gain of an amplifier is found to be 𝐺 = 20 ln(𝑉𝑜𝑢𝑡). The tasks are to find equations for: a) Draw a graph of Gain against Vout between 𝑉𝑜𝑢𝑡 = 1 and 𝑉𝑜𝑢𝑡 = 10 b) Determine the gradient of the graph at 𝑉𝑜𝑢𝑡 = 2 and 𝑉𝑜𝑢𝑡 = 5 c) Find the derivative 𝑑𝐺/ 𝑑𝑉𝑂𝑢𝑡 and calculate its value at 𝑉𝑜𝑢𝑡 = 2 and 𝑉𝑜𝑢𝑡 = 5 d) Compare your answers for part b) and part c) e) Find the second derivative 𝑑²𝐺/ 𝑑𝑉𝑂𝑢𝑡 ² 

Answer 1

Solution:

Given 𝐺 = 20 ln(𝑉_𝑜𝑢𝑡)

From 𝑉_𝑜𝑢𝑡 = 1 and 𝑉_𝑜𝑢𝑡 = 10.

Treat 𝑉_𝑜𝑢𝑡=x and G=y

(a):

then, the graph is:

(b):

Gradient At 𝑉_𝑜𝑢𝑡 =2= slope = $\dfrac yx= \dfrac{20 \ln(2)}{2}=6.931$

Gradient At 𝑉_𝑜𝑢𝑡 =5= slope = $\dfrac yx= \dfrac{20 \ln(5)}{5}=6.437$

(c):

$\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} = 20\times \dfrac{1}{𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{𝑉_{𝑜𝑢𝑡}}$

At 𝑉_𝑜𝑢𝑡 =2, $\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{2}=10$

At 𝑉_𝑜𝑢𝑡 =5, $\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{5}=4$

(d):

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

(e):

$\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{𝑉_{𝑜𝑢𝑡}} \\ \Rightarrow \dfrac{d^2G}{d 𝑉_{𝑜𝑢𝑡}^2} =\dfrac{-20}{𝑉_{𝑜𝑢𝑡}^2}$