Question #284578

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 


1
Expert's answer
2022-01-04T10:55:34-0500

Solution:

Given 𝐺 = 20 ln(𝑉𝑜𝑢𝑡)

From 𝑉𝑜𝑢𝑡 = 1 and 𝑉𝑜𝑢𝑡 = 10.

Treat 𝑉𝑜𝑢𝑡=x and G=y

(a):

then, the graph is:



(b):

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

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

(c):

d𝐺d𝑉𝑜𝑢𝑡=20×1𝑉𝑜𝑢𝑡=20𝑉𝑜𝑢𝑡\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} = 20\times \dfrac{1}{𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{𝑉_{𝑜𝑢𝑡}}

At 𝑉𝑜𝑢𝑡 =2, d𝐺d𝑉𝑜𝑢𝑡=202=10\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{2}=10

At 𝑉𝑜𝑢𝑡 =5, d𝐺d𝑉𝑜𝑢𝑡=205=4\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{5}=4

(d):

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

(e):

d𝐺d𝑉𝑜𝑢𝑡=20𝑉𝑜𝑢𝑡d2Gd𝑉𝑜𝑢𝑡2=20𝑉𝑜𝑢𝑡2\dfrac{d𝐺}{d 𝑉_{𝑜𝑢𝑡}} =\dfrac{20}{𝑉_{𝑜𝑢𝑡}} \\ \Rightarrow \dfrac{d^2G}{d 𝑉_{𝑜𝑢𝑡}^2} =\dfrac{-20}{𝑉_{𝑜𝑢𝑡}^2}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS