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VashaNatasha [74]
3 years ago
10

What are the desired characteristics or values for the following parameters of an ideal amplifier?o Phase change as a function o

f the log of the frequency o Common Mode Rejection Ratioo Input Resistanceo Output Resistance
Engineering
1 answer:
kvv77 [185]3 years ago
4 0

Answer and Explanation:

- Phase change as a function of the log of the frequency

The phase shift should be independent of the frequency of the input signal. It should have an Infinite bandwidth with zero phase shift

- Common Mode Rejection Ratio

An ideal amplifier has Infinite common-mode rejection ratio (CMRR). A perfect operational amplifier amplifies only the voltage difference between its two inputs, completely rejecting all voltages that are common to both.

- Input Resistance

Infinite input impedance and so, zero input current & Zero input offset voltage.

The differential input impedance of the operational amplifier is defined as the impedance between its two inputs; the common-mode input impedance is the impedance from each input to ground.

And the impedance between its inputs should be infinite.

- Output Resistance

Zero output impedance and Infinite output voltage range.

Output Resistance leads to voltage drop. Thee voltage drop across the output impedance effectively reduces the open loop gain. So, the optimal value for this should be zero.

N.B - For AC circuits such as this one being discussed, resistance is known as impedance.

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Answer:

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Explanation:

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3 years ago
A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st
Alenkasestr [34]

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

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5 0
1 year ago
The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery
Lynna [10]

Answer:

capacity  = 0.555 mAh

capacity  = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C  

capacity  = 2 × ampere × second   ...............1

put here value and we get

and 1 Ah = 3600 C

capacity  = \frac{2}{3600}

capacity  = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2   ...............2

capacity  = 1800 + 1800

capacity  = 3600 mAh

3 0
3 years ago
What are the different types of documents used to communicate engineering designs?
Ipatiy [6.2K]

Answer:

COMMON ENGINEERING DOCUMENTS

Inspection or trip reports.

Research, laboratory, and field reports.

Specifications.

Proposals.

Progress reports.

ect...

Explanation:

7 0
3 years ago
Write a naive implementation (i.e. non-vectorized) of matrix multiplication, and then write an efficient implementation that uti
erik [133]

Answer:

import numpy as np  

import time  

def matrixMul(m1,m2):      

   if m1.shape[1] == m2.shape[0]:  

       

       t1 = time.time()

       r1 = np.zeros((m1.shape[0],m2.shape[1]))

       for i in range(m1.shape[0]):

           for j in range(m2.shape[1]):

               r1[i,j] = (m1[i]*m2.transpose()[j]).sum()

       t2 = time.time()

       print("Native implementation: ",r1)

       print("Time: ",t2-t1)

       

       t1 = time.time()

       r2 = m1.dot(m2)

       t2 = time.time()

       print("\nEfficient implementation: ",r2)

       print("Time: ",t2-t1)

       

   else:

       print("Wrong dimensions!")

Explanation:

We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.

7 0
3 years ago
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