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3 years ago

Why is an integrated circuit (IC) referred to as a special device?" points more than 10 plz help

Engineering

1 answer:

sladkih [1.3K]3 years ago

5 0

Answer:

Because IC microcomputers are smaller and more versatile than previous control mechanisms, they allow the equipment to respond to a wider range of input and produce a wider range of output. They can also be reprogrammed without having to redesign the control circuitry

Explanation:

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Please answer fast. With full step by step solution.

lina2011 [118]

Let f(z) = (4z ² + 2z) / (2z ² - 3z + 1).

First, carry out the division:

f(z) = 2 + (8z - 2) / (2z ² - 3z + 1)

Observe that

2z ² - 3z + 1 = (2z - 1) (z - 1)

so you can separate the rational part of f(z) into partial fractions. We have

(8z - 2) / (2z ² - 3z + 1) = a / (2z - 1) + b / (z - 1)

8z - 2 = a (z - 1) + b (2z - 1)

8z - 2 = (a + 2b) z - (a + b)

so that a + 2b = 8 and a + b = 2, yielding a = -4 and b = 6.

So we have

f(z) = 2 - 4 / (2z - 1) + 6 / (z - 1)

f(z) = 2 - (2/z) (1 / (1 - 1/(2z))) + (6/z) (1 / (1 - 1/z))

Recall that for |z| < 1, we have

$\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n$

Replace z with 1/z to get

$\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}$

so that by substitution, we can write

$\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}$

Now condense f(z) into one series:

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}$

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}$

So, the inverse Z transform of f(z) is $\boxed{6+2^{2-n}}$ .

4 0

3 years ago

For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, aft

ivanzaharov [21]

Answer:

total time = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - $e^{(-kt^n)}$ ...........1

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}

solve it we get

k = 0.003786 = 37.86 × $10^{4}$

so here

y = 1 - $e^{(-kt^n)}$

1 - y = $e^{(-kt^n)}$

take ln both side

ln(1-y) = -k × $t^n$

t = $\sqrt[n]{-\frac{ln(1-y)}{k}}$ .............2

now we will put the value of y = 87% in equation with k and find out t

t = $\sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}$

total time = 304.21 s

7 0

3 years ago

A cylindrical specimen of a metal alloy 48.8 mm long and 9.09 mm in diameter is stressed in tension. A true stress of 327 MPa ca

Gnom [1K]

Answer:

The true stress necessary to plastically elongate the specimen is 406.69MPa

Explanation:

The Explanation or Solving is on the attached document.

5 0

3 years ago

1. Implement the k-means clustering algorithm either in Java or Python. • The program should be executable with at least 3 param

givi [52]

Answer:

The code for this Question in Python is as follows:

matplotlib inline

from copy import deepcopy

import numpy as np

import pandas as pd

from matplotlib import pyplot as plt

plt.rcParams['figure.figsize'] = (16, 9)

plt.style.use('ggplot')

# Importing the dataset

data = pd.read_csv('xclara.csv')

print(data.shape)

data.head()

# Getting the values and plotting it

f1 = data['V1'].values

f2 = data['V2'].values

X = np.array(list(zip(f1, f2)))

plt.scatter(f1, f2, c='black', s=7)

# Number of clusters

k = 3

# X coordinates of random centroids

C_x = np.random.randint(0, np.max(X)-20, size=k)

# Y coordinates of random centroids

C_y = np.random.randint(0, np.max(X)-20, size=k)

C = np.array(list(zip(C_x, C_y)), dtype=np.float32)

print(C)

# To store the value of centroids when it updates

C_old = np.zeros(C.shape)

# Cluster Lables(0, 1, 2)

clusters = np.zeros(len(X))

# Error func. - Distance between new centroids and old centroids

error = dist(C, C_old, None)

# Loop will run till the error becomes zero

while error != 0:

# Assigning each value to its closest cluster

for i in range(len(X)):

distances = dist(X[i], C)

cluster = np.argmin(distances)

clusters[i] = cluster

# Storing the old centroid values

C_old = deepcopy(C)

# Finding the new centroids by taking the average value

for i in range(k):

points = [X[j] for j in range(len(X)) if clusters[j] == i]

C[i] = np.mean(points, axis=0)

error = dist(C, C_old, None)

# Initializing KMeans

kmeans = KMeans(n_clusters=4)

# Fitting with inputs

kmeans = kmeans.fit(X)

# Predicting the clusters

labels = kmeans.predict(X)

# Getting the cluster centers

C = kmeans.cluster_centers_

fig = plt.figure()

ax = Axes3D(fig)

ax.scatter(X[:, 0], X[:, 1], X[:, 2], c=y)

ax.scatter(C[:, 0], C[:, 1], C[:, 2], marker='*', c='#050505', s=1000)

4 0

4 years ago

Electricity is the flow of electrons from a negatively charged

TiliK225 [7]

Yes.

electric current refers to directional negative-to-positive electrons from one atom to the next. Liquid conductors and gas conductors, electric current refers to electrons and protons flow in the opposite direction.

6 0

1 year ago