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3 years ago

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the frequencies 10, 12, 23 and 45 Hz. (a) What is the minimum sampling rate required to avoid aliasing? (b) If you sample at 40

sps, which frequency components will be aliased?

1 answer:

ValentinkaMS [17]3 years ago

3 0

Answer:

Sampling rate = 90

23 and 45 Hz

Explanation:

The Nyquist criteria states that the sampling frequency must be at least twice of the highest frequency component.

(a) We have 10, 12, 13, 23, and 45 Hz signals

The highest frequency component in this list is 45 Hz

So minimum sampling frequency needed to avoid aliasing would be

Sampling rate = 2*45 = 90 sps

Hence a sampling rate of 90 samples per second would be required to avoid aliasing.

(b) if a sampling rate of 40 sps is used then

40 = 2*f

f = 40/2 = 20 Hz

Hence 23 and 45 Hz components will be aliased.

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An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the $w_{p}$ to h₁, where $w_{p}$ is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute $w_{p}$ To computer

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water $v_{f}$ = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ = 0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

= 417.5 + 0.001043 (14900)

= 417.5 + 15.5407

= 433.04 kJ/kg

Find h₃

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = $s_{f}$ + $x_{4} s_{fg}$

$x_{4} = s_{4} -s_{f} /s_{fg}$

$x_{4}$ = 6.14 - 2.45 / 3.89

$x_{4}$ = 0.9497

The enthalpy h₄:

h₄ = $h_{f} +x_{4} h_{fg}$

= 908.4 + 0.9497(1889.8)

= 908.4 + 1794.7430

= 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅ = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = $s_{f}$ + $x_{6} s_{fg}$

$x_{6} = s_{6} -s_{f} /s_{fg}$

$x_{6}$ = 7.29 - 1.3028 / 6.0562

$x_{6}$ = 0.988

The enthalpy h₆:

h₆ = $h_{f} +x_{6} h_{fg}$

= 417.51 + 0.988 (2257.5)

= 417.51 + 2230.41

h₆ = 2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

$P_{p}$ is found by using:

mass flow rate = m = 1.74 kg/s

Volume of water = v₁ = 0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

$P_{p}$ = ( m ) ( v₁ ) ( p₂ - p₁ )

= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

= (1.74 kg/s) (0.001043 m³/kg) (14900)

= 27.04

$P_{p}$ = 27 kW

Compute heat added $q_{a}$ and heat rejected $q_{r}$ from boiler using computed enthalpies:

$q_{a}$ = ( h₃ - h₂ ) + ( h₅ - h₄ )

= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

= 2726 + 655

= 3381 kJ/kg

$q_{r}$ = h₆ - h₁

= 2648 kJ/kg - 417.5 kJ/kg

= 2232 kJ/kg

Compute net work

W $_{net}$ = $q_{a}$ - $q_{r}$

= 3381 kJ/kg - 2232 kJ/kg

= 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m = 1.74 kg/s

W $_{net}$ = 1150 kJ/kg

P = m * W $_{net}$

= 1.74 kg/s * 1150 kJ/kg

= 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

= 1.74 kg/s * 655

= 1140 kW

Compute Thermal efficiency of this system

μ $_{t}$ = 1 - $q_{r}$ / $q_{a}$

= 1 - 2232 kJ/kg / 3381 kJ/kg

= 1 - 0.6601

= 0.34

= 34%

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The acceleration (in m/s^2) of a linear slider (undergoing rectilinear motion) within a If the machine can be expressed in terms

Inga [223]

Answer:

47.91 sec

Explanation:

it is given that $\alpha =\frac{1}{4v^{2}}$

at t=0 velocity =0 ( as it is given that it is starting from rest )

we have to find time at which velocity will be 3.3 $\frac{m}{sec^{2}}$

we know that $\alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}$

$4v^{2}dv=dt$

integrating both side

$\frac{4v^{3}}{3}=t+c$ ---------------eqn 1

at t=o it is given that v=0 putting these value in eqn 1 c=0

so $\frac{4v^{3}}{3}=t$

when v= 3.3 $\frac{m}{sec^{2}}$

t= $\frac{4}{3}\times 3.3^{3}$

=47.91 sec

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Natasha2012 [34]

Answer:

Codes for each of the problems are explained below

Explanation:

PROBLEM 1 IN C++:

#include<iostream>

using namespace std;

//fib function that calculate nth integer of the fibonacci sequence.

void fib(int n){

// l and r inital fibonacci values for n=1 and n=2;

int l=1,r=1,c;

//if n==1 or n==2 then print 1.

if(n==1 || n==2){

cout << 1;

return;

}

//for loop runs n-2 times and calculates nth integer of fibonacci sequence.

for(int i=0;i<n-2;i++){

c=l+r;

l=r;

r=c;

cout << "(" << i << "," << c << ") ";

}

//prints nth integer of the fibonacci sequence stored in c.

cout << "\n" << c;

}

int main(){

int n; //declared variable n

cin >> n; //inputs n to find nth integer of the fibonacci sequence.

fib(n);//calls function fib to calculate and print fibonacci number.

}

PROBLEM 2 IN PYTHON:

def fib(n):

print("fib({})".format(n), end=' ')

if n <= 1:

return n

else:

return fib(n - 1) + fib(n - 2)

if __name__ == '__main__':

n = int(input())

result = fib(n)

print()

print(result)

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