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3 years ago

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Design a digital integrator using the impulse invariance method. Find and give a rough sketch of the amplitude response, and com

pare it with that of the ideal integrator. If this integrator is used primarily for integrating audio signals (whose bandwidth is 20 kHz), determine a suitable value for T.

1 answer:

mars1129 [50]3 years ago

8 0

Answer:

50 μsec

Explanation:

See the attached pictures for detailed answer.

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Inessa [10]

Answer:

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2 years ago

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The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s

Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

$lA=\sqrt{0.2^{2}+0.2^{2} }=0.2\sqrt{2}m$

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

$(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B}$ (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

$\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+ \frac{1}{2}k(l_{A}-l_{ul}) ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul}) ^{2}$

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

$\frac{m_{C}v_{B}^{2} }{R}-N_{B}-k(l_{B}-l_{ul})=0$

Replacing values and clearing NB:

NB = 694 N

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3 years ago

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User Location: Inside Viewport Viewport Location: Fresh Viewport in downtown San Fransisco, California Query: [Burger King] Resu

8_murik_8 [283]

Answer:

Steps should you take to rate this result are

Check the pin location.
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3 years ago

In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W

GrogVix [38]

Answer:

The shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

Explanation:

Given:

Temperature $T = 1273.15$ K

Initial Pressure $P_{1} = 1.8$ MPa

Final pressure $P_{2} = 0.1$ MPa

From the table superheated,

$h_{i} = 4635$ $\frac{K J}{Kg}$ and $h_{f} = 2706.54$ $\frac{K J}{Kg}$

Work done by shaft is,

$W = h_{f} - h_{i}$

$W = 2706.54 - 4635$

$W = -1928.46 \frac{kJ}{kg}$

But here efficiency is 0.56,

So work generated per kg is,

Work = $0.56 \times(- 1928.46)$

Work = $-1.3$ $\frac{MJ}{kg}$

Therefore, the shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

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3 years ago

Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 573 oC (this temperature is below the m

inn [45]

Answer:

$2.9\times 10^{-6}$

Explanation:

$Q_s$ = Energy for defect formation = 1.86 eV

T = Temperature = $573^{\circ}\text{C}=573+273.15=846.15\ \text{K}$

k = Boltzmann constant = $8.62\times 10^{-5}\ \text{eV/K}$

The fraction of lattice sites that are Schottky defects is given by

$\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}$

The required ratio is $2.9\times 10^{-6}$ .

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2 years ago