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Pavlova-9 [17]
2 years ago
8

A digital filter is given by the following difference equationy[n] = x[n] − x[n − 2] −1/4y[n − 2].(a) Find the transfer function

of the filter.(b) Find the poles and zeros of the filter and sketch them in the z-plane.(c) Is the filter stable? Justify your answer based on the pole-zero plot.(d) Determine the filter type (i.e. HP, LP, BP or BS) based on the pole-zeroplot.
Engineering
1 answer:
slega [8]2 years ago
6 0

Answer:

y(z) = x(z) - x(z) {z}^{ - 2}  -  \frac{1}{4} y(z) {z}^{ - 2}  \\ y(z) + \frac{1}{4} y(z) {z}^{ - 2} = x(z) - x(z) {z}^{ - 2} \\ y(z) (1 + \frac{1}{4}{z}^{ - 2}) = x(z)(1 - {z}^{ - 2}) \\  h(z) = \frac{y(z)}{x(z)}  =  \frac{(1 + \frac{1}{4}{z}^{ - 2})}{(1 - {z}^{ - 2})}

The rest is straightforward...

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The drag force, Fd, imposed by the surrounding air on a
Ad libitum [116K]

Answer:

a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

the power in hp

<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

             P=F.V

     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

           =12954.411*.001818

           =23.551 hp

      P_{a}=2769.29*102.67

           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

3 0
3 years ago
Chad is working on a design that uses the pressure of steam to control a valve in order to increase water pressure in showers. W
Natalija [7]

Answer:

C: Viscosity, the resistance to flow that fluids exhibit

Explanation:

Did it on Edge :)

8 0
2 years ago
Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr
mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

P _1= 15 psia

P _1= 100 psia

We know that work for isothermal process  

W=mRT\ln \dfrac{P_1}{P_2}

Lets take mass is 1 kg.

So work per unit mass

W=RT\ln \dfrac{P_1}{P_2}

We know that for air R=0.287KJ/kg.K

W=RT\ln \dfrac{P_1}{P_2}

W=0.287\times 310.92\ln \dfrac{15}{100}

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

\Delta S=-R\ln \dfrac{P_2}{P_1}

\Delta S=-0.287\ln \dfrac{100}{15}

ΔS = -0.544 KJ/Kg.K

3 0
3 years ago
The article provides information by using a list. What does it list? A. Thanksgiving food B. places where clams can be found C.
Gelneren [198K]

Answer:

C

Explanation:

7 0
3 years ago
Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per
Cloud [144]

Answer:

a) Re_{D} = 111896.745, b) Re_{D} = 1.119\times 10^{-7}

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}

Let assume that density and dynamic viscosity at 25 °C are 997\,\frac{kg}{m^{3}} 0.891\times 10^{-3}\,\frac{kg}{m\cdot s}, respectively. Then:

Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }

Re_{D} = 111896.745

b) The result is:

Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }

Re_{D} = 1.119\times 10^{-7}

6 0
3 years ago
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