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Pavlova-9 [17]
3 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]3 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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Which step in the engineering design phase is requiring concussion prevention from blows up to 40 mph an example of?
Charra [1.4K]

Answer:

Target your customers

Explanation:

took the test :)

6 0
3 years ago
What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
yanalaym [24]

Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

7 0
3 years ago
Is it better to do blue prints with paper and pencil or a computer program if you’re going to design a house? Why?
Hatshy [7]

Answer:

Computer program

Explanation:

I use Revit and its way better to do all that you can see 2D 3D the measurements and its super easy to use hope this helps

6 0
3 years ago
A packet weighs 40kg in air but when it is totally submerged into a 1mx1m square tank the weight of the packet is only 18kg. How
Irina18 [472]

Answer:

water  rise = 22 mm

Explanation:

weight of packet IN AIR = 40 *9.81 =392.4 N

weight of packet  IN WATER= 18 *9.81 =176.58 N

by Archimedi's principle

difference in weight = weight of displaced water

w_a - w_w = \rho_w v_d g

392.4 - 176.58 = 1000* v_d* 9.81

v_d = 0.022 m^3

v_d = A*H_rise

0.022 =1*H_rise

H_rise = 0.022 m = 22 mm

water  rise = 22 mm

5 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal
Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)

10*2=\sqrt{1+8F^{2} } -1

Solving for F:

F = 7.4162

v=F\sqrt{gd_{1} }

Here, g = gravity = 32.2 ft/s²

v=7.4162*\sqrt{32.2*1} =42.0833ft/s

b) The flow rate:

q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s

c) The Froude number:

v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s

F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345

d) The flow energy dissipated:

E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft

e) The critical depth:

d_{c} =(\frac{(\frac{q}{L})^{2}  }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2}  }{32.2} )^{1/3} =3.8030ft

4 0
3 years ago
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