Ask question

Ask question

All categories

brilliants [131]

2 years ago

13

How to measure the quality of the output signal in ADC?

1 answer:

hodyreva [135]2 years ago

8 0

Answer:

C. Signal-to-quantization-noise ratio

Explanation:

You might be interested in

The critical resolved shear stress for a metal is 39 MPa. Determine the maximum possible yield strength (in MPa) for a single cr

damaskus [11]

Answer:

78 MPa

Explanation:

Given that the critical resolved shear stress for a metal is 39 MPa, the maximum possible yield strength for a single crystal of this metal is twice the critical resolved shear stress for the metal. The maximum yield yield strength for a single crystal of this metal that is pulled in tension ( $\sigma_y$ ) is given as:

$\sigma_y=2*critical\ resolved\ shear\ stress(\tau_{css})\\\\\sigma_y=2*\tau_{css}\\\\\sigma_y=2*39\\\\\sigma_y=78\ MPa$

4 0

2 years ago

A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the g

olga55 [171]

Answer:

$P_2_{abs}=160\ psia$ (absolute).

Explanation:

Given that

Pressure ratio r

r=8

$r=\dfrac{P_2_{abs}}{P_1_{abs}}$

$8=\dfrac{P_2_{abs}}{P_1_{abs}}$ -----1

P₁(gauge) = 5.5 psig

We know that

Absolute pressure = Atmospheric pressure + Gauge pressure

Given that

Atmospheric pressure = 14.5 lbf/in²

P₁(abs) = 14.5 + 5.5 psia

P₁(abs) =20 psia

Now by putting the values in the above equation 1

$8=\dfrac{P_2_{abs}}{20}$

$P_2_{abs}=8\times 20\ psia$

$P_2_{abs}=160\ psia$

Therefore the exit gas pressure will be 160 psia (absolute).

7 0

3 years ago

Thick fluids such as asphalt and waxes and the pipes in which they flow are often heated in order to reduce the viscosity of the

Gala2k [10]

Answer:

what id the answer

Explanation:

7 0

2 years ago

A disk‑shaped part is to be cast out of aluminum. The diameter of the disk = 401 mm and its thickness = 25 mm. If the mold const

kupik [55]

Answer:

w

Explanation:

3 0

3 years ago

2. A thin vertical panel L = 3 m high and w = 1.5 m wide is thermally insulated on one side and exposed to a solar radiation flu

n200080 [17]

Answer: 383.22K

Explanation:

L = 3m, w = 1.5m

Area A = 3 x 1.5 = 4.5m2

Q' = 750W/m2 (heat from sun) ,

& = 0.87

Q = &Q' = 0. 87x750 = 652.5W/m2

E = QA = 652.5 x 4.5 = 2936.25W

T(sur) = 300K, T(panel) = ?

Using E = §€A(T^4(panel) - T^4(sur))

§ = Stefan constant = 5.7x10^-8

€ = emmisivity = 0.85

2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)

T(panel) = 383.22K

See image for further details.

5 0

3 years ago