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3 years ago

8

A baseband signal with a bandwidth of 100 kHz and an amplitude range of±1 V is to be transmitted through a channel which is cons

trained to a maximum transmission speed of 2 Mbps. Your task is to design a uni- form quantizer that introduces minimum quantization error. Determine the maximum number of levels L in the uniform quantizer. What is the maximum distortion introduced by the uniform quantizer? Assu Nyquist rate for sampling.

2 answers:

soldi70 [24.7K]3 years ago

7 0

Answer:

Baseband is a signal that has a near zero frequency range. In telecommunication and signal processing, baseband are transmitted without modulation.

Explanati

Verizon [17]3 years ago

6 0

Answer:

The maximum distortion is 0.49mV

Explanation:

See the pictures below

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"Carbon 14 (C-14), a radioactive isotope of carbon, has a half-life of 5730 ± 40 years. Measuring the amount of this isotope lef

igor_vitrenko [27]

Answer:

The age of the bones is approximately 14172 years.

Explanation:

The age of the bones can be determinated using the following decay equation:

$N_{(t)} = N_{0}e^{-\lambda t}$ (1)

<u>Where:</u>

N(t): is the quantity of C-14 at time t

No: is the initial quantity of C-14

λ: is the decay rate

t: is the time

First, we need to find λ:

$\lambda = \frac{ln(2)}{t_{1/2}}$

<u>Where:</u>

t(1/2): is the half-life of C-14 = 5730 y

$\lambda = \frac{ln(2)}{5730 y} = 1.21 \cdot 10^{-04} y^{-1}$

Now, we can calculate the age of the bones by solving equation (1) for t:

$t = \frac{-ln(\frac{N_{(t)}}{N_{0}})}{\lambda}$

We know that the bones have lost 82% of the C-14 they originally contained, so:

$N_{t} = (1 - 0.82)N_{0} = 0.18N_{0}$

$t = \frac{-ln(0.18)}{1.21 \cdot 10^{-04} y^{-1}}$

$t = 14172 y$

Therefore, the age of the bones is approximately 14172 years.

I hope it helps you!

3 0

3 years ago

Consider the problem of oxygen transfer from the interior lung cavity, across the lung tissue, to the network of blood vessels o

aalyn [17]

Answer:

See attached images

8 0

4 years ago

When _____ ,the lithium ions are removed from the_____ and added into the _____

bezimeni [28]

Answer:

b. Discharging; anode; cathode

Explanation:

When discharging , it means the battery is producing a flow electric current, the lithium ions are released from the anode to the cathode which generates the flow of electrons from one side to another. When charging Lithium ions are released by the cathode and received by the anode.

8 0

3 years ago

After testing a model of a fuel-efficient vehicle, scientists build a full-sized vehicle with improved fuel efficiency. Which st

Mazyrski [523]

B evaluate the solution. Sorry if I'm wrong.

4 0

3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

4 years ago