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Dima020 [189]
3 years ago
9

Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t

echnique is illustrated below where is the input, is the output and the ideal low-pass filter has a cutoff frequency of and a passband gain of 2. Determine . x(t) X(jω) = 0 g(t) = x(t)sin(2000πt) g(t) y(t) 2000π y(t) EENG 311: Signals and Systems Page 7 of 8 × lowpass filter y(t) cos(2000πt) g
Engineering
1 answer:
densk [106]3 years ago
6 0

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

g(t) = x ( t ) sin ( 2000 \pi t )

x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

next we apply Fourier transform

attached below is the remaining part of the solution

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2 years ago
Given a two-dimensional steady inviscid air flow field with no body forces described by the velocity field given below. Assuming
kolbaska11 [484]

Answer:

the pressure gradient in the x direction = -15.48Pa/m

Explanation:

  • The concept of partial differentiation was used in the determination of the expression for u and v.
  • each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.

4 0
3 years ago
A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
diamong [38]

Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}

\frac{3+r}{r}=3.095

thus r=1.43\ cm

Thus Electric field is zero at some distance r=1.43 cm right of q_2

3 0
3 years ago
[4 points] Does the green LED emit light when you connect the banana plug wires across it? What does that say about the threshol
castortr0y [4]

Yes, the green LED emits light when we connect the banana plug wires across it.

This indicates that the threshold voltage is lower for the green LED light than the blue.

<u>Explanation</u>:

A banana plug is named for its resemblance to the shape of a banana. They are wider in the middle of the plug, and narrower at the top and bottom. The banana plug can be easily plugged or unplugged into the ports of speaker or receiver.

LED lights emit photons when it is applied with electrical charge. LED lights are more efficient and last longer than incandescent light bulbs. Green light is commonly provides the calming effect. It is generally used in hyper-pigmentation treatment.

4 0
3 years ago
Design a plate and frame heat exchanger for the following problem:
qwelly [4]

Answer:

See explaination and attachment.

Explanation:

Iteration method is a repetitive method applied until the desired result is achieved.

Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below

x = pi(x)

Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).

Similarly for second, thrid and so on. approximation

x2 = pi(x1)

x3 = pi(x2)

x4 = pi(x3)

xn = pi(xn-1).

please go to attachment for the step by step solution.

8 0
3 years ago
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