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lesya692 [45]
3 years ago
10

Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find

the Fourier series coefficient of x(t). Then find its Fourier transform and sketch the magnitude and phase spectra.b)Sketch the magnitude and phase spectra of the output (i.e., |Y(????)|and∡Y(????)) if the system is a low-pass filter with H(????)={1|????|<3????020other????ise, where ????0=2πT0.c)Sketch the magnitude and phase spectra of the output(|Y(????)|and∡Y(????)) if the system is a high-pass filter with H(????)={1|????|>5????020other????ise, where ????0=2πT0.d)Sketch the magnitude and phase spectra of the outputif the system is a filter with H(????)=11+j????.

Engineering
1 answer:
Igoryamba3 years ago
4 0

Answer:

See explaination

Explanation:

The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by

|Y(w)| = |X(w)|

The phase spectrum of y(t) is given by

<Y(w) = -wto + <X(w)

please kindly see attachment for the step by step solution of the given problem.

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Answer:

Georgia Tech is committed to WGAR 53566 THE ANSWER IS JELLY IS KING AND THE JELLY IS KING AND  hope to improve the human condition in your community.

Explanation:

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3 years ago
A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is
Dafna1 [17]

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

v^2-u^2=2as

where,

s = distance covered by the object = 6.93 m

u = initial velocity  = 2.99 m/s

v = final velocity = ?

a = acceleration = 9.8m/s^2

Now put all the given values in the above equation, we get the final velocity of the ball.

v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)

v=12.03m/s

Thus, the final velocity of the ball is, 12.03 m/s

7 0
3 years ago
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

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<h3>What is PMCS?</h3>

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Read more on PMCS here: brainly.com/question/15720250

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Points for free
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Answer:

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Explanation:

3 0
3 years ago
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