(a) Power supplied to the rotorFor this problem, we are given that total power consumption of 72 kW, which means it's the total power consumed by the motor. Power consumed by the motor is a sum of stator copper losses (I2R) and rotor copper losses (I2R).I2R = (76)2 × 0.12 = 692.16 W (resistance between two stator terminals = 0.12)
Hence, power supplied to the rotor is:PR = 72,000 – 2,000 – 1,200 – 692.16PR = 68,107.84 W(b) Rotor I2R lossesRotor copper losses = PR – (2 kW + 1.2 kW) – Pcore – Pw-f-c = 68,107.84 W – (2 kW + 1.2 kW) – 692.16 W = 64,213.68 W(c) Mechanical power supplied to the load, in horsepowerWe know that, Power supplied to the load (mechanical power) = Power developed by the rotorPout = Pm = PR – (core + friction and windage losses) = 68,107.84 W – (2 kW + 1.2 kW) = 64,907.84 WNow, in order to convert watts to horsepower (hp),
we use the following conversion formula:1 hp = 746 WThus, the mechanical power supplied to the load, in horsepower is:Pm in horsepower = Pm/746 = 64,907.84 / 746 = 87.07 hp(d) Torque developed at 1728 r/minWe know that Power developed by the rotor, P = T × 2πN/60where N is rotor speed in rpm, T is torque developed and P is rotor power developed in watts.N = 1728 r/min, and P = 68,107.84 W.T = (P × 60) / (2πN) = (68,107.84 × 60) / (2π × 1728) = 219.6 N-m(e) EfficiencyEfficiency of the motor is given as the ratio of the output power.
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Prove the following logic expression:
i. AB+AB=A
ii. AB+BC+AC=AB+AC
The given logic expression AB + AB = A is true. The given logic expression AB + BC + AC = AB + AC is true.
The following logic expressions can be proven:
Proof of i. AB + AB = A
The given logic expression AB + AB = A is satisfied if we obtain A from both sides. For this, we shall use the Boolean algebraic identities.
Identify the left-hand side (LHS) and the right-hand side (RHS) of the given logic expression: LHS = AB + AB RHS = A
Let us apply Boolean algebraic identities to prove LHS = RHS: LHS = AB + AB= A (A + B) [Using A + A = A] = A.1 [Using A + A' = 1] = A [Using A.1 = A]
Therefore, LHS = RHS = A
Hence, the given logic expression AB + AB = A is true.
Proof of ii. AB + BC + AC = AB + AC
The given logic expression AB + BC + AC = AB + AC is satisfied if we obtain the same expressions on both sides.
For this, we shall use the Boolean algebraic identities.
Identify the LHS and the RHS of the given logic expression: LHS = AB + BC + ACRHS = AB + AC
Let us apply Boolean algebraic identities to prove LHS = RHS: LHS = AB + BC + AC= AB + AC + BC [Using A + BC = A + AC + AB] = AB + AC + B'C [Using B + B' = 1] = AB + AC [Using AB + B'C = AB + AC]
Therefore, LHS = RHS = AB + AC
Hence, the given logic expression AB + BC + AC = AB + AC is true.
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1. Which cable will most likely exhibit higher attenuation to RF signals? A. 0.5 inch diameter coaxial cable 00 B. 0.75 inch diameter coaxial cable 00 ANSWER: 2. What is the main advantage of a folded dipole over a half wave dipole antenna?
Attenuation refers to the reduction of signal strength as it travels through a medium such as a cable. The higher the attenuation, the more the signal is weakened.
The attenuation of a coaxial cable is primarily determined by the diameter of its inner conductor and the dielectric material between the inner conductor and outer shield. The attenuation of a cable increases as its diameter decreases, and therefore, the 0.5-inch diameter coaxial cable will exhibit higher attenuation to RF signals than the 0.75-inch diameter coaxial cable.
A folded dipole antenna is a type of antenna that is similar to a half-wave dipole antenna but has a folded section of wire in the middle. This folding increases the overall length of the antenna, which in turn increases its bandwidth.The bandwidth of an antenna refers to the range of frequencies over which it can operate effectively.
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In the small-signal equivalent circuit, the DC current source is replaced by a short circuit. Select one: True False Question 10 Not yet answered Marked out of \( 4.00 \) An npn transistor operates in
The statement is true. The small-signal equivalent circuit is used to determine the characteristics of a small-signal at the output without changing any parameters of the original circuit.
This model is used to simplify the analysis of the circuits that contain transistors. This equivalent circuit provides a simplified version of the circuit containing only those components required for small-signal analysis. In this, DC sources are replaced with short circuits and resistors are replaced with their small-signal equivalents.Explanation:This is true that in the small-signal equivalent circuit, the DC current source is replaced by a short circuit. The small-signal equivalent circuit is a simplified version of the circuit containing only those components required for small-signal analysis. This equivalent circuit provides a simplified version of the circuit that contains only those components required for small-signal analysis.
The equivalent circuit is formed by shorting out the DC voltage source and replacing the transistor with its small-signal model. The small-signal model is formed by analyzing the circuit for small variations in current and voltage from the bias point, which is the point at which the transistor is conducting just the right amount of current to produce the desired output voltage.
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A balanced delta – connected load has a phase current IAC = 10∠30° A.
a. Determine all line currents assuming that the circuit operates in the positive phase sequence.
b. Calculate the load impedance if the line voltage is VAB= 110∠0° V.
a) The line current that the circuit operates on is 10∠270°. b) The load impedance is 11∠330°.
Given data; A balanced delta – connected load has a phase current IAC = 10∠30° A.
The formula for calculating phase current (Iph) is:
Iph = IAC
If IAC = 10∠30°, then the phase current is:
Iph = 10∠30°.
a) Since the circuit is balanced, the line currents can be calculated by using the following formula;
Iab = Ica = Iph
Ibc = Iac = Iph∠-120°
Ica = Iab = 10∠30°
Ibc = 10∠(30°-120°)=10∠-90° = 10∠270°.
b) The formula for calculating line voltage (VL) is:
VL = √3 × VphIf
Vab = 110∠0°, then the line voltage is:
VL = √3 × Vph= √3 × 110 = 190.5V.
If Iab = 10∠30° and Vab = 110∠0°, then the load impedance can be calculated using the following formula;
Zab = Vab/Iab
Zab = 110∠0° / 10∠30°= 11∠-30° = 11∠330°
The load impedance is 11∠330°.
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The open-loop transfer function of a unity feedback system is Ke-0.1s G(s) = s(1 + 0.1s)(1+s) By use of Bode plot and/or Nichols chart, determine the following: (a) The value of K so that the gain margin of the system is 20 db. (b) The value of K so that the phase margin of the system is 60 deg. (c) The value of K so that resonant peak M, of the system is 1 db. What are the corresponding values of w, and a? (d) The value of K so that the bandwidth a of the system is 1.5 rad/sec.
The value of K so that the gain margin of the system is 20 dB is approximately 5.623.
To determine the value of K for a gain margin of 20 dB, we need to analyze the Bode plot or Nichols chart of the system. The gain margin represents the amount of gain that can be added to the system before it reaches instability. In other words, it quantifies the system's robustness against gain variations.
By examining the Bode plot or Nichols chart, we can find the frequency at which the magnitude of the open-loop transfer function is 0 dB (unity gain). At this frequency, the phase margin will be zero, and the system will be at the verge of instability.
To achieve a gain margin of 20 dB, we need to find the value of K that results in a magnitude of 0 dB at a certain frequency. By evaluating the transfer function Ke^(-0.1s)G(s) = s(1 + 0.1s)(1+s), we can determine that K ≈ 5.623 corresponds to the desired gain margin of 20 dB.
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Design a PI controller to drive the step response error to zero for the unity feedback system if G(s)= K/ (s+1)^2(s+10)
The system operates with a damping ratio of 0.6. Compare the performance of uncompensated and compensated systems
The comparison of the two step responses, it is observed that the compensated system has a faster rise time, a lower settling time, and zero steady-state error as compared to the uncompensated system.
In order to design a PI controller to drive the step response error to zero for the unity feedback system if G(s)= K/ (s+1)^2(s+10), the following steps must be followed:
Step 1: Find the error of the system in question.
For unity feedback system, the error is given by:
1/(1+G(s)).
Thus, the error is given as:
1/(1+G(s)) = 1/(1+ K/ (s+1)^2(s+10)) = (s+1)^2(s+10)/(s+1)^2(s+10) + K = (s+1)^2(s+10)/(s+1)^2(s+10) + K(s+1)^2(s+10)/(s+1)^2(s+10) = K(s+1)^2(s+10)+ (s+1)^2(s+10)/(s+1)^2(s+10) = (K(s+1)^2(s+10) + 1) / (s+1)^2(s+10)
Step 2: Determine the closed-loop transfer function, T(s) using the PI controller.
Since a PI controller is being designed, the transfer function is given as:
C(s) = Kp + Ki/sThe closed-loop transfer function T(s) is given as:T(s) = C(s)G(s) / (1+C(s)G(s))T(s) = [Kp + Ki/s]K/ (s+1)^2(s+10)[1/ (1 + [Kp + Ki/s]K/ (s+1)^2(s+10))]T(s) = [Kp + Ki/s]K/ (s+1)^2(s+10) + [Kp + Ki/s] / (s+1)^2(s+10) + [Kp + Ki/s]/(s+1)^2(s+10)[Kp + Ki/s]K/ (s+1)^2(s+10) + [Kp + Ki/s] + 1 = 0
Step 3: Determine the values of Kp and Ki using the given damping ratio. The damping ratio ζ and the natural frequency ωn are given as:
ζ = 0.6 = Kp / (2*ωn) => ωn = Kp / (2*ζ)
The natural frequency is given as:ωn = sqrt(Ki/K)Also, the steady-state error constant, Kp is given as:
Kp = lim(s → 0) sT(s) = Kp K / (1 + Kp K)
Thus, substituting the values of Kp and ωn into the transfer function, we have:
[Kp + Ki/s]K/ (s+1)^2(s+10) + [Kp + Ki/s] / (s+1)^2(s+10) + [Kp + Ki/s]/(s+1)^2(s+10)[Kp + Ki/s]K/ (s+1)^2(s+10) + [Kp + Ki/s] + 1 = 0[Kp + Ki/s] = s^3 + 20s^2 + 101s + K - Ki
Kp = lim(s → 0) sT(s) = Kp K / (1 + Kp K) => Kp = 0.1
The natural frequency ωn = Kp / (2*ζ) = 0.0833Ki = Kωn^2 = 5.2188
Comparing the performance of uncompensated and compensated systems.
The uncompensated transfer function, G(s) = K / (s+1)^2(s+10)
The compensated transfer function,
T(s) = [Kp + Ki/s]K / (s+1)^2(s+10) + [Kp + Ki/s] / (s+1)^2(s+10) + [Kp + Ki/s]/(s+1)^2(s+10)[Kp + Ki/s]K/ (s+1)^2(s+10) + [Kp + Ki/s] + 1 = 0
From the comparison of the two step responses, it is observed that the compensated system has a faster rise time, a lower settling time, and zero steady-state error as compared to the uncompensated system.
This means that the compensated system provides better performance as compared to the uncompensated system.
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in this context a verbal insight problem about a woman described as a ""muscular figure with a deep voice and a motorcycle"" would likely:
In this context, a verbal insight problem about a woman described as a "muscular figure with a deep voice and a motorcycle" would likely pertain to the challenge of accurately forming a mental image or perception of the woman based on the given description.
How would you reconcile the contrasting characteristics of a woman described as a "muscular figure with a deep voice and a motorcycle"?When presented with a verbal insight problem describing a woman as a "muscular figure with a deep voice and a motorcycle," the nature of the problem is likely to involve mental visualization and interpretation. The challenge lies in creating an accurate mental image or perception of the woman based on the provided description.
This may require mental flexibility and creativity to reconcile the seemingly contrasting characteristics mentioned, such as a muscular figure and a deep voice, with the presence of a motorcycle.
The problem may prompt individuals to explore various possibilities and use their cognitive abilities to form a coherent understanding of the described woman.
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An office with dimensions of 20 m (L) x 15 m (W) x 4 m (H) has 50 staff. A ventilation system supplying outdoor air to this office at a designed flow rate of 10 L/s/person. The outdoor CO₂ concentration is 300 ppm. The initial concentration of CO₂ in the office is 350 ppm and the CO₂ emission rate from each person is 0.01 L/s respectively. Determine the CO₂ concentration in ppm in the office at the end of the first 3 hours if it is full house.
The CO₂ concentration in the office at the end of the first 3 hours, considering a full house, would be approximately 540 ppm.
To determine the CO₂ concentration in the office after 3 hours, we need to consider the rate at which outdoor air is supplied, the CO₂ emission rate from each person, and the initial CO₂ concentration.
Calculate the total CO₂ emitted by all staff members.
CO₂ emission rate per person = 0.01 L/s
Number of staff members = 50
Total CO₂ emitted per second = CO₂ emission rate per person * Number of staff members
Total CO₂ emitted per second = 0.01 L/s * 50
Total CO₂ emitted per second = 0.5 L/s
Calculate the volume of the office.
Length (L) = 20 m
Width (W) = 15 m
Height (H) = 4 m
Volume of the office = Length * Width * Height
Volume of the office = 20 m * 15 m * 4 m
Volume of the office = 1200 m³
Step 3: Calculate the CO₂ concentration at the end of 3 hours.
Designed flow rate of outdoor air = 10 L/s/person
Number of staff members = 50
Total outdoor air supplied per second = Designed flow rate of outdoor air * Number of staff members
Total outdoor air supplied per second = 10 L/s/person * 50
Total outdoor air supplied per second = 500 L/s
CO₂ concentration change per second = (CO₂ emitted per second - CO₂ removed per second) / Volume of the office
CO₂ concentration change per second = (0.5 L/s - 500 L/s) / 1200 m³
CO₂ concentration change per second = -499.5 L/s / 1200 m³
CO₂ concentration change per hour = CO₂ concentration change per second * 3600 seconds
CO₂ concentration change per hour = -499.5 L/s / 1200 m³ * 3600 s/h
CO₂ concentration change per hour = -1498500 L/h / 1200 m³
CO₂ concentration at the end of 3 hours = Initial CO₂ concentration + CO₂ concentration change per hour * 3 hours
CO₂ concentration at the end of 3 hours = 350 ppm + (-1498500 L/h / 1200 m³) * 3 h
CO₂ concentration at the end of 3 hours ≈ 540 ppm
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which part of a synchronizer is splined to a shaft
The synchronizer sleeve or hub is splined to a shaft in a synchronization mechanism.
What component in a synchronization mechanism is typically splined to a shaft?A synchronizer is a device used in manual transmissions to enable smooth shifting between gears. It consists of several components, including the synchronizer sleeve or synchronizer hub, gear teeth, and blocking rings. The synchronizer sleeve or hub is a cylindrical component that slides along the length of a shaft, and it is responsible for engaging and disengaging the gear to be selected.
The synchronizer sleeve or hub is often splined to the shaft, which means it has grooves or ridges on its outer surface that match corresponding splines on the shaft. This spline connection allows the synchronizer sleeve or hub to rotate with the shaft while still being able to slide back and forth along its length. By splining the synchronizer component to the shaft, torque can be transmitted efficiently and synchronized with the rotational speed of the shaft, facilitating smooth gear engagement during shifting operations.
The precise design and configuration of synchronizers can vary depending on the specific transmission system and manufacturer. However, the splined connection between the synchronizer sleeve or hub and the shaft is a common feature in synchronizer designs to ensure effective and reliable gear shifting.
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iv. Draw the complete impulse generator circuit indicating values for each component.
To draw the complete impulse generator circuit indicating values for each component, we need to have a clear idea of what an impulse generator is and the components that make up the circuit.
The impulse generator circuit is a device that creates a high-voltage, short-duration electrical discharge that can be used for various purposes such as electrical testing or ignition in internal combustion engines. The circuit is made up of the following components:
1. Charging source (usually a capacitor)
2. Switching device (such as a spark gap)
3. Load (such as a spark plug)When the circuit is charged to a sufficient voltage, the switching device is triggered, causing the discharge to flow through the load. The value of each component depends on the desired output voltage and the load that the generator will be used to power.
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An LTI system is defined by its unit impulse response \( h(t)=u(t-2) \). If the input is \( x(t)=u(t+1) \) then the output \( y(t) \) is: Select one: None of these \( (t+1) u(t+1) \) \( (t-1) u(t-1) \
The unit impulse response of an LTI system is given as follows: h(t) = u(t-2), and the input is given as x(t) = u(t+1). We have to determine the output of the system, which is represented as y(t).
Solution:
When a system is linear and time-invariant (LTI), convolution can be used to calculate the output. The output of the LTI system can be calculated as:
y(t) = h(t) * x(t)
= ∫₋ₒ₊ₒ h(τ) x(t - τ) dτ
where h(τ) is the unit impulse response of the system.
Let us evaluate the above equation using the given values:
y(t) = ∫₋ₒ₊ₒ h(τ) x(t - τ) dτ
= ∫₋ₒ₊ₒ u(τ - 2) u(t - τ + 1) dτ
Here, we can note that the integration limits can be changed as follows:
∫₋ₒ₊ₒ u(τ - 2) u(t - τ + 1) dτ = ∫₋ₒ₊(t-1) u(τ - 2) u(t - τ + 1) dτ
Therefore, we can solve the above integral by splitting it into different intervals.
y(t) = ∫₋ₒ₊(t-1) u(τ - 2) u(t - τ + 1) dτ
= ∫₂₋ₒ u(τ - 2) u(t - τ + 1) dτ + ∫₋ₒ₊(t-1) u(τ - 2) u(t - τ + 1) dτ + ∫(t-1)₊ₒ u(τ - 2) u(t - τ + 1) dτ
We can evaluate the above three integrals separately.
Integral 1:
∫₂₋ₒ u(τ - 2) u(t - τ + 1) dτ = ∫₂₋ₒ 0 * u(t - τ + 1) dτ = 0
Integral 2:
∫₂₋ₒ u(τ - 2) u(t - τ + 1) dτ = ∫₂₋ₒ 1 * 1 dτ = t - 3
Integral 3:
∫(t-1)₊ₒ u(τ - 2) u(t - τ + 1) dτ = ∫(t-1)₊(t+2) u(τ - 2) u(t - τ + 1) dτ = ∫(t-1)₊(t+2) 0 * u(t - τ + 1) dτ = 0
The output of the system is given as:
y(t) = ∫₋ₒ₊ₒ u(τ - 2) u(t - τ + 1) dτ
= ∫₂₋ₒ u(τ - 2) u(t - τ + 1) dτ + ∫₋ₒ₊(t-1) u(τ - 2) u(t - τ + 1) dτ + ∫(t-1)₊ₒ u(τ - 2) u(t - τ + 1) dτ
= 0 + (t - 3)
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to prevent unwanted ground loops, instrumentation cable shielding is ___.
To prevent unwanted ground loops, instrumentation cable shielding is typically grounded at one end to minimize electromagnetic interference and maintain signal integrity by providing a low-impedance path for induced currents to flow.
To prevent unwanted ground loops, instrumentation cable shielding is typically grounded at one end. Grounding the shielding helps to minimize electromagnetic interference (EMI) by providing a low-impedance path for the induced currents to flow. When the shielding is grounded at only one end, it helps to eliminate potential differences between equipment and reduces the chances of ground loops forming.
Ground loops occur when there are multiple grounding points at different potentials, leading to circulating currents and unwanted noise in the system. By grounding the shielding at one end, any induced currents are directed away from the signal conductors and safely discharged to a single reference point, preventing interference and maintaining signal integrity.
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What is the range of the modulus of elasticity (GPa) and strength (MPa) ← of unidirectional composite of Carbon/Epoxy and Aramid/Epoxy, respectively?
The modulus of elasticity and strength of composites depend on many factors, including the orientation of the fibers in the composite.
The modulus of elasticity and strength of unidirectional composites of Carbon/Epoxy and Aramid/Epoxy, respectively are given as follows:
The modulus of elasticity of unidirectional Carbon/Epoxy composites range from 100 G Pa to 290 G Pa.
The modulus of elasticity of Aramid/Epoxy composites range from 70 G Pa to 110 G Pa.
The strength of unidirectional Carbon/Epoxy composites range from 500 MPa to 3000 MPa, while the strength of unidirectional Aramid/Epoxy composites range from 300 MPa to 2000 MPa.
These values may vary depending on the manufacturing process, the quality of the raw materials used, and other factors.
The values above are just a general guide to the range of modulus of elasticity and strength for these two types of composites.
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The inspector should establish a ____ method for conducting inspections in order to better identify unsafe conditions or behaviors. (702)
The inspector should establish a standardized method for conducting inspections in order to better identify unsafe conditions or behaviors.
What should the inspector establish to better identify unsafe conditions or behaviors during inspections?By implementing a consistent and systematic approach, the inspector can ensure that all relevant areas are thoroughly examined and evaluated.
This method can include predefined checklists, protocols, or procedures that guide the inspector's observations and assessments.
Having a standardized method helps to ensure that inspections are conducted consistently across different locations or situations, reducing the risk of overlooking potential hazards.
It also allows for easier comparison and analysis of inspection results over time, enabling the identification of patterns or trends that may indicate recurring safety issues.
Ultimately, establishing a standardized inspection method enhances the inspector's ability to identify and address unsafe conditions or behaviors effectively.
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Question 4 Rectifier type instruments generally use a PMMC movement along with a rectifier arrangement. Silicon diodes are preferred because of their low reverse and high forward current ratings. (a) Briefly describe about half-wave rectifier voltmeter and provide the required circuit design for half-wave rectifier voltmeter. (b) A basic D'Arsonval movement with a full scale deflection of 50 μA and internal resistance of 500 Q is used as a voltmeter. Determine the value of the multiplier resistance needed to measure a voltage range of 0-10 V. (c) Explain about the full-wave rectifier with the illustration of the diagram. [25 Mark]
Rectifier type instruments use a PMMC movement along with a rectifier arrangement. Silicon diodes are preferred due to their low reverse and high forward current ratings.
(a) A half-wave rectifier voltmeter is a type of instrument that measures the average value of a voltage by rectifying it to a unidirectional current and then converting it back to a voltage using a PMMC movement. The circuit design for a half-wave rectifier voltmeter involves connecting a series combination of a diode and a load resistor in parallel with the PMMC movement. The diode rectifies the input voltage, allowing only the positive half of the waveform to pass through. The PMMC movement, being sensitive to current, converts this rectified current into a corresponding voltage reading on the scale.
(b) To determine the value of the multiplier resistance needed for a voltage range of 0-10V using a D'Arsonval movement with a full-scale deflection of 50 μA and internal resistance of 500 Ω, we can use Ohm's law. The desired full-scale deflection current is 50 μA, and the maximum voltage to be measured is 10V. Using Ohm's law (V = I * R), we can rearrange the formula to solve for the multiplier resistance (Rm):
Rm = (Vmax / Ifsd) - Rint
Rm = (10V / 50 μA) - 500 Ω
Rm = 200 kΩ - 500 Ω
Rm = 199.5 kΩ
Therefore, a multiplier resistance of 199.5 kΩ is needed to measure the 0-10V voltage range.
(c) A full-wave rectifier is a circuit that converts an alternating current (AC) input into a unidirectional current using a bridge rectifier arrangement. It employs four diodes arranged in a bridge configuration to rectify both the positive and negative halves of the AC waveform. The AC input is applied to the bridge rectifier, and the diodes conduct alternately to convert the AC input into a pulsating DC output. This pulsating DC is then further filtered to obtain a smoother DC output.
The full-wave rectifier circuit eliminates the need for a center-tapped transformer, which is required in a half-wave rectifier circuit. It provides a more efficient utilization of the input power, resulting in a higher output voltage and reduced ripple. The full-wave rectifier is commonly used in applications where a more stable and smoother DC voltage is required, such as in power supplies and electronic devices.
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When the input to an LTI discrete-time system is a:[n] = 28[n. — 2), the output is y[n] = S[n − 1] + 8[n − 3). - (a). Find the impulse response h[n] of this system. (b). Is this system causal and stable? (c). Find the frequency response H(e) of the system. (d). Find the output of the system when the input is a[n] denotes the unit step sequence. = u[n], where u[n]
(a). Impulse response of the systemThe impulse response of the LTI discrete-time system is obtained by using the fact that when the input to an LTI discrete-time system is a unit impulse, the output is the impulse response, h[n]. Given, the input to the system is a[n] = 28[n - 2], the output is y[n] = s[n - 1] + 8[n - 3].
So, the input is written as the sum of shifted unit impulses as follows:a[n] = 28[n - 2] = 28δ[n - 2] + 28δ[n - 3] + 28δ[n - 4] + ...Thus, the output can be written as the sum of scaled and shifted impulse responses as follows:y[n] = s[n - 1] + 8[n - 3]= h[n - 1] + 8h[n - 3]Applying z-transform on both sides, we get,Y(z) = S(z) + 8z⁻³H(z)And the input can be expressed as a sum of shifted impulse responses as follows:A(z) = 28z⁻² + 28z⁻³ + 28z⁻⁴ + ...Therefore, we can write the output asY(z) = (28z⁻² + 28z⁻³ + 28z⁻⁴ + ...) H(z) + S(z)And
hence,H(z) = [Y(z) - S(z)] / [28z⁻² + 28z⁻³ + 28z⁻⁴ + ... + 28z⁻ⁿ + ...]From the given output expression, we see that the impulse response h[n] is h[n] = δ[n - 1] + 8δ[n - 3].(b). Causality of the systemA system is said to be causal if the output of the system does not depend on future values of the input, that is, the system does not "anticipate" the future inputs. From the impulse response expression, we see that the output at any time instant n depends only on the present and past input values, that is, a[n], a[n - 1], a[n - 2] and so on.
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Draw the root locus of the control system having open loop transfer function
G(s)H(s) = K(s+x) / s (s+4)(s+3)
The Root Locus plot is a method of finding the trajectories of the closed-loop poles of a system in the s-plane, given the system’s open-loop transfer function. In control system engineering, the Root Locus technique plays an essential role.
Let's draw the root locus of the control system having the open loop transfer function G(s)H(s) = K(s + x) / s (s + 4) (s + 3).Solution: Given that the open-loop transfer function is G(s)H(s) = K(s + x) / s (s + 4) (s + 3).The general transfer function of the control system is G(s) / (1 + G(s)H(s)).Let us consider the above open loop transfer function as the feedforward path of the control system, i.e., G(s) H(s).Therefore, the closed-loop transfer function T(s) will be: T(s) = G(s) H(s) / [1 + G(s) H(s)] Substituting G(s) H(s) in the above equation, we get: T(s) = K(s + x) / s (s + 4) (s + 3) + K(s + x)T(s) = K (s + x) / [s (s + 4) (s + 3) + K (s + x)]s (s + 4) (s + 3) + K (s + x) = 0s³ + (4 + 3K) s² + (3Kx + 4x + 12) s + 12x = 0Let us consider the denominator of the above equation as: D(s) = s³ + (4 + 3K) s² + (3Kx + 4x + 12) s + 12x.Now, the angle criterion of the Root Locus method can be applied. The necessary and sufficient conditions for a point to lie on the Root Locus are given as follows:1. The number of roots to the right of the point is equal to the number of poles of the system to the right of that point.2. The sum of the angles of departure of the Root Locus from the real axis, and the angles of arrival at a point is an odd multiple of 180°.
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In Bilinear Transformation Method, determine the ωo for a
second-order digital band-pass Butterworth filter with the
following specifications: upper cutoff frequency of 3600 Hz, lower
cutoff frequenc
Bilinear Transformation Method is a mathematical technique used for converting analog filters to their digital form. It preserves the location of the poles and zeros of the original analog filter in the digital domain.
To determine the ωo for a second-order digital band-pass Butterworth filter, with the given specifications: upper cutoff frequency of 3600 Hz, lower cutoff frequency, we will follow the following steps:
Step 1: Determine the analog filter transfer functionH (s) = K / (s^2 + ωo Qs + ωo^2 )whereK = gain constantωo = center frequencyQ = quality factor.
Step 2: Determine the transfer function for the low-pass filterHLP (s) = 1 / (s^2 + ωo s/Q + ωo^2 )
Step 3: Determine the transfer function for the band-pass filterHBP (s) = [s / (ωo Q)] / [(s/ωo)^2 + (s/ωoQ) + 1]
Step 4: Determine the digital filter transfer functionH (z) = HLP (s)|s = 2/T[(1 - z^-1) / (1 + z^-1)]^2
HBP (s)|s = 2/T[(1 - z^-1) / (1 + z^-1)] whereT = sampling periodThe sampling frequency Fs = 2 × 3600 Hz = 7200 HzSampling period T = 1/Fs = 1/7200 Hz = 1.389 × 10^-4 secondsSubstituting the values in the above formula we get H (z) = (0.00005806 z^2 + 0.0001161 z + 0.00005806) / (1.67 z^2 - 1.944 z + 0.7031) the center frequency ωo = 2π × 3600 Hz = 22619.47 rad/sThis is a second-order digital band-pass Butterworth filter with a center frequency of 22619.47 rad/s.
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Using MATLAB, draw Butterworth, Chebyshev, and Bessel filter frequency plots with the following:
n=512; % # of time samples
h=0.02; %sample interval
t=h*[0:n-1]; %time range
f=[0:n-1]/n/h; %frequency range
x=3*sin(2*pi*t)+sin(2*pi*t)+cos(2*pi**0.2*t); %signal
The parameters used in the filter design (e.g., filter order, cutoff frequency) are arbitrary in this example, and you may need to adjust them according to your specific requirements.
Here's the MATLAB code to draw Butterworth, Chebyshev, and Bessel filter frequency plots for the given parameters and signal:
```matlab
n = 512; % # of time samples
h = 0.02; % sample interval
t = h * [0:n-1]; % time range
f = [0:n-1] / (n * h); % frequency range
x = 3 * sin(2 * pi * t) + sin(2 * pi * t) + cos(2 * pi * 0.2 * t); % signal
% Butterworth filter
[butter_b, butter_a] = butter(4, 0.2, 'low');
butter_filtered = filter(butter_b, butter_a, x);
% Chebyshev filter
[cheby_b, cheby_a] = cheby1(4, 0.5, 0.2, 'low');
cheby_filtered = filter(cheby_b, cheby_a, x);
% Bessel filter
[bessel_b, bessel_a] = besself(4, 0.2, 'low');
bessel_filtered = filter(bessel_b, bessel_a, x);
% Plotting the frequency response
figure;
subplot(2, 2, 1);
plot(f, abs(fft(x)));
title('Original Signal');
xlabel('Frequency');
ylabel('Magnitude');
subplot(2, 2, 2);
plot(f, abs(fft(butter_filtered)));
title('Butterworth Filter');
xlabel('Frequency');
ylabel('Magnitude');
subplot(2, 2, 3);
plot(f, abs(fft(cheby_filtered)));
title('Chebyshev Filter');
xlabel('Frequency');
ylabel('Magnitude');
subplot(2, 2, 4);
plot(f, abs(fft(bessel_filtered)));
title('Bessel Filter');
xlabel('Frequency');
ylabel('Magnitude');
```
This code utilizes MATLAB's built-in filter design functions to design Butterworth, Chebyshev, and Bessel filters. The signal `x` is then filtered using each of these filters, and the frequency response is plotted using the Fast Fourier Transform (FFT). The resulting frequency plots for the original signal and each filtered signal are displayed in separate subplots.
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mar Instructions Show You have been asked to design a commercial website. Users will be able to browse or search for music and then download it to hard disk and any associated devices such as MP3 players. Briefly explain how you would identify the potential end users of such service, and then explain how you would conduct a summative evaluation for these users once the system had been built. 710 French Spanish B IEE words characters
To identify the potential end users of the commercial website for music downloading, you can follow these steps:
Market Research: Conduct market research to understand the target audience for the music service. This can include demographics, psychographics, and preferences related to music genres, devices, and technology usage.
User Surveys and Interviews: Create surveys and conduct interviews with potential users to gather their feedback and understand their needs and expectations from a music downloading service. Ask questions about their music preferences, downloading habits, preferred devices, and any pain points they have experienced with existing platforms.
User Personas: Based on the information collected from market research, surveys, and interviews, create user personas. User personas are fictional representations of different types of users who would use the music downloading service. Each persona should capture the characteristics, goals, motivations, and needs of a specific user segment.
User Testing: Conduct user testing sessions with prototypes or early versions of the website. Invite potential users from the identified user personas to interact with the system and perform common tasks such as browsing, searching, and downloading music. Observe their behavior, collect feedback, and note any issues they encounter or suggestions they provide.
Beta Testing: Release a beta version of the website to a limited group of potential end users. Encourage them to use the service and provide feedback on their experience. This can be done through feedback forms, surveys, or even user forums where they can share their thoughts, report issues, and suggest improvements.
Once the system has been built, a summative evaluation can be conducted to assess its overall effectiveness and gather insights for further improvements. Here's an approach for conducting a summative evaluation:
Define Evaluation Goals: Determine the specific goals and metrics you want to evaluate. These can include user satisfaction, ease of use, efficiency in finding and downloading music, and overall system performance.
Usability Testing: Conduct usability testing sessions with representative users. Provide them with specific tasks to perform on the website, such as searching for a specific song or downloading a playlist. Observe their interactions, collect quantitative and qualitative data, and identify any usability issues or bottlenecks.
Performance Testing: Evaluate the website's performance under different scenarios, such as high traffic or simultaneous downloads. Measure the system's response time, stability, and ability to handle user demands without significant delays or errors.
Surveys and Questionnaires: Administer surveys or questionnaires to a larger sample of users. Include questions about their overall satisfaction with the website, ease of use, quality of downloaded music, and any suggestions for improvements. Use Likert scales, open-ended questions, and structured response options to gather both quantitative and qualitative data.
Analyze Data and Feedback: Analyze the data collected from usability testing, performance testing, and surveys. Identify patterns, trends, and common themes in user feedback. Prioritize issues or areas of improvement based on the severity of impact and user feedback.
Iterate and Improve: Based on the findings from the summative evaluation, make necessary improvements to the website's design, functionality, performance, and user experience. Incorporate user feedback to enhance the system and address any identified issues or pain points.
By following these steps, you can identify potential end users and gather valuable feedback through user research, testing, and evaluation. This iterative approach helps ensure that the commercial website meets the needs and expectations of the target audience, providing an optimal music downloading experience.
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A (220+XY) Volts, 4-pole, Y-connected, three-phase induction motor has the following test data: Open load: Line current =2 A and input power =300 W. Blocked rotor: Current absorbed =(20+X)A and input power is =(700+YX)W (while the applied voltage is 30 Volts). Consider the friction and windage losses =(50−X)W, resistance between any two lines =0.2XΩ and compute the following equivalent circuit parameters of the motor:
An induction motor is a type of electric motor that converts electric energy into mechanical energy through the process of electromagnetic induction.
It works by applying a rotating magnetic field to the rotor, which causes it to spin.
The parameters of an induction motor can be determined by conducting various tests on it.
In this case, the test data for a three-phase induction motor is provided, and we need to calculate its equivalent circuit parameters.
The given test data is as follows:
Open load:
Line current = 2 A and
input power = 300 W
Blocked rotor:
Current absorbed = (20+X) A and
input power is = (700+YX) W (while the applied voltage is 30 Volts)
Friction and windage losses = (50−X) W
Resistance between any two lines = 0.2XΩ
Equivalent Circuit Parameters:
The equivalent circuit of a three-phase induction motor consists of three components:
resistance (R), reactance (X), and magnetizing reactance (Xm).
Rotor resistance (R2):
The rotor resistance is given by the ratio of blocked rotor input power to the square of the blocked rotor current.
R2 = Blocked rotor input power / (Blocked rotor current)^2
R2 = (700+YX) / (20+X)^2
Reactance (X2):
The reactance is given by the difference between the total impedance and the rotor resistance.
X2 = √[(Open circuit input power / (3*Open circuit current)^2) - R2^2]
X2 = √[(300 / (3*2)^2) - (700+YX) / (20+X)^2]^0.5
Magnetizing reactance (Xm):
The magnetizing reactance is the ratio of the open-circuit voltage to the no-load current.
Xm = Open circuit voltage / (3*Open circuit current)
Xm = (220+XY) / (3*2)
Therefore, the equivalent circuit parameters of the motor are Rotor resistance
(R2) = (700+YX) / (20+X)^2,
Reactance (X2) = √[(300 / (3*2)^2) - (700+YX) / (20+X)^2]^0.5,
and
Magnetizing reactance (X m) = (220+XY) / (3*2).
The answer has 193 words.
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Frequency modulated (FM) signal XFM (t) = 5.cos(1082zt + cos (4710³ t)) is given. (a) Find the carrier frequency (fe) (b) Find the modulation index (B) (c) Find the frequency (instantaneous frequency) of the FM signal (d) Find the message signal (m(t)).
The carrier frequency (fc) is given by:
[tex]fc = 1082z[/tex]Therefore,[tex]fc = 1082z = 1082 × 10 = 10820Hz[/tex](b) The modulation index (B) is given by:B = (maximum frequency deviation)/ message signal frequency.
The maximum frequency deviation (Δf) is given by:
[tex]Δf = kf[/tex] (maximum message amplitude)[tex]kf = (Δf)[/tex] / (maximum message amplitude)From the expression of the FM signal, we can see that the maximum amplitude is 5 Hence,[tex]Δf = 1/2(4710³) = 1.18 MHzkf = Δf / maximum message amplitudekf = 1.18 × 10⁶ / 5 = 2.36 × 10⁵B = 2.36 × 10⁵[/tex]
The instantaneous frequency of the FM signal (f) is given by
[tex]:f = fc + kfm(t)Where k = 2πk[/tex]
[tex]The message signal (m(t)) = cos(4710³ t)[/tex] Hence, [tex]kf = 2π × 2.36 × 10⁵[/tex]
Therefore, [tex]f = fc + kfm(t)f = 10820 + 2π × 2.36 × 10⁵ cos(4710³ t)Hz.[/tex]
To find the message signal (m(t)) , we can write the FM signal as:
[tex]XFM (t) = Acos(2πfct + 2πkf ∫m(t)dt)\\Let Y(t) = 2πkf ∫m(t)dt\\XFM (t) = Acos(2πfct + Y(t))[/tex]
Differentiating with respect to time, we get
[tex]:dXFM (t) / dt = - 2πAfcsin(2πfct + Y(t)) + 2πAkf (dm(t) / dt)Cos(2πfct + Y(t))[/tex]
Equating it to the given FM signal, we get:
[tex]dm(t) / dt = - sin(4710³ t) / 2πkf[/tex]
The message signal (m(t)) can be obtained by integrating dm(t) / dt over time:
[tex]m(t) = - 1 / (2πkf) cos(4710³ t) + constan[/tex]
tPutting the initial condition that message signal has zero amplitude at
[tex]t = 0,m(t)\\ = - 1 / (2πkf) cos(4710³ t)[/tex]
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Question 6 2 pts A three phase SCR rectifier supplies a resistive load with the parameters R = 2002. The rectifier is fed from a 415V (rms) 50Hz three phase AC source, and the SCR firing angle is set to 70°. Calculate the average voltage that is supplied to the load.
Given parameters: R = 200 Ω and SCR firing angle is 70°Frequency of AC source = 50HzVoltage of AC source = 415V (rms)We need to calculate the average voltage that is supplied to the load when a three-phase SCR rectifier supplies a resistive load with the above parameters
We know that the average voltage supplied to the load is given as:Vavg = Vm / π (1 + cos θ)Where,Vm = Maximum voltage of AC sourceθ = Firing angleπ = 3.1416First, we need to find the maximum voltage (Vm) of the AC source using the following relation Vm = √2 × Vrms Vm = √2 × 415Vm = 586.2 VNext, let's calculate the average voltage Vavg = Vm / π (1 + cos θ)Vavg = 586.2 / π (1 + cos 70°)Vavg = 104.6 V
The average voltage supplied to the load is 104.6 V, when a three-phase SCR rectifier supplies a resistive load with the above parameters ( R = 200 Ω, SCR firing angle is 70°). Hence, the answer is 104.6 and the is given in the above steps.
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I wish to transmit the message signal m(t) by DSB-SC modulating a carrier at 500 kHz. I have a variety of filters but do not have an oscillator that produces a cosine waveform at the frequency 500 kHz. However, I have two oscillators that produce cosine waveforms at 200 kHz and 300 kHz. I also have two identical non-linear devices with the same transfer characteristics y = 2². Illustrate the design of a circuit using block diagrams that will produce the required DSB-SC signal for me using only the devices I have. Clearly label each block, the inputs, and the outputs. Include trigonometric derivations to prove that your design generates the required signal.
Given, we wish to transmit the message signal m(t) by DSB-SC modulating a carrier at 500 kHz. And we have two oscillators that produce cosine waveforms at 200 kHz and 300 kHz.Let x1(t) = cos(2π(200)kt) and x2(t) = cos(2π(300)kt) be the inputs and we need to design a circuit using block diagrams that will produce the required DSB-SC signal for us using only these devices.
Now, the block diagram of the DSB-SC modulation technique is as follows:We need to remove the carrier component frequency from this circuit.The desired DSB-SC output can be obtained by multiplying the input message signal m(t) with a cosine signal at the same frequency as the carrier frequency. This can be achieved using the following equation: 2cos(2π(500)kt)cos(2π(500)kt) = cos(2π(1000)kt) + 1
First, the message signal m(t) is passed through a low-pass filter to remove any high-frequency components. The output of this filter is x(t).Now, x1(t) and x2(t) are mixed and then passed through a low-pass filter with cutoff frequency 100 Hz. The output of this filter is u(t).Now, u(t) is multiplied with x(t) to generate the desired DSB-SC signal. This can be achieved using a non-linear device with the transfer function y = 2². The output of this device is v(t).Finally, v(t) is passed through a low-pass filter with cutoff frequency 100 Hz to remove any high-frequency components. The output of this filter is the desired DSB-SC signal.
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the excerpt above is an example of the role of the media in partisan reporting. agenda setting. investigating corruption.
The excerpt above is an example of the role of the media in investigating corruption. In the excerpt, the media are highlighted to be exposing corrupt and unethical practices among state officials.
The description is an example of the media's investigative role and its commitment to ensuring that state officials act with integrity and transparency.In a corruption case, conduct a thorough interview of the primary subject, usually the suspected bribe recipient. Ask about his or her role in the suspect contract award and relevant financial issues, such as sources of income and expenditures.
Therefore, the excerpt is a clear illustration of the media's investigative role in society. By keeping an eye on state officials and exposing corrupt practices, the media plays a vital role in ensuring that the society is well governed.
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A six-step three-phase inverter has a 250V dc source and an output frequency of 50Hz.
A balanced Y-connected load consists of a series 25Ω resistance and 20mH inductance
in each phase.
Determine:
(a) Rms value of 50Hz component of load current
(b) THD of load current
You may consider harmonic order up to nth=17 for THD calculation.
Rms value of 50Hz component of load current: Given Data:
Output frequency (f) = 50 Hz Vdc
source = 250 V
Balanced Y-connected load25Ω resistance20mH inductance
Let’s calculate the inductive reactance of the given inductor as follows:
Reactance (X) = 2πFL
Reactance (X) = 2 × 3.14 × 50 × 20 × 10^-3
Reactance (X) = 6.28 ΩRMS
value of the current component can be calculated as follows:
VLine to Neutral = V p h RMS / √3 (where V p h RMS is the phase voltage)
The phase voltage can be calculated as follows:
V p h RMS = VLine to Neutral × √3VphRMS = 250 / √3VphRMS = 144.33 V
The inductor’s voltage is given as:
VL = XI Let's calculate the load current component:
IL = VL / XIL = V p h RMS / XLIL = 144.33 / 6.28IL = 22.96 A (Approximate)
the RMS value of the 50 Hz component of the load current is 22.96 A.
THD of load current:
In this case, the THD can be calculated as follows:
THD = (√(V^2n2 + V^2n3 + V^2n4 + … + V^2n17 ) / Vn1) × 100
Where Vn1 is the fundamental component, Vn2, Vn3…Vn17 are the second, third to 17th harmonic components respectively.
Vn1 is already calculated in part (a).
It is now necessary to calculate the remaining voltage components by considering the odd harmonics of the output frequency, starting with the third harmonic (the second harmonic is already considered in the inductor).
Let’s calculate the RMS value of the third harmonic component voltage:
V3 = (30 × VL) / πV3 = (30 × 6.28 × IL) / πV3 = 60.48 V
The RMS value of the fourth harmonic component voltage can be calculated as follows:
V4 = (20 × VL) / πV4 = (20 × 6.28 × IL) / πV4 = 40.32 V
The RMS value of the fifth harmonic component voltage can be calculated as follows:
V5 = (12 × VL) / πV5 = (12 × 6.28 × IL) / πV5 = 24.19 V
HD = ((60.48^2 + 40.32^2 + 24.19^2 + 12.56^2 + 6.99^2 + 3.65^2 + 1.79^2 + 0.81^2 + 0.35^2)^1/2) / 22.96THD = 28.53%
the THD of load current is 28.53%.
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A technique to search for small "nuggets" of information from the vast quantities of data stored in an organization's data warehouse, using technologies such as decision trees and neural networks, is called:
The technique to search for small "nuggets" of information from an organization's data warehouse using technologies such as decision trees and neural networks is called Data Mining.
Data Mining is a process of discovering patterns, relationships, and insights from large volumes of data. It involves applying various statistical and machine learning techniques to extract valuable information or knowledge that may not be readily apparent. Decision trees and neural networks are commonly used algorithms in data mining.
Decision trees are tree-like models that break down data into smaller and more manageable subsets based on different criteria or attributes. They can be used to classify data or make predictions by following a series of decision rules derived from the data.
Neural networks, on the other hand, are computational models inspired by the structure and function of the human brain. They consist of interconnected nodes or "neurons" that process and analyze input data to produce desired outputs. Neural networks are particularly effective for recognizing complex patterns and relationships within data.
By leveraging data mining techniques, organizations can uncover hidden patterns, correlations, and trends that can provide valuable insights for decision-making, optimization, and predictive analytics. It allows organizations to transform raw data into actionable knowledge and make data-driven decisions.
Data mining, employing techniques like decision trees and neural networks, enables organizations to extract meaningful information and discover valuable insights from vast amounts of data stored in data warehouses. By uncovering these "nuggets" of information, organizations can gain a competitive advantage, improve business processes, enhance customer experiences, and make more informed decisions based on data-driven evidence. Data mining plays a crucial role in various fields, including marketing, finance, healthcare, and manufacturing, helping organizations unlock the true potential of their data assets.
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__________ is the total sideband power when m-80% and carrier power is 5W. The total output current of an AM is _________ when the carrier power is 16 W with R = 70 ohms and percentage modulation of 70%
2.68 W is the total sideband power when m-80% and carrier power is 5W. The total output current of an AM is 1.6 A when the carrier power is 16 W with R = 70 ohms and percentage modulation of 70%
The given modulation index m = 0.8 and carrier power = 5 W in an AM circuit.
The formula to calculate the total sideband power is shown below:
Pₛ = (m²/2 + m) × P_c
Where, Pₛ = Total sideband power
P_c = Carrier power
m = Modulation index
By substituting the values in the formula, we get:
Pₛ = (0.8²/2 + 0.8) × 5= 2.68 W
Therefore, the total sideband power when m-80% and carrier power is 5W is 2.68 W.
The given carrier power P_c = 16 W, resistance R = 70 Ω and percentage modulation = 70%.
The formula to calculate the total output current is shown below: I_c = P_c / V_c
Where, I_c = Total output current
V_c = Carrier voltage
The formula to calculate the carrier voltage is shown below: V_c = V_m cosωct
By substituting the given values, we get the following:
V_m = 4.8 volts (Peak value) = (70% × 1.414 × 10) = 9.97 V (RMS value)
ωc = 2π × f_c = 2 × 3.14 × 10^6 Hz = 62.8 × 10^6 rad/s
V_c = 9.97 × cos (62.8 × 10^6 × 0) = 9.97The total output current is:
I_c = P_c / V_c= 16 W / 9.97 V= 1.6 A
Rounded to one decimal place, the total output current of an AM is 1.6 A when the carrier power is 16 W with R = 70 ohms and percentage modulation of 70%.
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Determine wether signals is periodic. or not a) X₁ (t) = 2e ³ (t + 7/4), u(t) b) x₂ [n] = u[n] + u[n]
a) X₁(t) is not periodic.
b) x₂[n] is periodic.
a) X₁(t) = 2e^(3(t + 7/4)), u(t)
To determine whether a signal is periodic or not, we need to check if it repeats itself after a certain time interval. In the case of X₁(t), we have an exponential function multiplied by a unit step function. The exponential function grows exponentially with time, and the unit step function ensures that the signal is only active for positive values of t. Since the exponential function does not repeat itself and keeps growing indefinitely, X₁(t) does not exhibit any periodicity. Therefore, X₁(t) is not periodic.
b) x₂[n] = u[n] + u[n]
In this case, we have a signal x₂[n] that is the sum of two unit step functions. The unit step function u[n] has a value of 1 for n ≥ 0 and 0 for n < 0. Adding two unit step functions results in a signal that has a value of 2 for n ≥ 0 and 0 for n < 0. This signal repeats itself every two units of n. Hence, x₂[n] exhibits periodicity with a period of 2.
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FILL THE BLANK.
Only cars made after _____________ are required by the NHTSA to have a dual front airbags.
Only cars made after September 1, 1997 are required by the NHTSA to have a dual front airbags. What are the dual front airbags Dual front airbags, also known as "driver-side airbag" and "front-passenger airbag," are an automotive safety feature that deploys during a high-speed collision to safeguard drivers and passengers.
Airbags are designed to prevent the human body from colliding with hard surfaces within the vehicle and reduce the risk of severe injuries. The National Highway Traffic Safety Administration (NHTSA), the federal agency in charge of regulating and ensuring vehicle safety standards, mandated the use of dual front airbags in cars made after September 1, 1997.
Your front airbags are dual-stage airbags. This means they have two inflation stages that can be ignited sequentially or simultaneously, depending on crash severity. In a crash, both stages will ignite simultaneously to provide the quickest and greatest protection.
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