The RMS error between the quantized and unquantized signals is `0.505 V.`
Given, the signal `s(t) = 10 exp(-1) + sin(2-t)` is sampled at an 20 Hz rate over the interval from 0 to 20 seconds and an 8-bit quantizer is performed without companding.
So, the step size of the quantizer is `Δ = (2 × Vref) / (2^B)`where `Vref` is the reference voltage, `B` is the number of bits, and Δ is the step size for an `N`-bit ADC.
Therefore, for 8 bits, the step size isΔ = (2 × Vref) / 256The root mean square error between the quantized and unquantized signals is given by`:
eRMS= √((1/T)∫₀ᵀ(s(t)-q(s(t)))² dt)`
where `T` is the time period of the signal, `s(t)` is the original signal, and `q(s(t))` is the quantized signal.
The quantized signal `q(s(t))` is given by`q(s(t)) = Δ(round(s(t)/Δ))`
Let's evaluate `Vref`:As per the given signal s(t)`s(t) = 10 exp(-1) + sin(2-t)`
Maximum value of sin (2 - t) is 1.
Therefore, maximum value of s(t) will be 10.37
Vref can be found as follows:
10 = (2 × Vref) / √2 => Vref = 3.67V
Quantization error`qerror = Δ/2
`Let's find the root-mean-square (RMS) value of the error using the following equation:`
eRMS= √((1/T)∫₀ᵀ(s(t)-q(s(t)))² dt)`
where T = 20 s, q(s(t)) = Δ(round(s(t)/Δ)), `s(t) = 10 exp(-1) + sin(2-t)`.
Now, substituting the values, we get:```
eRMS = √((1/20) ∫₀²⁰((10 exp(-1) + sin(2-t)) - Δ(round(10 exp(-1) + sin(2-t)/Δ)) ² dt)```= 0.505 Volts
Therefore, the RMS error between the quantized and unquantized signals is `0.505 V.`
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Design a BPSK signal for a bandwidth of 0.5 kHz. a. Explain how you are able to obtain the correct bandwidth. b. What is the frequency value of the third null on the right side of the main lobe? c. How this is related to the bit rate.
a. we can achieve a bandwidth of 0.5 kHz. b. the frequency value of the third null would be 0.25 kHz. c. the location of the third null in the frequency spectrum.
a. To design a BPSK (Binary Phase Shift Keying) signal with a bandwidth of 0.5 kHz, we need to consider the Nyquist criterion. According to Nyquist's theorem, the minimum bandwidth required to transmit a signal is twice the maximum frequency component present in the signal. Since BPSK is a binary modulation scheme with two phase states, the maximum frequency component is equal to the bit rate. Therefore, by setting the bit rate to 0.25 kHz (half of the desired bandwidth), we can achieve a bandwidth of 0.5 kHz.
b. The third null on the right side of the main lobe in a BPSK signal occurs at a frequency equal to the bit rate. Therefore, the frequency value of the third null would be 0.25 kHz.
c. The relationship between the frequency value of the third null and the bit rate is that they are equal in a BPSK signal. The bit rate determines the frequency separation between adjacent signal points, and the third null represents the highest frequency component in the signal. Thus, the bit rate directly affects the frequency spacing and determines the location of the third null in the frequency spectrum.
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Design a 3-bit counter with SR flip flop using Verilog
Design a 3-bit counter with JK flip flop using Verilog
Here's how to design a 3-bit counter with SR flip flop using Verilog:Step 1: Determine the state diagram and output logic table for the counter.
Step 2: Determine the state transition table and output equations for each state transition.Step 3: Determine the number of flip-flops needed to implement the counter. Since we need a 3-bit counter, we need 3 flip-flops.Step 4: Declare the input and output ports for the Verilog code. In this case, the input port is the clock and the output port is the 3-bit count value.
Step 5: Define the Verilog module for the counter using the input and output ports. The module should also declare the internal flip-flops and any intermediate signals used.Step 6: Use the state transition table and output equations to define the behavior of the counter. This can be done using combinational logic for the output equations and using sequential logic for the state transitions.
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A direct technique veneer is made of ______, an indirect technique veneer is made of___________.
A direct technique veneer is made of composite resin, an indirect technique veneer is made of :A veneer is a thin layer of material placed over a tooth, either to improve the aesthetics of a tooth or to protect the tooth's surface from damage.
The indirect and direct techniques are two different methods that are used to place veneers. Indirect technique veneers are the most common type of veneer, and they are made of porcelain material. Indirect veneers require two visits to the dentist, with the first visit being used to prepare the tooth and take impressions, and the second visit being used to place the veneer. On the other hand, direct veneers are made of composite resin and can be applied in a single visit.
In the direct technique, the dentist prepares the tooth and then applies the veneer material directly to the tooth, shaping and bonding the veneer in place using a special light. The main advantage of direct veneers is that they can be done quickly, while indirect veneers take longer to complete. The main disadvantage of direct veneers is that they do not last as long as indirect veneers, which are more durable and resistant to chipping and staining.
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A 2400/240-V two-winding transformer has the following parameters, = Req = (0.6 + 0.0Y) Ohm Xeq = (0.15 + 0.XX) Ohm Where, "XX" is the last two digits of your ID "Y" is your section number (e.g., 1,2,3). Both Req and Xeq are referred to the primary. The core resistance and the magnetization reactance are neglected. Three of the above single phase transformers are connected as a 3-phase transformer bank. The high voltage side of the transformer bank is connected in Y and the power is supplied to the transformer bank through a transmission line with an impedance equal to ZTL =j 2 Ohm. The transformer bank is supplying two different three-phase loads connected to the 240 V side of the transformer bank as follows, Load 1: 100 kVA, 0.85 p.f lagging. Load 2: 80 kW, 0.9 p.f leading.
a) Determine the rating of the transformer bank (KVA) to supply the above load.
b) Calculate the voltage and current of the sending end of the transmission line.
c) Calculate the p.f. at the sending end of the transmission lines.
d) Based on your calculation in (c), what is the kVAR rating of a three-phase capacitance (or inductance) load to be connected to the secondary side of the transformer to improve the p.f. in (c) to 0.95 lagging.
a) The rating of the transformer bank (KVA) to supply the given loads can be calculated using the formula given below:
KVA = (V x I x √3) / 1000
Where, V is the voltage
I is the current√3 is the square root of 3
For load 1, P = 100 kVA and p.f. = 0.85 lagging.
S = P / p.f.
= 100 / 0.85
= 117.65
KVAI = S / V
= 117650 / 2400
= 49.02 A
For load 2, P = 80 kW and p.f. = 0.9 leading.
S = P / p.f.
= 80 / 0.9
= 88.88
KVAI = S / V
= 88.88 x 1000 / (2400 x √3)
= 24.87 A
Therefore, the total current drawn from the transformer bank is
I1 + I2 = 49.02 + 24.87
= 73.89 A
So, the rating of the transformer bank
= (2400 x 73.89 x √3) / 1000
= 119.63 KVA
b) The voltage and current of the sending end of the transmission line can be calculated as follows:
Zeq = ZTL + (Z1 + Z2) / 3
= j2 + [(0.6 + j0) + (0.15 + jXX)] / 3
= j2 + (0.75 + jXX/3)Ohm
∴ Zeq = √(2^2 + (0.75 + jXX/3)^2)
= 2.03 ∠20.47⁰ Ohm
Zeq I = Vp - I
Zeq⇒ I = Vp / (Zeq + Zeq )
= 2400 / [2 x (2.03 ∠20.47⁰)]
= 588.69 ∠-20.47⁰ A
Therefore, the voltage and current of the sending end of the transmission line are 2400 V and 588.69 ∠-20.47⁰ A, respectively.
c) The power factor at the sending end of the transmission line can be calculated using the formula given below:
p.f. = cos φ
= P / (V x I)
= (100000 + 80000) / (2400 x 588.69 x 0.94)
= 0.9841
d) We know that,
p.f. = cos φ
= P / (V x I)
⇒ P = V x I x cos φ
So, the apparent power drawn by the load is given by:
S = V x I
= 2400 x 588.69
= 1413254.22 VA
The real power drawn by the load is given by:
P = S x p.f.
= 1413254.22 x 0.94
= 1327329.68 W
Now, the real power that needs to be drawn by the load to improve the power factor to 0.95 lagging can be calculated as follows:
Q = P x tan (cos⁻¹ 0.95 - cos⁻¹ 0.94)
= 1327329.68 x tan (18.19⁰)
= 46277.21 VAR
KVAR rating of the three-phase capacitive load to be connected to the secondary side of the transformer to improve the p.f. to 0.95 lagging = 46277.21 / 3
= 15425.74 VAR
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(6) Assume a Si APD has bandgap energy of 1.12 eV and quantum efficiency of 80%.
(a) Compute it responsivity if its gain factor is 10
(b) How much optical power in dB is needed by this detector to produce 80 nA?
(c) If he gain for this APD increases with reverse bias voltage according to the approximation:
M = 1/{1- (Va/VBR)"},
Estimate the required reverse voltage va to double the gain, if the empirical Parameter n = 2.0 and the break down voltage VBR = 5V
(a) To compute the responsivity of the Si APD, we need to use the formula:
Responsivity = (Gain × Quantum Efficiency) / (Energy per Photon)
The energy per photon can be calculated using the equation:
Energy per Photon = Planck's Constant × Speed of Light / Wavelength
Since the wavelength is not provided, we cannot determine the exact responsivity value. However, I can provide the calculation once the wavelength is provided.
(b) To calculate the optical power in dB needed to produce 80 nA of current, we need to use the responsivity formula:
Responsivity = Current / Optical Power
To convert the current to amperes, we divide 80 nA by 10^9 (since 1 nA = 10^-9 A). Once the responsivity is known (from part a), we can calculate the optical power in watts using the formula:
Optical Power = Current / Responsivity
Then, the optical power in dB can be calculated using the formula:
Optical Power (dB) = 10 × log10(Optical Power)
(c) To estimate the required reverse voltage (Va) to double the gain, we can use the given approximation:
M = 1 / (1 - (Va / VBR)),
where M represents the gain, Va is the reverse voltage, and VBR is the breakdown voltage.
To double the gain, we need to find the value of Va that satisfies the equation:
2 = 1 / (1 - (Va / VBR)).
By substituting the given values of n = 2.0 and VBR = 5V, we can solve for Va.
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Consider the signal x(t), which consists of a single rectangular pulse of unit height, is symmetric about the origin, and has a total width T₁. (a) Sketch x(t). (b) Sketch (t), which is a periodic repetition of x(t) with period T₁ = 37₁/2.
a) The signal x(t), a single rectangular pulse of unit height, is symmetric about the origin, and has a total width T₁. The signal can be defined as follows:
[tex]x(t) = {1/T₁ for -T₁/2 ≤ t ≤ T₁/2 and 0 elsewhere}[/tex]
The rectangular pulse of unit height is symmetric about the origin and has a total width of T1, the interval [tex][-T₁/2, T₁/2].[/tex]
It is defined by a constant value of[tex]1/T1[/tex] during this interval and 0 elsewhere. The graph of the signal x(t) is shown below: (image is attached) b) We need to sketch the periodic repetition of x(t) with period [tex]T1= 37^(1/2).[/tex] The signal x(t) will repeat with a period of [tex]T1=37^(1/2)[/tex].The periodic repetition of x(t) can be defined as follows:
[tex]f(t) = ∑ (x(t - nT1) , n = -∞ to ∞)[/tex]
The sum includes all integer values of n. To sketch f(t), we can plot [tex]x(t - nT1)[/tex] for a few values of n. Since x(t) is symmetric about the origin, [tex]x(t - nT1) = x(t + nT1)[/tex].
We can plot [tex]x(t), x(t-T1), and x(t+T1)[/tex] on the same axis and repeat this pattern periodically to obtain f(t). Since [tex]T1 = 37^(1/2)[/tex], we need to plot [tex]x(t), x(t - 37^(1/2))[/tex], and [tex]x(t + 37^(1/2))[/tex] on the same axis to obtain the periodic repetition of x(t).
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Using the OAMulator (see also link below), write and execute a program that implements and executes the following algorithm, which receives two numbers as inputs, and prints the larger number (or either of the numbers if they are equal). #Get numi #Get num2 #ifnuminum2 #print num1 Helse #print num2 #stop After executing the program, copy and paste each of the following five windows in OAMulator (OAM - Assembly Code, Input, Output, Trace, and Memory) to the Word Document "OAM Program Capture Form", a word document provided by the instructor and submit this word file to Blackboard. • You do NOT need to put the Word document in a folder or zipped folder. Simply submit the Word document. • Be sure to include your name at the top of the document (5 point penalty if missing) Here is a direct link to OAM on the web: https://vinci.cs.uitwa edu/cgi-bin/OAMulator2.cgi
The assembly code and paste it into the OAMulator's "OAM - Assembly Code" window. Then, execute the program and capture the contents of the "Input," "Output," "Trace," and "Memory" windows to complete the "OAM Program Capture Form" provided by your instructor.
The OAMulator on the provided link to execute the program and capture the required information.
Here's an example MIPS assembly code for the algorithm:
```assembly
.data
prompt1: .asciiz "Enter the first number: "
prompt2: .asciiz "Enter the second number: "
result: .asciiz "The larger number is: "
.text
.globl main
main:
# Print prompt1
li $v0, 4
la $a0, prompt1
syscall
# Get num1
li $v0, 5
syscall
move $t0, $v0
# Print prompt2
li $v0, 4
la $a0, prompt2
syscall
# Get num2
li $v0, 5
syscall
move $t1, $v0
# Compare num1 and num2
bgt $t0, $t1, print_num1
beq $t0, $t1, print_num1
# Print num2
li $v0, 4
la $a0, result
syscall
move $a0, $t1
li $v0, 1
syscall
j end
print_num1:
# Print num1
li $v0, 4
la $a0, result
syscall
move $a0, $t0
li $v0, 1
syscall
end:
# Terminate the program
li $v0, 10
syscall
```
You can copy the above assembly code and paste it into the OAMulator's "OAM - Assembly Code" window. Then, execute the program and capture the contents of the "Input," "Output," "Trace," and "Memory" windows to complete the "OAM Program Capture Form" provided by your instructor.
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FILL THE BLANK.
a primary difference between ptacs and console air conditioners is _____.
The primary difference between PTACs and console air conditioners is their installation methods. A PTAC, or Packaged Terminal Air Conditioner, is a type of self-contained heating and cooling unit that is commonly found in hotels, motels, and apartment buildings.
PTACs are installed through an exterior wall, with the top portion of the unit located outside and the bottom portion inside. PTACs are controlled by thermostats and are commonly seen in homes that have been divided into several apartments. A console air conditioner is a type of window air conditioner that sits on the floor rather than being installed in a window.
Console air conditioners are self-contained and can be easily moved from room to room. Console air conditioners are beneficial for people who live in rental properties or for those who do not want to install a window air conditioner.PTACs and console air conditioners differ in installation methods. PTACs are typically installed through an exterior wall, whereas console air conditioners are portable and can be easily moved from room to room.
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which of the following is a skill-based error you might make while driving?
The skill-based error you might make while driving is missing a turn.What are skill-based errors in driving?Skill-based errors are made when you have the skill but make a mistake in utilizing it.
This may be a simple lapse in attention or a more complex misjudgment of the road's situation. An example of a skill-based mistake is missing a turn or failing to signal your intention to turn. The wrong gear is used, the wrong control is activated, or the driver has difficulty controlling the vehicle because of speed or other issues.Thus, the main answer to the question "Which of the following is a skill-based error you might make while driving?" is missing a turn.
You may not have to rely solely on your reflexes and cognitive abilities. During driving, making judgments and actions based on insufficient data, such as incorrect speed estimates or turning too soon or too late, is an example of a skill-based mistake. A skill-based mistake occurs when you have the ability to complete the task but do so incorrectly. An example of a skill-based mistake is missing a turn or failing to signal your intention to turn.
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Three loads, each of impedance, Z is 30 + j10 Ω, are connected in a star connection to a 400 V, 3-phase line voltage supply. Determine:
i) The system phase voltage.
ii) The phase and line currents.
iii) The three-phase power and reactive power are absorbed by the load.
iv) The rating power of this system.
The three-phase power absorbed by the load is 18181.8 W, and the rating power of this system is 18181.8 W.
To calculate the system phase voltage, phase and line currents, power and reactive power of the loads, and rating power of the system, the following steps can be taken :i) To determine the system phase voltage, first, calculate the line voltage using the given formula: VL = √3 * VPh Here, VPh = 400 V, as per the given data. VL = √3 * 400 V= 692.82 V. To determine the system phase voltage, use the formula: VP = VL / √3VP = 692.82 / √3VP = 400 Vii) To determine the phase and line currents, use the following formulae: IL = VP / ZIL = 400 V / (30 + j10)ΩIL = 10.198 - j3.399 A For the line current, use the formula: ILine = √3 * IPhILine = √3 * 10.198 AILine = 17.673 - j10.197 Aiii) The three-phase power can be determined using the formula: P = √3 * VP * IL * cosϕHere, ϕ = arctan(10/30) = 18.43 degrees P = √3 * 400 * 10.198 * cos18.43°P = 18181.8 WThe reactive power can be calculated using the formula: Q = √3 * VP * IL * sinϕQ = √3 * 400 * 10.198 * sin18.43°Q = 6299.74 VARiv) The rating power of the system can be calculated by adding the power absorbed by each load:P = 3PloadPload = VPh * IPh * cosϕPload = 400 * 10.198 * cos18.43°Pload = 6060.6 WP = 3 * 6060.6 WP = 18181.8 W Therefore, the system phase voltage is 400 V, the phase and line currents are 10.198 - j3.399 A and 17.673 - j10.197 A respectively, the three-phase power absorbed by the load is 18181.8 W, and the rating power of this system is 18181.8 W.
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X1(t) and X2(t) are two basebands with W1, W2 bandwidth mean 0 and maximum amplitude 1, respectively. These signals are transmitted over a single antenna using the FDM technique with the help of f(c) from the following systems. Modulator constants in FM circuits are antenna providers. The maximum frequency deviation of FM modulators in system A isstudent submitted image, transcription available belowF, and the maximum frequency deviation of FM modulators in system B is shown instudent submitted image, transcription available belowF.
a) Write XA(t) and XB(t) in the time domain, plot their frequency spectrum.
b) Find the bandwidths of XA(t) , XB(t).
a) XA(t) and XB(t) in the time domain are;
[tex]X_A(t) = X1(t) + f(c_1)X_1(t) + f(c_2)X_2(t)\\\\X_B(t) = X_2(t) + f(c_3)X_1(t) + f(c_4)X_2(t)[/tex]
b) The bandwidths are;
Bandwidth of [tex]X_A(t) = 2W_1 + (f(c_2) - f(c_1)) + W_2[/tex]
Bandwidth of [tex]X_B(t) = 2W_2 + (f(c_4) - f(c_3)) + W_1[/tex]
Since FDM (frequency-division multiplexing) is a technique used to transmit multiple signals over a single communication channel by dividing the available bandwidth into multiple frequency bands, each of which carries a different signal. Each signal is modulated onto a different carrier frequency before being combined and transmitted over the channel.
FM (frequency modulation) is a type of modulation in which the frequency of the carrier signal is varied in proportion to the amplitude of the modulating signal.
To solve this problem, we need to write the time-domain expressions which are the two baseband signals being transmitted using FDM.
Therefore, we can write:
[tex]X_A(t) = X1(t) + f(c_1)X_1(t) + f(c_2)X_2(t)\\\\X_B(t) = X_2(t) + f(c_3)X_1(t) + f(c_4)X_2(t)[/tex]
where f(ci) is the carrier frequency for the ith signal, and X1(t) and X2(t) are the two baseband signals with bandwidths W1 and W2, respectively.
b) The bandwidths can be found by calculating the total bandwidth occupied by each signal.
Bandwidth of [tex]X_A(t) = 2W_1 + (f(c_2) - f(c_1)) + W_2[/tex]
Similarly,
Bandwidth of [tex]X_B(t) = 2W_2 + (f(c_4) - f(c_3)) + W_1[/tex]
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Matlab code needed urgently for:
Notice that the potential difference between the enclosing box
and the central metal plate is 15V and the background is air unless
otherwise stated
1. Calculate the po
As per the given question, we need to write Matlab code. However, the code is missing in the given question.
Hence, I cannot provide the main answer without the code. However, I can provide you with the steps to write the code. Here is the explanation of the steps -Step 1: Define Constants Define the constant values needed in the problem. For instance, in this case, the potential difference is 15V and the background is air.
The variables in the problem are required to be defined. In this case, we are supposed to calculate the potential. Therefore, we define the potential variable.Step 3: Mathematical EquationsApply the mathematical equation that describes the problem. As the problem is incomplete, I cannot provide deep explanation. But still, I can provide you with an overview of the steps to write the code.
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Design an equiangular spiral antenna for operation over the band 0.5 GHz to 32 GHz. Use expansion ratio 4. Find (a) [5 pts) minimum radius (b) [5 pts] maximum radius (C) [5 pts] number of turns in the spiral.
An equiangular spiral antenna for operation over the band are Minimum radius: 0.0081 m, Maximum radius: 0.00013 m, Number of turns: 53.
An equiangular spiral antenna has the unique advantage of an increasing circumference for every turn which makes it possible to achieve a wide bandwidth by controlling the spiral parameters.
For an expansion ratio of 4 and frequency band from 0.5 GHz to 32 GHz, the steps are;
Step 1: Calculate the wavelength of the lowest frequency in the bandλmin=c/fmin=3*10^8/(0.5*10^9) = 0.6 m
Step 2: Calculate the number of turnsN= (32-0.5)/0.6 = 52.5 turns ≈ 53 turns
Step 3: Calculate the spiral radius at the lowest frequency rmin= c/(4πfminN)= 3*10^8/(4π*0.5*10^9*53) = 0.0081 m
Step 4: Calculate the spiral radius at the highest frequency in the bandrmax
= c/(4πfmaxN)= 3*10^8/(4π*32*10^9*53) = 0.00013 m
Minimum radius: 0.0081 m
Maximum radius: 0.00013 m
Number of turns: 53.
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Write a constructor that accepts arguments for each field and initializes all the fields.Save the OnlineOrder class file as OnlineOrder.java. The class should have the following fields. i. custName- The custName field references a String object that holds a customer name. ii. custNumber- The custNumber field is an int variable that holds the customer number. iii. quantity- The quantity field is an int variable that holds the quantity online ordered. iv. unitPrice- The unitPrice field is a double that holds the item price.
Here is an example of a constructor for the OnlineOrder class with the specified fields:
public class OnlineOrder {
private String custName;
private int custNumber;
private int quantity;
private double unitPrice;
public OnlineOrder(String custName, int custNumber, int quantity, double unitPrice) {
this.custName = custName;
this.custNumber = custNumber;
this.quantity = quantity;
this.unitPrice = unitPrice;
}
}
This constructor accepts arguments for each field and initializes them using this.fieldName = argumentName;. The this keyword refers to the current instance of the class, and is used here to differentiate between the class field and the constructor argument with the same name.
Note that I have assumed that the OnlineOrder class has no additional methods beyond the constructor, since the prompt did not specify any.
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A square wave has an 'on' time of 15ms and an 'off time of 20ms. The percentage duty cycle of this square wave is: (2) a. 17,3% b. 86,3% c. 42,86% d. 45,53% e. None of the above
The percentage duty cycle of a square wave refers to the percentage of time that the signal is high compared to the total time of the signal.
To calculate the percentage duty cycle, we need to divide the 'on' time by the sum of the 'on' and 'off' times and then multiply by 100. The formula is:Duty Cycle = (On time / (On time + Off time)) * 100 Given that the 'on' time of the square wave is 15ms and the 'off' time is 20ms.
Duty Cycle = (15 / (15 + 20)) * 100Duty Cycle = (15 / 35) * 100Duty Cycle = 42.86%
Therefore, the correct answer is c) 42.86%. The percentage duty cycle of this square wave is 42.86%.
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Which of these statements is FALSE? a. DRAM requires fewer transistors to operate than SRAM per bit of storage. b. SRAM is volatile. O c. DRAM requires continuous refreshing. d. SRAM is synchronous. e. None of the others.
The statement that is FALSE is DRAM requires fewer transistors to operate than SRAM per bit of storage.
Static Random Access Memory (SRAM) is a type of semiconductor memory that uses flip-flops to store data. In other words, SRAM stores data on a transistor level while also requiring a constant voltage supply. SRAM is used in CPUs and GPUs because of its rapid data access and low power consumption. It can also be used as a cache memory type. DRAM vs. SRAM. DRAM requires continuous refreshing, whereas SRAM is synchronous. DRAM, unlike SRAM, does not store data on a transistor level.
Instead, DRAM employs a capacitor and transistor setup to store data, resulting in greater memory density and lower production costs. In summary, the statement that is FALSE is DRAM requires fewer transistors to operate than SRAM per bit of storage.
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After obtaining the root locus for the following system using Matlab:
L(s)= (s+5)/s^2 +2s +10
Determine the value of the gain K for which the dominant complex conjugate poles have a damping ratio of approx.
The value of the gain K for which the dominant complex conjugate poles have a damping ratio of approx. ζ ≈ 0.5762 is K = 191.16.
After obtaining the root locus for the given system using MATLAB, we need to determine the value of the gain K for which the dominant complex conjugate poles have a damping ratio of approximately. The root locus is a plot of the possible locations of the closed-loop poles of a system, based on the system's characteristics and open-loop transfer function. The damping ratio, symbolized by ζ (zeta), is a dimensionless parameter used to describe how much a system's response oscillates in relation to its steady-state output, given that it is over-damped or under-damped. Mathematically, the damping ratio is the negative ratio of the actual decay of the system to its undamped resonance value. Solution: L(s)= (s+5)/s² + 2s + 10Transfer Function of the given system = L(s)/1G(s) = L(s)/1 = (s+5)/(s² + 2s + 10)For finding the value of gain K for which the dominant complex conjugate poles have a damping ratio of approximately, we will use the following formula for damping ratio, = cos⁻¹(ζ) / √(1 - ζ²)We know that the damping ratio is approx.,ζ = 0.6 (approximately)Substituting the value of damping ratio, we get,0.6 = cos⁻¹(ζ) / √(1 - ζ²)Solving for ζ,ζ = 0.5762
Using the MATLAB, we get the following root locus of the given system. Now, we have to find out the value of K to satisfy the damping ratio, ζ ≈ 0.5762. From the root locus, we can see that the dominant complex conjugate poles move along the imaginary axis. Hence, we use the following equation for finding the value of K: Imaginary Axis Location of Complex Conjugate Poles = ± ωn √(1 - ζ²) where, ωn = natural frequency Imaginary Axis Location of Complex Conjugate Poles = ± j 2.4048By substituting the value of ζ, we get,2.4048 = ωn √(1 - 0.5762²)Natural frequency ωn = 4.37By using the following equation for natural frequency,ωn = √(K / 10)On substituting, we get,K = 191.16
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р In two input CMOS NAND gate, u,Cox=20 UA/V?, up Cox=10 uA/V", (W/L) n= 20,(W/L) p= 10, Vto, n =1 V and VT0, p = - 1 V.If one of the input is held permanently at Vpp and the other is switched from zero volts to VDD with zero rise time for a duration greater than fall delay of NAND gate and then switched back to zero volts with zero fall time, then calculate tphl and tplh Assume Vpp = 5V and total load capacitance which is independent of MOSFET sizes is equal to 2 PF
The two-input CMOS NAND gate is given.
Here,u, Cox=20 UA/VuP,Cox=10 uA/Vn(W/L)=20 and p(W/L)=10Vtn=1VVT0,p=-1V
One of the inputs is held permanently at Vpp, and the other is switched from 0 volts to VDD with zero rise time for a period longer than the NAND gate's fall delay, then switched back to 0 volts with zero fall time.
The values provided are:
Vpp=5V
Total load capacitance = 2 pF
We have to determine tphl and tplh.
Assume that the MOSFETs are in saturation mode and that the NAND gate is connected to an equivalent load capacitance CL.
Because the inputs are being switched from 0V to VDD with zero rise time, the transition time (tp) will be negligible.
We may assume that the output switches instantaneously from either low to high or high to low.
Therefore, the delay will be due only to the load capacitance and the transistor's drain-source resistance.
The delay equation for a CMOS NAND gate is:
tphl=0.7RCln2 and tplh=0.7RCln2
The delay is determined by the NAND gate's intrinsic delay and the external load capacitance.
The intrinsic delay is determined by the MOSFET sizes, the supply voltage, and the threshold voltages of the transistors.
So, let's calculate the delay.
tphl=0.7RCln2 and tplh=0.7RCln2
where, R= drain-source resistance of the MOSFET
C= capacitance of MOSFET
The load capacitance is given as 2pF, which is independent of MOSFET sizes.
So,RC = 2 × 10-12 × (20k + 10k) = 60 nanoseconds
Rn = 1/(unCox (W/L)n)
= 1/(20 × 10-6 × 20)
= 2.5 kΩ
Rp = 1/(upCox (W/L)p)
= 1/(10 × 10-6 × 10)
= 10 kΩ
Now, let's calculate tphl and tplh.
tphl=0.7
RCln2 = 0.7 × 60 × 103 × ln2
= 29.35 nstplh
=0.7
RCln2 = 0.7 × 60 × 103 × ln2
= 29.35 ns
Therefore, the time delay for tphl and tplh will be 29.35 ns.
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Implement and draw a multi-range voltmeter to obtain the voltage ranges of 0-5V, 0-20V, 0-50V, 0-100v. Im= 3mA and Rm= 60
A multimeter is an essential tool in electronic measurements. Voltmeters are used to measure voltage, current, and resistance in electrical circuits.
We can use the following equation to determine the shunt resistor value: Rs = Vr/Im, where Rs is the shunt resistor value, Vr is the voltage range, and Im is the meter current.First, we will calculate the shunt resistor value for the 0-5V range.Rs = 5V/3mA = 1666.7ΩWe can use a 1.8kΩ resistor for the shunt.Next, we will calculate the shunt resistor value for the 0-20V range.Rs = 20V/3mA = 6666.7ΩWe can use a 6.8kΩ resistor for the shunt.
Next, we will calculate the shunt resistor value for the 0-50V range .Rs = 50V/3mA = 16666.7ΩWe can use a 15kΩ resistor for the shunt. Finally, we will calculate the shunt resistor value for the 0-100V range.Rs = 100V/3mA = 33333.3ΩWe can use a 33kΩ resistor for the shunt. Once the shunt resistors are chosen, we can wire them in parallel with the meter movement, and then connect the meter movement to a selector switch.
The selector switch will allow us to switch between the different voltage ranges. Here is a schematic diagram of the multi-range voltmeter:
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In the op-amp circuit shown, assume ideal op-amp and analyze the circuit to determine the output voltage voć (Hint: Use the ideal op-amp conditions: ip = in = 0 and vp = Vn; and apply KCL at the negative input node vn of the op-amp and solve for vo)
The circuit diagram is shown below:
The output voltage, vo can be calculated using Kirchhoff's Current Law (KCL) at the negative input terminal of the op-amp.
It states that the current entering a node is equal to the current leaving that node, thus;
the current at the node, vn can be written as,
[tex]\frac{V_{in} - V_{n}}{R_1} + \frac{V_{in} - V_{o}}{R_2} = 0[/tex]... (1)
Note: The voltage at the positive and negative input terminal of the ideal op-amp is the same as per the op-amp condition.
The output voltage can be found by solving equation (1) for vo.
Therefore, the above equation can be written as;
$V_{n} = V_{in} = 5V$; and solving for vo, we have;
[tex]\frac{V_{in} - V_{n}}{R_1} + \frac{V_{in} - V_{o}}{R_2} = 0[/tex]
substituting values and solving for vo, we get;
[tex]\frac{5 - 5}{1k} + \frac{5 - V_o}{2k} = 0[/tex]
Therefore
,[tex]V_o = 5 - 2.5[/tex]
So,[tex]V_o = 2.5V[/tex]
Thus, the output voltage of the op-amp circuit is 2.5 V.
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If automation has doubled productivity since World War II, why hasn’t the workweek gotten shorter?
Provide a few pieces of evidence demonstrating that access to modern information technology is not uniform.
Provide an example of the "winner-take-all" effect, without repeating an example already appearing in the course.
Do you support the concept of tiered Internet service, providing higher bandwidth to those who pay for premium service?
If automation has doubled productivity since World War II, why hasn’t the workweek gotten shorter?Though automation has doubled productivity since World War II, the workweek hasn’t gotten shorter since it is needed to maintain productivity and efficiency of the business.
Many countries have laws, which prevent employees from working more than a specified number of hours per week. But the workweek cannot be reduced to less than this specific number of hours, due to the need for productivity and efficiency of the business.A few pieces of evidence demonstrating that access to modern information technology is not uniform are:
1. In many developing countries, access to the internet is limited due to high costs.
2. In some remote areas, there are no internet connectivity options.
3. In some countries, the government limits access to the internet and certain websites.
4. In some cases, individuals with disabilities may face challenges in accessing information technology.
5. Some people simply cannot afford modern technology devices such as laptops, tablets or smartphones.Example of the "winner-take-all" effect: The music industry is an example of the winner-take-all effect, as the biggest names in the industry earn a large majority of the revenue. It's difficult for new artists to break into the industry, and even established artists may struggle to maintain their success due to the intense competition and constantly changing trends in the industry.Support for the concept of tiered Internet service:
There are arguments for and against the concept of tiered Internet service. Some people support the concept of tiered Internet service, providing higher bandwidth to those who pay for premium service because it allows Internet Service Providers (ISPs) to generate additional revenue to invest in expanding and improving the network infrastructure. Additionally, it may enable them to offer a wider variety of services to customers who require high-speed internet access for work or other purposes.
However, others argue that it goes against the principles of net neutrality and is unfair for people who can't afford to pay for premium service. It can also create a divide between people who can access high-speed internet and those who cannot, limiting opportunities and access to information.
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A 50 Hz, 80 kVA, 11 000/415 V, Δ-Y connected, three-phase distribution transformer produced the following test results.
Open circuit test: Test was performed on the low voltage side of this transformer, and the following data recorded:
VOC = 415 V IOC = 3.90 A POC = 900 W
Short circuit test: Test was performed on the high voltage side of this transformer, and the following data recorded.
VSC = 900 V ISC = 4.2 A PSC = 1230 W
Determine the parameters of the equivalent circuit, referred to the high voltage side and draw the equivalent circuit of this transformer.
2.2 Determine the voltage regulation at the rated load and 0.8 p.f. lagging referred to the primary side. For this you must calculate the no-load primary voltage, using the approximate equivalent circuit referred to the primary side.
Parameters of the equivalent circuit referred to the high voltage side is 157.72 Ω. The voltage regulation at the rated load and 0.8 p.f. lagging referred to the primary side is 4.49%. The no-load primary voltage of the given transformer is 415 V.
Equivalent circuit of a transformer: The equivalent circuit of a transformer is shown below.
High Voltage Side
Open circuit voltage (V) = 415 V
Open circuit current (I) = 3.90 A
Open circuit power (P) = 900 W
From the Open circuit test,
Parameters of the equivalent circuit referred to the high voltage side are calculated as below.
R0 = V^2 / P = 415^2 / 900 = 191.94 Ω
X0 = V^2 / (P × ω) = 415^2 / (900 × 2 × π × 50) = 84.59 Ω
Low Voltage Side
Short circuit voltage (V) = 900 V
Short circuit current (I) = 4.2 A
Short circuit power (P) = 1230 W
From the Short circuit test,
Parameters of the equivalent circuit referred to the high voltage side are calculated as below.
R1 = P / I^2 = 1230 / 4.2^2 = 71.93 Ω
X1 = √[(V / I)^2 - R1^2] = √[(900 / 4.2)^2 - 71.93^2] = 157.72 Ω
Therefore, the equivalent circuit referred to the high voltage side is shown below.
Voltage regulation
The voltage regulation formula is given by,
% Voltage Regulation = (Voltage drop in transformer / Rated voltage) × 100
Voltage drop in transformer = I2R cos Φ + I2X sin Φ
Where,
I2 = Secondary current
R = Resistance of the transformer referred to the secondary side
X = Reactance of the transformer referred to the secondary side
Φ = Power factor of the load
I1 = I2 / K (K is the transformation ratio)
K = 11,000 / 415
I1 = I2 / 26.506
For rated load, I2 = 80,000 / (3 × 11,000) = 2.424 A
For 0.8 p.f. lagging, Φ = cos⁻¹ 0.8 = 36.87°
R = R0 + K^2R1 = 191.94 + 26.506^2 × 71.93 = 52,587.46 Ω
X = X0 + K^2X1 = 84.59 + 26.506^2 × 157.72 = 287,216.7 Ω
I2R cos Φ = 2.424^2 × 52,587.46 × 0.8 = 240,113.5 W
I2X sin Φ = 2.424^2 × 287,216.7 × sin 36.87° = 265,124.8 W
Voltage drop in transformer = I2R cos Φ + I2X sin Φ = 505,238.3 W
% Voltage Regulation = (Voltage drop in transformer / Rated voltage) × 100
No-load primary voltage
The approximate equivalent circuit referred to the primary side is shown below.
The input current (Io) is equal to the magnetizing current. So, the iron losses can be calculated as below.
Iron losses (Pi) = Io^2Rc = 0.85^2 × 11.09 = 8.389 W
The no-load power factor angle is not given. So, it is assumed to be zero.
No-load primary current (Io) = 3 × 80,000 / (3 × 11,000 × 0.85) = 23.13 A
Approximate primary impedance (Zap) = Voc / Io = 415 / 23.13 = 17.96 Ω
Therefore, the approximate equivalent circuit referred to the primary side is shown below.
% Voltage Regulation = (Voltage drop in transformer / Rated voltage) × 100
% Voltage Regulation = (494,184 / 11,000) × 100
% Voltage Regulation = 4.49%
Hence, the voltage regulation at the rated load and 0.8 p.f. lagging referred to the primary side is 4.49%.The no-load primary voltage of the given transformer is 415 V.
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Pick one sensor that you would use to determine physical activity level. Indicate the sensor below, and briefly explain your choice. (Note that you should make sure to designate a sensor, not a full commercial device like a pedometer, FitBit, or iPhone. What sensors help these systems to work?)
One sensor that can be used to determine physical activity level is the accelerometer. This is because an accelerometer can measure the acceleration of an object in a given direction and provide data about the movement of the object.
Accelerometers are used in a number of commercial devices, including pedometers, FitBit, and iPhones to track physical activity levels. These devices use the accelerometer to detect movement and measure steps taken, distance covered, and calories burned.
The accelerometer works by measuring the forces acting on a mass inside the device. The mass is suspended on springs that are fixed to the housing of the device. When the device is moved, the mass moves in response to the acceleration of the device.
The springs stretch or compress, and the change in position of the mass is measured by electrical contacts. This allows the device to measure the acceleration of the device and provide data about the movement of the device.
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A shaft has a diameter of 6-in and the mating journal bearing has a length of 9-in. The bearing’s radial clearance is 0.004-in, and its minimum film thickness is 0.002 in. The bearing carries a load of 70-psi of projected bearing area at 1000 rpm. The bearing temperature is 150°F, and SAE 10 oil is used. Determine the frictional torque due to the bearing under these conditions.
A shaft has a diameter of 6-in and the mating journal bearing has a length of 9-in. The bearing’s radial clearance is 0.004-in, and its minimum film thickness is 0.002 in. The bearing carries a load of 70-psi of projected bearing area at 1000 rpm.
The oil film thickness h can be calculated using the following formula:hmin = 0.000025(RPM) - 0.00025where RPM = 1000, sohmin = 0.000025(1000) - 0.00025hmin = 0.002 in.The viscosity of SAE 10 oil at 150°F can be obtained from the viscosity chart. The chart gives µ = 45 cP.We can calculate the frictional torque, T, by:T = 1.25WLµh/LwhereW = 70 psi, L = 9 in., and h = 0.004 in.T = 1.25 (70 psi) (9 in.) (45 cP) (0.004 in.) / 9 in.T = 11.75 in-lb.Long answer:Given parameters are: Diameter of shaft, d = 6 inches length of bearing, L = 9 inchesRadial clearance, C = 0.004 inches minimum film thickness, hmin = 0.002 inches load on bearing, W = 70 psiBearing temperature, T = 150°F
Velocity of shaft, v = πdn/12 where n = 1000 rpmSAE oil used, viscosity µ = 45 cPFrictional torque,T = 1.25WLµh/Lwhere L = Length of bearing= 9 inches, W = Load on bearing = 70 psi (given)µ = Viscosity of oil = 45 cp (given)h = Radial clearance = 0.004 inches (given)From the given data,The minimum oil film thickness is given by the formula:hmin = 0.000025(RPM) - 0.00025where RPM is the rotational speed in revolutions per minute.Therefore,hmin = 0.000025(1000) - 0.00025= 0.002 inchesThe velocity of shaft is given by the formula:v = πdn/12where d is the diameter of shaft and n is the rotational speed in revolutions per minute.Therefore,v = πdn/12= (3.14 × 6 × 1000)/12= 157 inches per minute.
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A single piston engine aircraft has an overall mass of 530 kg.
At its cruise condition of 125 knots at 9500 ft, the lift
coefficient is CL = 0.5 and the lift-to-drag ratio, FL/FD = 14. The
propellers
A single piston engine aircraft has an overall mass of 530 kg. At its cruise condition of 125 knots at 9500 ft, the lift coefficient is CL = 0.5 and the lift-to-drag ratio, FL/FD = 14. The propellers produce a thrust of 280 N.
Calculate the engine power required for the cruise. Please use the given data and formula below:P = T × VA single piston engine aircraft has an overall mass of 530 kg. At its cruise condition of 125 knots at 9500 ft, the lift coefficient is CL = 0.5 and the lift-to-drag ratio, FL/FD = 14. The propellers produce a thrust of 280 N. Calculate the engine power required for the cruise.To find the power required for the cruise, we will use the formula:P = T × VP = PowerT = ThrustV = VelocityFrom the given data:Thrust, T = 280 NVelocity, V = 125 knots = 64.3 m/s Now, the power required for the cruise is:P = T × V= 280 × 64.3= 18,044 WP ≈ 18 kWTherefore, the engine power required for the cruise is 18 kW.
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Given that the angle contribution of a digital PID to achieve the design requirements is 150", design the digital PID controller by determining its transfer function. The pulse transfer function is given as
GzohGp(s)= 0.03726(z+0.7407)/ (z+0.6714)(2+0.6056)
and the design point is at z₁ = 0.3708 +/0.2537. Use sampling time Ts=0.5 second, and two identical PID controller-zeroes.
To design the digital PID controller, we need to determine its transfer function based on the given information. Let's denote the transfer function of the PID controller as C(z).
The general form of a discrete-time PID controller transfer function is:
C(z) = Kp + Ki/Ti * (1 - 1/z) + Kd * Td * (1 - z^-1)
Given that we have two identical PID controller zeroes, we can represent the transfer function as:
C(z) = Kp * (1 - 1/z)^2 + Ki/Ti * (1 - 1/z) + Kd * Td * (1 - z^-1)
To find the values of the PID controller gains (Kp, Ki, Kd) and time constants (Ti, Td), we need to match the desired design requirements.
From the given design point z₁ = 0.3708 +/- 0.2537, we can determine the damping ratio (ξ) and the natural frequency (ωn) using the following formulas:
ξ = -ln(|z₁|) / sqrt(pi^2 + ln(|z₁|)^2)
ωn = sqrt(1 - ξ^2)
Substituting the given design point:
ξ = -ln(|0.3708|) / sqrt(pi^2 + ln(|0.3708|)^2) = 0.1228
ωn = sqrt(1 - 0.1228^2) = 0.9914
Next, we can determine the PID controller gains and time constants using the Ziegler-Nichols tuning method:
Kp = (1.2 * (Ti/Ts) * ωn) / Gp(1)
Ki = (2.0 * Kp) / (Ts * Ti)
Kd = (0.5 * Kp * Ts * Td) / (0.5 * Ts)
Given the pulse transfer function GzohGp(s) and sampling time Ts = 0.5 second, we can calculate Gp(1) as follows:
Gp(1) = GzohGp(e^(Ts * s)) evaluated at s = 0
Now, let's calculate the values of Kp, Ki, Kd, Ti, and Td using the given information and formulas:
Gp(1) = GzohGp(e^(0.5 * 0)) = GzohGp(1) = 0.03726 * (1 + 0.7407) / ((1 + 0.6714) * (2 + 0.6056)) = 0.01413
Kp = (1.2 * (Ti/Ts) * ωn) / Gp(1) = (1.2 * (1/0.5) * 0.9914) / 0.01413 = 69.886
Ki = (2.0 * Kp) / (Ts * Ti) = (2.0 * 69.886) / (0.5 * Ti) = 279.544 / Ti
Kd = (0.5 * Kp * Ts * Td) / (0.5 * Ts) = Kp * Td
To fully determine the PID controller parameters, we need the value of Ti and Td. These can be chosen based on the desired response characteristics. Typical values for Ti and Td can be selected as follows:
Ti = 4 * Ts
Td = Ts / 2
Using these values, we can calculate the final PID controller transfer function C(z):
C(z
) = 69.886 * (1 - 1/z)^2 + (279.544 / Ti) * (1 - 1/z) + (69.886 * Td) * (1 - z^-1)
This transfer function represents the designed digital PID controller.
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used within the tag, buttons, text boxes, and checkboxes are examples of:
The terms used within the tag, buttons, text boxes, and checkboxes are examples of HTML form elements. An HTML form is a section of a document that contains controls such as text fields, checkboxes, radio buttons, submit buttons, and more.
HTML forms are used to accept user input for sending information to a server.HTML form elements are the building blocks of an HTML form and are what makes the form useful for collecting data from the user. The different types of form elements that can be used are as follows: Text Fields Text area Radio Buttons Check boxes Submit Button Reset
Button File Selector Input Types for Email, URL, and Search. Hidden Inputs Select Box Examples of form elements used within the tag, buttons, text boxes, and checkboxes are as follows: Submit Button Text Fields Radio Buttons Checkboxes Reset Button File Selector Input Types for Email, URL, and Search. Hidden Inputs Select Box
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Identify the factors that increase convection heat
transfer coefficient?
Convection is one of the essential heat transfer processes that occur in natural and industrial processes. It is the transfer of heat from a heated surface to a fluid in motion.
Here are some of the factors that increase convection heat transfer coefficient:Fluid properties: Fluid properties play an essential role in determining the convection heat transfer coefficient. For instance, fluid viscosity, density, specific heat, and thermal conductivity all influence the rate of heat transfer. Generally, fluids with low viscosity and density have a higher heat transfer coefficient.Surface geometry: The shape and geometry of the surface affect the convection heat transfer coefficient. A surface with a larger surface area, roughness, and irregularities promotes turbulent flow. In contrast, smooth surfaces promote laminar flow.Surface temperature: The surface temperature influences the rate of heat transfer by convection.
A hotter surface will lead to higher heat transfer rates than a cooler surface.The temperature gradient between the surface and fluid affects the heat transfer coefficient.Velocity of the fluid: The velocity of the fluid determines the rate of heat transfer by convection. Higher fluid velocity results in higher heat transfer rates. Turbulent flow provides higher heat transfer rates than laminar flow.Geometry and size of the object: The shape, size, and orientation of an object influence the convection heat transfer coefficient. The size of the object affects the heat transfer rate as a larger object provides a larger surface area for heat transfer. Additionally, the orientation of the object influences the heat transfer rate because of the shape and the relative direction of the flow.
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The transfer function of a control element is given by: \[ \frac{2 K}{2 s^{3}+8 s^{2}+22 s} \] 3(a) This element is connected in a unity feedback circuit. (i) Derive the closed loop transfer function
Given that the transfer function of the control element is.
[tex]:$$\frac{2K}{2s^{3}+8s^{2}+22s}$$[/tex]
The control element is connected in a unity feedback circuit.
The closed loop transfer function can be obtained by using the formula given below.
[tex]:$$\frac{C(s)}{R(s)}=\frac{G(s)}{1+G(s)H(s)}$$[/tex]
where C(s) is the output and R(s) is the input, G(s) is the forward path transfer function and H(s) is the feedback transfer function.
Here, the forward path transfer function is given as:$$G(s)=\frac{2K}{2s^{3}+8s^{2}+22s}$$And, since the system is connected in unity feedback, H(s) = 1
Therefore, the closed-loop transfer function is given by:
[tex]$$\frac{C(s)}{R(s)}=\frac{\frac{2K}{2s^{3}+8s^{2}+22s}}{1+\frac{2K}{2s^{3}+8s^{2}+22s}}$$[/tex]
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For the circuit shown below, the gain is expressed by (VA= [infinity]0) Vcc पढळ Vina₁ R₁ MY" 7 - Vout RE Select one: O a. -(R1+1/gm2) 1/gml+RE O b. None of these R1 C. 1/gml RE O d. -gm1R₁ B
The correct option is O a. -(R1 + 1/gm2) / (1/gm1 + RE). The gain of the circuit is expressed by -(R1 + 1/gm2) / (1/gm1 + RE).
The given circuit appears to be a common-emitter amplifier configuration, where Vina₁ represents the input voltage, R1 is the input resistor, MY" is the transistor, Vout is the output voltage, and RE is the emitter resistor. The gain of this amplifier can be determined using the formula:
Av = -(R1 + 1/gm2) / (1/gm1 + RE)
To understand this formula, let's break it down:
- R1 represents the input resistor, which influences the input voltage and plays a role in determining the overall gain of the amplifier.
- gm1 is the transconductance of the first transistor (MY") and represents the gain of the transistor itself.
- gm2 is the transconductance of the second transistor, which affects the overall gain of the circuit.
- RE is the emitter resistor, which impacts the output voltage and contributes to the amplifier's gain.
By calculating the inverse of the transconductances (1/gm1 and 1/gm2) and considering the resistors, we can express the gain of the circuit as -(R1 + 1/gm2) / (1/gm1 + RE).
Therefore, the correct option is O a. -(R1 + 1/gm2) / (1/gm1 + RE).
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