Answer:
• it charges banks more interest
• it sells more securities
• it decreases the money supply
In response to high inflation, the Fed charges banks more interests and pays the banks less interests. It also sells not securities.
Answer:
The MATLAB Code for this PI Controller will be:
Kp = 350;
Ki = 300;
Kd = 50;
C = pid(Kp,Ki,Kd)
T = feedback(C*P,1);
t = 0:0.01:2;
step(T,t)
Explanation:
When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.
Obtain an open-loop response and determine what needs to be improved
Add a proportional control to improve the rise time
Add a derivative control to reduce the overshoot
Add an integral control to reduce the steady-state error
Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.
The further explanation is attached in the Word File.
<h3><u>CSMA/CD Protocol:
</u></h3>
Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.
But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.
There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.
<u>Example:
</u>
, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.
12:00 AM A will see collisions
Pocket Size to detect the collision.
CSMA/CD is widely used in Ethernet.
<u>Efficiency of CSMA/CD:</u>
- In the previous example we have seen that in worst case time require to detect a collision.
- There could be many collisions may happen before a successful completion of transmission of a packet.
We are given number of collisions (contentions slots)=4.
Distance = 1km = 1000m
Location of the class depends on satiation
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV