Energy Signals
Вставка
- Опубліковано 17 жов 2024
- Signal and System: Energy Signals
Topics Discussed:
1. The definition of Energy Signals.
2. Condition for a signal to be an energy signal.
3. Why power is zero when energy is finite?
4. Energy signal examples.
5. Properties of energy signal.
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Music:
Axol x Alex Skrindo - You [NCS Release]
#SignalAndSystemByNeso #Signal&System #EnergySignals
6:13 ah finally heard the explanation of the part where I confused for these few days!! Thank you so much!!!
2nd day of my signal and system, hopefully I can finish in 1 week. I must say, these lectures are very well explained and laid-out, you have to watch the all the 178 lectures to get a great understanding of the subject. Thank you so much neso academy!!!!
What's your chess elo rating?
Thank you sir🙏🙏
Your explanation is extraordinary better than our lectures.
This video was very energetic and powerful, and it definitely was *not* average. Thank you so much!
what is the reason for assuming the value of resistance 1 ohm during the derivation of equations for energy and power signals?
Thank you very much for your great work !!
Good explanation for the energy and power signals and why they are 0 and infinte energy and power signals|thank you sir 😊
Sir ye integration Kya hota Hai aur differentiation I know but don't no what is this please make a video sir
sir please upload all lectures of signal and system
THANK YOU
Why power is not absolute integration like energy? 3:47
its good to have straight A's
Goood
bro,when u wil upload the complete lectures regarding Fourier series n Fourier transform
Sir, my query is whether impulse signal is an energy signal or power signal? And How?
Neither Energy ( because infinite energy)
Nor Power ( because not finite power)
What is normalised energy and normalised power? How they are different from average energy and power??Plzz provide a suitable link fir the above bcoz I have searched and results are wrt yo athletic
Smith chart idea hai or Nahi oo be similar normalisation means reducing to small quantity
Thanks for a clear understanding lecture.
what is difference between avarage value and dc value?
I think dc value is an even as shown our previous lecture of an even and odd
Unit of energy is V^2*s or I^2*s depending upon signal is voltage or current
What a phenomenal video.
Great video on the topic, thanks. However I can not agree with you on your first conclusion. A finite energy signal (energy signal) must not always be absolutely integrable. Take x(t)=(1/t)*heaviside(t-1) for example.
is energy finite or infinite?! If energy is finite/infinite then universe is finite/ infinite, but we don't really know. So in signals and systems for some odd reason we define energy = finite! Why? Maybe to avoid problems with other formulas/concepts in signals&systems?
Why should T be +infinity that may be -infinity please clarify my doubt sir
for the 1st property you left the power 2 on E for finite
Thanks neso
I have no word to say for your explanations
Great sir...
Thank you sir
Thank you sir for teaching us
If x(t) is u(t)+5u(t-1)+2u(t-2).
Is their Any condition that energy signal must be period or aperiod??....?
We always calculate Energy of a non-periodic signal, because it can be finite( provided it does exist in a certain range, and not from -inf to +inf), so that we can compare withthe other signals. For a periodic signal like sine signal, Energy is infinite and can't compare directly
if we are integrating inst. power from -infinity to +infinity then how we r getting finite value ? sir plz clear my doubt..
Because from minus infinity to plus infinity area is calculated for a specific shape which we apply in a formula to calculate energy...Anything other than that will have zero area so zero energy
(mistake at 3.26) formula for average power is [total energy/total time]
where total time is negative infinity to positive infinity. The boundary is set as such or else you will always have positive values for power signals.
Power =energy / time , we know that AVG marks=total marks/no of subject,like that he written ...
Thank u vry much
Please provide the last video link also
How to know whether it is finite or infinite
Countable is finite
t is from -∞ to ∞ means what???
Why integration of power is zero but integration of energy is finite
when calculating x(t)^2 you don't need mod[x(t)]. It's already positive.
this is because the order can be reversed .you may start by mod then square
This is because in case if we have a complex signal, So of course you need to get the mod and squre it.
clc;
clear;
close all;
% Q1:Find the main dimensions and number of poles of 1000 KW 500 V 300 RPM DC
% machine. Assume the specific magnetic loading Bav= 0.7 W/m², ampere
% conductor per meter = 40,000 , square pole face and ratio of pole
% Arc to pole pitch is 0.7 efficiency of generator is 92%.
power = 1000; %powe, in KW
V = 500; %input voltage, in Volt
Bav = 0.7; %average value of flux density in air gap, in T ( Tesla)
N = 300; %speed in rpm
n = 0.92; %efficiency of generator
f = 50; %frequency in Hz
P = (120*f)/N; % number of poles
psi = 0.7; %L/tau ratio of pole arc to pole per meter
q = 400001; %specific electrical loading
Pa = power/n; %power in armature coil
D = (Pa/((1.64*10^-4*Bav*q*N)*(psi*pi/P)))^(1/3); %diameter of armature
L = psi*pi*D/P; %length of armature coil
disp(P);
disp(Pa);
disp(D);
disp(L);
clc;
clear;
close all;
% Q2:-Calculate the armature diameter and core length for 7.5kW,4pole,
%1000rpm,and 220V shunt motor. Assume: Full load efficiency = 0.83,field
%current is 2.5% of rated current. The maximum efficiency occurs at full load.
power = 7500; %powe, in W
V = 220; %input voltage, in Volt
Bav = 0.6;%(Assume)average value of flux density in air gap, in T ( Tesla)
N = 1000; %speed in rpm
n = 0.83; %full load efficiency of motor
f = 50; %frequency in Hz
P = 4; % number of poles
psi = 1; %(Assume)L/tau ratio of pole arc to pole per meter
q = 30000; %(Assume) specific electrical loading
Pin = power/n; %Electrical input to the motor
Ish = (0.025*Pin)/220; %shunt current
losses = Pin - power; %losses at full load
cl = losses/2; %constant loss
Fcl = V*Ish; %field copper loss
Pa = Pin - (cl+Fcl);%Power developed in armature
D = (Pa/((1.64*10^-4*Bav*q*N)*10^3*(psi*pi/P)))^(1/3); %diameter of armature
L = psi*pi*D/P; %length of armature coil
disp(P);
disp(Ish);
disp(Pa);
disp(D);
disp(L);
disp(losses);
disp(cl);
disp(Fcl);
clc;
clear;
close all;
%Q 3: - The diameter and length of a 500Kw, 500V, 455 rpm, 6 pole, and D.C.
%generator are 84 cm and 35 cm respectively. If it is lap wound with 660
%conductors. Estimate the specific electric and magnetic loading.
p = 500000; %power rating in watt
V = 500; %volatage in V
N = 455; %speed of motor in rpm
P = 6; %number of pole
D = .84; %diamerer in meater
L = .35; %length in meater
Z = 660; %number of conductor
A = P; %in lap winding no of parallel path(A) is equal to no of pole
phai = (V*60*A)/(Z*N*P); %flux generated
Bav = (P*phai)/(pi*D*L); %specific magnetic loading in wb/m^2
Ia = p/V; % Armature current = (power rating/volatge)
%Iz = Ia/A; %Armature current/no of parallel path
ac = (Ia*Z)/(A*pi*D); %specific electrical loading in A-conductor/m
disp(phai);
disp(Ia);
disp(Bav);
disp(ac);
clc;
clear;
close all;
%A design is required for a 50Kw, 4-pole, 600rpm, and D.C. shunt generator
%with full load terminal voltage being 200V. If the max air gap flux
%density is 0.83 wb/m2 and armature ampere conductor per meter is 30000.
%Calculate the suitable dimensions of armature core to give square pole
%face. Assume full load armature voltage drop 3% of the rated terminal
%voltage and the field current 1% of rated full load current, ratio of
%pole face to pole pitch is 0.67.%
p = 50000; %rated power
P = 4; %number of poles
N = 600; %speed of generator
V = 200; %voltage in V
Bg = 0.83; %air gap flux in wb/m^2
ac = 30000; %specific electrical loading
%b = 1; %due to square pole face
%L = 1; %due to square pole face
x = 0.03; %full load armature voltage drop (Ia*Ra)
y = 0.01; %field current 1% of rated full load current
sai = 0.67; %ratio pole face to pole pitch (sai*Bg = Bav)
Bav = sai*Bg; %specific magnetic loading
Co = 1.64*10^-4*Bav*ac; %cofficent
E = V+(x*V); %full load voltage
Il = (p/V); %full load current
If = y*Il; %field current
Ia = Il+If; %armature current
Pa = Ia*E; %power in armature
DDL = (Pa*(10^-3)/(Co*N)); % Pa = Co*D^2*L*N; %Armature power take in kw
r = sai*(pi/4); % (r = L/D)
D = (DDL/r)^(1/3); % diameter
L = r*D; %lenght
disp(Co)
disp(E)
disp(Il)
disp(If)
disp(Ia)
disp(Pa)
disp(DDL)
disp(r)
disp(D)
disp(L)
clc;
clear;
close all;
%Q5:-Using MATLAB calculate the main dimension detail of a 10KVA,2000/400V,
%50Hz, single phase shell type oil immersed self-cooled transformer.
%Assume voltage per turn 10V, flux density 1.1wb/m^2, current density
%2A/mm^2, window space factor 0.33, the ratio of window height to window
%width is 3 and the ratio of core depth to width of central limbs is 2.5,
%the stacking factor is 0.9.
F=input('Enter the value of frequency in Hz:');
Q=input('Enter the value of output power in KVA:');
X=input('Enter the value of primary voltage V1:');
V2=input('Enter the value of secondary voltage V2:');
Et=input('Enter the value of voltage per turn Et:');
B=input('Enter the flux density in wb/m^2 B:');
j=input('Enter the value of current density j:');
S=input('Enter the value of stacking factor S:');
K=input('Enter the value of window space factor K:');
R=input('Enter the value of ratio of core depth to central limb,R:');
Hr=input('Enter the value of ratio of height and width of window, Hr:');
Ai=Et/(4.44*F*B); %Net iron area
Agi=Ai/S; %Gross iron area
a=(Agi/10)^1/2; %Width of central limb
b=2*a*R; %Core depth
Ay=Agi/2; %Gross area of yoke
Dy=b; %Depth of yoke
Hy=Ay/Dy; %Height of yoke
Aw=Q/(2.22*F*B*K*j*Ai*10^-3); %Area of window
Hw=(Aw*Hy)^1/2; %Height of window
Ww=Aw/Hw; %Width of window
H=Hw+2*Hy; %Height of frame
overall=2*Hw+4*a; %0verall length of frame
Tp=X/Et; %H.V winding turns
Ts=V2/Et; %L.V winding turns
Ip=Q/X;%H.V winding current
Is=Q/V2;%L.V winding current
Ap=Ip/j;%H.V winding conductor area
As=Is/j; %L.V winding conductor area
ds=(As/3.14)^1/2; %L.V winding conductor diameter
dp=(Ap/3.14)^1/2; %H.V winding conductor diameter
disp(Ai);
disp(Agi);
disp(b);
disp(a);
disp(Ww);
disp(Aw);
disp(Hw);
disp(ds);
disp(dp);
disp(Hy);
disp(H);
disp(overall);
disp(Tp);
disp(Ts);
disp(Ip);
disp(Is);
disp(Ap);
disp(As);
disp(Dy);
clc;
clear;
close all;
%Q 6: - Determine the number of poles, main dimensions, pole pitch and
%armature mmf/pole of a 92Kw, 220V, 1480 rpm D.C. motor whose full load
%efficiency, specific magnetic loading and specific electric loading,
%the pole arc to pole pitch ratio are 89.76%, 0.545T, 32750 Ac/m and 0.67
%respectively. Assume square pole face.
p = 92000; %rated power
N = 1480; %speed of generator
V = 220; %voltage in V
n = 0.8976; %full load efficincy
Bav = 0.545; %specific magnetic loading in wb/m^2
ac = 32750; %specific electrical loading
sai = 0.67; %ratio pole face to pole pitch (sai*Bg = Bav)
f = 50; %frequency in Hz
P = (120*f)/N; %number of pole
PP = round(P);
Pa = p/n; %Power in armature
Co = 1.64*10^-4*Bav*ac; %cofficent
DDL = (Pa*(10^-3)/(Co*N)); % Pa = Co*D^2*L*N; Armature power take in kw
r = sai*(pi/4); % (r = L/D)
D = (DDL/r)^(1/3); % diameter
L = r*D; %lenght
disp(PP)
disp(Pa)
disp(Co)
disp(DDL)
disp(r)
disp(D)
disp(L)
clc;
clear;
close all;
%Q 7: - Determine the main dimensions of a 10 H.P ,400 V, 1500 r.p.m ,
%dc shunt motor, the average gap density is 0.45 tesla and ampere
%conductors/meter are 20,000. The max efficiency is 85%. Assume shunt
%field current to be 0.9A and the diameter to length ratio is 2.7.
%The pole arc to pole pitch is 0.7.
p = 10*746; %rated power
N = 1500; %speed of generator
V = 400; %voltage in V
n = 0.85; %full load efficincy
Bav = 4.5; %specific magnetic loading in wb/m^2
ac = 20000; %specific electrical loading
Ish = 0.9; %shunt field current
Rsh = V/Ish; %shunt field resistance
r = 1/2.7; % r = L/D
pin = p/n; %Input power
pf = ((pin-p)/2)-(Ish^2*Rsh);%power loss due to friction
Pa = p+pf; %total power
Co = 1.64*10^-4*Bav*ac; %cofficent
DDL = (Pa*(10^-3)/(Co*N)); % Pa = Co*D^2*L*N; Armature power take in kw
D = (DDL/r)^(1/3); % diameter
L = r*D; %lenght
disp(Rsh)
disp(r)
disp(pin)
disp(pf)
disp(Pa)
disp(Co)
disp(DDL)
disp(D)
disp(L)
clc;
clear;
close all;
%Q 8: - A 5Kw, 250V, 4-pole, 1500 rpm, shunt generator is designed to have
%a square pole face. The loadings are: Average flux density in the gap is
%0.42 wb/m2 and ampere conductor per meter is 15000. Find the main
%dimension of the machine. Assume load efficiency is 87% and ratio of pole
%arc to pole pitch is 0.66.
p = 5000; %rated power
P = 4; %number of pole
N = 1500; %speed of generator
V = 250; %voltage in V
n = 0.87; %full load efficincy
Bav = 0.42; %specific magnetic loading in wb/m^2
ac = 15000; %specific electrical loading
sai = 0.66; %ratio pole face to pole pitch (sai*Bg = Bav)
r = sai*(pi/4); % (r = L/D)
Pa = p/n; %Power in armature
Co = 1.64*10^-4*Bav*ac; %cofficent
DDL = (Pa*(10^-3)/(Co*N)); % Pa = Co*D^2*L*N; Armature power take in kw
D = (DDL/r)^(1/3); % diameter
L = r*D; %lenght
disp(r)
disp(Pa)
disp(Co)
disp(DDL)
disp(D)
disp(L)
goor3efvgrghthn
Krishna
Thanks neso