Assignment-3
1. Taking first constant term of fourier series.As T is large higher order terms are approximately zero.
F= A*τ/T ,
Where τ=T-del(T),
4sqrt(2)=5* {T-delta(T)}/T,
Delta(T)/T = 0.12
- (a)-By superposition theorem,
Current by first source,
I1= |V1/Z1 |= |(120./sqrt((20)^2 + (2.pi.60.*0.1)^2) |
Current by second source-
I2=V2/Z2= |V2./sqrt((20)^2 + (2.pi.50.*0.1)^2)|
Total current,
I=I1+I2,
Given Irms=5;
Solving ,
V2rms=76.91 volts
(b) Power factor,geometric mean due to both sources–
Sqrt(Cos(phi1)Cos(phi2))=sqrt((R/Z1)(R/Z2))=400/sqrt(1400*1840) = 0.25
3.(a) F(x)=x^2 +2 ;
F(-x)=(-x)^2 +2=F(x) ,
This is an even function.
(b) F(x)=(x^2+2)tan(x)^2 ;
F(-x)=((-x)^2 +2)tan(-x)^2=F(x);
This is an even function.
(c) F(x)= (x^2+2)sin(x)tan(x)^2 ;
F(-x)= ((-x)^2+2)sin(-x)tan(-x)^2 = – F(x) ;
This is an odd function.
4.For fourier series we will evaluate the constants, T=(0-(-2)+(2)-0)=4;
So that a0=2/4=1/2;
Where w0=2.pi/T=2.pi/4=pi/2
So that an=-1 ;
So that bn=2/n.pi;
So the fourier series is,
5.(a) Inverse fourier transform is,
Ft=F^(-1) [F(w)];
Where F(w)= | F(w)| exp(j*theta(w));
Where | F(w)|= (alpha^2 ) + ((beta)^2);
Theta = tan^(-1) (beta/alpha) ;
The code for solving this is ,
alpha=input(‘alpha=’);
beta=input(‘beta=’);
Fw=((alpha.^2)+(beta.^2))
theta=atan(beta./alpha)
Hence
After integrating between limits (-infinity) to (0) and then (0) to (infinity) we will have Ft.It is difficult to solve above equation.This is the difficulty in this problem.
Assignment-4
1.(a)The damping ratio is the ratio of real and imaginary parts,
So that, ζ = -σ/ω , σ is the x-axis ,j ω is the y-axis, in s plane ,
We have to analysis the graph in s plane,in which these formulas can be applied.But the graph is in V,t plane.This is the difficulty.If we can convert this V,t graph to s-plane graph than damping ratio can be find by above formula.
(c) the undamped frequency of oscillation is,
(b)= Damped frequency is , ω= ω(undamped).sqrt(1- ζ^2)
2. (a) £ (20)= 20/s ;
£ (20sin(5t) =100/(s^2+25) ;
£ (20exp(-5t) = 20/(s+5) ;
£ (20 exp(-5t)cos(5t))= 100/((s+5)^2 + 25) ;
£ (20t^2) = 2/s^2 ;
(b) £^(-1) [1/(2s+4)] =0.5 exp(-2t) ;
£^(-1) [5s/(16+s^2)] = 5.cos(4t) ;
£^(-1) [10/s^2] = 10t ;
£^(-1) [4/((s-2)^2 +16)] = exp(2t)* sin(4t) ;
£^(-1) [ (9s+23)/(s+2)(s+3)] = £^(-1) [( 9(s+2) + 5)/(s+2)(s+3)]
= 9exp(-3t) + 5(exp(-2t)+ exp(-3t))
- (1)£^(-1) [1/(s^2+2s+2)] = £^(-1) [1/((s+1)^2 + 1)] = exp(-t)sin(t)
(2) £^(-1) [1/(s^2+s+2)] = £^(-1) [ (1)/(s+1/2)^2 + 7/4)]
= (2/sqrt(7))exp((-1/2)t)sin(sqrt(7)t/2)
(3) £^(-1) [1/(s^2+0.5*s+2)] = (4/sqrt(31))exp((-1/4)t)sin(sqrt(31)t/4) ;
(4) £^(-1) [1/(s^2+0.4*s+4)] =
We compare coefficients and match the graphs—
We will check how much decrease in volts with decrease in frequency—
4 – D
3 – B
2- C
1—A
- (a) Laplace equation will be, tou*s*deltaT(s) + deltaT(s) = (Tw-Ta)
deltaT(s) = (Tw- Ta)/tou*(s + 1/tou)
(b) Transient response is,
deltaT(s)= ((Tw- Ta)/tou)exp(-t/tou) ,
where tou= mc/Ah
(c) Plot is ,
Ta=20;Tw=100;m=2.5.*10.^(-3);A=10.^(-3);c=0.2;h=0.78;
tou= m.*c./(A.*h) ;
t=0:tou:4*tou;
deltaT= ((Tw- Ta)./tou).*exp(-t./tou) ;
plot(t,deltaT);
xlabel(‘t’);
ylabel(‘deltaT’);
- (a)
Vo= Vi* (RXc/(R+Xc))/(XL +(RXc/(R+Xc)) = ((100/0.001)(100*s)/(100+s./0.001))/((100*s*0.5)+(100.*s/0.001)/(100 + s/0.001))
Vo= (10^(5)*s/(1+10s))/(50*s + (10^3)s/(1+10s));
Vo= (10^(5))*s/(500*s^2+1050*s)
Vo=(10^5)/(500*s+ 1050);
(b) Vo= .01*200/(s+ 2.1);
V(t)= 2exp(-2.1t) ;
t=0:1:100;
Vo=2.*exp(-2.1.*t);
plot(t,Vo);
xlabel(‘t’);
ylabel(‘Vo’);
NOTE—Answer 5 of assignment3 and answer 1 of assignment 4 are not exact solution but hints .
LC03
But you can order it from our service and receive complete high-quality custom paper. Our service offers Science essay sample that was written by professional writer. If you like one, you have an opportunity to buy a similar paper. Any of the academic papers will be written from scratch, according to all customers’ specifications, expectations and highest standards.”
