matlab assignment jk institute
TRANSCRIPT
SOL 1 :
clear
clc
F=0;
format short
F = input('Enter temperature in Fahrenheit : ');
K = (((F-32)*5/9) + 273); %The expression converts the temperature
fprintf('Temperature in Kelvin would be : %f\n',K)
OUTPUT :
Enter temperature in Fahrenheit : 32
Temperature in Kelvin would be : 273.000000
>>
SOL 2 :
clear
clc
format compact
format short
A(1)=0;
A(2)=1;
n= input ('Enter the value of n (n>2) : ');
i=3;
while i<=n
A(i)= A(i-1)+ A(i-2);
i=i+1;
end
disp(A)
OUTPUT :
Enter the value of n (n>2) : 10
0 1 1 2 3 5 8 13 21 34
>>
SOL 3 :
clear
clc
y=input('Enter the year (in YYYY format) : ','s');
m=input('Enter the numeric value for month (in MM format) : ','s');
d=input('Enter the date (in DD format) : ','s');
date=strcat(m,'/',d,'/',y);
[N,S]=weekday(date,'long');
fprintf ('The day is %s \n',S)
OUT PUT :
Enter the year (in YYYY format) : 1990
Enter the numeric value for month (in MM format) : 02
Enter the date (in DD format) : 10
The day is Saturday
>>
SOL 4 :
clear
clc
x(1)=0;
i=1;
n=input(' Enter no of elements in the array : ');
disp(' Now enter array elements one by one')
while i<=n
x(i)=input('');
i=i+1;
end
m=mean(x);
s=std(x);
disp(' You entered the array as :')
disp(x)
fprintf('\n The mean of the elements is : %f\n The Standard Deviation of the
elements is : %f\n',m,s)
OUTPUT :
Enter no of elements in the array : 4
Now enter array elements one by one
2
3
5
9
You entered the array as :
2 3 5 9
The mean of the elements is : 4.750000
The Standard Deviation of the elements is : 3.095696
>>
SOL 5 (i) :
clear
clc
format short
i=1;
for t=-6*pi:pi/10:6*pi
x(i)=t;
y(i)=sin(t);
if y(i)<=0
y(i)=0;
end
i=i+1;
end
R(:,1)=x;
R(:,2)=y;
disp(' t f(t)')
disp(R)
SOL 5 (ii) :
clear
clc
format short
x=-6*pi:pi/10:6*pi;
y= sin(x);
w= y>0;
R(:,1)=x;
R(:,2)=w(1,:) .* y(1,:);
disp(' t f(t)')
disp(R)
OUTPUT :
t f(t) -18.8496 0.0000 -18.5354 0.3090 -18.2212 0.5878 -17.9071 0.8090 -17.5929 0.9511 -17.2788 1.0000 -16.9646 0.9511 -16.6504 0.8090 -16.3363 0.5878 -16.0221 0.3090 -15.7080 0 -15.3938 0 -15.0796 0 -14.7655 0 -14.4513 0 -14.1372 0 -13.8230 0 -13.5088 0 -13.1947 0 -12.8805 0 -12.5664 0.0000 -12.2522 0.3090 -11.9381 0.5878 -11.6239 0.8090 -11.3097 0.9511 -10.9956 1.0000 -10.6814 0.9511 -10.3673 0.8090 -10.0531 0.5878 -9.7389 0.3090 -9.4248 0 -9.1106 0 -8.7965 0 -8.4823 0 -8.1681 0 -7.8540 0 -7.5398 0 -7.2257 0 -6.9115 0 -6.5973 0 -6.2832 0.0000 -5.9690 0.3090 -5.6549 0.5878 -5.3407 0.8090 -5.0265 0.9511 -4.7124 1.0000
-4.3982 0.9511 -4.0841 0.8090 -3.7699 0.5878 -3.4558 0.3090 -3.1416 0 -2.8274 0 -2.5133 0 -2.1991 0 -1.8850 0 -1.5708 0 -1.2566 0 -0.9425 0 -0.6283 0 -0.3142 0 0 0 0.3142 0.3090 0.6283 0.5878 0.9425 0.8090 1.2566 0.9511 1.5708 1.0000 1.8850 0.9511 2.1991 0.8090 2.5133 0.5878 2.8274 0.3090 3.1416 0.0000 3.4558 0 3.7699 0 4.0841 0 4.3982 0 4.7124 0 5.0265 0 5.3407 0 5.6549 0 5.9690 0 6.2832 0 6.5973 0.3090 6.9115 0.5878 7.2257 0.8090 7.5398 0.9511 7.8540 1.0000 8.1681 0.9511 8.4823 0.8090 8.7965 0.5878 9.1106 0.3090 9.4248 0.0000 9.7389 0 10.0531 0 10.3673 0 10.6814 0 10.9956 0 11.3097 0
11.6239 0 11.9381 0 12.2522 0 12.5664 0 12.8805 0.3090 13.1947 0.5878 13.5088 0.8090 13.8230 0.9511 14.1372 1.0000 14.4513 0.9511 14.7655 0.8090 15.0796 0.5878 15.3938 0.3090 15.7080 0.0000 16.0221 0 16.3363 0 16.6504 0 16.9646 0 17.2788 0 17.5929 0 17.9071 0 18.2212 0 18.5354 0 18.8496 0 >>
SOL 6 :
clear
clc
i=1;
for k = -9:0.5:9
T(i)=k;
if k<0
Y(i) = (-3*(k^2)) + 5;
else
Y(i) = (3*(k^2)) + 5;
end
i=i+1;
end
R(:,1) = T;
R(:,2) = Y;
disp(' T Y(T)')
disp(R)
OUTPUT :
T Y(T)
-9.0000 -238.0000
-8.5000 -211.7500
-8.0000 -187.0000
-7.5000 -163.7500
-7.0000 -142.0000
-6.5000 -121.7500
-6.0000 -103.0000
-5.5000 -85.7500
-5.0000 -70.0000
-4.5000 -55.7500
-4.0000 -43.0000
-3.5000 -31.7500
-3.0000 -22.0000
-2.5000 -13.7500
-2.0000 -7.0000
-1.5000 -1.7500
-1.0000 2.0000
-0.5000 4.2500
0 5.0000
0.5000 5.7500
1.0000 8.0000
1.5000 11.7500
2.0000 17.0000
2.5000 23.7500
3.0000 32.0000
3.5000 41.7500
4.0000 53.0000
4.5000 65.7500
5.0000 80.0000
5.5000 95.7500
6.0000 113.0000
6.5000 131.7500
7.0000 152.0000
7.5000 173.7500
8.0000 197.0000
8.5000 221.7500
9.0000 248.0000
>>
SOL 7(i) :
clear
clc
format short
i=1;
for x=-1:0.1:3
X(i)=x;
Y(i)=(x^2)-(3*x)+2;
i=i+1;
end
R(:,1)=X;
R(:,2)=Y;
disp(' X Y(x)')
disp(R)
plot(X,Y,'--r','linewidth',3)
xlabel('x-axis'); ylabel('y-axis');
SOL 7(ii) :
clear
clc
format short
X=-1:0.1:3;
Y= (X.^2)-(3.*X)+2;
R(:,1)=X;
R(:,2)=Y;
disp(' X Y(x)')
disp(R)
plot(X,Y,'--r','linewidth',3)
xlabel('x-axis'); ylabel('y-axis');
OUTPUT : X Y(x) -1.0000 6.0000 -0.9000 5.5100 -0.8000 5.0400 -0.7000 4.5900 -0.6000 4.1600 -0.5000 3.7500 -0.4000 3.3600 -0.3000 2.9900 -0.2000 2.6400 -0.1000 2.3100 0 2.0000 0.1000 1.7100 0.2000 1.4400 0.3000 1.1900 0.4000 0.9600 0.5000 0.7500 0.6000 0.5600 0.7000 0.3900 0.8000 0.2400 0.9000 0.1100 1.0000 0 1.1000 -0.0900 1.2000 -0.1600 1.3000 -0.2100 1.4000 -0.2400 1.5000 -0.2500 1.6000 -0.2400 1.7000 -0.2100 1.8000 -0.1600 1.9000 -0.0900 2.0000 0 2.1000 0.1100 2.2000 0.2400 2.3000 0.3900 2.4000 0.5600 2.5000 0.7500 2.6000 0.9600 2.7000 1.1900 2.8000 1.4400 2.9000 1.7100 3.0000 2.0000 >>
SOL 8 :
function f = fact(x)
f=1;
if x<0
disp('Entered value is negative')
return
end
if x~=ceil(x) | x~=floor(x)
disp('Entered value is not an integer')
return
end
for k = x:-1:0
if k==0
k=1;
end
f = f * k;
end
OUTPUT :
>> fact(-1)
Entered value is negative
>> fact(12.90)
Entered value is not an integer
>> fact(7)
ans =
5040
>>
SOL 9 :
Clear
clc
x=0;
while x < 1
x = input('Enter value for x : ');
if x >= 1
disp('The entered value is out of domain of function so returning back . . .')
return
end
y = log(1/(1-x));
disp(y)
end
OUTPUT :
Enter value for x : -1
-0.6931
Enter value for x : 0
0
Enter value for x : -12
-2.5649
Enter value for x : 1
The entered value is out of domain of function so returning back...
>>
SOL 10:
clear
clc
F=[75 100 125];
T = (5/9*(F-32))+273;
Io=0.000002;
q= 1.6 * (10^(-19));
k= 1.38 * (10^(-23));
Vd= -1.0 : 0.1 : 10.6;
qvd= q * Vd;
KT1= k * T(1);
KT2= k * T(2);
KT3= k * T(3);
I1=Io*(exp(qvd/KT1)-1);
I2=Io*(exp(qvd/KT2)-1);
I3=Io*(exp(qvd/KT3)-1);
plot(Vd,I1,'--b',Vd,I2,'-r',Vd,I3,':k');
legend('75F','100F','125F');
xlabel('Vd (Volts)');
ylabel('I (in Amp)');
OUTPUT :
>>
SOL 11 :
clear
clc
ln1=input('Enter the no of elements in the I sequence : ');
xn1(1,:)=zeros(1,ln1);
disp('Enter the values for the elements ')
for r=1:ln1
xn1(r)=input('');
end
ln2=input('Enter the no of elements in the II sequence : ');
xn2(1,:)=zeros(1,ln2);
disp('Enter the values for the elements ')
for r=1:ln2
xn2(r)=input('');
end
xk2=zeros(1,ln2);
xk1=zeros(1,ln1);
%To find the DFT of the sequence
for k=0:ln1-1
for n=0:ln1-1
xk1(k+1)=xk1(k+1)+(xn1(n+1)*exp((1*2*pi*k*n / ln1)*1i));
end
end
for k=0:ln2 - 1
for n=0:ln2 - 1
xk2(k+1)=xk2(k+1)+(xn2(n+1)*exp((1*2*pi*k*n / ln2)*1i));
end
end
disp('DFT of I sequence')
disp(xk1)
disp('DFT of II sequence')
disp(xk2)
OUTPUT :
Enter the no of elements in the I sequence : 2
Enter the values for the elements
1
3
Enter the no of elements in the II sequence : 3
Enter the values for the elements
2
4
6
DFT of I sequence
4.0000 -2.0000 + 0.0000i
DFT of II sequence
12.0000 -3.0000 - 1.7321i -3.0000 + 1.7321i
>>