Instructor
SRA
SRA
These are for supportive ideas only.
Do not misuse these.
Do not misuse these.
Assignment-I
Course Name : Signals and Systems
Course Code : EEE303
Assignment No : 01
Answer to the question no. 1
Code for plotting x1 vs. t:
clear all
t=0:4000;
t=t/100;
x1=sin(2*pi*t/5);
plot (t,x1);
Figure: Plot of x1(t) vs. t
Code for plotting x2 vs. t:
clear all
t=0:4000;
t=t/100;
x2=sin(2*pi*t/pi);
plot (t,x2);
Figure: Plot of x2(t) vs. t
Code for plotting x3 vs. t:
clear all
t=0:4000;
t=t/100;
x3=[.5*sin(2*pi*t/5)+.5*sin(2*pi*t)];
plot (t,x3);
Figure: Plot of x3(t) vs. t
Answer to the question no. 2
Code for plotting x4 vs. t:
clear all
t=0:4000;t=t/100;
x1=sin(2*pi*t/5);
x2=sin(2*pi*t/pi);
x3=[.5*sin(2*pi*t/5)+.5*sin(2*pi*t)];
x4=x1+x2;
plot (t,x4);
Figure: Plot of x4(t) vs. t
Code for plotting x5 vs. t:
clear all
t=0:4000;t=t/100;
x1=sin(2*pi*t/5);
x2=sin(2*pi*t/pi);
x3=[.5*sin(2*pi*t/5)+.5*sin(2*pi*t)];
x5=x2+x3;
plot (t,x5);
Figure: Plot of x5(t) vs. t
Answer to the question no. 3
From the wave shapes of the above diagram, it seems that x4=x1+x2 and x5=x1+x3 both are periodic. We will see if it is really periodic or not in the mathematical measurement.
Mathematically,
x4=x1+x2
=sin (2πt/5)+sin(2πt/π)
So,
T1=2π/(2π/5)=5
T2=2π/(2π/π)=π
Now,
T1/T2=5/π
This is not a rational number.
So, x4 is not periodic.
And,
x5=x1+x3
= sin (2πt/5)+sin(6πt/5)cos (4πt/5)
= sin (2πt/5)+0.5*2sin(6πt/5)cos (4πt/5)
= sin (2πt/5)+0.5sin(2πt)+0.5sin (2πt/5)
= (3/2)*sin (2πt/5)+0.5sin(2πt)
So,
T1=2π/(2π/5)=5
T2=2π/(2π)=1
Now,
T1/T2=5
This is a rational number.
So, x5 is periodic
Further analysis of wave shapes:
Now if we again look at the waves shapes of x4, we can see that , though wave shapes seems to be repeated bur their magnitude is not same. So they are not actually repeated.
But the wave shapes of x5 are really repeated with perfect magnitudes.
Course Name : Signals and Systems
Course Code : EEE303
Assignment No : 01
Answer to the question no. 1
Code for plotting x1 vs. t:
clear all
t=0:4000;
t=t/100;
x1=sin(2*pi*t/5);
plot (t,x1);
Figure: Plot of x1(t) vs. t
Code for plotting x2 vs. t:
clear all
t=0:4000;
t=t/100;
x2=sin(2*pi*t/pi);
plot (t,x2);
Figure: Plot of x2(t) vs. t
Code for plotting x3 vs. t:
clear all
t=0:4000;
t=t/100;
x3=[.5*sin(2*pi*t/5)+.5*sin(2*pi*t)];
plot (t,x3);
Figure: Plot of x3(t) vs. t
Answer to the question no. 2
Code for plotting x4 vs. t:
clear all
t=0:4000;t=t/100;
x1=sin(2*pi*t/5);
x2=sin(2*pi*t/pi);
x3=[.5*sin(2*pi*t/5)+.5*sin(2*pi*t)];
x4=x1+x2;
plot (t,x4);
Figure: Plot of x4(t) vs. t
Code for plotting x5 vs. t:
clear all
t=0:4000;t=t/100;
x1=sin(2*pi*t/5);
x2=sin(2*pi*t/pi);
x3=[.5*sin(2*pi*t/5)+.5*sin(2*pi*t)];
x5=x2+x3;
plot (t,x5);
Figure: Plot of x5(t) vs. t
Answer to the question no. 3
From the wave shapes of the above diagram, it seems that x4=x1+x2 and x5=x1+x3 both are periodic. We will see if it is really periodic or not in the mathematical measurement.
Mathematically,
x4=x1+x2
=sin (2πt/5)+sin(2πt/π)
So,
T1=2π/(2π/5)=5
T2=2π/(2π/π)=π
Now,
T1/T2=5/π
This is not a rational number.
So, x4 is not periodic.
And,
x5=x1+x3
= sin (2πt/5)+sin(6πt/5)cos (4πt/5)
= sin (2πt/5)+0.5*2sin(6πt/5)cos (4πt/5)
= sin (2πt/5)+0.5sin(2πt)+0.5sin (2πt/5)
= (3/2)*sin (2πt/5)+0.5sin(2πt)
So,
T1=2π/(2π/5)=5
T2=2π/(2π)=1
Now,
T1/T2=5
This is a rational number.
So, x5 is periodic
Further analysis of wave shapes:
Now if we again look at the waves shapes of x4, we can see that , though wave shapes seems to be repeated bur their magnitude is not same. So they are not actually repeated.
But the wave shapes of x5 are really repeated with perfect magnitudes.
Assignment-02
EEE 303
Signals and Systems
Code: x(t)
clear all
t=linspace (-1,3, 1000);
x= (-t+1).*(-1<t & t<0) + t.*(0<t & t<2) +2.*(2<t & t<3);
plot (t,x)
Plot 01: Showing the plot of function x(t) in the time axis t
Code: x(-t)
clear all
t=linspace (-1,3,100000);
x=(-t+1).*(-1<t & t<0)+t.*(0<t & t<2)+2.*(2<t & t<3);
x1=fliplr (x);
t1=-fliplr (t);
scale =2
t2=t1/scale
plot(t2,x1)
Plot 02: Showing the plot of function x(-t) in the time axis t
Code: x (3-t)
clear all
t=linspace(-1,3,1000);
x=(-t+1).*(-1<=t & t<=0)+t.*(0<=t & t<2)+2.*(2<=t & t<3);
x1=fliplr(x);
t1=-fliplr(t);
shift=(3);
t2=t1+shift;
plot(t2,x1)
Plot 03: Showing the plot of function x(3-t) in the time axis t
Code: x(-2t)
clear all
t=linspace (-1,3,1000);
x=(-t+1).*(-1<t & t<0)+t.*(0<t & t<2)+2.*(2<t & t<3);
x1=fliplr(x);
t1=-fliplr(t);
scale=2
t2=t1/scale
plot(t2,x1)
Plot 03: Showing the plot of function x(-2t) in the time axis t
Code: x (3-2t)
clear all
t=linspace(-1,3,1000);
x=(-t+1).*(-1<t & t<0)+t.*(0<t & t<2)+2.*(2<t & t<3);
x1=fliplr(x);
t1=-fliplr(t);
scale=2;
t2=t1/scale;
shift=(3/2);
t3=t2+shift;
plot(t3,x1)
Plot 04: Showing the plot of function x(3-2t) in the time axis t
EEE 303
Signals and Systems
Code: x(t)
clear all
t=linspace (-1,3, 1000);
x= (-t+1).*(-1<t & t<0) + t.*(0<t & t<2) +2.*(2<t & t<3);
plot (t,x)
Plot 01: Showing the plot of function x(t) in the time axis t
Code: x(-t)
clear all
t=linspace (-1,3,100000);
x=(-t+1).*(-1<t & t<0)+t.*(0<t & t<2)+2.*(2<t & t<3);
x1=fliplr (x);
t1=-fliplr (t);
scale =2
t2=t1/scale
plot(t2,x1)
Plot 02: Showing the plot of function x(-t) in the time axis t
Code: x (3-t)
clear all
t=linspace(-1,3,1000);
x=(-t+1).*(-1<=t & t<=0)+t.*(0<=t & t<2)+2.*(2<=t & t<3);
x1=fliplr(x);
t1=-fliplr(t);
shift=(3);
t2=t1+shift;
plot(t2,x1)
Plot 03: Showing the plot of function x(3-t) in the time axis t
Code: x(-2t)
clear all
t=linspace (-1,3,1000);
x=(-t+1).*(-1<t & t<0)+t.*(0<t & t<2)+2.*(2<t & t<3);
x1=fliplr(x);
t1=-fliplr(t);
scale=2
t2=t1/scale
plot(t2,x1)
Plot 03: Showing the plot of function x(-2t) in the time axis t
Code: x (3-2t)
clear all
t=linspace(-1,3,1000);
x=(-t+1).*(-1<t & t<0)+t.*(0<t & t<2)+2.*(2<t & t<3);
x1=fliplr(x);
t1=-fliplr(t);
scale=2;
t2=t1/scale;
shift=(3/2);
t3=t2+shift;
plot(t3,x1)
Plot 04: Showing the plot of function x(3-2t) in the time axis t