PROBLEM 1


>> load('handel.mat')

t0=cputime;
Y=fft(y);
yTime=cputime-t0

yTime =

    0.4910

>> hallelujah=y(1:65536);
t0=cputime;
Y=fft(hallelujah);
yTime=cputime-t0

yTime =

    0.0770

>> [y_size, t]=size(y);
y_size

y_size =

       73113

>> % Since the original size of y is 73113x1 and hallelujah is 65536x1
>> % hallelujah is smaller than y and hence requires less cpu time.



PROBLEM 2
>> load('handel.mat')
i=sqrt(-1);
F4=[1 1 1 1; 1 i -1 -i; 1 -1 1 -1; 1 -i -1 i]
R=[1 0 0 0; 0 0 0 1; 0 0 1 0; 0 1 0 0]

% Since F is symmetric, F* is the same as conj(F4)
R_HerF=R*(conj(F4))

F4minusRHerF=F4-R_HerF

F4 =

   1.0000             1.0000             1.0000             1.0000          
   1.0000                  0 + 1.0000i  -1.0000                  0 - 1.0000i
   1.0000            -1.0000             1.0000            -1.0000          
   1.0000                  0 - 1.0000i  -1.0000                  0 + 1.0000i


R =

     1     0     0     0
     0     0     0     1
     0     0     1     0
     0     1     0     0


R_HerF =

   1.0000             1.0000             1.0000             1.0000          
   1.0000                  0 + 1.0000i  -1.0000                  0 - 1.0000i
   1.0000            -1.0000             1.0000            -1.0000          
   1.0000                  0 - 1.0000i  -1.0000                  0 + 1.0000i


F4minusRHerF =

     0     0     0     0
     0     0     0     0
     0     0     0     0
     0     0     0     0

>> % Note that F-RF*=0, which means F=RF*, or F=R*conj(F), since F* is the same as conj(F) 
% due to the fact that F is always symmetric, so F=F' always. Since s=Fx is the Fourier
% transformation of a real sequence x and s is the complex conjugate of s,
% we have the same case as F-RF*.  Thus, we can conclude that s-R*conj(s)=0, which 
% means s=R*conj(s).

% Since s=R*conj(s) and power(s)=power(conj(s)), power(s)=Power(R*conj(s))=Power(R*s).  
% Therefore, the only way that the power(s) can equal to Power(R*s) is if the graph is
% symmetric.