signal Analysis Important Questions

1.1)  Consider the following signals having the same frequency but different amplitudes    and phase components.  h(t)= 1.2cos(500 πt-π/5)  and g(t)= cos( 500 πt+π/3).

(a) Test whether the signal s(t)= 1.2+ 0.5 h(t) is periodic signal or not. If it is periodic find the  fundamental period and its power / energy.

  Ans:

Given h(t)=1.2cos(500πt-π/5) and g(t)=cos(500πt+π/3)

a)     s(t)=1.2+0.5h(t)

      =1.2+0.5(1.2cos(500πt-π/5))

      =1.2+0.6(cos(500πt-π/5))

 Since s(t)=Acos(Ωₒt+Ø)

    A=0.6

    Ωₒ=500π then Fₒ=250Hz

    Ø=-π/5

           Then s(t+Tₒ)=1.2+0.6(cos(500π(t+Tₒ)-π/5))

 Tₒ=1/ Fₒ=1/250

     = 1.2+0.6cos(500 πt+2π- π/5)

     =1.2+0.6cos(500 πt- π/5)

     =s(t)

Therefore s(t) is a periodic signal

Fundamental period=1/250

Power=

           =

            =1.62

   (b) Let y(t)=h (t )+g (t ). Express this signal in terms of y(t)=Acos(2πFοt+ϕ) . Find the  values of ‘ A’, ‘ Fο ’ and ‘ϕ ’. Determine the power of  y(t) . Is this power is equal to the sum of powers of h(t) and g(t)? Comment.

   Ans:

b)y(t)=h(t)+g(t)

  = 1.2cos(500πt-π/5)+cos(500πt+π/3)

   =>f1=250                  ;f2=250   =>T1=1/250           ;T2=1/250

  T1/T2=1

 no periodic function

  (c) Let z(t)= h(t).g(t) . Plot the signal  z(t). Is it a periodic signal? If it is periodic, find the  fundamental period. Also find the its energy / power of this signal z(t). 

Ans

:z(t)=h(t)g(t)

                  =1.2cos(500πt-π/5)* cos(500πt+π/3)

      =0.6(cos(500πt-π/5+500πt+π/3)+cos(500πt-π/5-500πt+π/3))

      =0.6(cos(1000πt-2π/15)+cos(8π/15))

      =0.6(cos(1000πt-2π/15)-0.6

      Here  Ωₒ=1000π

        Fₒ= 500

        Tₒ =1/500

   Z(t+Tₒ)=0.6(cos(1000π(t+ Tₒ)-2π/15)-0.6

               =0.6(cos(1000πt+1000π/500) -2π/15)-0.6

               =0.6(cos(1000πt+2π-2π/15)-0.6

               =0.6(cos(1000πt-2π/15)-0.6

               =z(t)

Therefore  z(t) is periodic signal

               Power(P)=500

 (d)  Identify the amplitude, frequency and phase components of the following   signals(1)p(t)=g(2t),and(2)q(t)=g(t/2)and(3)r(t)=g^2(t).

  Ans:  p(t)=g(2t)

    1)

              g(t)=cos(500πt+π/3);

             g(2)=cos(500π(2t)+π/3)

        =cos(1000πt+ π/3)

Here  Amplitude=1

Frequency=500Hz

Phase angle= π/3

       2)

           q(t)=g(t/2)

           g(t/2)= cos(500π(t/2)+π/3)

           g(t/2)=cos(250πt+π/3)

             Here  amplitude=1

                        Frequency=125

                        Phase angle= π/3

     3) r(t)= (t)

               =cos(500πt+π/3)* cos(500πt+π/3)

                =2/2( cos(500πt+π/3)* cos(500πt+π/3))

               =1/2[cos((500πt+π/3+500πt+π/3) +cos(500πt+π/3-500πt-π/3)]

              =1/2[cos(1000πt+2π/3)]

       Here Amplitude=1/2

                 Frequency=500

                 Phase angle=2π/3

 (e) Find the power / energy of the above signals p( t) ,q( t )and r(t).  

 Ans:

  1)  Since p(t+Tₒ)=p(t)  and it is periodic signal

        Power=1/T  dt

                     =500 dt                

                     = 500  dt

                      =500[1/2(1/500)+sin(2000π(1/500)+2π/3)/2*2000π]

                      =500[1/1000+sin(2π/3)/2*2000π]

                      =1/500[0.001+0.000068]

                      =500[0.0016]

                      =0.8

     2) Since q(t+Tₒ)=q(t)  and therefore it is periodic signal

             Power of q(t)=1/T  dt

                                     =125 )/2dt

                                   =125(1/2(1/125)+sin(500sinπ*1/125+2π/3)/500*π*2)

                                   =125[0.002+0.00027]

                                   =0.28

     3)  r(t)=1/2[cos(1000πt+2π/3)]

           Since r(t+Tₒ)=r(t) and therefore it is periodic signal

              Power=1/t  dt

                          =125  dt

                          =125[1/2*1/500+sin(2000π*1/500+4π/3)/2*2000*π)]

                          =125[0.001-0.000068]

                          =125[0.00093]

                           =0.11

 (f) If the signal g(t )is shifted by 0.6 msec in time, find the corresponding phase    shift in the frequency domain. Plot the shifted signal overlapping on g(t)  and find the phase shift from  this plot. Comment.    

Ans: Given g(t)=cos(500πt+π/3)

       here  Amplitude=A=1

                  Ω=500π

                  F=250

                 T=1/250

From the question we know that g(t-0.6m)

   Therfore g(t-0.6m)=cos(500π(t-0.6)+π/3)

                                    =cos(500πt-300π+ π/3)

                                    =cos(500πt-899π/3