Analog electronics

:   

:    Any function f(t) can be expanded  using Fourier series, which is expressed as the summation of sinusoidal (sine, cosine or both) terms  as given by:

                                                   ∞                                           ∞

f t =	A0 + ∑ A n Sin (nωt ) + ∑ B n cos(               n=1                                                             n=1	nωt )	------- (3.1)
where	1 T A0 = ∫ f t.dt T 0 T		------ (3.2)

2

                                    A_n                                  dt                                       ------ (3.3)

2 T

                                   B_n dt                                     ------ (3.4)

 

Example 3.1 Find the first four coefficients of the half wave rectified output, using Fourier series expansion. Characteristics of the wave are:

 f t = E_mSinωt  when 0 ≤ t ≤ T / 2  f t = 0 T / 2 ≤ t ≤ T

 

Solution:     The coefficients of Fourier series of the half wave rectified output wave are given as:

                                                             ₁ T                                     ₁ T / 2                                  T              

                          A₀ = ∫E_mSin(ωt).dt = ^ ∫E_mSin(ωt dt) + ∫(0).dt

                                                            T 0                                     T  0                                   T / 2           

E

= m  π		------ (3.5)
2 T / 2                                                                                   An =  ∫EmSin(ωt).Sin(nωt).dt		------ (3.6)

                                                          ^T  0                                                  

2^E_m T / 2

                                          ⁼  ∫[Cos(n −1)ωt −Cos(n +1)ωt dt]

                                                               T     0

                         = 2E2π_mωSinn(n−−1)1)ωωt ₋ Sin(n(n++1)1)ωωt ^ ^T₀ ^{/ 2}

 (

E Sin (n −1)π Sin (n + 1)π

                                                 (n −1) − (n + 1)                    

                                          = 0            (for all values of n, except for n = 1)

For n = 1 denominator of the above equation will be inderminent. So we calculate A₁ by putting n = 1 directly in equation (3.6) as:

2 T / 2

                                                A₁ ⁼  ∫ E_m Sinωt.Sinωt dt.           T ₀

                                                               2Em T / 2           2                          2Em T /2

                                           =  ∫Sin ωt dt. =  ∫[1−Cos2ωt dt]       

                                                                 T     0                                         2T    0

                                                              E^m T    Sin2ωt^{T / 2}     E^m T      

                                   = T 2 − _2ω  0  = _T ₂ − 0

E^m

                                           =                                                                       ------ (3.7)

2

B_n may be calculated as:

2 T / 2

 B_n ⁼  ∫E_mSin( . ).ωt Cos(nω. ).t dt T ₀

2Em ^{T / 2} −Sin(n−1) .ωt+Sin(n+1) .ωt

                                          =      ∫0                 2                     dt

T

                                                           EmωCos(n−1)ωt    Cos(n +1)ωt^{T / 2}

                                         = 2              −1)ω − (n +1)ω  0       

π  (n

                                                              Em Cos(n −1)π Cos(n +1)π       1             1 

                                         =                   −1) − (n +1) − (n −1) + (n +1)

                                                            2π     (n

                                                                      Em  1              1            1             1 

If n is odd, then                   Bn = π −1) − (n +1) − (n −1) + (n +1) = 0  ------ (3.8)

2 (n

                                                                    Em        1            1            1             1 

If n is even, then                           Bn = 2 − −1) + (n +1) − (n −1) + (n +1) π (n

                                                                      2Em        1             1 

                                              = 2π − (n −1) + (n +1)

                                                                    Em − n −1+ n −1          2Em

                                                 =              2                           =− 2                                     ------- (3.9)

                                                                       π       (n −1)             (n −1)π

                                                (where n is even i.e. n = 2,4,6,8…..)

So by putting the values of A’s and B’s in equation 3.1 we get the required function for half wave rectified output as:   

           E^m           E^m                              2E^m                                    2E^m                                   2E^m

E = + Sinωt − Cos2ωt − Cos4ωt − Cos6ωt........      ------- (3.10) π 2 3π 15π 35π

 

Example 3.2 Consider a full wave rectified signal of peak value E_m and period T. Find its coefficients in the Fourier series expansion.

Solution:   The coefficients of Fourier series of the full wave rectified output wave are given as:

 

                             ₁ T                                        ₁ T / 2                                  T                                      

A₀ ⁼ ∫E_mSin(nωt).dt = ^ ∫E_mSin(ωt dt) ⁺ ∫− E_mSin( . )ωt dt T 0 T  0 T / 2 

                                                  ^m                                                ------ (3.11)

                            ₂ T / 2                                                       T                                                   

        A_n = ^ ∫E_mSin(ωt).Sin(nωt).dt ⁺ ∫E_mSin(ωt).Sin(nωt).dt      

                            T  0                                                        T / 2                                                

                             4Em T / 2                                          

     =   ^ ∫Sin(ωt).Sin(nωt).dt        ^T  0   

4E^m T / 2

              =  ∫[Cos(n −1)ωt −Cos(n +1)ωt dt]

                               2T    0

                               2EmωCos (n −1)ωt     Cos (n + 1)ωt ^{T / 2}

                = 2π         (n −1)ω − (n + 1)ω  0         

                              Em Sin(n −1)π− Sin0    Sin(n +1)π+ Sin0

              = π          (n −1)          −          (n +1)        

E^m Sin(n −1)π Sin(n +1)π

                = π ^   ₋₁₎ − (n ₊₁₎ ^ = 0                    ------ (3.12)       _(n

           

2x2 T / 2

           B_n ⁼  ∫E_mSin( . ).ωt Cos(nω. ).t dt

                                T     0

4E_m ^{T / 2} − Sin(n−1) .ωt + Sin(n+1) .ωt

               =      ∫0                 2                     dt

T

                             2E_mωCos(n −1)ωt    Cos(n +1)ωt^{T / 2}

                   = 2π  (n −1)_ω − ₍n +1)_ω ₀          

                                Em Cos(n −1)π Cos(n +1)π       1             1 

                  =                     −1) − (n +1) − (n −1) + (n +1) π  (n

 

                                                   E^m  1              1            1             1 

If n is odd, then   B_n = π (n −1) − (n +1) − (n −1) + (n +1) = 0      ----- (3.13)

                                                                           Em        1            1            1             1 

If n is even, then                                  Bn =      − −1) + (n +1) − (n −1) + (n +1) π  (n

                                                                      2Em        1             1 

                                                       = 2            − −1) + (n +1) π (n

                                                                    2Em − n −1+ n −1          4Em

 =  2  =− 2      ------- (3.14) π  (n −1)  (n −1)π

                                                                       (where n is even i.e. n = 2, 4, 6, 8…..)

So by putting the values of A’s and B’s in equation 3.1, the required function for full wave rectified output is obtained as:   

           2E^m           4E^m                                   4E^m                                    4E^m

E = − Cos2ωt − Cos4ωt − Cos6ωt........                ------- (3.15) π 3π 15π 35π

 

Example 3.3 Consider a symmetrical triangular wave of peak value E_m and period T as shown in figure (3.1). Find its coefficients in the Fourier series.

                               

                                                     Fig. 3.1  

 

 

Solution:   let  f(t) = X + Y.t    at t = 0           f(t) = 0  So X = 0               at t = T         f(t) = E_m

                      E_m = Y T            or    Y = (E_m)/T

E^m                         f t = (       ).t is the equation of the triangular wave. So the coefficients of

T

Fourier series of this wave are given as:

                                                   1 T Em                        Em T 2      Em

                              A₀ ⁼ ∫     . .t dt ⁼ ₂       =                           ------ (3.16)

                                                   T ₀ T               T  2        2

   A_n = 2 T∫ Em . .t Sin(nωt).dt

                                                 T  0 T                        

                                               2E    ^T

                         =                  t dt

                                                T     0

                                              2Em −tCos(nωt)     ^T 1.Cos(nωt) 

                             = ₂     +∫    .dt T             nω     ₀ nω     

                                               2E_m −tCos(nωt)     Sin(nωt)^T

                          = 2          nω      + n 2ω2  0

T

 

                                            2Em −TCos(2nπ)     Sin(2nπ)

                       = T 2         nω       + n 2ω2 

 

                                                   2Em TCos(2nπ)        2Em

                         =− T 2         nω  =− nω Cos2nπ 

E^m

                              =−       (since Cos 2nπ=1 for all values of n)    nπ

             

                   B_n ⁼ 2 T∫ Em . .t Cos(nωt).dt

                                                 T  0 T                         

                                               2E^m tSin(nωt)      ^T 1.Sin(nωt) 

                                   = ₂                   −∫           .dt

                                                T         nω       ₀              nω       

                                               2Em tSin(nωt)    Cos(nωt)^T

                                 = 2          nω + n 2ω2  0

T

                                               2Em TSin(2nπ)     Cos(2nπ)         1 

                                 = 2             nω + n 2ω2 − n 2ω2 

T

                                    = ₂2E₂^m ₂ [Cos(2nπ) −1]= ^2E₂ ^m ₂ [1−1]= 0         ------ (3.17) n ω T            2n π

So by putting the values of A’s and B’s in equation 3.1, the required Fourier series of the

E^m                      E^m                E^m                E^m triangular wave is as:    E =         −          Sinωt −           Sin2ωt −         Sin3ωt........                  

                                                                  2 π               2π             3π

 

Example 3.4 Find the Fourier series expansion of the square wave shown in figure (3.2).

                              

                                                 Fig. 3.2 

 

Solution:   The coefficients of Fourier series of the square wave are defined as:

                                          1 T                       1 T / 2                T             

                   A₀ = ∫ f t.dt = ^ ∫E_mdt + ∫(0)dt

                                          T 0                       T  0                 T / 2           

                                           E m T      E m

                           =       2  = 2

T

           2 T∫/ 2                                  T∫                2E^m [−Cos(nωt)]^T₀ ^{/ 2}

A_n = ^ E_m.Sin(nωt).dt ⁺ (0).Sin(nωt).dt=

             T  0                                       T / 2                                  nωT

2E^m

   = ( )1[ −Cos(nωT / 2)] nωT

2E^m

   = (  )[1−Cos(nπ)] = 0 for even values of n nπ

2E^m

 = (    )    for odd values of n nπ

 

             ₂ T / 2                                        T                                  

B_n = ^ ∫E_m.Cos(nω. ).t dt ⁺ ∫(0)Cos(nωt).dt T  0              T / 2          

              2E^{m                                      T / 2}             2E^m                   nωT       E^m

     = ( )[Sin(nωt)]_o = (     )Sin(     ) =       Sin(nπ) = 0  for all values of n. nωT   nωT     2  nπ

 

   So by putting the values of A’s and B’s in this equation, the required Fourier series of the Square wave is given as:     

E^m           2E^m            2E^m            2E^m            2E^m  f t =      +           Sin(ωt) +  Sin(3ωt) +           Sin(5ωt) +           Sin(7ωt) +....

                      2        π                3π                5π                 7π