Unit 4

Analysis of Fourier Methods

Q1) Explain Fourier Series

A1)

To represent any periodic signal x(t) Fourier developed an expression called Fourier series. This is in terms of an infinite sum of sines and cosines or exponentials.

A periodic signal is one which repeats itself periodically over -∞ < t < ∞

A sinusoidal signal x(t) = A sin Ω0t is periodic with period T = 2π/ Ω0.

Let us consider a signal x(t) is a sum of sine and cosine functions whose frequencies are integer multiple of Ω0.

x(t) = a0 + a1 cos (Ω0t) + a2 cos (2 Ω0t) + a3 cos ( 3 Ω0t) ……. + ak cos(k Ω0t) + b1 sin(Ω0t) + b2 sin (2Ω0t) + b3 sin(3Ω0t) + ……….. + bk sin( k Ω0t) ------------- (1)

= a0 +

Where a0,a1……. Ak and b1,b2………bk are constants and is the fundamental frequency.

If the signal x(t) has to be periodic then it should satisfy the condition

x(t + T) = x(t)

x(t + T) = a0 + ---------------- (2)

= a0 +   Here

              = a0 +     -------------- (3)

            = x(t)

Evaluation of Fourier Coefficients

The constants a0,a1,a2 ……… an , b1,b2…….bn are called Fourier coefficients. To evaluate ao we have to integrate both sides of eq(3) over one period (t0,t0+T)n of x(t) from an arbitrary time t0.

Thus    +  -------(4)

+ ------(5)

= a T + ------(6)

Each of the integrals in the summation in the above equation(6) is zero since the net areas of sinusoids over the complete periods are zero for any nonzero integer n .

Therefore

= aoT----------(7)

a0 = 1/T ------(8)

To evaluate an and bn 

= 0 for m≠n when m=n ≠0 = T/2.-------(9)

= 0 for values of m and n-------(10)

= 0 when m≠n when m=n ≠0 = T/2------(11)

To find an multiply eq(3) by cos(and integrate over one period. That is

+t  -----------(12)

The first integral on the RHS of the above equation yields to zero because we are integrating over one period.

Therefore

= am T/2-----(13)

Am = 2/T  or

An = 2/T

Similarly

Bn=2/T

The Fourier series of functions is used to find the steady-state response of a circuit.

Q2) Explain the functional symmetry condition

A2)

There are different types of symmetry that can be used to simplify the process of evaluating the Fourier coefficients.

• Even function symmetry

A function is  defined to be even if and only if

f(t) = f(-t)    

If a function satisfies the equation then it is said to be even because polynomial functions with only even exponents have this type of behavior.

Av = 2/T

Ak = 4/T          bk=0  for all values of k.

The function is even if the coefficients of b are zero.

When t0 = -T/2 then

Av = 1/T

Av     =  1/T   + 

When t= -x we observe that f(-x) =f(x) since it is even function.

   =   =   

Which does show that integrating from -T/2 to 0 is the same as integrating from 0 to T/2

Therefore

Ak = 2/T    + 2/T

= =-

Similarly

= =

To find the Fourier coefficients the integration lies between 0 and T/2.

• Odd function symmetry

A periodic function is defined to be odd if f( t) = -f(t) due to the fact that polynomial functions with only odd exponents behave this way.

The expression is

Av= 0 ; ak = 0 for all k;

Bk = 4/T

The evenness (oddness) of a function can be dismantled by shifting the periodic function along the time axis.             

• Half-wave symmetry:

A periodic signal satisfying the condition

x(t) = -x(t ± T/2)

Is said to half symmetry . The Fourier series expansions of such type of periodic signals contains odd harmonics only.

• Quarter-wave symmetry

If a function has half-wave symmetry and symmetry about the midpoint of the positive and negative half-cycles, the periodic function is said to have quarter--wave symmetry. 

Q3) Explain the Exponential Fourier Series

A3)

The exponential Fourier series is another form of Fourier series. Using Euler’s identity we can write

x(t) = A0 + ej(Ω0nt + n) –e- j(Ω0nt + n)]

       =  A0 +  ej(Ω0nt ) ej n) –e- j(Ω0nt ) . ej n)]

= A0  + ejn ) ej(Ω0nt )  + (An/2  e-j n )   e- j(Ω0t )   ]   -------- (1)

Let n=-k

x(t) = A0  + ejn ) ej(Ω0nt ] + ej k )   e j(Ω0kt ) ------- (2)

Comparing (1) and (2) we get

An = Ak (-n ) = k     n>0  k<0-------------------(3)

Let us define  c0 = A0  ; cn =An/2 ejn   for n>0

By changing the index from k to n and combining into one equation we get

x(t) = A0 +  ejn ) ej(Ω0nt)   + ejn) ej(Ω0nt)]

x(t) = n ej(Ω0nt)  ]

The above series is known as Exponential Fourier Series.

To develop the coefficients of the the exponential Fourier series

We know that

x(t) = n ej(Ω0nt)  ]  where  Ω0 = 2π/T

Multiply  e-jk Ω0 t and integrate over one period .Then

   e-jk Ω0 t   dt = cn ej(Ω0nt) ] e-jk Ω0 t   dt

= cn  ej(Ω0nt) e-jk Ω0 t   dt

Substituting the relation   e-jk Ω0 t   dt       = 0 for k ≠n  and T when k=n

e-jk Ω0 t   dt  = T ck

Therefore

Ck = 1/T   e-jk Ω0 t   dt

Or

Cn = 1/T   e-jk Ω0 t   dt

Where cn are the Fourier series coefficients of exponential Fourier series.

Q4) Explain Fourier Integral

A4)

Fourier Integral

The formula for the decomposition of a non- periodic function into harmonic components whose frequencies range over a continuous set of values.

If a function f(x) satisfies the Dirichlet condition on every finite interval and if the integral

converges then

F(x) = 1/π           --------- (1)

The formula was first introduced by J. Fourier in connection with the solution of certain heat conduction problems but was proved later by other mathematicians.

Formula (1) can also be given in the form

f(x) =   ------- (2)

Where

a(u) = 1/π  

b(u) = 1/π  

In particular for even functions

f(x) =   where

a(u) = 2/π

By taking the limits of Fourier series for functions with period 2T as T -> ∞

Then a(u) and b(u) are analogues of the Fourier coefficients of f(x)

Using complex numbers, we can replace formula (1) with

f(x) = 1/2π    e ju(x-t) f(t)

This is called Fourier Integral.

Q5) Explain Fourier transform

A5)

Fourier Transform

Consider a periodic signal f(t) with period T. The complex Fourier series representation of f(t) is given as

f(t) = k ejkw0t

     = k ej2π/T0kt     -------- (1)

Let 1/T0 = f   then equation (1) becomes

f(t) = k ej2πkft    --------------- (2)

But you know that

Ak = 1/ T0  e-jkw0t  dt

Substitute in eq(2)

f(t) =     e-jkw0t  dt ej2πk ft

Let to = T/2 then

e-j2πkt     dt  ej2πkt   ft  f

As lim T-> ∞ f  approaches differential df, k f becomes continuous variable hence summation becomes integration 

f(t) = lim T-> ∞  {    e-j2πk ft dt] ej2πk ft  f

= e-j2π ft  dt] ej2π ft df

f(t)=   ejwt dw

Where F(w) = e-j2π ft  dt

Fourier transform of a signal is given by

f(t) = F(w)  =    e-jwt dt

And Inverse Fourier transform is given by

f(t) =   ejwt dw

Find the Fourier integral of

f(x) = |sin x|               |x| ≤ π

       = 0                          |x| ≥ π

Q6) Deduce that π +1/ 1 - 2  cos (π/2) d = π/2

A6)

f(x) = 2/π

=t   =

=

= 1/ 1- 2 [cos π + 1]

=π +1/ 1 - 2  cos (π/2) d = π/2

Find the Fourier Integral of

f(x) = 1        |x| ≤ 1

f(x) = 1/π   (t-x) dt d

      = 1/π  dt d

   =  1/π /   ] -1 1     d

=  1/π  - / d

= 1/π – sin ]/ d

= 2/π   / d       =  π/2  when |x| < 1 and 0 when |x| >1

By setting x=0

= /    d  = π/2.

Q7) Explain the multiplication effect

A7)

If x1(t) and x2(t) are both periodic with period T and the Fourier series co-efficient of x1(t) and x2(t) are cn and dn respectively then

FS[x1(t) x2(t) ] = dn-l   

Proof:

FS[x1(t) x2(t) ] = 1/T x2(t) e – jnΩot dt ------------------------(1)

Using synthesis equation, we write

x 1 (t) = l  e jlΩot   --------------------------------------(2)

x 2(t) = k   e jkΩot   ---------------------------------------(3)

Substituting (2) and (3) in (1) we get

FS[x1(t) x2(t) ] = 1/T l  e jlΩot   k   e jkΩot   e – jnΩot dt-----------------(4)

= 1/T = 1/T l  e jlΩot  d k   e jkΩot   e – jnΩot dt --------------------(5)

= l    d k    ∂ (n –(k+l)) -------------------------------(6)

= l  d n-l   ------------------------------------------(7)

Q8) Explain DTFT

A8)

Discrete Fourier Transform

The discrete-time Fourier transform (DTFT) or the Fourier transform of a discrete–time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence ejωn.

The DTFT sequence x[n] is given by

X(w) =   e-jwn ---------------(1)

Here X(w) is a complex function of real frequency variable w and can be written as

X(w) = Xre (w) + j X img(w)

Where  Xre (w) , j X img(w) are real and Imaginary parts of X(w)

And | X(w)|   can be represented as

.

Inverse Discrete Fourier Transform is given by

Q9) Find the four point DFT of the sequence

x(n) = {0,1,2,3}

A9)

Here N=4. W40 = e-j2πn/4 = e-j π/ 2 = cos 0 – j sin  = 1   for n=0

W41 = e-j2 π/4 = cos π/2 – j sin π /2 = -j

W42  = e-j π = cos π – j sin π = -1

W43    = e-j2.3 π/4 = cos 3 π/2 – j sin 3 π/2 = j

For k=0

X(k) = e-j2 π nk/N

X(0) = 

X(0) = x(0)+ x(1)+x(2) + x(3) = 0 +1+2+3 = 6

X(1) = e-j2 π nk/N

X(1) = e-j2 π n/4

= x(0) e0 + x(1) e –j2 π /4+ + x(2) e-j4 π/4+ x(3) e- j 6 π/4

= 0 + 1 –j + 2( -1) + 3(j)

= -2+ 2j

X(2) = e-j2 π n2/4

X(2) =   e-j π n

X(2) = x(0) 1+ x(1) e-j π + x(2) e-j2 π  +  x(3) e-j3 π 

X(2) = -2

X(3) = e-j2 π n3/4

X(3) = x(0) e0 + x(1) e-j3 π/2   + x(2) e-j3 π  + x(3) e-j9 π/2

X(3) = -2-2j.

DFT = { 6, -2+2j,-2,_2-2j}

Q10) Explain Parseval’s theorem

A10)

Consider two periodic signals x1(t) and x2(t) with equal period T. If the Fourier series co-efficient of these two signals are cn and dn then

1/T    x2(t)   =  1/T n  e j n Ωot [ m e jmΩot  ]    dt---------------------(1)

= 1/T     n d *m           e j(n-m)Ωot  dt ------------------------------(2)

= 0     n≠ m

= n d *n   n=m --------------------------------(3)

If x1(t) = x2(t) = x(t) then eq(3) becomes

1/T     2 =   2     --------------------------------(4)

The above equation can be written as

  2     = c0 2 +   2    

                                          n≠0      

                             = c0 2 + n c *n     

                                          n≠0 

a0 2 + [Re(c 2 n ) + Im (cn ) 2]

= a0 2 + 2 n /2 + b 2 n /2 ---------------------------(5)