3.1 Trigonometric and exponential Fourier series: Discrete spectra and symmetry of waveform

Any arbitrary periodic function can be represented by an infinite frequencies. This representation is known as Fourier series representation. The fourier representation of periodic function is extended to non periodic function as a function of frequency is called Fourier transform. The Fourier series representation can be obtained only for periodic function, but the Fourier transform technique can be applied to both periodic and non-periodic function to obtain frequency domain representation of functions. The information about magnitude and phase of various frequency components can be obtained by Fourier analysis.

The plot of magnitude versus frequency is called magnitude spectrum. • The plot of phase versus frequency is called phase spectrum.

Trigonometric Fourier series Representation of Periodic Signals.

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.

Condition for symmetry

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.             

Example:

For the following diagram explain whether it is even or odd.

Solution:

The signal is neither even nor odd.

To make the signal even there is a slight shift to be applied as shown in the figure.

To make the signal odd

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.

Example:

Show that the signal x(t) that satisfies half wave symmetry contains Fourier coefficients  with odd harmonics only.

A signal with half-wave symmetry that satisfies the condition x(t+T/2) = - x(t)

The Fourier coefficients of the signal are

Cn = 1/T Ω0     dt

= 1/T [ Ω0     dt      +   Ω0     dt

= 1/T Ω0     dt      +   Ω0     dt

= Ω0     dt      -     Ω0     dt                

Here e-jnπ = 1 for n Even.

Hence Cn=0

Ω0 = 2π/T

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. 

Problems:

Find the trigonometric Fourier series for the periodic signal x(t) as shown in the figure.

For the given signal the period T =4.

For our convenience let us choose one period signal from t= -1 to t=3

That is t0=-1 and t0+ T =3.

The fundamental frequency

= 2 π/T =2 π/4=  π/2.

x(t) = { 1 for -1 ≤t ≤1

-1 for 1≤t≤3}

a0= 1/T 

= ¼ = ¼[ + ]

= ¼ [ t |-1 1 + (-t)| 1 3 ]

= ¼ [ 1 – (-1) – (3-1)] = 0

a0 = 0.

An = 2/T

½  

= ½  +

= ½ [ 2/ nπ sin(nπ/2t) | -1 1  - 2/nπ sin(nπ/2t)|1 3  

= ½ [ 8/nπ sin(nπ/2)] = 4/nπ sin(nπ/2)

An=4/nπ sin(nπ/2)

Bn =  1/2

= ½ +

= ½ [ -2 /nπ cos(nπ/2t) | -1 1 + 2/nπ cos(nπ/2t) | 1 3

= -2/nπ(cosnπ/2 – cos nπ/2) + 2/nπ (cos 3nπ/2 – cosnπ/2)

= 0

Therefore 4/nπ sinnπ/2 = {  0        n even

4/nπ  = 1,,9,13…..

-         4/nπ    3,7,11,15…….

x(t) =

= 4/π cos (π/2 t) – 4/3π cos (3π/2  t)+ 4/5π  cos(5π/2 t) - …………..

= 4/π [ cos(π/2  t) -1/ 3 cos( 3π/2 t ) + 1/5 cos (5 π/2 t) – 1/7 cos(7π/2 t) + ………..

Exponential Fourier Series

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

An cos(Ω0nt + n ) = An [ ej(Ω0nt + n) –e -j(Ω0nt + n)

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.

Problems:

The time period of the signal x(t) isT=4.

Ω0 = 2π/T = 2 π/4= π/2

C0 = 1/T = ¼ [ +

C0 = ¼ [ 2+1 ] = ¾

Cn = 1/T [    e-jnΩot dt = ¼[  e-jnπ/2t dt + e-jn π/2t dt

= ½. 1/ -jn π/2 [e-jn π/2t ] 0 1 + ¼ 1/-jn π/2[ e-jn π/2t ]  1 2

= - 1/jn π[e-jn π/2  – 1] – 1/ 2jn π (e-jn π – e-jn π/2 )

= 1/jn π[ 1 - e-jn π/2 ] – 1/ 2 e-jn π  + ½ e-jn π/2 )

= 1/jn π [ 1- ½ (-1) n -1/2 e-j n π/2 ]

Key takeaway

In Fourier series

a0 = 1/T

An = 2/T

Similarly

Bn=2/T

For exponential Fourier Series

Cn = 1/T   e-jk Ω0 t   dt

3.2 Steady state response of a network to non-sinusoidal periodic inputs, power factor, effective values

Average Value:

Fig 1 Sinusoidal waveform

The arithmetic mean of all the value over complete one cycle is called as average value

=

For the derivation we are considering only hall cycle.

Thus varies from 0 to ᴫ

i = Im   Sin

Solving

We get

Similarly, Vavg=

The average value of sinusoid ally varying alternating current is 0.636 times maximum value of alternating current.

RMS value: Root mean square value

Fig 2 Sinusoidal waveform

The RMS value of AC current is equal to the steady state DC current that required to produce the same amount of heat produced by ac current provided that resistance and time for which these currents flows are identical.

I rms =  

Direction for RMS value:

Instantaneous current equation is given by

i = Im Sin

But

I rms =  

=  

=

=

Solving

=

=

Similar we can derive

V rms= or 0.707 Vm

the RMS value of sinusoidally alternating current is 0.707 times the maximum value of alternating current.

Reactance

It is opposition to the flow of an AC current offered by inductor.

XL = ω L    But     ω = 2 ᴫ F

XL = 2 ᴫ F L

It is measured in ohm

XL∝FInductor blocks AC supply and passes dc supply zero

2.     Capacitive Reactance (Xc)

It is opposition to the flow of ac current offered by capacitor

Xc =

Measured in ohm

Capacitor offers infinite opposition to dc supply 

Impedance (Z)

The ac circuit is to always pure R pure L and pure C it well attains the combination of these elements. “The combination of R1 XL and XC is defined and called as impedance represented as 

Z = R +i X

Ø = 0

  only magnitude

R = Resistance, i = denoted complex variable, X =Reactance XL or Xc

Polar Form

Z   = L I

Where =

Measured in ohm

Power factor (P.F.)

It is the cosine of angle between voltage and current

If Ɵis –ve or lagging (I lags V) then lagging P.F.

If Ɵ is +ve or leading (I leads V) then leading P.F.

If Ɵ is 0 or in phase (I and V in phase) then unity P.F.

Steady state behavior of RLC circuit with excitation

The series RLC circuit is shown below. Let current i(t) be sinusoidal. The value of instantaneous voltage across R is in phase with current. The instantaneous voltage across L leads current by 900. The instantaneous voltage across C lags current by 90o.

Writing the loop equations, we get

Vs-VR-VL-VC=0

VS-IR-L- = 0

VS= IR + L +

The voltage triangle will be

Vs=

VR = iRsint

VL= iXLsint+90)

VC= iXcsint-90)

Z = =

Impedance Z=

Key takeaway

V rms= or 0.707 Vm

the RMS value of sinusoidally alternating current is 0.707 times the maximum value of alternating current.

Examples

Q) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate total impedance?

Sol: Impedance Z=

Z= = 18.27ohm

Q) For a series RLC circuit having R=12ohms, L= 0.2H, C=60F. They are connected across 100v 50Hz supply. Calculate circuit current?

Sol: I=

Z=

Z= = 13.89ohm

I = 100/13.89 =7.2A

Q) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate power factor?

Sol: cosφ =

Impedance Z=

Z= = 18.27ohm

Cosφ = =

φ = 56.81o lagging

Sinusoidal Steady State Analysis:

1)     The method used is phasor method.

2)     Take Laplace transform of R, L and C in the circuit.

3)     Make all initial conditions zero.

4)     Derive the expression for desired voltage and current.

5)     Replace S by jω where ω is the frequency at which the circuit is operating.

Que) If V1=6cos2t, find i2 at steady state.

Sol: Apply KVL

6-sI1(s)-2I2(s)=0

-2I2(s)-I2(s)+ I1(s)=0

I2(s)= I1(s)

I1(s)=[I2(s)

Substituting I1(s) in first equation

6-s[}I2(s)- 2I2(s)=0

Solving above equation we get

I2(s)=

Replace s=jω, s=2j

I2(jω)===

=

Que) If V1=2sin2t, Va at steady state will be?

Sol: Zeq=

After Laplace Transform the above circuit will be

By voltage divider rule Va=

Replace s by jω , s=2j

Va==

Va==

Va=

3.3 Fourier transform and continuous spectra

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)

f(x)= lim  1/π    sin    dt  x-t  

This is called Fourier Integral.

Problem:

1.Find the Fourier integral of

f(x) = |sin x|               |x| ≤ π

= 0                          |x| ≥ π 

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

Solution:

f(x) = 2/π 

=t   =

=

= - cost(1-) ]0 π  - cost(1+)   ]0 π

2(1-)                 2(1+)

= 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.

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

Fourier transform are classified into

Continuous Fourier Transform

Fourier series was defined for periodic signals. A periodic signals can be considered as a periodic signal with fundamental period

Consider a periodic square wave:

x(t) =  1 for  |t| < T1

= 0   T1< |t|   < T0/2   

The fourier series co-efficient is

Ak = 2 sin(kwoT1)

k wo T0 

To.ak  =  2 sin(wT1)     | w=kw0

W

Key takeaway

3.4 Three phase unbalanced circuit and power calculation.

Three Phase

In three phase the windings are separated by 1200 each. The voltage produced in those windings are 1200 apart from each other. Below shown is one coil RR’ and two more coils YY’ and BB’ each having phase shift of 1200.

The instantaneous value of voltages is given as

VRR’ = Vmsinωt

VYY’ = Vmsin(ωt-120)

VBB’ = Vmsin(ωt-240)

The three phase voltages are of same magnitude and frequency.

Phase sequence

The change in voltage is in order VRR’- VYY’- VBB’. So, the three-phase are changed in that order and are called as phase change.

VRR’ = Vmsinωt

VYY’ = Vmsin(ωt-120)

VBB’ = Vmsin(ωt-240)

Phasor Diagram

We are getting resultant line current IR  by subtracting  2 phase currents IRY and IBR   take phase currents at reference as shown

Cos 300  =  

=

IL =Iph

Phase current IYB lags behind VYB  which is phase voltage as the load is inductive

Star Connection:

In this type similar ends are connected to common point called as neutral and having a star shape. These connections are used in case of unbalanced current flowing in the three-phase. To avoid any kind of damage we use this connection.

Line voltage VL = Vphase

Line current IL = Iphase

Delta Connection:

There are three wires with no neutral. They are used for short distance due to unbalanced current in circuit.

Line voltage VL = Vphase

Line current IL = Iphase

Power relation for delta load star power consumed per phase

PPh  = VPh IPh  Cos Ø

For 3 Ø total power is

PT= 3 VPh  IPh  Cos Ø …….①

For star

VL and IL = IPh  (replace in ①)

PT  = 3    IL  Cos Ø

PT  = 3   VL IL  Cos Ø – watts

For delta

VL  = VPh  and IL =      (replace in ①)

PT = 3VL    Cos Ø

PT VL IL Cos Ø – watts

Total average power

P = VL IL  Cos Ø – for ʎ and load

K (watts)

Total reactive power

Q = VL IL Sin Ø – for star delta load

K (VAR)

Total Apparent power

S = VL IL   – for star delta load

K (VA)

Three Phase Unbalanced Circuits

If the load connected across the three phases are not identical to each other, which means they have different magnitudes and power factors. These kinds of loads are known as unbalanced loads. The phase currents in delta and phase or line current in star connections differ in unbalanced loading, giving rise to flow of current in neutral wire.

IN = IR+IY+IB

Unbalanced Delta-connected loads

Fig 3 Unbalance Delta connected load

For delta load

VL = Vph

Phase currents are given by

Unbalanced three wire star connected load

Fig 4 Unbalanced three wire star connected load

If neutral N of the load is not connected to the neutral point O of the supply, there exists a voltage between the supply neutral point and the load neutral point. The load phase voltage is not the same as the supply phase voltage. There are many methods to solve such networks. Most commonly used are

In star transformation, the star connected loads are replaced by equivalent delta connected loads.

The problem is then solved as an unbalanced delta connected load. The line current so calculated are equal to the line currents of the original star connected load.

An unbalanced three wire star connected load can also be solved by Millman’s Theorem. Let VRO, VYO and VBO be the phase voltages of supply, which are equal in magnitude but differ in phase from each other by 120o. Let VRN, VYN and VBN be the load phase voltages, which are unequal in magnitude as well as differ in phase by unequal angles. The voltages between the neutral points, N and O, VNO is calculated using Millman’s theorem.

Where YR, YB and YY are the admittances of the branches of the unbalanced star connected load. The voltage across phase impedance or load phase voltage can be calculated as

The phase currents in the load are given as

For star load line current is equal to phase current.

Solved Examples

Q1) Three similar resistors are connected in star across 400V 3-phase lines. Line current is 4A. Calculate the value of each resistor.

Sol: For star connection:

IL=Iph=4A

Vph=VL/ = 400/ = 231V

Rph= 231/4= 57.75ohm

For Delta Connection:

IL=4A

Iph= IL/

=4/ ==2.30A

Zph=400/2.30=173.9ohm

Rph= 173.9/3 = 57.97ohm

Q2) Three identical impedances are connected in delta 3-phase supply of 400V. The line current is 30A and total power taken from the supply is 10kW. Calculate the resistance and reactance value of each impedance?

Sol: VL=Vph=400V

IL=30A

Iph=IL/= 30/ =17.32A

Zph=Vph/Iph= 400/17.32=23.09ohm

P=VLIL Cos Ø

Cos Ø = 10000/ 400x30 = 0.48

Sin Ø =0.88

Rph=Zph Cos Ø= 23.09x0.48=11.08ohm

Xph=Zph Sin Ø = 23.09x0.88=20.32ohm

Q3) A star connected alternator supplies a delta connected load. The impedance of the load branch is 6+j5 ohm/phase. The line voltage is 230V. Determine the current in the load branch and power consumed by the load.

Sol: Zph= = 7.8ohm

VL=Vph=230V

Iph=Vph/Zph=230/7.8=29.49A

Iph=IL/

IL= Iph=x29.49=51.07A

P=VLIL Cos Ø = x 230x51.07x0.768=15.62kW

Q4) The load connected to a 3-phase supply comprise three similar coils connected in star. The line currents are 25A and the kVA and kW inputs are 18 and 10 respectively. Find the line and phase voltage, the kVAR input resistance and reactance of each coil?

Sol: IL= 25A

P= 10000W

Cos Ø = 10/18 = 0.56

P=VLIL Cos Ø

10000= x VLx25x0.56

VL =412.39V

Vph= VL/ = 412.39/=238.09V

KVAR= = 14.96

Zph=238.09/25=9.52ohm

Rph=Zph Cos Ø= 9.52x0.56=5.33ohm

Xph=Zph Sin Ø = 9.52x0.83=7.88ohm

Q5) A balanced delta connected load consisting of three coils draws 8 A at 0.5 p.f from 100V 3-phase ac supply. If the coils are reconnected in star across the same supply. Find the line current and total power consumed?

Sol: For Delta connection:

IL=8A

Iph= IL/= 8A

Vph=100V

Zph=100/8=12.5ohm

Rph=Zph Cos Ø=12.5x0.5 = 6.25ohm

Xph=Zph Sin Ø = 12.5x0.866=10.825ohm

P=VLIL Cos Ø

= x 100x 8x0.5=1200W

For Star Connection:

Vph= VL/= 100/V=57.73V

Zph=100/8=12.5ohm

Iph=57.73/12.5=4.62A

P=VLIL Cos Ø

= x 100x 4.62x0.5

P= 400W

Q6) A 3-phase supply with a line voltage of 250V has an unbalanced delta connected load as shown in figure

Find (a) Phase current (b) line current?

Sol: From above figure we can write

The phase currents are

The Line currents are

Q7) Find vc(t) for circuit shown in the figure?

Sol: Let the current in circuit be i(t). From KVL we have

Taking F.T we get

As i(0+) =0 hence I(0)=0

Takin IFT we get

Q8) Find Fourier transform of the waveform given below?

Sol:

Q9) Find the Fourier series coefficients for the given saw tooth wave(in exponential form) as shown in figure

Sol:

Q10) A continues time periodic signal x(t) has a fundamental period T=8. The nonzero Fourier series coefficients for x(t) are: 𝑎1 = 𝑎−1 = 2. 𝑎3 = 𝑎∗−3 = j4

Express x(t) in the form of-

Sol:

Here,

Therefore,

Q11) For the circuit shown in figure, Determine the voltage response 𝑣𝑜 𝑡 to a current source excitation i(t)=2𝑒−𝑡𝑢(𝑡), using Fourier transform?

Sol: We apply KCL

Taking FT of both sides we have

Therefore,

By partial fraction of above equation, we get

Taking IFT of above equation we get

Q12) In the Circuit of fig., A 400 V, 50 Hz, three phase supply of phase sequence ABC is supplied to a delta connected load consisting of a 100 Ohm resistor between lines A and B, a 378 mH inductor between lines B and C, and a 37.8 micro farad capacitor between lines C and A. Determine Phase and line currents.

Sol: VL = 400V, f=50Hz, R= 100 Ohm, L= 318mH, C=31.8 micro farad

Phase Current

Line Currents

References: