fourier series important questions and solutions pdf Fourier Series Representation of signal | Trigonometrical Fourier Series and Exponential Fourier Series formula examples ?
Fourier Series
Fourier Series Fourier transform provides effective reversible link between frequency domain and time domain representation of the signal. non-periodic signals can be represented with the help of Fourier transform.
Fourier Series Representation of signal
1. Trigonometrical Fourier Series
2. Exponential Fourier Serise
M = A|H (joo)
0 = L01 + <H (joo)
CF = Complementary function (free response) transient
PI = Forced response (steady steady response)
y(t) = CF + PI
Trigonometrical Fourier Series
if x(t) is time domain continuous periodic signal.
x(t) = a0 + a1 cos oot + a2 cos 2 oot + ………+ b1 sin oot + b2 sin 2 oot + …………
x(t) = a0 + an cos not + bn sin not
ao = DC component of the signal
an = cos not = cosine component of signal
bn sin = sine component of signal
for n = 1 components are called fundamental signal.
for n > 1 components are called harmonics of the signal.
n > 1 n = even = even harmonic
n > 1 n = odd = odd harmonic
DC value, ao = 1/T x(t) dt
average of x(t) or average value
an = 2/T x(t) cos nt dt
bn = 2/T x(t) sin nt dt
= 2 {average of x(t) cos nt}
= 2 {average of x(t) sin nt}
where, T = fundamental time period
oo = 2/T
Smallest possible period is called fundamental time period (T)
Special Cases
(i) Even symmetry x(-t) = x(t)
bn = 0
Sine component must be absent odd component must be absent.
(ii) Odd symmetry
Either x(-t) = – x(t) or signal is anti-symmetry.
ao as well as an would be zero.
DC and even (cosine) component of the signal would absent.
Half-wave odd symmetry
x(t + TO/2) = – x(t)
since, these are odd means ao = 0, an = 0
bn = {4/T x(t) sin nt dt, for odd values of n
Half-wave even symmetry
x(t + to/2) = x(t)
bn = 0
an = {4/t x(t) sin nt dt, for even values of n
ao = purely real
an = Real
bn = Real
Advantage,
a0, an and bn = Real
a0 <0o, an <0o, bn <0o
a0 + a1 cos t + a2 cos 2 t +……….+ b1 sin t + b2 sin 2 t +………….
Never (00t + 0) in this because a0, an and bn are real and there is 0 = 0 but
x(t) = cn ejnot
Example 1.
Odd + half-wave odd symmetry
a0 = 0, an = 0
bn = {4/T x(t) sin nt dt
bn = 4/T x(t) sin nt dt
bn = 8A/Tnoo sin2 n/2
= 4/T A/noo [1 – (-1)n ]
sin n = 0
cos n = (-1)n for odd (-1) and for even (+1)
sin 2n = 0, cos 2n = 1
bn = [1 – (-1)n] 2A
bn =4A/n for odd values of n
x(t) = 4A/ sin + 4A/3 sin 3 +…………..
(b) Exponential Fourier Series Representation
if x(t) is continuous time periodic signal
x(t) = ao + an cos not + bn sin not
it can be represented in exponential form
x(t) = cnejnot
where, cn = an – jbn/2
cn = 1/2 [2/t x(t) cos not dt – j 2/t x(t) sin not dt]
cn = 1/t x(t) e-jnot dt
for n = 0, co = a0 (DC value)
Properties of cn
(i) if signal is even signal bn = 0, cn is pure real quantity. only magnitude representation required.
(ii) if signal is odd signal an = 0; cn = purely imaginary
Example 2. (i) what is value of cn?
(ii) What is fourier series representation of even signal?
(iii) Represent cn in spectrum form.
Sol. x(t) = n = – oo cn ejnot
cn = an – jbn/2
signal is even, bn = 0
cn = an/2 = Real quantity
for one time period (t),
x(t) = {A, – d/2 < t < d/2
cn = 1/t [t/2 x(t) e-jont dt
cn = 1/t (A) e- jont dt
cn = 2A/ont sin ond/2
cn = Ad/t sin (ond/2)/(ond/2)
ond/2
n = T/d
Example 3.
if signal is odd.
1. What is the value of cn ?
2. Represent exponential fourier series.
3. Draw line spectrum of cn magnitude and phase spectrum.
Sol. Half-wave odd symmetry
cn = – bn/2
bn = 4/t x(t) sin ont dt
bn = 4/t (A) sin ont dt
= 4/t (A) [-cos ont/on]t/2
= 4A/T [cos on t/2 + 1/on]
= 4A/onT [1 – cos on T/2]
= 4A/2 [1 – (-1)N]
= 2A /n [1 – (-1)n]
cn = j/2. 2A/n [1 – (-1)n]
cn = -jA/n [1 – (-1)n]
x(t) = cn ejnot
Phase Spectrum
x(t) = cn ejnot
x(t) = ………+ c_5 e-j5oot + c_3e-3 joot + ……c-1e-joot + c1ejoot + c3e3joot + ………
x(t) = …………- 2A/5 e-5 joot – 2A/3 e3joot + ………..+ …..2A ejoot + 2A/3 e3joot+………….
x(t) = A/sin oot + A/3 sin 3 oot + A/5 sin 5 oot + …………
Trigonometry – Exponential
Previous questio
cn = Ad/t {sin on d/2/ond/2}
00 = 2/t, d= 7/4
cn = A/4 sin n/4/n/4
n/4 = n = 4
Distance between spectrum = 2/4 – 4 = 4
d/dt {r(t)} = u(t)
d/dt {u(t)} = (t)
Dira-Delta (00) area = 1
(t) dt = u(t)
u(t) dt = r(t)
Properties of Fourier Series Coefficient
(i) if x(t) is continuous time periodic signal
fourier coefficient = cn
(ii) fourier coefficient of x(t – to)
cn = cn e-jont0
(iii) fourier coefficient of d/dt x(t)
cn = d/dt cn
Circuit Transients (RL, RG, and RLC)
{t = 0+ initial condition
{t = oo, final condition
. in time domain
. in s-domain
initially, at t = 0 the capacitor behaves as short-circuit and inductor acts as open-circuit.
As t – oo or in the steady state condition the capacitor is fully charged and therefore, acts as a open-circuit where as the inductor acts as a short-circuit.
With initial condition at t = 0, the capacitor acts as a constant voltage source and the inductor acts as a constant current source.
Representation of L, C with initial Condition in the s-domain
Capacitor
I(t) = Cdv(t)/dt
taking laplace on both sides
I(s) = sCV (s) – CVC (0+) ………….(i) fig. 14 (a)
KCL (parallel)
V (s0 = 1/SC I(S) + VC (0+)/S …………..(ii)
KVL (serise) fig. 14(b)
Inductor
V(t) = L d/dt I (t)
V (S) = sLI (s) – LIL (0+)
KVL (series) fig. 16(a)
I(S) = 1/sL (S) + 1/s IL (0+)
KCL (parallel) fig. 16(b)
We follow the following steps in solving problems of first order RL and RC circuits.
Step 1 We consider a steady state before the switch changes its position at t = 0, i.e., for t = 0, we apply the steady state conditions.
Note for Step 1
(a) DC circuits are in a steady state since the current and voltages do not change in time we also use the term steady state to describe currents and voltages that display repetitive temporal waveform. thus, systems in which the current and voltages can be descrided by constant amplitude constant frequency sinusoidal functions are also considered to be in a steady state. and the term transient refers to the behaviour of the voltage or current when it is in transition between one steady state to another.
(b) At steady state capacitor behaves as open circuit.
IC = CdVC/dt for steady state
IC – 0 hence open- circuit and inductor behaves as short-curcuit
VL = L dI/dt
for steady state – t
VL = 0, hence short-circuit
Step – 2 Then for the changed position of switch, we again consider the steady state conditions.
In step 1. We calculate the value f(0-) where f(0-) can be IL(0-) (current through inductor) or VC(0-) (voltage through capacitor)
Add in Step 2 We calculate the value f(00) where f(00) can again be described as IL and VC(00)
Then, at any time t, the response can be described as
f (t) = f (00) – {f (00) – f(0-)} e-t
where, f(t) can be current through inductor and voltage through capacitor at any time t and = time constant.
For RC circuit,
Req. C – REQ C
Where, Req = resistance seen across terminals of capacitor by replacing all sources with their internal resistances.
For RL circuit,
L/Req
Req = resistance seen across terminals of inductor by replacing all sources with their internal resistances.
Example 4. consider the following circuit:
Switch S is closed at t = 0
(i) Find VC as function of time.
(ii) Find VR as function of time
Sol. VC (t) = VC (00) – (VC (00) – VC (0-)) e-t/Req. c ……………..(i)
Step 1 to calculate VC (0-)
For t < 0, the circuit at steady state position will be
VC (0-) = V4
= 4/4 + 2 x 5 = 10/3 V
Step 2 For t < 0, again taking the steady state position VC (00) = voltage across parallel combination 4 and 4 = 2
VC (00) = (2/2 + 2) x 5 V = 2.5 V
REQ (4||4||2) in series with 1 = 2 C = 3 F
from EQ. (i)
VC (t) = 2.5 – (2.5 – 3.3) e-t/reqc
VC (t) = 2.5 + 0.8 e-t/6
IC = CdVC /dt = C {-0.8/6 e-t/6}
IC = -0.4 e-t/6
VR = ICR = – 0.4 e-t/6
t = 0+, Vr = – 0.4
but at t = 0-, VR = 0
Example 5. (RL network)
to find I (0+), di/dt (0+)
in the given network, the switch is initially open. it is closed at t = 0 find. the initial value of the current and initial rate of change of current through the circuit.
Sol. The switch is initially open. hence, at t = 0-, switch is open
so, I (0-) = 0
At t = 0+ (when switch is closed at t = 0)
L is open-circuit
I (0+) = 0
Conclusion,
IL (0+) = IL (0-)
VC (0+) = VC (0-)
General equation for t > 0
Applying KVL to the circuit of fig. (c),
V0 = RI (t) + Ldi (t)/dt
t = 0+
VO = RI (0+) + L dI (0+) / dt
dI (0+)/ dt = + V0/L A/s
Do you Know?
1. the current through the inductor and voltage across the capacitor do not change instantaneously.
2. These values remain same just before and just after the switch has been closed.
Example 6. to find I (t) for t > 0. in the given network the switch is initially at position ‘a’ for a long time till the steady state is reached. at t = 0, the switch is moved to position ‘b’. find a general expression for I (t) for t < 0.
Sol. the switch is initially at position a so at t = 0-,
the circuit is in steady state condition. inductor is short-circuited I (0-) = VO/R1
Since, I(0-) = I (0+) = VO/R1
t > 0 I (0+) = VO/R1
(When switch is at position ‘b’)
Applying KVL to the circuit,
I(s) [R1 + R2 + SL] – V0L/R1 = 0
I(S) = V0L/R1 x 1/L[s + R1 + R2/L]
a1 = R1 + R2 /L = a
I (S) = VO/R1 1/(s + a)
Taking inverse laplace transform,
I (t) = VO/R1 e-at
I (t) = VO/R1 e (L/R1 + R2)
T = 1/a = L/REQ , REQ = R1 + R2
T = time constant
At t = t
I(t) = VO/R1 e-1
I (t) = 0.377 VO/R1
I (t) = 0.377 I (0+)
T = L/REQ
Do you Know ?
1. the time constant of an RL network represents a time at which the current through the network is decreased to 37.7% it’s initial value.
2. the time constant controls the rate of decay of the current through the RL network and depends upon the equivalent values of R and L present in the network.
Summary
RL Circuit
I (t) = I (0+) e-t
Where, I(0+) ………Initial wave
L/REQ = time constant
Summary
Given, Any network
Find initial condition in L and C.
Sinusoidal Steady State Analysis
The sinusoid is a distinctive form of signal waveform. if a sinusoid source is connected to a network of linear, passive elements, then every voltage and current in that network will be sinusoid in the steady state, differing from the source waveform only in amplitude and phase angle.
It is observed that,
(i) The sinusoid may be repeatedly differentiated or integrated and still be a sinusoid of the same frequency.
(ii) The sum of a number of sinusoid of one frequency but of different ampliyude and phase in a sinusoid of same frequency.
Signal described by exponential function est has a steady state value for s = joo and s = – joo.
Sinusoid and e+joot
A sinusoidal waveform shown in figure can be expressed as
V = V0 sin (oot + 0)
VO = amplitude of sine wave
oo = frequency rad/s
0 = phase angle of V w.r.t V’ (reference waveform)
Time period of the waveform changes from oot = 0 to oot = 2
eq. (i) can be written as
V = VO cos [oot + 0 – 2]
V = VO cos [oot + B]
V = VO cos B cos (oot + B) + (-V sin B) sin
Now, the laplace transform of sine and cosine are
L [sin ] = 00/s2 + 002
L [sin] = s/s2 + oo2
It indicates sinusoidal signals having transformation with poles and zeros restructed to imaginary axis shown in figure.
From the above mentioned equation, we have
sin = ejot – e-jot
cos = ejot + e-jot/2
The term ejot can be interpreted as unit rotating phasor in the positive (or couter-clockwise) direction and e-jot is interpreted as negative (or clockwise) rotating direction.
The rotating phasors are shown in figure.
Procedure to Find Steady State Response
The steady state response of the excitation VO cos or VO sin may be found by determining the response to the excitation ejot, multiplying this response by V = VO ej0 and then determining the real part for VO cos and imaginary part for VO sin of this product.
Phasor and Phasor Diagram
Consider a network of L independent loops excited by sinusoidal sources, each of same frequency. in order to determine the steady state response in this network L- loop equations can be written in transform form which are of the form
V = Z(joo) I
Then for the given network, we have
Z11 (joo) I1 + Z12(joo) I2 +………..+ (Z1L (joo) IL = V1
Z21 (joo) I1 + Z22 (joo) I2 +…………+ Z2L (joo) IL = V2
ZLI (joo) I1 + ZL2 (joo) I2 +……….+ ZLL (joo) IL = VL
The general method to find steady state reponse involves the following steps
(i) express the network equation in transform form and let s = joo in all the network function.
(ii) all initial conditions are set to zero.
(iii) all source voltages are described in terms of cosine function VI (T) = VI cos for which the phasor is VI
(iv) the system of equations that results in the form of eq. (i) are solved by usual methods.
(v) the phasor IK yields the required phase and magnitude information for solution to be as IK (t) = IK cos. so solution is complete.
Locus Diagrams
If one of the circuit elements is variable, then depending upon its values, the circuit characteristics varies. the locus of the extremity of the current phasor, obtained for various values of a variable element is called a locus diagram.
Intro Exercise – 5
1. in the RC circuit shown in figure, VC (0) = 10 V, the VC(t) is
(a) e-t V
(b) 5e-t V
(c) 10 e-t V
(d) 10 + e-t V
2. In the circuit of figure VC (0-) = 5 V. IF VC (0.15) = 5/E V at = 0.1 s, the value of C is
(a) 2.5 uf
(b) 4 uf
(c) 250 uf
(d) 400 uf
3. In the circuit of figure ….., if IC(0) = – 2 mA, at t = 2 ms. the voltage VC (2ms) is
(a) 6.23 V
(b) 1.56 mV
(c) – 1.56 mV
(d) 0.15 V
4. In the RL circuit of figure if IL (at 50 us) = 9.2 mA and IL (150 us) = 1.24 mA, the initial condition IL (0) is
(a) 43.3 mA
(b) 25 mA
(c) 86.7 mA
(d) – 86.7 mA
5. In the circuit of figure ………., VC (at 1 ms) = 18.39 V and VC (at 2 ms) = 6.77 V, the value of R is
(a) 120
(b) 210.5
(c) 6
(d) 2.3 k
6. In the circuit of figure……IL (0) = 10 mA. the VR (t) for t > 0 is
(a) -32 e-4 x 104t V
(b) 10 e-4 x 104t V
(c) -10 e-4 x 104t V
(d) 10 e-4 x 104t mV
7. In the circuit of figure, if IL (0) = 160 mA, the current IL (t) is
(a) 160 e-500t mA
(b) -160 e-500t mA
(c) 160 e-t/500 mA
(d) -60 e-t/500 mA
8. In the circuit of figure the voltage VC (at 5 ms) is
(a) 11.43 V
(b) 19.86 V
(c) 6.45 V
(d) VC (0) is required
9. The system function H(S) = 1/(s + 1) for an input signal cos t, the steady state response is
(a) 1/2 cos (t – 4)
(b) cos t
(c) cos (t – 4)
(d) 1/2 cos t
10. In the given figure, the switch was closed for a long time before opening at t = 0. the voltage VX at t = 0+ is
(a) 25 v
(b) 50 v
(c) -50 v
(d) zero
11. After closing the switch S at t = 0, the current I(t) at any instant t in the network shown in the given figure will be
(a) 10 + 100 e100t
(b) 10 – 10 e100t
(c) 10 + 10 e-100t
(d) 10 – 10 e-100t
12. The natural response of an RLC circuit is described by the differential equation d2v(t)/dt + 2dv (t)/dt + V (t) = 0, V(0) = 10, dv(0)/dt = 0, the v(t) is
(a) 10 (1 + t) e-t V
(b) 10 (1 – t) e-t V
(c) 10 e-t V
(d) 10 te-t V
13. In the circuit of figure for positive time constant the range of a is
(a) a > 3.33
(b) a < 3.33
(c) a < 0.03
(d) a > 0.03
14. In the circuit of figure the equivalent resistance seen by the capacitor
(a) – 470
(b) 470
(c) -90
(d) VC(0) is required
15. In the series RC circuit shown in the figure, the voltage across C starts increasing when the DC source is switched-on. the rate of increase of voltage across C at the instant just after switch is closed (i,e., at t = 0+) will be
(a) zero
(b) infinty
(c) RC
(d) 1/RC
16. A sinusoidal voltage V and frequency f is connected to a series circuit at variable resistance R and a fixed reactance X. the locus of the tip of the current phasor I, as R is varied from 0 to oo is
(a) a semicircle with a diameter V/X
(b) a straight line with a slope R/X
(c) an ellipse with V/R as major-axis
(d) a circle at radius R/X and origin at (0, V/2)
17. Find the volue of I (t) in the figure given below.
(a) 6 – (6 cos 500t + 6 sin 5000t) x e-50t mA
(b) 8 – (8 cos 500t + 0.06 sin 5000t) e-50t mA
(c) 6 – (6 cos 5000t + 0.06 sin 5000t) e-50t mA
(d) 6e-50t sin 5000t mA
18. The step response of an RLC series is given by d2 I (t) /dt2 + 2di (t) /dt + 5i(t) = 10, i(0)+ = 2, di (0+)/dt = 4. the i(t) is
(a) 1 + e-4t cos 4t A
(b) 4 – 2e-4t cos 4t A
(c) 2 + e-4t sin 4t A
(d) 10 + e-4t sin 4t A
19. Find the value of VC (t) for t > 0 in the figure given below
(a) 4e-1000t – e-2000t V
(b) (3 + 6000t) e-2000t V
(c) 2e-1000t + e-2000t V
(d) (3 – 6000t) e-2000t V
20. What is the value of I(t) in the given figure ?
(a) 20 cos (300t + 68.20) A
(b) 20 cos (300t – 68.20) A
(c) 2.48 cos (300t – 68.20) A
(d) 2.48 cos (300t – 68.20) A
Answers with Solutions
1. (c)
VC (t) = VC (OO) + [VC (0) – VC (00)]e-1/rc
VC (0) = 10 V
VC(00) = 0
VC(T) = 0 + [10 – 0]
VC(T) = 10e-t
2. (a)
VC(t) = VC(00) = [VC(0) – VC(00)]e-t/RC
VC (OO) = 0
VC(0) = 5 V
R = 40 K
t = 0.1 s
VC(0.1s) = 5/e V
5/e = 0 + (5 – 0)e-0.1/40xc
0.1/40C = 1 C = 2.5 uf
3. (a)
VC(0) = 2 x 4
VC(0+) = 8 V = VC(0+)
VC(t) = VC(00) + [VC(0) – VC(00)] e-t/rc
VC (t) = 0 + [8 – 0] et/8 x 10-3
VC (2ms) = 8e = 6.23
4. (b)
IL(t) = IL(00) + [IL(0+) – IL(OO) e
IL (OO) = 0
t = 50 us
IL(50 us) = 9.2 m
IL(50 u) = 9.2 m = 0 + [IL(a+) – 0]e
8.2 m = IL (0+) = e
t = 150 us
IL (150 = 1.24m = IL (0+) e
9 x 23 = I3L (0+) e IL (0+) = 25.06 mA
5. (b)
VC (t) = VC (00) + [VC(0+) -VC(00)] e-1/rq
18.39 = VC (0+) e1
6.77 = VC(0+) x e-2
2.72 = e 1ma
(R(4)/R + 4) (5) = 1 ms
R = 210.5 K
11. (d)
it can be verified by checking the value of the value of the current at t = 0 and t = oo which should be zero and 10/1 = 10 A, respectively.
12. (a)
d2V/dt2 + 2dv/dt + V = 0
taking laplace transform
s2 + 2s + 1 = 0
(s + 1) (s + 1) = 0
s = – 1, – 1
we know that V(t) = (AI + A2t)e-t
V(0) = 10 V, dV (0)/dt = 0
A1 = A2 = 10
V(t) = 10(1 + t) e-t
13. (c)
Vtest = 100 + 50 (1 – a 100)
VTEST = RTH > 0
100 + 50 (1 – a 100) > 0
3 > a 100 a < 0.03
14. (b)
IX = 1 A
VTEST = 120 + 70 x 5
VTEST = 470 V
REQ = VTEST = 470
15. (d)
we kow that voltage across the capacitor at any time t,
VC = V (1 – e-t/rc)
V = 1 V
VC = 1 – e-t/rc
dVC/dt = 1/RC e-t/rc
At t = 0, dvc/dt = 1/RC
16. (a)
I = V/R + jX = V (R – JX)/R2 + X2)
IX = VR/R2 + X2
IV = – VX/R2 + X2
I2X + I2X = V2/R2 + X2 = V (V/R2 + X2) = VIY/X
IX2 (IY + V/2X)2 = V2/4X2
17. (c)
a = 1/2RC = 1/2 x 2 x 54 = 50
1/LC = 1/8 x 5 = 5000 rad/s
a < 0, under damped response
s = – 50 + 502 – 50002 = – 50 + j5000
I (t) = 6 + (A cos 5000t + B sin 5000t) e-50t mA
I (0) = 6 = 6 + A, A = – 6
dI(0)/dt = – 50 A + 5000 B = 0
B = – 0.06
So, I (t) = 6 – (6 cos 5000 + 0.06 sin 5000t) e-50t mA
18. (c)
d2i(t)/dt2 + 2di (t)/dt = 5I (t) = 0
taking laplace transform
s2 + 2s + 5 = 0 (s2 = – 1 + j4)
I (t) if + (A1 cos 4t + A2 sin 4t)e-t(4)
IF = 10/5 = 2 A
I (0+) = 2 + A1 = 2 A1 = 0
di (0+)/dt = 4 = – 4 A2 A2 = 1
I (t) = 2 + (0 cos 4t + (1) sin 4t) e-4t
I (t) = 2 + e-t(4) sin 4t
19. (b)
VC(0+) = 30 x 100 = 3 V
dVC (0+)/dt = IL(0-) = 0 IL (0+) = (dVC (0+)/dt
s2 + 100/ 25 x 10-3 s + 1/25 x 10-3 x 10 x 10-6 = 0
s = – 2000, – 2000
VC(t) = (A1 + A2t) e-2000 t
dVx(t)/dt = A2e-2000 + (A1 + A2T) e-2000t (-2000)
VC (0+) = A1 = 3, dvc(0)/dt = A2 – 2000 x 3 = 0
A2 = 6000
VC(t) = (3 + 6000t) e-2000t V
20. (d)
Z = 3 + j(25 m) (300)
Z = 3 + J(7.5)
Z = 8.06 <68.20
I = 20<00/8.08 <68.20 = 2.48 < – 68.20 A
I (t) = 2.48 cos (300t – 68.20) A