Unit - 3
Discrete Time Fourier Transform
Q1) For the following diagram explain whether it is even or odd.
A1) 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.
Q2) Show that the signal x(t) that satisfies half wave symmetry contains Fourier coefficients with odd harmonics only.
A2) 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
Q3) Compute the Exponential series of the following series.
The time period of the signal x(t) isT=4.
A3) 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) + ………..
Q4) 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
A4)
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
Q5) Find the four point DFT of the sequence x(n) = {0,1,2,3}
A5) 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}
Q6) Let N =4 and
How to calculate the DFT of x
A6)
Q7) Compute the N-point DFT of x(n)=3δ(n).
A7)
Q8) Compute the N-point DFT of x(n)=7(n−n0)
A8) We know that,
Substituting the value of x(n),
Q9) Find the Fourier Integral of
f(x) = 1 |x| ≤ 1
0 |x| ≥ 1
A9)
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.
Q10) The open loop transfer function of a system with unity feedback gain G( S ) = 20 / S2 + 5S + 4. Determine the ξ, Mp, tr, tp.
A10)
Finding closed loop transfer function,
C( S ) / R( S ) = G( S ) / 1 + G( S ) + H( S )
As it is unity feedback so, H(S) = 1
C(S)/R(S) = G(S)/1 + G(S)
= 20/S2 + 5S + 4/1 + 20/S2 + 5S + 4
C(S)/R(S) = 20/S2 + 5S + 24
Standard equation for second order system,
S2 + 2ξWnS + Wn2 = 0
We have,
S2 + 5S + 24 = 0
Wn2 = 24
Wn = 4.89 rad/sec
2ξWn = 5
(a). ξ = 5/2 x 4.89 = 0.511
(b). Mp% = e-∏ξ / √1 –ξ2 x 100
= e-∏ x 0.511 / √1 – (0.511)2 x 100
Mp% = 15.4%
(c). tr = ∏ - φ / Wd
φ = tan-1√1 – ξ2 / ξ
φ= tan-1√1 – (0.511)2 / (0.511)
φ = 1.03 rad.
tr = ∏ - 1.03/Wd
Wd = Wn√1 – ξ2
= 4.89 √1 – (0.511)2
Wd = 4.20 rad/sec
tr = ∏ - 1.03/4.20
tr = 502.34 msec
(d). tp = ∏/4.20 = 747.9 msec
Q11) A second order system has Wn = 5 rad/sec and is ξ = 0.7 subjected to unit step input. Find (i) closed loop transfer function. (ii) Peak time (iii) Rise time (iv) Settling time (v) Peak overshoot.
A11) The closed loop transfer function is
C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2
= (5)2 / S2 + 2 x 0.7 x S + (5)2
C(S)/R(S) = 25 / S2 + 7s + 25
(ii). tp = ∏ / Wd
Wd = Wn√1 - ξ2
= 5√1 – (0.7)2
= 3.571 sec
(iii). tr = ∏ - φ/Wd
φ= tan-1√1 – ξ2 / ξ = 0.795 rad
tr = ∏ - 0.795 / 3.571
tr = 0.657 sec
(iv). For 2% settling time
ts = 4 / ξWn = 4 / 0.7 x 5
ts = 1.143 sec
(v). Mp = e-∏ξ / √1 –ξ2 x 100
Mp = 4.59%
Q12) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1 + ST)
Calculate the value by which k should be multiplied so that damping ratio is increased from 0.2 to 0.4?
A12) C(S)/R(S) = G(S) / 1 + G(S)H(S) and H(S) = 1
C(S)/R(S) = K/S(1 + ST) / 1 + K/S(1 + ST)
C(S)/R(S) = K/S(1 + ST) + K
C(S)/R(S) = K/T / S2 + S/T + K/T
For second order system,
S2 + 2ξWnS + Wn2
2ξWn = 1/T
ξ = 1/2WnT
Wn2 = K/T
Wn =√K/T
ξ = 1 / 2√K/T T
ξ = 1 / 2 √KT
Forξ1 = 0.2, for ξ2 = 0.4
ξ1 = 1 / 2 √K1T
ξ2 = 1 / 2 √K2T
ξ1/ ξ2 = √K2/K1
K2/K1 = (0.2/0.4)2
K2/K1 = 1 / 4
K1 = 4K2
Q13) Consider the transfer function C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2
Find ξ, Wn so that the system responds to a step input with 5% overshoot and settling time of 4 sec?
A13)
Mp = 5% = 0.05
Mp = e-∏ξ / √1 –ξ2
0.05 = e-∏ξ / √1 –ξ2
Cn 0.05 = - ∏ξ / √1 –ξ2
-2.99 = - ∏ξ / √1 –ξ2
8.97(1 – ξ2) = ξ2∏2
0.91 – 0.91 ξ2 = ξ2
0.91 = 1.91 ξ2
ξ2 = 0.69
(ii). ts = 4/ ξWn
4 = 4/ ξWn
Wn = 1/ ξ = 1/ 0.69
Wn = 1.45 rad/sec
Q14) Find the initial value for the function f(t) = 2u(t)+3cost u(t)?
A14)
f(t) = 2u(t)+3cost u(t)
F(s) =
SF(s) = 2+
By initial value theorem
= f(0+)
= 2+ = 5 = f(0+)
Hence, initial value of the function is 5
Q15) Find final value of the function F(s) =
A15)
F(s) =
SF(s) =
By final value theorem
=
= = 0.1
So, final value of the function is 0.1
Q16) Find fourier Sine transform of
(i) (ii)
A16)
(i) By definition, we have
(ii) By definition, we have
Q17) Find Fourier transform of
And hence evaluate
A17)
Fourier transform of f(x) is given by
Taking inverse fourier transform of (1)
Substituting in (2)
Equating real parts on both sides, we get
Putting on both sides
is an even function of
Now by property of definite integrals
Q18) Prove that Fourier transform of is self reciprocal.
A18)
Fourier transform of f(x) is given by
By putting z=x-i
being even function of z.
Put
Hence we see that Fourier transform of is given by . Variable x is transformed to . We can say that fourier transform of is self reciprocal.
Q19) Draw and explain all the standard input test signals for time domain analysis?
A19) The Impulse signal, Ramp signal, unit step and parabolic signals are used as the standard test signals. All these signals are explained below.
Impulse Signal:
This signal has zero amplitude everywhere except at the origin. Fig below shown the representation of Impulse signal.
Fig: Unit Impulse Signal
The mathematical representations
A(t) = 0 for t ≠0
dt = A e
Where A represent energy or area of the Laplace Transform of Impulse signal is
L [A (t)] = A
UNIT IMPULSE SIGNAL:
If A = 1
(t) = 0 for t ≠0
L [(t)] = 1
The transfer function of a linear time invariant
System is the Laplace transform of the impulse response of the system. If a unit impulse signal is applied to system then Laplace transform of the output c(s) is the transfer function G(s)
As we know G(s) = c(s)/R(S)
r(t) = (t)
R(s) = L [(t)] = 1
:. G(S) = C(s)
(b) Step signal:
Step signal of size A is a signal that change from zero level to A in zero time and stays there forever.
Fig: Unit Step Signal
r(t)= A t >=0
=0 t<0
L[r(t)] = R(s) = A/s
UNIT STEP SIGNAL: - If the magnitude of the slip signal is I then it is called unit step signal.
u(t) = 1
t>=0
t<0
L[u(t)] = 1/s
(c) Ramp Signal: -
The vamp signal increase linearly with time from initial value of zero at t= 0 as shown in fig is below
Fig: Ramp Signal
r(t) = At t>=0
=0 t<0
A is the slope of the line The Laplace transform of ramp signal is
L[r(t)] = R(s) = A/s2
(d) Parabolic Signal: -
The instantaneous value of a parabolic signal varies as square of the time from an initial value of zero t=0. The signal representation in fig 14 below.
Fig 4 Parabolic Signal
r(t) At2 t>=0
=0 t<0
Then Laplace Transform is given as
R(s) = L[At2] = 2A/s3
If no error then E(t) =0, :. R(t) = c(t), output is tracking the input.
Steady state Errors signal (ess): - (t )
Ess = t e(t)
Using final values, the theorem
= ess = S.E (s)
ess= S[R(s)/1+G/(s)]
Q20) What are the specifications of transient response of second order systems?
A20)
Fig: Transient Response of second order system
Specifications:
1) Rise Time (tp): The time taken by the output to reach the already status value for the first time is known as Rise time.
C(t) = 1-e-wnt/1-2 sin (wdt+ø)
Sin (wd +ø) = 0
Wdt +ø = n
tr =n-ø/wd
For first time so ,n=1.
tr = -ø/wd
T=1/
2) Peak Time (tp)
The peak value attained by the output is called peak time. The time required by the output to reach this value is lp.
d(cct) /dt = 0 (maxima)
d(t)/dt = peak value
tp = n/wd for n=1
tp = wd
3) Peak Overshoot Value:
Maximum deviation of output from steady state value is called peak overshoot value (Mp).
(ltp) = 1 = Mp
(Sin(Wat + φ )
(Sin( Wd∏/Wd + φ)
Mp = e-∏ξ / √1 –ξ2
Condition 3 ξ = 1
C( S ) = R( S ) Wn2 / S2 + 2ξWnS + Wn2
C( S ) = Wn2 / S(S2 + 2WnS + Wn2) [ R(S) = 1/S ]
C( S ) = Wn2 / S( S2 + Wn2 )
C( t ) = 1 – e-Wnt + tWne-Wnt
The response is critically damped.
(4). Settling Time (ts):
ts = 3 / ξWn ( 5% )
ts = 4 / ξWn ( 2% )