Module9
Fourier Transform of Standard Signals
Objective:To find the Fourier transform of standard signals like unit impulse, unit step etc. and
any periodic signal.
Introduction:
The Fourier transform of a finite duration signal can be found using the formula
๐ ๐ = ๐ฅ(๐ก)๐โ๐๐๐ก ๐๐กโ
โโ
This is called as analysis equation
The inverse Fourier transform is given by
๐ฅ(๐ก) = ๐ ๐ ๐๐๐๐ก ๐๐โ
โโ
This is called as synthesis equation
Both these equations form the Fourier transform pair.
Description:
Existence of Fourier Transform:
The Fourier Transform does not exist for all aperiodic functions. The condition for a function
x(t) to have Fourier Transform, called Dirichletโs conditions are:
1. x t is absolutely integrable over the interval -โ to +โ,that is
x(t) dt
โ
โโ
< โ
2. x t has a finite number of discontinuities in every finite time interval. Further, each of
these discontinuities must be finite.
3. x t has a finite number of maxima and minima in every finite time interval.
Almost all the signals that we come across in physical problems satisfy all the above
conditions except possibly the absolute integrability condition.
Dirichletโs condition is a sufficient condition but not necessary condition. This means,
Fourier Transform will definitely exist for functions which satisfy these conditions. On the
other hand, in some cases , Fourier Transform can be found with the use of impulses even for
functions like step function, sinusoidal function,etc.which do not satisfy the convergence
condition .
Fourier transform of standard signals:
1. Impulse Function ๐ ๐ญ
Givenx t = ฮด t ,
ฮด t = 1 for t = 00 for t โ 0
Then
X ฯ = x t eโjฯtdt = ฮด t eโjฯt dt โ
โโ
โ
โโ = eโjฯtโธt=0 = 1
โด F ฮด t = 1 or ฮด t FT 1
Hence , the Fourier Transform of a unit impulse function is unity.
X ฯ = 1 for all ฯ
X ฯ = 0 for all ฯ
The impulse function with its magnitude and phase spectra are shown in below figure:
Similarly,
F ฮด t โ to = ฮด t โ to eโjฯtdt = eโjฯt0 i. e. ฮด(t โ to)FT eโjฯto
โ
โโ
2. Single Sided Real exponential function ๐โ๐๐ญ๐ฎ(๐ญ)
Given x t = eโat u t , u(t) = 1 fort โฅ 00 fort < 0
Then
X ฯ = x t eโjฯtdt = eโat u t eโjฯtdt
โ
โโ
โ
โโ
= eโat eโjฯtdt = eโ(a+jฯ)tdt = e(โ a +jฯ t
โ a + jฯ
โ
0
โ
0 0
โ
= eโโ โ e0
โ(a + jฯ)
=0 โ 1
โ(a + jฯ)=
1
a + jฯ
โด F eโat u t =1
a+jฯ or eโat u t
FT
1
a+jฯ
Now, X ฯ =1
a+jฯ=
aโjฯ
a+jฯ (aโjฯ)
=aโjฯ
a2+ฯ2 = a
a2+ฯ2 โ jฯ
a2+ฯ2 = 1
a2+ฯ2 โ tanโ1 ฯ
a
โด X(ฯ) =1
a2 + ฯ2 , X ฯ = โ tanโ1
ฯ
a forallฯ
Figure shows the single-sided exponential function with its magnitude and phase spectra.
3.Double sided real exponential function ๐โ๐ ๐ญ
Given x t = eโa t
โด x t = eโa t = eโa โt =eat fort โค 0eโat = eโat fort โฅ 0
= eโa โt u โt + eโat u t
= eat u โt + eโat u(t)
X ฯ = x t eโjฯtdt
โ
โโ
= eat eโjฯtdt + eโat eโjฯtdt = e aโjฯ tdt + eโ a+jฯ tdt
โ
0
0
โโ
โ
0
0
โโ
= eโ aโjฯ tdt + eโ a+jฯ tdt = โ
0
โ
0
eโ aโjฯ t
โ(aโjฯ)
0
โ
+ eโ a +jฯ t
โ(a+jฯ)
0
โ
= eโโ โeโ0
โ(aโjฯ)+
eโโ โeโ0
โ(a+jฯ)=
1
aโjฯ+
1
a+jฯ=
2a
a2+ฯ2
โด F eโa t = 2a
a2 + ฯ2oreโa t
FT
2a
a2 + ฯ2
โด X ฯ = 2a
a2 + ฯ2forallฯ
And X ฯ = 0 forallฯ
A Two sided exponential function and its amplitude and phase spectra are shown in figures
below:
4. Complex Exponential Function๐๐ฃ๐๐๐ญ :
To find the Fourier Transform of complex exponential function ejฯ0t , consider finding the
inverse Fourier transform of ฮด(ฯ โ ฯ0). Let
X ฯ = ฮด(ฯ โ ฯ0)
โด x t = Fโ1 X ฯ = Fโ1[ฮด(ฯ โ ฯ0)] =1
2ฯ X ฯ ejฯtdฯ
โ
โโ
=1
2ฯ ฮด(ฯ โ ฯ0)ejฯtdฯ
โ
โโ=
1
2ฯejฯ0t
โด Fโ1 ฮด ฯ โ ฯ0 =ejฯ0t
2ฯorFโ1 2ฯฮด ฯ โ ฯ0 = ejฯ0t
= F ejฯ0t = 2ฯฮด(ฯ โ ฯ0)
Or ejฯ0tFT 2ฯฮด(ฯ โ ฯ0)
5. Constant Amplitude (1)
Let x t = 1 โ โ โค t โค โ
The waveform of a constant function is shown in below figure .Let us consider a small section of
constant function, say, of duration ๐.If we extend the small duration to infinity, we will get back
the original function.Therefore
x t = Lt tโโ
[rect t
ฯ ]
Where rect t
ฯ =
1 forโฯ
2โค t โค
ฯ
2
0 elesewhere
By definition, the Fourier transform of x(t) is:
X(ฯ) = F[x(t)] = F Lt tโโ
rect t
ฯ = Lt
tโโF rect
t
ฯ
= Lt tโโ
1 ฯ 2
โฯ 2 eโjฯt dt = Lt
tโโ
eโjฯ t
โjฯ โฯ 2
ฯ 2
= Lt tโโ
eโjฯ (ฯ 2 )โejฯ (ฯ 2 )
โjฯ = Lt
tโโ
2sin [ฯ ฯ
2 ]
ฯ = Lt
tโโ ฯ
sin [ฯ ฯ
2 ]
ฯ(ฯ
2)
= Lt tโโ
ฯ sa(ฯฯ
2) = 2ฯ Lt
tโโ
ฯ 2
ฯsa(
ฯฯ
2)
Using the sampling property of the delta function i. e. Lt tโโ
ฯ 2
ฯsa(
ฯฯ
2) = ฮด(ฯ) , we get
X(ฯ) = F Lt tโโ
rect t
ฯ = 2ฯฮด ฯ
6.Signum function sgn(t)
The signum function is denoted by sgn(t) and is defined by
sgn(t) = 1 for t > 0
โ1 for t < 0
This function is not absolutely integrable. So we cannot directly find its Fourier transform.
Therefore, let us consider the function eโaโนtโธsgn(t) and substitute the limit aโ0 to obtain the
above sgn(t)
Given x(t) = sgn(t) = Lt aโ0
eโaโนtโธ sgn(t) = Lt aโ0
[ eโat u t โ eโat u โt
โด X(ฯ) = F[sgn(t)] = Lt aโ0
[ eโat u t โ eโat u โt eโjฯtโ
โโdt
= Lt aโ0
eโat eโjฯtu t dt โ eat eโjฯtu โt dt โ
โโ
โ
โโ
= Lt aโ0
eโ(a+jฯ)tdt โ e(aโjฯ)tdt 0
โโ
โ
0 = Lt
aโ0 eโ(a+jฯ)tdt โ eโ(aโjฯ)tdt
โ
0
โ
0
= Lt aโ0
eโ(a +jฯ )t
โ(a+jฯ)
0
โ
โ eโ(aโjฯ )t
โ(aโjฯ)
0
โ
= Lt aโ0
1
a+jฯโ
1
aโjฯ =
1
๐๐โ
1
โ๐๐ =
2
๐๐
F[sgn(t)] = 2
๐๐
sgn(t)๐น๐
2
๐๐
โด โนX(โต)โธ = 2
๐ and ๐(โต) =
๐
2๐๐๐ ๐ < 0 ๐๐๐ โ
๐
2๐๐๐ ๐ > 0
Figure below shows the signum function and its magnitude and phase spectra
7. Unit step function u(t)
The unit step function is defined by
u(t) = 1 for t โฅ 00 for t < 0
since the unit step function is not absolutely integrable, we cannot directly find its Fourier
transform. So express the unit step function in terms of signum function as:
u(t) = 1
2+
1
2 ๐ ๐๐ ๐ก
x(t)= u(t) = 1
2[1 + ๐ ๐๐ ๐ก ]
X(๐) = F[u(t)] = F 1
2[1 + ๐ ๐๐ ๐ก ]
= 1
2 ๐น 1 + ๐น[๐ ๐๐ ๐ก ]
We know that F[1] = 2๐๐ฟ(๐) and F[sgn(t)] = 2
๐๐
F[u(t)]= 1
2 2๐๐ฟ ๐ +
2
๐๐ = ๐๐ฟ ๐ +
1
๐๐
u(t)๐น๐ ๐๐ฟ ๐ +
1
๐๐
โด โนX(โต)โธ=โ at โต=0 and is equal to 0 at โต=โโ and โต=โ
Figure shows the unit step function and its spectrum.
8. Rectangular pulse ( Gate pulse) ๐ญ
๐ or rect
๐ญ
๐
Consider a rectangular pulse as shown in below figure. This is called a unit gate function and is
defined as
x(t) = rect t
ฯ =
t
ฯ =
1 ๐๐๐โน tโธ <ฯ
2
0 ๐๐ก๐๐๐๐ค๐๐ ๐
Then X(โต) = F[ x(t)] = F ๐ก
๐ =
t
ฯ
โ
โโ๐โ๐๐๐ก ๐๐ก
= 1 ๐โ๐๐๐ก ๐๐ก๐ 2
โ๐ 2 =
๐โ๐๐๐ก
โ๐๐ ๐ 2
๐ 2
= ๐โ๐๐ (๐ 2) โ ๐ ๐๐ (๐ 2)
โ๐๐
= ๐
๐(๐ 2) ๐ ๐๐ (๐ 2) โ ๐โ๐๐ (๐ 2)
2๐ = ๐
sin ๐(๐ 2)
๐(๐ 2)
= ๐ ๐ ๐๐๐ ๐(๐ 2)
โด F ๐ก
๐ = ๐ ๐ ๐๐๐ ๐(๐ 2) , that is
rect t
ฯ =
t
ฯ
๐น๐ ๐ ๐ ๐๐๐ ๐(๐ 2)
Figure shows the spectra of the gate function
The amplitude spectrum is obtained as follows:
At ๐ = 0, sinc(๐๐/2)=1. Therefore, โนX(๐)โธ ๐๐ก ๐ = 0 is equal to ๐. At (๐๐/2)=ยฑ๐๐, i.e.
at
๐ = ยฑ2๐๐
๐ , n = 1, 2, โฆโฆsinc(๐๐/2) =0
The phase spectrum is: ๐(๐) = 0 if sinc(๐๐/2)> 0
= ยฑ๐ if sinc(๐๐/2) < 0
The amplitude response between the first two zero crossings is known as main lobe and the
portions of the response for ๐ < โ 2๐
๐ and ๐ >
2๐
๐ are known as side lobes. From the
amplitude spectrum, we can find that majority of the energy of the signal is contained in the main
lobe. The first zero crossing occurs at ๐ = 2๐
๐ or at f =
1
๐ Hz. As the width of the rectangular
pulse is made longer, the main lobe becomes narrower. The phase spectrum is odd function of ๐.
If the amplitude spectrum is positive, then phase is zero, and if the amplitude spectrum is
negative, then the phase is โ ๐ ๐๐ ๐.
9. Triangular Pulse โ t
ฯ
Consider the triangular pulse as shown in below figure. It is defined as:
x(t) = โ t
ฯ =
1
ฯ 2 t +
ฯ
2 = 1 + 2
t
ฯ for โ
ฯ
2< ๐ก < 0
1
ฯ 2 t โ
ฯ
2 = 1 โ 2
t
ฯ for 0 < ๐ก <
ฯ
2
0 elsewhere
i.e. as x(t) = โ t
ฯ = 1 โ
2โนtโธ
ฯ for โนtโธ <
ฯ
2
0 otherwise
Then X(โต) = F[ x(t)] = F โ t
ฯ = โ
t
ฯ
โ
โโ๐โ๐๐๐ก ๐๐ก
= 1 +2t
ฯ
0
โ๐ 2 ๐โ๐๐๐ก ๐๐ก + 1 โ
2t
ฯ
๐ 2
0๐โ๐๐๐ก ๐๐ก
= 1 โ2t
ฯ
๐ 2
0๐๐๐๐ก ๐๐ก + 1 โ
2t
ฯ
๐ 2
0๐โ๐๐๐ก ๐๐ก
= ๐๐๐๐ก๐ 2
0๐๐ก โ
2t
ฯ
๐ 2
0๐๐๐๐ก ๐๐ก + ๐โ๐๐๐ก๐ 2
0๐๐ก โ
2t
ฯ
๐ 2
0๐โ๐๐๐ก ๐๐ก
= [๐๐๐๐ก +๐ 2
0๐โ๐๐๐ก ]๐๐ก โ
2
ฯ ๐ก[๐๐๐๐ก +
๐ 2
0๐โ๐๐๐ก ]๐๐ก
= 2 cos ๐๐ก๐ 2
0๐๐ก โ
2
ฯ 2๐ก cos ๐๐ก
๐ 2
0๐๐ก
= 2 sin ๐๐ก
๐
0
๐ 2
โ 4
๐ ๐ก
sin ๐๐ก
๐
0
๐ 2
+ cos ๐๐ก
๐2 0
๐ 2
= 2
๐ sin ๐
๐
2 โ
4
๐๐ ๐
2sin
๐๐
2 โ
4
๐2๐ cos
๐๐
2โ 1
=4
๐2๐ 1 โ cos
๐๐
2 =
4
๐2๐ 2 ๐ ๐๐2 ๐๐
4
= 8
๐2๐ ๐๐
4
2 sin 2 ๐๐
4
๐๐
4
= ๐
2๐ ๐๐๐2
๐๐
4
F โ t
ฯ =
๐
2๐ ๐๐๐2
๐๐
4
Or โ t
ฯ
๐น๐
๐
2๐ ๐๐๐2
๐๐
4
Figure shows the amplitude spectrum of a triangular pulse.
10. Cosine wave cos โต0๐ก
Given x(t) = cosโต0๐ก
Then X(โต) = F[ x(t)] = F[cosโต0๐ก] = F 1
2 ๐๐โต0๐ก + ๐โ๐โต0๐ก
= 1
2 ๐น ๐๐โต0๐ก + ๐น ๐โ๐โต0๐ก =
1
2 2๐๐ฟ โต โ โต๐ + 2๐๐ฟ โต + โต๐
= ๐ ๐ฟ โต โ โต๐ + ๐ฟ โต + โต๐
โด F[cos โต0๐ก] = ๐ ๐ฟ โต โ โต๐ + ๐ฟ โต + โต๐ or cos โต0๐ก๐น๐ ๐ ๐ฟ โต โ โต๐ + ๐ฟ โต + โต๐
Below Figure shows the cosine wave and its amplitude and phase spectra.
11. Sine wave sin โต0๐ก
Given x(t) = sinโต0๐ก
Then X(โต) = F[ x(t)] = F[sinโต0๐ก] = F 1
2๐ ๐๐โต0๐ก โ ๐โ๐โต0๐ก
= 1
2๐ ๐น ๐๐โต0๐ก โ ๐น ๐โ๐โต0๐ก =
1
2๐ 2๐๐ฟ โต โ โต๐ โ 2๐๐ฟ โต + โต๐
= โ๐๐ ๐ฟ โต โ โต๐ โ ๐ฟ โต + โต๐
โด F[cosโต0๐ก] = โ๐๐ ๐ฟ โต โ โต๐ โ ๐ฟ โต + โต๐ or cos โต0๐ก๐น๐ โ๐๐ ๐ฟ โต โ โต๐ โ ๐ฟ โต + โต๐
Below Figure shows the sine wave and its amplitude and phase spectra.
Fourier transform of a periodic signal
The periodic functions can be analysed using Fourier series and that non-periodic
function can be analysed using Fourier transform. But we can find the Fourier transform of a
periodic function also. This means that the Fourier transform can be used as a universal
mathematical tool in the analysis of both non-periodic and periodic waveforms over the entire
interval. Fourier transform of periodic functions may be found using the concept of impulse
function.
We know that using Fourier series , any periodic signal can be represented as a sum of
complex exponentials. Therefore, we can represent a periodic signal using the Fourier integral.
Let us consider a periodic signal x(t) with period T. Then, we can express x(t) in terms of
exponential Fourier series as:
x(t) = ๐ถ๐ ๐๐๐ โต0๐กโ
๐=โโ
The Fourier transform of x(t) is:
X(โต) = F[x(t)] = F ๐ถ๐ ๐๐๐ โต0๐กโ
๐=โ
= ๐ถ๐ ๐น ๐๐๐ โต0๐ก โ๐=โ
Using the frequency shifting theorem, we have
๐น 1๐๐๐ โต0๐ก = ๐น 1 โนโต=โตโ๐โต0 = s๐๐ฟ โต โ ๐โต0
X(โต) = 2๐ ๐ถ๐ ๐ฟ โต โ ๐โต0 โ๐=โ
Where ๐ถ๐ ๐ are the Fourier coefficients associated with x(t) and are given by
๐ถ๐ = 1
๐ ๐ฅ(๐ก)
๐ 2
โ๐ 2 ๐โ๐๐ โต0๐ก๐๐ก
Thus, the Fourier transform of a periodic function consists of a train of equally spaced impulses.
These impulses are located at the harmonic frequencies of the signal and the strength of each
impulse is given as 2๐๐ถ๐ .
Solved Problems:
Problem 1:Find the Fourier transform of the signals e3tu(t)
Solution:
Given x(t) = e3tu(t)
The given signal is not absolutely integrable. That is e3tu t = โโ
โโ. Therefore, Fourier
transform of x(t) = e3tu(t) does not exist.
Problem 2: Find the Fourier transform of the signals cos ฯotu(t)
Solution:
Given x(t) = cos ฯot u(t)
i.e. = ejฯo t +eโjฯo t
2 u(t)
โด X(โต) = F[cos ฯot u(t)] = ejฯo t +eโjฯo t
2 u(t)
โ
โโeโjฯt dt
= 1
2 eโj(ฯโฯo )tโ
0dt + eโj(ฯ+ฯo )tโ
0 dt
= 1
2
eโj(ฯโฯo )t
โj(ฯโฯo )+
eโj(ฯ+ฯo )t
โj(ฯ+ฯo )
0
โ
= 1
2
โe0
โj(ฯโฯo )+
โe0
โj(ฯ+ฯo )
With impulses of strength ฯ at ฯ = ฯo and ฯ = โฯo
โด X(โต) = 1
2
1
j(ฯโฯo )+
1
j(ฯ+ฯo )+ ฯฮด ฯ โ ฯo + ฯฮด(ฯ + ฯo)
= 1
2
j2ฯ
(jฯ)2+ ฯo2 + ฯฮด ฯ โ ฯo + ฯฮด(ฯ + ฯo)
= jฯ
(jฯ)2+ ฯo2 +
1
2 ฯฮด ฯ โ ฯo + ฯฮด(ฯ + ฯo)
Problem 3: Find the Fourier transform of the signals sin ฯot u(t)
Solution:
Given x(t) = sin ฯot u(t)
i.e. = ejฯo tโ eโjฯo t
2j u(t)
โด X(โต) = F[sin ฯot u(t)] = ejฯo tโeโjฯo t
2j u(t)
โ
โโeโjฯt dt
= 1
2j eโj(ฯโฯo )tโ
0dt โ eโj(ฯ+ฯo )tโ
0 dt
= 1
2j
eโj(ฯโฯo )t
โj(ฯโฯo )โ
eโj(ฯ+ฯo )t
โj(ฯ+ฯo )
0
โ
= 1
2j
โe0
โj(ฯโฯo )โ
โe0
โj(ฯ+ฯo )
With impulses of strength ฯ at ฯ = ฯo and ฯ = โฯo
โด X(โต) = 1
2j
1
j(ฯโฯo )โ
1
j(ฯ+ฯo )+ ฯฮด ฯ โ ฯo โ ฯฮด(ฯ + ฯo)
= 1
2j
j2ฯo
(jฯ)2+ ฯo2 + ฯฮด ฯ โ ฯo โ ฯฮด(ฯ + ฯo)
= ฯo
(jฯ)2+ ฯo2 โ j
ฯ
2 ฮด ฯ โ ฯo + ฮด(ฯ + ฯo)
Problem 4: Find the Fourier transform of the signals eโtsin 5t u(t)
Solution:
Given x(t) = eโtsin 5t u(t)
x(t) = eโt ej5tโ eโj5t
2j u(t)
โด X(โต) = F[eโt sin 5t u(t)] = F eโt ej5tโ eโj5t
2j u(t)
= 1
2j [
โ
โโeโt(ej5t โ eโj5t)u(t)] eโjฯt dt
= 1
2j
eโ[1+j ฯโ5 ]t
โ[1+j ฯโ5 ]โ
eโ[1+j ฯ+5 ]t
โ[1+j ฯ+5 ]
0
โ
= 1
2j
1
[1+j ฯโ5 ]โ
1
[1+j ฯ+5 ]
= 5
[1+j ฯโ5 ][1+j ฯ+5 ] =
5
(1+jฯ)2 + 25 [neglecting impulses]
Problem 5: Find the Fourier transform of the signals eโ2tcos 5t u(t)
Solution:
Given x(t) = eโ2tcos 5t u(t)
x(t) = eโ2t ej5t + eโj5t
2 u(t)
โด X(โต) = F[eโ2t cos 5t u(t)] = F eโ2t ej5tโ eโj5t
2 u(t)
= 1
2 [
โ
โโeโ2t(ej5t โ eโj5t)u(t)] eโjฯt dt
= 1
2[ eโ 2+j ฯโ5 tโ
0 dt + eโ 2+j ฯ+5 tโ
0 dt ]
= 1
2
eโ[2+j ฯโ5 ]t
โ[2+j ฯโ5 ]โ
eโ[2+j ฯ+5 ]t
โ[2+j ฯ+5 ]
0
โ
= 1
2
1
[1+j ฯโ5 ]โ
1
[1+j ฯ+5 ]
= 1
2
2(2+jฯ)
(2+jฯ)2 + 25 =
2+jฯ
(2+jฯ)2 + 25 [neglecting impulses]
Problem 6: Find the Fourier transform of the signals eโโนtโธsin 5โนt โธfor all t
Solution:
Given x(t) = eโโนtโธsin 5โนt โธfor all t
i.e. x(t) = ๐๐ก๐ ๐๐5 โ๐ก = โ๐๐ก๐ ๐๐5๐ก ๐๐๐ ๐ก < 0
๐โ๐ก๐ ๐๐5 ๐ก = ๐๐ก๐ ๐๐5๐ก ๐๐๐ ๐ก > 0
i.e. x(t) = โ๐๐ก๐ ๐๐5๐ก ๐ข(โ๐ก) + ๐๐ก๐ ๐๐5๐ก ๐ข(๐ก)
โดX(๐) = 1
2j [โ
โ
โโet ej5t โ eโj5t u โt + eโt(ej5t โ eโj5t)u(t)] eโjฯt dt
= โ1
2j [
0
โโe[1โj ฯโ5 ]t โ e[1โj ฯ+5 ]t]dt +
1
2j [
โ
0eโ[1+j ฯโ5 ]t โ eโ[1+j ฯ+5 ]t]dt
= โ1
2j [
โ
0eโ[1โj ฯโ5 ]t โ eโ[1โj ฯ+5 ]t]dt +
1
2j [
โ
0eโ[1+j ฯโ5 ]t โ eโ[1+j ฯ+5 ]t]dt
= โ1
2j
eโ[1โj ฯโ5 ]t
โ[1โj ฯโ5 ]โ
eโ[1โj ฯ+5 ]t
โ[1โj ฯ+5 ]
0
โ
+ 1
2j
eโ[1+j ฯโ5 ]t
โ[1+j ฯโ5 ]โ
eโ[1+j ฯ+5 ]t
โ[1+j ฯ+5 ]
0
โ
= โ1
2j
1
[1โj ฯโ5 ]โ
1
[1โj ฯ+5 ] +
1
2j
1
[1+j ฯโ5 ]โ
1
[1+j ฯ+5 ]
= 5
(1โjฯ)2 + 25+
5
(1+jฯ)2 + 25 [neglecting impulses]
Problem 7: Find the Fourier transform of the signals eat u(-t)
Solution:
Given x(t) = eat u(-t)
โด X(๐) = F[eat u(-t)] = eat u(โt)โ
โโeโjฯt dt
= e(aโjฯ)t0
โโ dt = eโ(aโjฯ)tโ
0 dt =
eโ(aโjฯ )t
โ(aโjฯ)
0
โ
= 1
aโjฯ
Problem 8: Find the Fourier transform of the signals teat u(t)
Solution:
Given x(t) = teat u(t)
โด X(๐) = F[teat u(t)] = teat u(t)โ
โโeโjฯt dt
= teโ(a+jฯ)tโ
โโ dt = ๐ก
eโ(a +jฯ )t
โ(a+jฯ)
0
โ
โ eโ(a +jฯ )t
โ(a+jฯ)
โ
0 dt = ๐ก
eโ(a +jฯ )t
โ(a+jฯ)
0
โ
โ eโ(a +jฯ )t
(a+jฯ)2 0
โ
=
1
(a+jฯ)2
Problem 9: Find the inverse Fourier transform of X(๐) = ๐๐
(2+๐๐ )2
Solution:
We know that F[teโat u(t)] = 1
(๐+๐๐ )2
โด F[teโ2t u(t)] = 1
(2+๐๐ )2
Let teโ2t u(t) = x1(t)
Then 1
(2+๐๐ )2 = X1(๐)
Using differentiation in time domain property [ i.e. ๐
๐๐กx(t)
๐น๐ j๐X(๐)], we have
F ๐
๐๐ก๐ฅ1(๐ก) = j๐๐1(๐)
๐นโ1 j๐๐1(๐) = ๐
๐๐ก ๐ฅ1(๐ก) =
๐
๐๐ก teโ2t u(t)
Problem 10: Find the Fourier transform of the Gaussian signal x(t) = ๐โ๐๐ก2
Solution:
Given x(t) = ๐โ๐๐ก2
The Fourier transform of the given signal is:
X(๐) = F[๐โ๐๐ก2] = ๐โ๐๐ก2
๐โ๐๐๐กโ
โโdt = ๐โ(๐๐ก2+๐๐๐ก )โ
โโdt = ๐โ๐2/4๐ ๐
โ[๐ก ๐+ ๐๐
2 ๐ ]2โ
โโdt
Let p = ๐ก ๐ + ๐๐
2 ๐
โด dp = ๐ dt
โด X(๐) = ๐โ๐ 2/4๐
๐ ๐โ๐2โ
โโdp
= ๐โ๐ 2/4๐
๐ ๐ โต ๐โ๐2โ
โโdp = ๐
= ๐
๐๐โ๐2/4๐
โด F[๐โ๐๐ก2] =
๐
๐๐โ๐2/4๐ or ๐โ๐๐ก2 ๐น๐
๐
๐๐โ๐2/4๐
The graphical representation of Gaussian signal and its spectrum are shown in below figure:
Assignment Problems:
1. Find the Fourier transform of the following:
(a) eโat u(t) (b) teโ2tu(t) (c) ej2tu(t)
2. Find the Fourier transform of the following:
(a) ej2tcos๐๐๐ก u(t) (b)ej2tsin๐๐๐ก u(t)
3. Find the Fourier transform of the Gaussian modulated signal x(t) =๐โ๐๐ก2cos๐๐ t
4. Find the Fourier transform of the signal x(t) = ๐โ๐โน๐กโธsgn(t).
5. Find the inverse Fourier transform of X(๐) = ๐๐
(3+๐๐ )2
6. Find the inverse Fourier transform of X(๐) = ๐โ4๐๐(๐)
7. Find the Fourier transform of x(t-4)+x(t+4).
8. Find the Fourier transform of ๐ฟ ๐ก + 4 + ๐ฟ ๐ก + 2 + ๐ฟ ๐ก โ 2 + ๐ฟ ๐ก + 4
9. Find the Fourier transform of the signal x(t) shown in below figure:
10. Determine the inverse Fourier transform of the spectrum given in below figure:
Simulation:
The function โfftโ is used to find the fourier transform of discrete samples of any signal x(n) and
its magnitude is plotted using the function โabsโ. Fourier transform of a gate pulse is sinc
function.
Program:
Fs = 150; % Sampling frequency
t = -0.5:1/Fs:0.5; % Time vector of 1 second
w = .2; % width of rectangle
x = rectpuls(t,w); % Generate Square Pulse
nfft = 512; % Length of FFT
% Take fft, padding with zeros so that length(X) is equal to
nfft
X = fft(x,nfft);
% FFT is symmetric, throw away second half
X = X(1:nfft/2);
%Take the magnitude of fft of x
mx =abs(X);
% Frequency vector
f = (0:nfft/2-1)*Fs/nfft;
% Generate the plot, title and labels.
figure(1);
plot(t,x);
title('Square Pulse Signal');
xlabel('Time (s)');
ylabel('Amplitude');
figure(2);
plot(f,mx);
title('Magnitude Spectrum of a Square Pulse');
xlabel('Frequency (Hz)');
ylabel('Power');
OUTPUT:
References:
[1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, โSignals & Systemsโ, Second
edition, Pearson Education, 8th
Indian Reprint, 2005.
[2] M.J.Roberts, โSignals and Systems, Analysis using Transform methods and MATLABโ,
Second edition,McGraw-Hill Education,2011
[3] John R Buck, Michael M Daniel and Andrew C.Singer, โComputer explorations in Signals
and Systems using MATLABโ,Prentice Hall Signal Processing Series
[4] P Ramakrishna rao, โSignals and Systemsโ, Tata McGraw-Hill, 2008
[5] Tarun Kumar Rawat, โSignals and Systemsโ, Oxford University Press,2011
[6] A.Anand Kumar, โSignals and Systemsโ , PHI Learning Private Limited ,2011