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

𝑋 𝜔 = 𝑥(𝑡)𝑒−𝑗𝜔𝑡 𝑑𝑡∞

−∞

This is called as analysis equation

The inverse Fourier transform is given by

𝑥(𝑡) = 𝑋 𝜔 𝑒𝑗𝜔𝑡 𝑑𝜔∞

−∞

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 Dirichlet‟s conditions are:

1. x t is absolutely integrable over the interval -∞ to +∞,that is

x(t) dt

∞

−∞

< ∞

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.

Dirichlet‟s 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 𝛅 𝐭

Givenx t = δ t ,

δ t = 1 for t = 00 for t ≠ 0

Then

X ω = x t e−jωtdt = δ t e−jωt dt ∞

−∞

∞

−∞ = e−jωt⎸t=0 = 1

∴ F δ t = 1 or δ t FT 1

Hence , the Fourier Transform of a unit impulse function is unity.

X ω = 1 for all ω

X ω = 0 for all ω

The impulse function with its magnitude and phase spectra are shown in below figure:

Similarly,

F δ t − to = δ t − to e−jωtdt = e−jωt0 i. e. δ(t − to)FT e−jωto

∞

−∞

2. Single Sided Real exponential function 𝐞−𝐚𝐭𝐮(𝐭)

Given x t = e−at u t , u(t) = 1 fort ≥ 00 fort < 0

Then

X ω = x t e−jωtdt = e−at u t e−jωtdt

∞

−∞

∞

−∞

= e−at e−jωtdt = e−(a+jω)tdt = e(− a +jω t

− a + jω

∞

0

∞

0 0

∞

= e−∞ − e0

−(a + jω)

=0 − 1

−(a + jω)=

1

a + jω

∴ F e−at u t =1

a+jω or e−at u t

FT

1

a+jω

Now, X ω =1

a+jω=

a−jω

a+jω (a−jω)

=a−jω

a2+ω2 = a

a2+ω2 − jω

a2+ω2 = 1

a2+ω2 – tan−1 ω

a

∴ X(ω) =1

a2 + ω2 , X ω = − tan−1

ω

a forallω

Figure shows the single-sided exponential function with its magnitude and phase spectra.

3.Double sided real exponential function 𝐞−𝐚 𝐭

Given x t = e−a t

∴ x t = e−a t = e−a −t =eat fort ≤ 0e−at = e−at fort ≥ 0

= e−a −t u −t + e−at u t

= eat u −t + e−at u(t)

X ω = x t e−jωtdt

∞

−∞

= eat e−jωtdt + e−at e−jωtdt = e a−jω tdt + e− a+jω tdt

∞

0

0

−∞

∞

0

0

−∞

= e− a−jω tdt + e− a+jω tdt = ∞

0

∞

0

e− a−jω t

−(a−jω)

0

∞

+ e− a +jω t

−(a+jω)

0

∞

= e−∞ −e−0

−(a−jω)+

e−∞ −e−0

−(a+jω)=

1

a−jω+

1

a+jω=

2a

a2+ω2

∴ F e−a t = 2a

a2 + ω2ore−a t

FT

2a

a2 + ω2

∴ X ω = 2a

a2 + ω2forallω

And X ω = 0 forallω

A Two sided exponential function and its amplitude and phase spectra are shown in figures

below:

4. Complex Exponential Function𝐞𝐣𝛚𝟎𝐭 :

To find the Fourier Transform of complex exponential function ejω0t , consider finding the

inverse Fourier transform of δ(ω − ω0). Let

X ω = δ(ω − ω0)

∴ x t = F−1 X ω = F−1[δ(ω − ω0)] =1

2π X ω ejωtdω

∞

−∞

=1

2π δ(ω − ω0)ejωtdω

∞

−∞=

1

2πejω0t

∴ F−1 δ ω − ω0 =ejω0t

2πorF−1 2πδ ω − ω0 = ejω0t

= F ejω0t = 2πδ(ω − ω0)

Or ejω0tFT 2πδ(ω − ω0)

5. Constant Amplitude (1)

Let x t = 1 − ∞ ≤ t ≤ ∞

The waveform of a constant function is shown in below figure .Let us consider a small section of

constant function, say, of duration 𝜏.If we extend the small duration to infinity, we will get back

the original function.Therefore

x t = Lt t→∞

[rect t

τ ]

Where rect t

τ =

1 for−τ

2≤ t ≤

τ

2

0 elesewhere

By definition, the Fourier transform of x(t) is:

X(ω) = F[x(t)] = F Lt t→∞

rect t

τ = Lt

t→∞F rect

t

τ

= Lt t→∞

1 τ 2

−τ 2 e−jωt dt = Lt

t→∞

e−jω t

−jω −τ 2

τ 2

= Lt t→∞

e−jω (τ 2 )−ejω (τ 2 )

−jω = Lt

t→∞

2sin [ω τ

2 ]

ω = Lt

t→∞ τ

sin [ω τ

2 ]

ω(τ

2)

= Lt t→∞

τ sa(ωτ

2) = 2π Lt

t→∞

τ 2

πsa(

ωτ

2)

Using the sampling property of the delta function i. e. Lt t→∞

τ 2

πsa(

ωτ

2) = δ(ω) , we get

X(ω) = F Lt t→∞

rect t

τ = 2πδ ω

6.Signum function sgn(t)

The signum function is denoted by sgn(t) and is defined by

sgn(t) = 1 for t > 0

−1 for t < 0

This function is not absolutely integrable. So we cannot directly find its Fourier transform.

Therefore, let us consider the function e−a⎹t⎸sgn(t) and substitute the limit a→0 to obtain the

above sgn(t)

Given x(t) = sgn(t) = Lt a→0

e−a⎹t⎸ sgn(t) = Lt a→0

[ e−at u t − e−at u −t

∴ X(ω) = F[sgn(t)] = Lt a→0

[ e−at u t − e−at u −t e−jωt∞

−∞dt

= Lt a→0

e−at e−jωtu t dt − eat e−jωtu −t dt ∞

−∞

∞

−∞

= Lt a→0

e−(a+jω)tdt − e(a−jω)tdt 0

−∞

∞

0 = Lt

a→0 e−(a+jω)tdt − e−(a−jω)tdt

∞

0

∞

0

= Lt a→0

e−(a +jω )t

−(a+jω)

0

∞

− e−(a−jω )t

−(a−jω)

0

∞

= Lt a→0

1

a+jω−

1

a−jω =

1

𝑗𝜔−

1

−𝑗𝜔 =

2

𝑗𝜔

F[sgn(t)] = 2

𝑗𝜔

sgn(t)𝐹𝑇

2

𝑗𝜔

∴ ⎹X(⍵)⎸ = 2

𝜔 and 𝑋(⍵) =

𝜋

2𝑓𝑜𝑟 𝜔 < 0 𝑎𝑛𝑑 −

𝜋

2𝑓𝑜𝑟 𝜔 > 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 ≥ 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 𝑠𝑔𝑛 𝑡

x(t)= u(t) = 1

2[1 + 𝑠𝑔𝑛 𝑡 ]

X(𝜔) = F[u(t)] = F 1

2[1 + 𝑠𝑔𝑛 𝑡 ]

= 1

2 𝐹 1 + 𝐹[𝑠𝑔𝑛 𝑡 ]

We know that F[1] = 2𝜋𝛿(𝜔) and F[sgn(t)] = 2

𝑗𝜔

F[u(t)]= 1

2 2𝜋𝛿 𝜔 +

2

𝑗𝜔 = 𝜋𝛿 𝜔 +

1

𝑗𝜔

u(t)𝐹𝑇 𝜋𝛿 𝜔 +

1

𝑗𝜔

∴ ⎹X(⍵)⎸=∞ at ⍵=0 and is equal to 0 at ⍵=−∞ and ⍵=∞

Figure shows the unit step function and its spectrum.

8. Rectangular pulse ( Gate pulse) 𝐭

𝛕 or rect

𝐭

𝛕

Consider a rectangular pulse as shown in below figure. This is called a unit gate function and is

defined as

x(t) = rect t

τ =

t

τ =

1 𝑓𝑜𝑟⎹ t⎸ <τ

2

0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

Then X(⍵) = F[ x(t)] = F 𝑡

𝜏 =

t

τ

∞

−∞𝑒−𝑗𝜔𝑡 𝑑𝑡

= 1 𝑒−𝑗𝜔𝑡 𝑑𝑡𝜏 2

−𝜏 2 =

𝑒−𝑗𝜔𝑡

−𝑗𝜔 𝜏 2

𝜏 2

= 𝑒−𝑗𝜔 (𝜏 2) − 𝑒 𝑗𝜔 (𝜏 2)

−𝑗𝜔

= 𝜏

𝜔(𝜏 2) 𝑒 𝑗𝜔 (𝜏 2) – 𝑒−𝑗𝜔 (𝜏 2)

2𝑗 = 𝜏

sin 𝜔(𝜏 2)

𝜔(𝜏 2)

= 𝜏 𝑠𝑖𝑛𝑐 𝜔(𝜏 2)

∴ F 𝑡

𝜏 = 𝜏 𝑠𝑖𝑛𝑐 𝜔(𝜏 2) , that is

rect t

τ =

t

τ

𝐹𝑇 𝜏 𝑠𝑖𝑛𝑐 𝜔(𝜏 2)

Figure shows the spectra of the gate function

The amplitude spectrum is obtained as follows:

At 𝜔 = 0, sinc(𝜔𝜏/2)=1. Therefore, ⎹X(𝜔)⎸ 𝑎𝑡 𝜔 = 0 is equal to 𝜏. At (𝜔𝜏/2)=±𝑛𝜋, i.e.

at

𝜔 = ±2𝑛𝜋

𝜋 , n = 1, 2, ……sinc(𝜔𝜏/2) =0

The phase spectrum is: 𝑋(𝜔) = 0 if sinc(𝜔𝜏/2)> 0

= ±𝜋 if sinc(𝜔𝜏/2) < 0

The amplitude response between the first two zero crossings is known as main lobe and the

portions of the response for 𝜔 < − 2𝜋

𝜏 and 𝜔 >

2𝜋

𝜏 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 𝜔 = 2𝜋

𝜏 or at f =

1

𝜏 Hz. As the width of the rectangular

pulse is made longer, the main lobe becomes narrower. The phase spectrum is odd function of 𝜔.

If the amplitude spectrum is positive, then phase is zero, and if the amplitude spectrum is

negative, then the phase is – 𝜋 𝑜𝑟 𝜋.

9. Triangular Pulse ∆ t

τ

Consider the triangular pulse as shown in below figure. It is defined as:

x(t) = ∆ t

τ =

1

τ 2 t +

τ

2 = 1 + 2

t

τ for −

τ

2< 𝑡 < 0

1

τ 2 t −

τ

2 = 1 − 2

t

τ for 0 < 𝑡 <

τ

2

0 elsewhere

i.e. as x(t) = ∆ t

τ = 1 −

2⎹t⎸

τ for ⎹t⎸ <

τ

2

0 otherwise

Then X(⍵) = F[ x(t)] = F ∆ t

τ = ∆

t

τ

∞

−∞𝑒−𝑗𝜔𝑡 𝑑𝑡

= 1 +2t

τ

0

−𝜏 2 𝑒−𝑗𝜔𝑡 𝑑𝑡 + 1 −

2t

τ

𝜏 2

0𝑒−𝑗𝜔𝑡 𝑑𝑡

= 1 −2t

τ

𝜏 2

0𝑒𝑗𝜔𝑡 𝑑𝑡 + 1 −

2t

τ

𝜏 2

0𝑒−𝑗𝜔𝑡 𝑑𝑡

= 𝑒𝑗𝜔𝑡𝜏 2

0𝑑𝑡 −

2t

τ

𝜏 2

0𝑒𝑗𝜔𝑡 𝑑𝑡 + 𝑒−𝑗𝜔𝑡𝜏 2

0𝑑𝑡 −

2t

τ

𝜏 2

0𝑒−𝑗𝜔𝑡 𝑑𝑡

= [𝑒𝑗𝜔𝑡 +𝜏 2

0𝑒−𝑗𝜔𝑡 ]𝑑𝑡 −

2

τ 𝑡[𝑒𝑗𝜔𝑡 +

𝜏 2

0𝑒−𝑗𝜔𝑡 ]𝑑𝑡

= 2 cos 𝜔𝑡𝜏 2

0𝑑𝑡 −

2

τ 2𝑡 cos 𝜔𝑡

𝜏 2

0𝑑𝑡

= 2 sin 𝜔𝑡

𝜔

0

𝜏 2

− 4

𝜏 𝑡

sin 𝜔𝑡

𝜔

0

𝜏 2

+ cos 𝜔𝑡

𝜔2 0

𝜏 2

= 2

𝜔 sin 𝜔

𝜏

2 −

4

𝜔𝜏 𝜏

2sin

𝜔𝜏

2 −

4

𝜔2𝜏 cos

𝜔𝜏

2− 1

=4

𝜔2𝜏 1 − cos

𝜔𝜏

2 =

4

𝜔2𝜏 2 𝑠𝑖𝑛2 𝜔𝜏

4

= 8

𝜔2𝜏 𝜔𝜏

4

2 sin 2 𝜔𝜏

4

𝜔𝜏

4

= 𝜏

2𝑠𝑖𝑛𝑐2

𝜔𝜏

4

F ∆ t

τ =

𝜏

2𝑠𝑖𝑛𝑐2

𝜔𝜏

4

Or ∆ t

τ

𝐹𝑇

𝜏

2𝑠𝑖𝑛𝑐2

𝜔𝜏

4

Figure shows the amplitude spectrum of a triangular pulse.

10. Cosine wave cos ⍵0𝑡

Given x(t) = cos⍵0𝑡

Then X(⍵) = F[ x(t)] = F[cos⍵0𝑡] = F 1

2 𝑒𝑗⍵0𝑡 + 𝑒−𝑗⍵0𝑡

= 1

2 𝐹 𝑒𝑗⍵0𝑡 + 𝐹 𝑒−𝑗⍵0𝑡 =

1

2 2𝜋𝛿 ⍵ − ⍵𝑜 + 2𝜋𝛿 ⍵ + ⍵𝑜

= 𝜋 𝛿 ⍵ − ⍵𝑜 + 𝛿 ⍵ + ⍵𝑜

∴ F[cos ⍵0𝑡] = 𝜋 𝛿 ⍵ − ⍵𝑜 + 𝛿 ⍵ + ⍵𝑜 or cos ⍵0𝑡𝐹𝑇 𝜋 𝛿 ⍵ − ⍵𝑜 + 𝛿 ⍵ + ⍵𝑜

Below Figure shows the cosine wave and its amplitude and phase spectra.

11. Sine wave sin ⍵0𝑡

Given x(t) = sin⍵0𝑡

Then X(⍵) = F[ x(t)] = F[sin⍵0𝑡] = F 1

2𝑗 𝑒𝑗⍵0𝑡 − 𝑒−𝑗⍵0𝑡

= 1

2𝑗 𝐹 𝑒𝑗⍵0𝑡 − 𝐹 𝑒−𝑗⍵0𝑡 =

1

2𝑗 2𝜋𝛿 ⍵ − ⍵𝑜 − 2𝜋𝛿 ⍵ + ⍵𝑜

= −𝑗𝜋 𝛿 ⍵ − ⍵𝑜 − 𝛿 ⍵ + ⍵𝑜

∴ F[cos⍵0𝑡] = −𝑗𝜋 𝛿 ⍵ − ⍵𝑜 − 𝛿 ⍵ + ⍵𝑜 or cos ⍵0𝑡𝐹𝑇 −𝑗𝜋 𝛿 ⍵ − ⍵𝑜 − 𝛿 ⍵ + ⍵𝑜

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) = 𝐶𝑛 𝑒𝑗𝑛 ⍵0𝑡∞

𝑛=−∞

The Fourier transform of x(t) is:

X(⍵) = F[x(t)] = F 𝐶𝑛 𝑒𝑗𝑛 ⍵0𝑡∞

𝑛=∞

= 𝐶𝑛 𝐹 𝑒𝑗𝑛 ⍵0𝑡 ∞𝑛=∞

Using the frequency shifting theorem, we have

𝐹 1𝑒𝑗𝑛 ⍵0𝑡 = 𝐹 1 ⎹⍵=⍵−𝑛⍵0 = s𝜋𝛿 ⍵ − 𝑛⍵0

X(⍵) = 2𝜋 𝐶𝑛 𝛿 ⍵ − 𝑛⍵0 ∞𝑛=∞

Where 𝐶𝑛 𝑠 are the Fourier coefficients associated with x(t) and are given by

𝐶𝑛 = 1

𝑇 𝑥(𝑡)

𝑇 2

−𝑇 2 𝑒−𝑗𝑛 ⍵0𝑡𝑑𝑡

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 2𝜋𝐶𝑛 .

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 = ∞∞

−∞. Therefore, Fourier

transform of x(t) = e3tu(t) does not exist.

Problem 2: Find the Fourier transform of the signals cos ωotu(t)

Solution:

Given x(t) = cos ωot u(t)

i.e. = ejωo t +e−jωo t

2 u(t)

∴ X(⍵) = F[cos ωot u(t)] = ejωo t +e−jωo t

2 u(t)

∞

−∞e−jωt dt

= 1

2 e−j(ω−ωo )t∞

0dt + e−j(ω+ωo )t∞

0 dt

= 1

2

e−j(ω−ωo )t

−j(ω−ωo )+

e−j(ω+ωo )t

−j(ω+ωo )

0

∞

= 1

2

−e0

−j(ω−ωo )+

−e0

−j(ω+ωo )

With impulses of strength π at ω = ωo and ω = −ωo

∴ X(⍵) = 1

2

1

j(ω−ωo )+

1

j(ω+ωo )+ πδ ω − ωo + πδ(ω + ωo)

= 1

2

j2ω

(jω)2+ ωo2 + πδ ω − ωo + πδ(ω + ωo)

= jω

(jω)2+ ωo2 +

1

2 πδ ω − ωo + πδ(ω + ωo)

Problem 3: Find the Fourier transform of the signals sin ωot u(t)

Solution:

Given x(t) = sin ωot u(t)

i.e. = ejωo t− e−jωo t

2j u(t)

∴ X(⍵) = F[sin ωot u(t)] = ejωo t−e−jωo t

2j u(t)

∞

−∞e−jωt dt

= 1

2j e−j(ω−ωo )t∞

0dt − e−j(ω+ωo )t∞

0 dt

= 1

2j

e−j(ω−ωo )t

−j(ω−ωo )−

e−j(ω+ωo )t

−j(ω+ωo )

0

∞

= 1

2j

−e0

−j(ω−ωo )−

−e0

−j(ω+ωo )

With impulses of strength π at ω = ωo and ω = −ωo

∴ X(⍵) = 1

2j

1

j(ω−ωo )−

1

j(ω+ωo )+ πδ ω − ωo − πδ(ω + ωo)

= 1

2j

j2ωo

(jω)2+ ωo2 + πδ ω − ωo − πδ(ω + ωo)

= ωo

(jω)2+ ωo2 − j

π

2 δ ω − ωo + δ(ω + ωo)

Problem 4: Find the Fourier transform of the signals e−tsin 5t u(t)

Solution:

Given x(t) = e−tsin 5t u(t)

x(t) = e−t ej5t− e−j5t

2j u(t)

∴ X(⍵) = F[e−t sin 5t u(t)] = F e−t ej5t− e−j5t

2j u(t)

= 1

2j [

∞

−∞e−t(ej5t − e−j5t)u(t)] e−jωt dt

= 1

2j

e−[1+j ω−5 ]t

−[1+j ω−5 ]−

e−[1+j ω+5 ]t

−[1+j ω+5 ]

0

∞

= 1

2j

1

[1+j ω−5 ]−

1

[1+j ω+5 ]

= 5

[1+j ω−5 ][1+j ω+5 ] =

5

(1+jω)2 + 25 [neglecting impulses]

Problem 5: Find the Fourier transform of the signals e−2tcos 5t u(t)

Solution:

Given x(t) = e−2tcos 5t u(t)

x(t) = e−2t ej5t + e−j5t

2 u(t)

∴ X(⍵) = F[e−2t cos 5t u(t)] = F e−2t ej5t− e−j5t

2 u(t)

= 1

2 [

∞

−∞e−2t(ej5t − e−j5t)u(t)] e−jωt dt

= 1

2[ e− 2+j ω−5 t∞

0 dt + e− 2+j ω+5 t∞

0 dt ]

= 1

2

e−[2+j ω−5 ]t

−[2+j ω−5 ]−

e−[2+j ω+5 ]t

−[2+j ω+5 ]

0

∞

= 1

2

1

[1+j ω−5 ]−

1

[1+j ω+5 ]

= 1

2

2(2+jω)

(2+jω)2 + 25 =

2+jω

(2+jω)2 + 25 [neglecting impulses]

Problem 6: Find the Fourier transform of the signals e−⎹t⎸sin 5⎹t ⎸for all t

Solution:

Given x(t) = e−⎹t⎸sin 5⎹t ⎸for all t

i.e. x(t) = 𝑒𝑡𝑠𝑖𝑛5 −𝑡 = −𝑒𝑡𝑠𝑖𝑛5𝑡 𝑓𝑜𝑟 𝑡 < 0

𝑒−𝑡𝑠𝑖𝑛5 𝑡 = 𝑒𝑡𝑠𝑖𝑛5𝑡 𝑓𝑜𝑟 𝑡 > 0

i.e. x(t) = −𝑒𝑡𝑠𝑖𝑛5𝑡 𝑢(−𝑡) + 𝑒𝑡𝑠𝑖𝑛5𝑡 𝑢(𝑡)

∴X(𝜔) = 1

2j [−

∞

−∞et ej5t − e−j5t u −t + e−t(ej5t − e−j5t)u(t)] e−jωt dt

= −1

2j [

0

−∞e[1−j ω−5 ]t − e[1−j ω+5 ]t]dt +

1

2j [

∞

0e−[1+j ω−5 ]t − e−[1+j ω+5 ]t]dt

= −1

2j [

∞

0e−[1−j ω−5 ]t − e−[1−j ω+5 ]t]dt +

1

2j [

∞

0e−[1+j ω−5 ]t − e−[1+j ω+5 ]t]dt

= −1

2j

e−[1−j ω−5 ]t

−[1−j ω−5 ]−

e−[1−j ω+5 ]t

−[1−j ω+5 ]

0

∞

+ 1

2j

e−[1+j ω−5 ]t

−[1+j ω−5 ]−

e−[1+j ω+5 ]t

−[1+j ω+5 ]

0

∞

= −1

2j

1

[1−j ω−5 ]−

1

[1−j ω+5 ] +

1

2j

1

[1+j ω−5 ]−

1

[1+j ω+5 ]

= 5

(1−jω)2 + 25+

5

(1+jω)2 + 25 [neglecting impulses]

Problem 7: Find the Fourier transform of the signals eat u(-t)

Solution:

Given x(t) = eat u(-t)

∴ X(𝜔) = F[eat u(-t)] = eat u(−t)∞

−∞e−jωt dt

= e(a−jω)t0

−∞ dt = e−(a−jω)t∞

0 dt =

e−(a−jω )t

−(a−jω)

0

∞

= 1

a−jω

Problem 8: Find the Fourier transform of the signals teat u(t)

Solution:

Given x(t) = teat u(t)

∴ X(𝜔) = F[teat u(t)] = teat u(t)∞

−∞e−jωt dt

= te−(a+jω)t∞

−∞ dt = 𝑡

e−(a +jω )t

−(a+jω)

0

∞

− e−(a +jω )t

−(a+jω)

∞

0 dt = 𝑡

e−(a +jω )t

−(a+jω)

0

∞

− e−(a +jω )t

(a+jω)2 0

∞

=

1

(a+jω)2

Problem 9: Find the inverse Fourier transform of X(𝜔) = 𝑗𝜔

(2+𝑗𝜔 )2

Solution:

We know that F[te−at u(t)] = 1

(𝑎+𝑗𝜔 )2

∴ F[te−2t u(t)] = 1

(2+𝑗𝜔 )2

Let te−2t u(t) = x1(t)

Then 1

(2+𝑗𝜔 )2 = X1(𝜔)

Using differentiation in time domain property [ i.e. 𝑑

𝑑𝑡x(t)

𝐹𝑇 j𝜔X(𝜔)], we have

F 𝑑

𝑑𝑡𝑥1(𝑡) = j𝜔𝑋1(𝜔)

𝐹−1 j𝜔𝑋1(𝜔) = 𝑑

𝑑𝑡 𝑥1(𝑡) =

𝑑

𝑑𝑡 te−2t u(t)

Problem 10: Find the Fourier transform of the Gaussian signal x(t) = 𝑒−𝑎𝑡2

Solution:

Given x(t) = 𝑒−𝑎𝑡2

The Fourier transform of the given signal is:

X(𝜔) = F[𝑒−𝑎𝑡2] = 𝑒−𝑎𝑡2

𝑒−𝑗𝜔𝑡∞

−∞dt = 𝑒−(𝑎𝑡2+𝑗𝜔𝑡 )∞

−∞dt = 𝑒−𝜔2/4𝑎 𝑒

−[𝑡 𝑎+ 𝑗𝜔

2 𝑎 ]2∞

−∞dt

Let p = 𝑡 𝑎 + 𝑗𝜔

2 𝑎

∴ dp = 𝑎 dt

∴ X(𝜔) = 𝑒−𝜔 2/4𝑎

𝑎 𝑒−𝑝2∞

−∞dp

= 𝑒−𝜔 2/4𝑎

𝑎 𝜋 ∵ 𝑒−𝑝2∞

−∞dp = 𝜋

= 𝜋

𝑎𝑒−𝜔2/4𝑎

∴ F[𝑒−𝑎𝑡2] =

𝜋

𝑎𝑒−𝜔2/4𝑎 or 𝑒−𝑎𝑡2 𝐹𝑇

𝜋

𝑎𝑒−𝜔2/4𝑎

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) e−at u(t) (b) te−2tu(t) (c) ej2tu(t)

2. Find the Fourier transform of the following:

(a) ej2tcos𝜔𝑜𝑡 u(t) (b)ej2tsin𝜔𝑜𝑡 u(t)

3. Find the Fourier transform of the Gaussian modulated signal x(t) =𝑒−𝑎𝑡2cos𝜔𝑐 t

4. Find the Fourier transform of the signal x(t) = 𝑒−𝑎⎹𝑡⎸sgn(t).

5. Find the inverse Fourier transform of X(𝜔) = 𝑗𝜔

(3+𝑗𝜔 )2

6. Find the inverse Fourier transform of X(𝜔) = 𝑒−4𝜔𝑈(𝜔)

7. Find the Fourier transform of x(t-4)+x(t+4).

8. Find the Fourier transform of 𝛿 𝑡 + 4 + 𝛿 𝑡 + 2 + 𝛿 𝑡 − 2 + 𝛿 𝑡 + 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 „fft‟ is used to find the fourier transform of discrete samples of any signal x(n) and

its magnitude is plotted using the function „abs‟. 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, “Signals & Systems”, Second

edition, Pearson Education, 8th

Indian Reprint, 2005.

[2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”,

Second edition,McGraw-Hill Education,2011

[3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals

and Systems using MATLAB”,Prentice Hall Signal Processing Series

[4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008

[5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011

[6] A.Anand Kumar, “Signals and Systems” , PHI Learning Private Limited ,2011