Convolution and Fourier Transform

Dr. Hariharan Muthusamy,

School of Mechatronic Engineering,

Universiti Malaysia Perlis.

Convolution Consider a linear system whose behaviour is specified by the impulse

response h(n). To find the output signal be y(n), resolve the signal x(n) into a weighted sum of impulses, and the linearity and time shift property of the LTI system.

The above expression gives the response y(n) of the LTI system as a function of the input signal x(n) and impulse response h(n) is called the convolution sum.

Steps involved in finding out the convolution sum

1. Folding: Fold the signal h(k) about the origin, ie at k=0;

2. Shifting: Shift h(-k) to the right by no,if no is positive or shift h(-k) to the left by no , if no is negative to obtain h(no – k).

3. Multiplication: Multiply x(k) by h(no – k) to obtain the product sequence

yo(k) = x(k) h(no – k)

k

k)x(k)h(ny(n)

4. Summation: Sum all the values of the product sequence yo(k) to obtain the value of the output at time n = no

Find the convolution of two finite duration sequences

otherwise

1n1

0

1x(n)

otherwise

1n1

0

1h(n)

2. Find the convolution of the two signals x(n) = u(n) and h(n) = anu(n), |a|<1, n>=0.

Fourier Transform When the repetition period T becomes infinity, the waveform f(t) become

non-periodic, the separation between two adjacent harmonics will be zero.

Assuming f(t) is initially periodic, we have

dtef(t)Tc)F(jn

Let

,ecTTf(t)

dtef(t)T1

c

where

,ecf(t)

T/2

T/2

tjnno

tjnn

T/2

T/2

tjnn

tjnn

o

o

o

o

ω

ω

ω

ω

ω

As T ->∞, the fundamental frequency becomes infinitely small, hence ωo -> dω

dtef(t))F(j

de)F(j21

f(t)

e)F(jn21

ecTT1

f(t)

tj

tj

otjn

o

tjnn

o

o

ω

ω

ω

ω

ω

ωωπ

ωωπ

Properties of Fourier TransformLinearity

FT obeys superposition and homogeneity principles. According to this property FT of the sum of two signals is equal to sum of FT of individual signals.

F[f1(t)]= F1(jω) F[f2(t)]= F2(jω)

F[a1f1(t)+b1f2(t)] = a1 F1(jω) + b1F2(jω)

Where a1 and b1 are constants.

Time Shifting Time Reversal

F[f(t)]= F(jω) F[f(t)]= F(jω)

F[f(t-t0) = e-jωt0 F(jω) F[f(-t)]= F(-jω)

Frequency Shifting

According the this property multiplication of the signal ejωot with f(t) shifts the frequency spectrum by ωo

F[f(t)]= F(jω)

F[f(t) ejωot ]= F[j(ω- ωo)]

Time Scaling:

F[f(t)]= F(jω) F[f(at)]= 1/|a| F(j ω/a), when a>1 f(at) is a compressed version of the signal f(t) by a factor a.

When a < 1, f(at) is an expanded version of the signal x(t) by a factor a

Differentiation in time

Differentiating a signal results in a multiplication of the Fourier

transform by jω, F[f(t)]= F(jω) then F[d/dt(f(t))]= jω F(jω)

Differentiation in FrequencyMultiplication of signal x(t) with time t results in differentiation of frequency spectrum F(jω)F[f(t)]= F(jω) then F[t f(t)]= j d/dω F(jω)

Time IntegrationIntegration of a signal results in a division of Fourier transform by jω. However to account for the dc or average value that can result from integration, we must add the term πF(0)δ(ω)F[f(t)] = F(jω), then

)F(j)(jdt

f(t)dF n

n

n

ωω

πF(0)δ(ω)F(jω(jω1

f(τ(τ)Ft

Conjugation F[f(t)]= X(jω), F[f*(t)] = F*(-jω)

Fourier transform of complex and real functionsf(t) =fR(t)+jfI(t)

Fourier transform of f(t) is given by

F[f(t)]=

Auto-correlationF[Rff(τ)] = F[f(t) * f*(-t)] = τff(jω) = |F(jω|2

DualityF[f(t)]= F(jω)

F[f(t)]= 2πf(-jω)

dtef(t))F(j tj

ωω

Convolution The convolution of two signals results in the multiplication of their Fourier

transforms in the frequency domain.

F[f(t)*h(t)]= F(jω) H(jω)

The output of a system can be obtained from the convolution of input signal and the system impulse response

τττ d))h(tf(y(t)

Fourier transform of Single Gate Function

otherwise

T/2tT/2for

0

1,f(t)

2T

Tsinc

2T2T

sinT.

ej

1

dt.e1

dtef(t))F(j

T/2

T/2tj

tjT/2

T/2

tj

ω

ω

ωω

ω

ω

ω

ω

The amplitude spectrum is shown in Fig. At ω = 0, sin c(ωT/2) = T, therefore F(jω) at ω = 0 is equal to T.

At ωT/2 = ± nπ, sin c(ωT/2) = 0,

Fourier Transform of Rectangular Pulse

otherwise

Tt0for

0

1,f(t)

2T2T

sineT

2jee2e

eej

e

ej

1

dt.e1

dtef(t))F(j

T/2j-

T/2jT/2jT/2j

T/2jT/2jT/2j

T

0tj

tjT

0

tj

ω

ω

ω

ω

ω

ω

ω

ωωω

ωωω

ω

ω

ω

2T

sincTe T/2j ωω

Fourier transform of a rectangular pulse 2 seconds long with a magnitude of 10 volts

ωωω

ω

ω

ω

ω

ω

ω

ω

ωωω

ωωω

ω

ω

ω

ω

csine20

sine20

2j

eee20

eej

e10

j

1e10

j

e10

dt0.e1

dtef(t))F(j

j

j

jjj-

jjj

j2-

tj

tjT/2

T/2

tj

Exponential Pulse

0 t for

0tfor

0

,ef(t)

-at

atan

a

1

ja1

dte

0tfor0f(t)since,dtee0

dtef(t)dtef(t)

dtef(t))F(j

1

22

0

t)j(a

0

tjat

0

tj0

tj

tj

ωω

ω

ω

ω

ω

ωω

ω

Fourier transform for the double exponential pulse

0 t for

0tfor

e

,ef(t)

at

-at

22

ω0

ω

ωω

ωω

ω

)(

ωω

ω

ajF a2

ja1

ja1

dtedte

dteedtee

dtef(t)dtef(t)

dtef(t))F(j

0

t)j(a

-

t)j(a

0

tjat0

tjat

0

tj0

tj

tj

Fourier transform of triangular pulse

f(t) = A (1+ 2t/T), -T/2<t<=0 f(t) = A(1-2t/T) 0<=t <=T/2

T/2

0

tj0

T/2

tj0

T/2

T/2

0

tjtj

tj0

T/2

T/2

0

tj

0

T/2

T/2

0

tjtj

T/2

T/2

tjtj

dtetT2A

dtetT2A

dteAdteA

dtetT2

1AdtetT2

1A

dtef(t)dtef(t)

dtef(t)dtef(t))F(j

ωωωω

ωω

ωω

ωωω

4ωTsin2

Tω4A

2ωTcos1

Tω4A

dtω

sinωiω

2ωTsin

2T

T4A

ω2

ωTsin2A

dt2tcosωtT2Adtcosωo2A

dtetdtetT2AdtedteAF ︵jω ︵

obtain weintegrals,thirdandfirstthein t-tot Changing

22

2

T/2

0

T/2

0

T/2

0

T/2

0

jωωT/2

0

jωωT/2

0

T/2

0

jωωjωω

Impulse Function (Unit Impulse)The impulse function, which has an infinite amplitude and is infinitely narrow. This is defined as δ(t) = 0 for all values except at t = 0The Fourier transform or the impulse function δ(t) is obtained as

The frequency spectrum of the impulse function δ(t) has a constant amplitude and extends over positive and negative frequencies.The inverse Fourier transform of the unit impulse is given by

1dte(t))F(j tj

ωω

)(21.

ωπ

ωωδπ

ωωδπ

ωω

Therefore21

f(t)

t21

d)(2

ede)(

21

f(t)tj

tj

Hence, the Fourier transform of the constant function is an impulse at the origin with an area equal to 2π

Signum Function

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

0tif

0tif

0tif

1

0

1

f(t)

ωωω

ωω

ω

ωω

ωω

ω

j2

j1

j1

je

je

dte(1)dte1)(

dtesgn(t))F(j

0

tj0tj

0

0

tjtj

tj

Unit Step Function

The unit step function is obtained by suddenly closing a switch of a DC circuit. For easier analysis, the waveform of the unit step function split into two component waveforms.

The first waveform is similar to the signum function with half amplitude. Therefore, the Fourier transform function is given by

F1(jω) = ½ (2/jω) = 1/jωThe second waveform is related to the unit impulse function and hence its

Fourier transform is given by

F2(jω) = ½ [2π δ(ω)] = πδ(ω)Therefore, the Fourier transform of the step function becomesF(jω) = F1(jω) + F2(jω)

=1/jω + πδ(ω)

Sinusoidal FunctionsThe Fourier transforms of the sinusoidal functions cos ωot and sin ωot are

obtained as given below.

F[cos ωot] =

Using the transform pair

F[cos ωot] =π [δ(ω – ωo) + δ(ω+ωo)]

Similarly

F[sin ωot] =π [δ(ω + ωo) + δ(ω - ωo)]

2ee

Ftjtj oo ωω

getwe),(2e otj o ωωπδω

FOURIER TRANSFORM OF POWER AND ENERGY SIGNALS The average power of a signal x(t) over a single period (t1, t1+T)

is given by

Where x(t) is a complex periodic signal.

A signal f(t) is called a power signal, if the average power expressed by

If x(t) is bounded, P∞ is finite. Every bounded and periodic signal is a power signal. But it is true that a power signal is not necessarily a bounded and periodic signal.

dtx(t)T1

P2Tt

tav

1

i

T

T

2

tx x(t)

2T1

LtP

Energy Signal

A signal x(t) is called an Energy signal, if its total energy over the interval (- ∞, ∞) is finite, that is

For a digital signal, the energy is defined by

As n ->∞, the energy of period signals becomes infinite, whereas the energy of aperiodic pulse-signals have a finite value.

dtx(t)LtE

T

T

2

Tx

2

n

x(n)E

Determine the signal energy and signal power for (a). f(t) = e-3|t|, (b). f(t) = e-3t

31

e62

dte2

dtedte

dteE

06t

0

6t

0

6t0

6t

2

|t|3

The signal power P∞ = 0, since E∞is finite. Hence, the signal f(t) is an energy signal

T

T

T

TT

e6

2

6T

T6t-

3t-

e61

dte

dteE

As T -> ∞, ET approaches infinity.

Its average power is

Hence, e-3t is neither an energy signal nor a power signal.

TeeLtE

TLtP

TT

TTT 821 66

Compute the signal energy for x(t) = e-4t u(t)

Find the Fourier transform of the signal f(t) shown in Fig.

f(t) = (A/T) t, for 0 < t < T, A , for T < t < 2T

Tj2T/2jT/2jT/2j2

Tj2Tj2

2T

T

tjT

0

2

tjtj

2T

T

tjT

0

tj-

eA

jeeeT

A

eA

j1eT

A

je

A)j(

ej

et

TA

dte AdtetTA

E

ωωωω

ωω

ωωω

ωω

ωω

ωω

ωωω

Tj2T/2j

Tj2T/2j

Tj2T/2j2

e2T

sincejA

eA

j2T

sinceA

j

eA

j2T

sineT2A

j

ωω

ωω

ωω

ωω

ωω

ω

ωω

ω

Obtain the Fourier transform of the trapezoidal pulse is shown in Fig.

f(t) = At/(tp – ta) + Atp/(tp – ta), for –tp < t < -ta

= A, for –ta < t < ta

= Atp/(tp – ta) - At/(tp – ta), for tp < t < -ta

Therefore,

p

a

p

a

a

p

a

p

a

p

t

t ap

tjt

t ap

tjp

t

t ap

tj

t

t ap

tjp

t

t ap

tj

dttt

Atedt

tt

etAdt

ttAte

dttt

eAtdt

ttAte

)F(j

ωωω

ωω

ω

ωω

ωωω

ωωω

ωωωω

ωωωω

ωωωωωω

ωωω

ωωωω

ωωω

ωω

jee

Aj

eett

At

ee1

ee1

jee

ttt

A

je

Aj

ej

ett

At

ej

teej

tett

A

dteAdtedtett

At

dtetdtettt

A

aaaa

ppaa

aa

a

a

p

a

a

p

p

a

a

p

a

a

a

p

p

a

a

p

p

a

tjtjtjtj

ap

p

tjtj

2tjtj

2

tjtj

aap

t

t

tjt

t

tjt

t

tj

ap

p

t

t2

tjtjt

t2

tjtj

ap

t

t

tjt

t

t

t

tjtj

ap

p

t

t

t

t

tjtj

ap

pa2

pa2ap

p

ap

pa

aaap

p

p2a2aa

ap

tcostcos2

tcostcos2

tt

t

tt

t1tsin

2A

t2sinA

t2sin1

tt

At

t2cos1

t2cos1

tsin2t

ttA

ωωω

ωωω

ωω

ωω

ωω

ωω

ωω

ωω