1

The Fourier Series for Discrete-Time Signals

Suppose that we are given a periodic sequence Suppose that we are given a periodic sequence with period N. The Fourier series with period N. The Fourier series representation for x[n] consists of N representation for x[n] consists of N harmonically related exponential functionsharmonically related exponential functionseej2j2kn/Nkn/N, k = 0, 1,2,…….,N-1, k = 0, 1,2,…….,N-1and is expressed as and is expressed as

1N

0k

N/kn2jkec]n[x

where the coefficients ck can be computed as:

0n

N/kn2jk e]n[x

N

1c

2

Example 2: Determine the spectra of the following signals:(a) x[n] = [1, 1, 0, 0], x[n] is periodic with period 4 (b) x[n] = cosn/3(c) x[n] = cos(2)n

Solution: (a) x[n] = [1, 1, 0, 0]Solution: (a) x[n] = [1, 1, 0, 0]

3

0n

N/kn2j1N

0n

N/kn2jk e]n[x

4

1e]n[x

N

1c

Now 2

10011

4

1]3[x]2[x]1[x]0[x

4

1]n[x

4

1c

3

0n0

00e]1[x]0[x4

1e]n[x

4

1e]n[x

4

1c 2/j

3

0n

2/nj3

0n

4/n2j1

j14

1j01

4

1sinjcos11

4

122

3

j3

0n

nj3

0n

4/n22j2 e.11

4

1e]n[x

4

1e]n[x

4

1c

0sinjcos14

1

j14

1j01

4

1)2/3sin(j2/3cos1

4

1e]n[x

4

1c

3

0n

4/3n2j3

The magnitude spectra are:

2

1c0

4

2c1 0c2

4

2c3

and the phase spectra are:

00 41

undefined2 43

4

(b) x[n] = cosn/3

Solution: In this case, fSolution: In this case, f00 = 1/6 and hence x[n] is = 1/6 and hence x[n] is periodic with fundamental period N = 6.periodic with fundamental period N = 6.

NowNow

5

0n

3/knj6/kn2j5

0n

5

0n

N/kn2jk e

3

ncos

6

1e

3

ncos

6

1e]n[x

6

1c

5

0n

k1jk1j3/knj3/nj3/nj5

0n

3n

3n

ee12

1eee

2

1

6

1

0coscoscoscoscos0cos6

1

3

ncos

6

1

3

ncos2

12

1c

35

34

33

32

3

5

0n

5

0n0

Similarly, c2 = c3 = c4 = 0, c1 = c5 = ½.

5

(c) Cos(2)n

Solution: The frequency fSolution: The frequency f00 of the signal is of the signal is

1/1/2 Hz. Since f2 Hz. Since f00 is not a rational number, is not a rational number,

the signal is not periodic. Cosequently, this the signal is not periodic. Cosequently, this signal cannot be expanded in a Fourier signal cannot be expanded in a Fourier series.series.

6

Power density Spectrum of Periodic Signals

The average power of a discrete time periodic signal with period N is The average power of a discrete time periodic signal with period N is

1N

0n

2

x )n(xN

1P

The above relation may also be written as

1N

0n

1N

0n

N/kn2*k

1N

0n

*x ec]n[x

N

1]n[x]n[x

N

1P

or

1N

0n

221N

0kk

1N

0n

N/kn2j1N

0n

*kx

]n[xN

1c

e]n[xN

1cP

This is Parseval’s Theorem for Discrete-Time Power Signals.

7

Similarly, for discrete time energy signals, the Parseval’s Theorem may be stated as Similarly, for discrete time energy signals, the Parseval’s Theorem may be stated as follows:follows:

21N

0kk

21N

0nx cN]n[xE

If the signal x[n] is real, [i.e. x*[n] = x[n]], then we can easily show that

|c-k| = |ck| (even symmetry)-c-k = ck (odd symmetry)|ck| = |cN-k| (Periodicity)ck = cN-k (periodicity)

8

More specifically, we haveMore specifically, we have

|c|c00| = |c| = |cNN| | cc00 = - = - ccNN

|c|c11| = |c| = |cN-1N-1|| cc11 = - = - ccN-1N-1

|c|cN/2N/2| = |c| = |cN/2N/2| | ccN/2N/2 = 0 if N is even = 0 if N is even

|c|c(N-1)/2(N-1)/2| = |c| = |c(N+1)/2(N+1)/2| | cc(N-1)/2(N-1)/2 = = (N+1)/2(N+1)/2 if N if N

is is odd odd

9

Example: Determine the Fourier Series Coefficients and the Power Density Spectrum of the following periodic signal.

Solution:Solution:

-N L N

AX[n]

n

1N

0n

N/kn2j1N

0n

N/kn2jk Ae

N

1e]n[x

N

1c

k = 0, 1, 2, …., N-1

1N,...,2,1k,

0k,e

N

Ac

N/k2j

N/kL2j

e1e1

NA

NALn1L

0n

N/k2jk

10

But But

)N/ksin(

)N/kLsin(e

ee

ee

e

e

e1

e1

N/)1L(kj

N/kjN/kj

N/kLjN/kLj

N/kj

N/kLj

N/k2j

N/kL2j

Therefore,

otherwiseN/ksin

N/kLsin

0kc 2

2

NA

2

NAL

2

k

11

The Fourier Transform of Discrete-Time Aperiodic Signals

The Fourier Transform of a finite energy discrete The Fourier Transform of a finite energy discrete time signal x[n] is defined astime signal x[n] is defined as

n

jwne]n[x)w(X

X(w) may be regarded as a decomposition of x[n] into its Frequency components. It is not difficult toVerify that X(w) is periodic with frequency 2.

The Inverse Fourier Transform of X(w) may be defined as

dwe)w(X2

1]n[x

2

jwn

12

Energy Density Spectrum of Aperiodic Signals

Energy of a discrete time signal x[n] is defined asEnergy of a discrete time signal x[n] is defined as2

nx ]n[xE

Let us now express the energy Ex in terms of the spectral characteristic X(w). First we have

n n

jwn*x dwe)w(X

2

1]n[x]n[x]n[xE

If we interchange the order of integration and summation in the above equation, we obtain

dw)w(X2

1dwe]n[x)w(X

2

1E

2

n

jwnx

13

Therefore, the energy relation between x[n] and Therefore, the energy relation between x[n] and X(w) isX(w) is

dw)w(X2

1]n[xE

22

nx

This is Parseval’s relation for discrete-time aperiodic signals.

14

Example: Determine and sketch the energy density spectrum of the signal x[n] = anu[n], -1<a<1

Solution: Solution:

0njw

njwjwn

0n

n

n

jwn

ae1

1aeeae]n[x)w(X

The energy density spectrum (ESD) is given by

jwjw

2

xx ae1ae1

1)w(X)w(X)w(X)w(S

2awcosa21

1

0 w

X(w)

a = 0.5 a= -0.5

15

Example: Determine the Fourier Transform and the energy density spectrum of the sequence

Solution:Solution:

otherwise,0

1Ln0,A]n[x

)2/wsin(

)2/wLsin(Ae

e1

e1AAee]n[x)w(X )1L)(2/w(j

jw

jwL1L

0

jwn

n

jwn

The magnitude of x[n] is

otherwise,A

0w,LA)w(X

)2/wsin()2/wLsin(

and the phase spectrum is

)2/wsin()2/wLsin(

2w )2L(A)w(X

The signal x[n] and its magnitude is plotted on the next slide. The Phase spectrum is left as an exercise.

16

x[n]

|X(w)|

17

Properties of Discrete Time Fourier Transform (DTFT)

Symmetry Properties:Symmetry Properties: Suppose that both the signal x[n] and its Suppose that both the signal x[n] and its transform X(w) are complex valued. Thentransform X(w) are complex valued. Then

x[n] = xx[n] = xRR[n] + jx[n] + jxII[n] (1)[n] (1)

X(w) = XX(w) = XRR(w) + jX(w) + jXII[w] (2)[w] (2)

n

jwne]n[x)w(X

Substitution of (1) and (2) gives

wnsinjwncos]n[x]n[xe]n[x]n[x)w(jX)w(Xn

IRjwn

nIRIR

Separating the real and imaginary parts, we have