Fourier Transform of a Periodic Signal Properties of the Continuous time Fourier Transform

Properties of the Continuous time Fourier Transform Fourier Transform of a Periodic Signal ?

Fourier Transform of a Periodic Signal

let the signal x(t) be periodic with period to. such signal can be expressed by exponential fourier series as

x(t) = x(k) ejkt          [here,  = 2/to]

where,   x(k) = 1/t   x(t) e-jkt dt

taking fourier transform on both sides,

x(0) = f [x (k) ejkt  dt]

x(0) = f{x (k) ejkt}

x(0) = x(k) f {ejkt},

x(0) = x(k) 2 (0 – ko)

x(0) = 2 x(k) (0 – ko)

Properties of the Continuous time Fourier Transform

x(t) = 1/2   x(j)ejt d

x(j) = x(t) e-jt dt

Inverse Fourier Transform

The inverse fourier transform can be obtained by the basic definition. it can also be obtained with the help of partial fraction expansions. following fourier transform pairs are used for partial fraction expansions.

e-at u(t) = 1/a + j

te-at  u(t)  = 1/(a + j)2

Example 5.  Determine inverse fourier transform of

(i) x(j) = 2j + 1/(j + 2)2 by partial fraction expansions.

(ii) x(j) = 1/(a + j)2 by convolution property.

Sol.   (i)  x(j) =  a/jt2 +  b/(j + 2)2

x(j) = 2/(j + 2) – 3/(j + 2)2

taking inverse FT using standard FT pairs,

x(t) = 2e-2t  u(t) – 3t e-2t u(t)

(ii) x(j) = 1/(a + j)2 = 1/(a + j)(a + j)

= x1(j). x2(j)

here, x1 (j) = 1/a + j and x2 (j) = 1/a + j

x1(t) = e-at u(t) and x2(t) e-at u(t)

using convolution property,

x(t) = x1(t)* x2(t)

= [e-a u. e-a(t-) u(t-) d

here, u = 0 for < 0

and    u(t-) = 1 for < t

x(t) = [e-a . e-a(t-) d

x(t) = e-a. e-at. ea d

x(t) = e-at  d = e-at

x(t) = te-at u(t)

te-at  u(t) = 1/(a + j)2

The Discrete-Time Fourier Transform

In the previous chapter, we used the time domain representation of the signal. given any signal {x[n]} we can write it as linear combination of basic signals {[n – k}]. Another representation of signals that has been found very useful is frequency domain representation. in the mid 1960s as algorithm for calculation of the fourier transform was discovered, known as the fast fourier transform (FFT) algorithm. this spurred the development of digital signal processing in many areas.

The fourier representation of signals derives its importance from the fact that exponential signals are eigen functions for the discrete time LTI systems. what we mean by this is that if {zn} is input signal to an LTI system then output is given by H(z) {zn}. let us consider an LTI system with impulse response {h[n]}. then output is given by

y[n] = h[k] x [n – k]

h[k] zn – k

= zn   h[k] z-k

= h (z) zn

where, h(z) = h[k] z-k assuming that the summation in right-hand side converges. thus, output is same exponential sequence multiplied by a constant that depends on the value of z.

The constant h(z) for a specified value of z is the eigen value associated with eigen function (zn).

In the analysis of LTI system, the usefulness of decomposing a more general signal in terms of eigen functions can be seen from the following example. let {x[n]} correspond to a linear combination of two exponential

{x[n]} = a1 {z1n} + a2 {z2n}

From the eigen function property and superposition property, the response {y[n]} is given by

{y[n]} = a1 H (z1){z1n} + a2 H (z2) {z2n}

more general if

{x[n]} = ak {zkn}

then, {y[n]} = an H(zn) {zkn}

Thus, if input signal can be represented by a linear combination of exponential signals, the output can also be represented by a linear combination of same exponentials, moreover the coefficients of the linear combination in the output is obtained by multiplying ak, the coefficient in the input representation by corresponding eigen value H(zk). the procedure outlined above is useful, if we can represent a large class of signals in terms of complex exponentials. in this chapter, we will consider representation of aperiodic signals in terms of signal {ejon}.

Representation of Aperiodic Signals : The Discrete Time Fourier Transform (DTFT)

Here, we take the exponential signals to be {ejon}, where o is a real number. the representation is motivated by the harmonic analysis, but instead of following the historical development of the representation we give directly the definig equation.

let {x[n]} be discrete time signal such that |x[n]| < 0 that is {x[n]} sequence is absolutely summable.

The sequence {[n]} can be represented by a fourier and integral of the form

x[n] = 1/2 [x (ejo) ejon d …………(1)

x (ejo) = x[n] e-jon …………………….(2)

Eqs. (1) and (2) give the fourier representation of the signal {x[n]}. eq. (1) is referred as synthesis equation or the inverse discrete time fourier transform (IDTFT) and eq. (2) is fourier transform in the analysis equation.

Fourier transform of a signal is general in a complex valued function, we can write

x(ejo) = xr (ejo) + j x1 (ejo) …………..(3)

where, xr (ejo) is the real part of x(ejo) and x1 (ejo) is imaginary part of the function x(ejo). we can also use a polar form

x(ejo) = |x(ejo)| ex (ejo) ………………….(4)

where, |x(ejo)| is magnitude and x(ejo) is the phase of x(ejo). we also use the term fourier spectrum or simply, the spectrum to refer to x(ejo). thus, |x(ejo)| is called the magnitude spectrum and x(ejo) is called the phase spectrum.

From eq. (2) we can see that x(ejo) is a periodic function with period 2, i.e., x (ejo+2) = x (ejo) we can interpret (1) as fourier coefficients in the representation of the periodic function x(ejo). in the fourier series analysis our attention is on the periodic function, here we are concerned with the representation of the signal {x[n]}. so, the roles of the two equation are interchanged compared to the fourier series analysis of periodic signals.

Now, we show that if we put eq. (2) in eq. (1) we indeed get the signal {x[n]}.

let,  x[n] = 1/2  (x[m]e-jom) e+jom d

where, we have substituted x (ejo) from eq. (2) into eq.

(1) and called the result as x[n].

Since, we have need n as index on the left hand side we have used m as the index variable for the sum defining the fourier transform. under our assumption that {x[n]} sequence is absolutely summable we can interchange the order of integration and summation. thus,

x[n] = x[m] = (1/2 e+jo(n – m) d)…………………….(5)

the integral with the parenthesis can be evaluated as

if m = n then ejo(n-m) = 1

1/2  1 . d = 1

if m = n then ejo(n – m) = cos (n – m) + j sin (n – m)

1/2  ejo(n – m) do = 1/2  cos (n – m) d + j/2  sin (n – m) d

= 1 sin (n – m)/2 (n – m)   j cos (n – m)/2   n – m

= 0

Thus, in eq. (5) there is only one non-zero term in RHS, corresponding to m = n and we get x[n] = x[n]. this result is true for all values of n and so eq. (1) is indeed a representation of signal {x[n]} in terms of eigen functions {ejom).

In above demonstration we have saaumed that {x[n]} is absolutely summable. determining the class of signals which cas be represented by eq. (1) is equivalent to considering the convergence of the infinite sum in equation (2). if we fix a value of then, RHS of equation (2) is a complex valued series whose partial sum is given by

Xn (ejo) = x[n] e-jon

The limit as n – o. if the partial sum xn (ejo0) exists if the series is absolutely summable.

|xn (ejo 0) |=| x[n] e-jo n|

<  |x[n]e-jo n |by triangle inequality

= |x[n]|

since, the limit |x[n]| exists by our assumption the limit n – xn (ejo 0) exists for every. furthermore it can be shown that the series converges uniformly to a continuous function of .

If a sequence has only finitely many non-zero terms then it is absolutely summable and so the fourier transform exists. since, a stable sequence is by definition, an absolutely summable sequence, its fourier transform also exists.

for example, let {x[n]} = {an u[n]}

fourier transform of this sequence will exist if it is absolutely summable. we have

|x[n]| = |a|n

This is a geometric series and sum exists if |a|<1 in that case

|a|n = 1/1 – |a| < + 0

Thus, the fourier transform of the sequence {an u[n]} exists, if |a| < 1. the fourier transform is

x(ejo)  e-jon

= (ae-jo)n

= 1/1 – ae-jo ………………….(6)

where, the last equality follows from sum of a geometric series, which exists if |ae-jon | < 1, i.e., |a| < 1.

Absolute summability is a sufficient condition for the existence of a fourier transform. fourier transform also exists foe square summable sequence.

|x[n]|2 < 0

For such signals the convergence is not uniform. this has implications in the design of discrete system for filtering.

We also deal with signals that are neither so absolutely summable nor square summable. to deal with some of these signals we allow impulse function as a fourier transform. the impulse function is defined by the following properties

(a)  (0) d = 1

(b)  x (ejo) (0 – 00) do = x (ejo 0), if x (ejo) is continuous at o = oo ; (shifting or convolution property)

(c) x (ejo) (0) = x (ej0) (0), if x (ej0) is continuous at 0 = 0

since, x (ejo) is a periodic function, let us consider

x (ejo) = 2 (0 + 2k) ……………(7)

if we substitute this in eq. (1), we get

x[n] = 1/2 (0 + 2h) ejon d

= d(0) ejon d

since, there is only one impulse in the interval of integration

x[n] = 1

thus, we can say that eq. (7) represents fourier transform of a signal such that x[n] = 1 for all n.

As a generalization of the above example consider a sequence {x[n]} whose fourier transform is

x (ejo) = 2 (0 – 00 + 2k), – < 0o <

substituting this in eq. (1), we get

x[n] = 1/2   2 (0 – 00 + 2h) ejon d …………..(8)

= (0 – 0o) ejon d

As only one term corresponding to k = 0 will be there in the interval of the integration

x[n] ejo 0n

So, the signal is {ejo 0n} when fourier transform is given by eq. (8). more generally is x[n] is sum of an arbitrary set if complex exponentials

{x[n]} = a1 {ejo 1n} + a2 {ajo 2n} +……+ am {ejo mn}

Thus, its fourier transform is

x{ejo) = a1  2 (0 – 01 + 2k) + a2  2 (0 – 02 + 2k) +. ………+ am  2 (0 – 0m + 2k) ………………..(9)

Thus, x(ejo) is a periodic impulse train, with impulses located at the frequencies 01, 02 ………..0m of each of the complex exponentials and at all points that are multiples of 2 from these frequencies. An interval of 2 contains exactly one impulse from each of the summation of RHS of eq. (9).

let  {x[n]} = {cos 0o n}

for example,     x[n] = 1/2 ejo 0n + 1/2 e-jo 0n


x(ejo) = [(0 – 00 + 2k) + (0 + 0o + 2k)]