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Fourier Series –

Fourier Series  joseph fourier developed the theory for the study of their sinusoidal representation. named after him as fourier analysis, it is extensively useful in various branches of science and engineering, as it offers an insight into the frequency content a signal.

When the french mathematician joseph’ fourier (1768 – 1830) was trying to solve a problem in heat conduction, he needed to express a function f as an infinite series of sine and cosine functions

f(x) = ao + (an cos nx + bn sin nx)

= ao + a1 cos x + a2 cos 2x + a3 cos 3x + ……….+ b1 sin x + b2 sin 2x + b3 sin 3x + ……(1)

Earlier, Daniel bernoulli and leonard euler had used such series while investigating problems concerning vibrating strings and astronomy.

The series in eq. (1) is called a trigonometric series or fourier series and it turns out that expressing a function as a fourier series is sometimes more advantageous than expanding it as a power series. in particular, astronomical phenomena are usually periodic, as are heartbeats, tides, and vibrating strings, so it makes sense to express them in terms of periodic functions.

We star by that the trganoneure scrrcs converges and has a continuous function f(x) as its sum on the interval that is.

f(x) = ao + (an cos nx + bn sin nx), – < x<

Our aim is to find formulas for the coefficients an and bn terms of f. recall that for a power series f(x) = cn (x – a)n, we found a formula for the coefficients in terms of derivatives cn = f(n) (a)/n here we use integrals.

If we integrate both sides of eq. (2) and assume that it’s permissible to integrate the series term-by-term, we get

f(x) dx  ao dx + (an cos nx + bn sin nx) dx

= 2 ao + an cos nx dx + bn  sin nx dx

cos nx dx = 1/n sin nx

= 1/n [sin n – sin(-n)] = 0

because n is integer, similarly, sin nx dx = 0.

f(x) dx = 2ao

and solving for ao, gives

ao = 1/2  f(x) dx ………………….(3)

Note that, ao is the average value of f over the interval

To determine an for n > 1, we multiply both sides of eq. (2) by cos mx (where, m is an integer and m > 1) and integrate term-by-term from – to

f(x) cos mx dx

= [ao + (an cos nx + bn sin nx)] cos mx dx

= ao  cos mx dx + an cos nx cos mx dx + bn sin nx cos mx dx …………………(4)

We have seen that the first integral is zero. with the help of the table of integrals, it is not hard to show that

sin nx cos mx dx = 0,   for all n and m

cos nx cos mx dx = [0, for n = m   , for n = m

so, the only non-zero term in eq. (4) is am and we get

f(x) cos mx dx = am

Solving for am and then replacing m by n, we have

an = 1  f(x) cos nx dx,     n = 1,2,3  ……………(5)

Similarly, if we multiply both sides of eq. (2) by sin mx and integrate from -to we get

bn = 1  f(x) sin nx dx,   n = 1,2,3 …………..(6)

We have derived formulas 3, 5 and 5 assuming f is a continuous function such that eq. (2) holds and for which the term-by-term integration is legitimate. but we can still consider the fourier series of a wider class of functions. A piecewise continuous function on [a.b] is continuous except perhaps for a finite number of removable or jump discontinuities. (in other words, the function has no infinite discontiuities.

Definition

Let f be piecewise continuous function on then the fourier series of f is the series

ao + (an  cos nx + bn sin nx) ………………….(7)

where, the coefficient an and bn in this series are defined by

ao = 1/2 f(x) dx

an = 1 f(x) cos nx dx

bn = 1 f(x) sin nx dx

and are called the fourier coefficient of f.

note that, in definition that we are not saying, f(x) is equal to its fourier series. later we will discuss conditions under which that is actually true. for now we are just saying that associated with any piecewise continuous function f on is a certain series called a fourier series.

x(t) = akejkot   = ak ejk(2 /t) t

ak = 1/t x(t) e-jk ot  dt = 1/t x(t) e-jk(2 /t)t dt

Discrete Time Fourier Series

We now start  considering discrete time signals. A discrete time signal is a function (real or complex valued) whose argument runs over the integers, rather than over the real line, we shall use square brackets as in x[n], for discrete time signals and round parenthesis, as in x(t), for continuous time signals. this is the notation used in EECE 359 and EECE 369. discrete-time signals arise in two ways. firstly, the signal could really be representing a discrete sequence of values. for example, x[n] could be the nth digit in a string of binary digits being transmitted along some data, bus in a computer or it could be the maximum temperature for day number n. secondly, a discrete time signal could arise from sampling a continuous time signal at a discrete sequence of times.

Periodic Signals

Just as in the continuous time case, discrete time signals may or may not be periodic. we start by considering the periodic case. imagine an application in which we have to measure some function x(t), that is periodic of period 2l. and compute its fourier coefficients from the measurements. we can think of x(t) as the amplitude of some periodic signal at time t. because we can only make finitely many measurements, we cannot determine x(t) for all values of t. suppose that we measure x(t) as n equally spaced values of t covering the full period 0 < t < 2l. say at

t = 0, 2l/n, 2 2l/n ………..(n – 1) 2l/n . because we do not know x(t) for all t, we cannot compute the complex fourier coefficient

ck = 1/2l  x(t) e-ik dt ……………..(1)

exactly. but we can get a riemann sum approximation to it using only t’s for which x(t) is known. all we need to do is divide the domain of integration up into n intervals each of length 2l/n, for t in the interval n2l/n < t < (n + 1) 2l/n,  we approximate the integrand x(t) e-lk/1 t by its value at t = n 2l/n, which is x(n2l/n) eik2l/n = x(n 2l/n) -2 i kn/n. so, we approximate the integral over n 2l/n < t < (n + 1) 2l/n by the area of a rectangle of height x(n 2l/n) e-2i kn/n and width 2l/n. this gives

ck ~ ck(n) = 1/2l  x(n 2l/n) e-2i kn/n) 2l/n

= 1/n   x(n 2l/n) e-2i kn/n

To save writing for what follows set x[n] = x(n 2l/n) and x[k] = c(n)k. then

x[k] = 1/n  x[n] e2i kn/n

note that x[n] and x[k] are both periodic of period n. that is

x[k + n] = 1/n  x[n] e-2 i (k + n)n/n

= 1/n   x[n] e-2 i kn/n e-2i nn/n

= 1/n x[n] e-2 i kn/n because e-2 ni = 1

= x[k]

x[n + n] = (n + n) 2l/n) = x(n 2l/n + 2l)

= x (n 2l/n) = x[n]

the vector (x[k] k = 0,1,2 ……n – 1, defined by

x[k] = 1/n   x[n] e-2 i kn/n ……………………(2)

is called the discrete fourier series (or by some people the discrete fourier transform) of the vector [x(j) j = 0 , 1, 2 ………n – 1. one of the main facts about discrete fourier series is that we can recover all of the (n different) x[n]’s exactly form x[0], x[1],………….x[n – 1] or any other n consecutive x[k] using the inverse formula

x[n] = x[k] e2 nk/n ……………….(3)

Proof  we need to show that if x[k] is defined by (2), than (3) is true. to verify this we just substitute the definition of x[k] into the right hand side of (3), taking care to rename the summation variable to ensure that we don’t use n to stand for two different quantities in the same formula.

e2 i nk/n  x[k] = e2 i nk/n 1/n   e-2 i n’k/n x[n]

= 1/n  e2 i (n – n)/n  x[n]

= 1/n  x[n]  e2 i k (n – n)/n

= 1/n   x[n]  (e2 i n – n/n)

= 1/n  x[n] rk

with  r = e 2 i n – n/n

for one value of n’ namely n’ = n, r = 1 and

e2 i k(n – )/n  = 1 = n

for all other values of n’ we can use the standard formula

1 + r + r2 + ………….ro = 1 – rp + 1/1 – r

(which you can check by multiplying out (1 – r) (1 + r +…….+ r0) and getting (1 – rp + 1) with

p = n – 1 and r = e2 (n -n)/n

= 1 – (e2 i (n – n)/n)n/1 – e2 i (n – n)/n  = 1 – e2 i (n – n)/1 – e2 i (n – n)/n

=  0

because n – n is an integer. substituting the values, we have just found for the k sums gives

e2 i nk/n x[k] = 1/n  x[n]   e2 i k (n – n)/n

= 1/n  x[n] {n  , if n = n  0, if n = n as desired.

The Fast fourier Transform

The fast fourier transform does not refer to a new or different type of fourier transform. it refers to a very efficient  type of fourier transform. it refers to a very efficient algorithm (made popular by a publication or J. W. cooley and J.W. tukey in  1965, but actually known to gauss in about 1805) for computing the discrete-time fourier and inverse fourier sums (2) and (3). we will not be covering this algorithm in this course, though it is not particularly sophisticated. the main idea behind it, is explained in the supplementary notes the fast fourier transform.

You have access to the fast fourier transform through the MATLAB commands FFT and IFFT. but a little care must be exercised when using FFT and IFFT because they implement different conventios than ours. if the input vector x is of length N, the x = (x) FFT (x) is a vector xhat with the N elements

xhat (k) = x(n) e-2i(k – 1) (n – 1)/N , 1 < k < N

and if the input vector xhat is of length N, then x = IFFT (xhat) is a vector x with the N elements

x(n) = 1/n  that  (k) e2 i (k – 1) (n – 1)/n   1 < n < N.

in contrast to eq. (2) and (3) the factor of 1/n appears in

IFFT. the reason for the funny looking exponents is that, in MATLAB, vector indices start with 1 rather than 0.

So, given any complex numbers x[0], ………….x[n – 1], the vector x given by eq. (2) above can be computed, in MATLAB, as follows

x = [x[0], x[1], x[2],……….x[n – 1]]

notice, the factor of 1/N in the second line. the resulting vector xhat will have n entries, corresponding to [x[0], x[1], ……x[n – 1]]. but notice that MATLAB’s subscripting rules require

x[0] = xhat (1), x[1] = xhat (2), ………………x[n – 1] ……….xhat (n).

similarly, if complex numbers x[0], …….x[n – 1] are given, the vector x in eq. (3) above can be found using these MATLAB commands

xhat = [x[0], x[1], ……………x[n- 1]

x = n * IFFT (xhat)

here, again the factor of n in the second line corrects for a different system of conventions between these notes and the MATLAB software system.

Aperiodic Signals

We now develop a frequency expansion for non-periodic discrete time functions using the same strategy as we did in the continuous time case.

Again, for simplicity we’ll only develop the expansions for functions x[n] that are zero for all sufficiently larger |n|. Again, our conclusions will actually apply to a much broader class of functions. let N be an even integer that is sufficiently large that x[n] = 0 for all |n| > 1/2 N. we can get a discrete-time fourier series expansion for the part of x[n]. with |n|< 1/2 N by using the periodic extension trick.

Define x(n) [n] to be the unique discrete time function determined by the requirements that

(i) x(n) [n] = x[n] for – n/2 < n n/2

(ii) x(n) [n] is periodic of period n

then, for -n/2 < n < n/2,

x[n] = x(n) [n] = -n/2 < k< n/2x(n)  [k] e2i nk/n

where, x(n) [k] = 1/n -n/2 < n< n/2 x[n]e-2 i nk/n ………………(4)

here, we have exploited the fact that since, both x(n) [n] and x(n) [k] are periodic of period n. we may choose the range of summation in eqs. (2) and (3) to be any consecutive set of n integers. we have chosen  – n/2 < k < n/2 and -n/2 < n< n/2 because they will lead to nice limits when we send n .

To get a representation of x[n] that is valid for all n’s not just those in a finite interval n/2 < n < n/2 , we take limit  n – oo. to evaluate this limit we again interpret the sum over k in eq. (4) as a reimam sum approximation to a certain integral. for each integer k, define the kth frequency to be ook = 2  k/n . also use two successive frequencies and define x(oo) = x[n]ejon. since, x[n] = 0 for all |n| n/2 ,

xn [k] = 1/n -n/2 < n < n/2 x[n]e-2 i nk/n

=  1/n n = – oo x[n] e-i2 k/n – n

= 1/n n = – oo x[n] e-jokn = 1/2 x(0k)

in this notation,

x[n] = x(n)[n] = -n/2 < n < n/2 1/2 x(0k) e2 i nk/n

= 1/2  x(0k) eiokn

for any -n/2 < n < n/2. note that we have multiplied the summation restriction – n/2 < k < n/2 by 2/n to get the equivalent restriction as we let n the restriton – n/2 < n < n/2 disappears and the right hand side, which is exactly a riemann sum approximation sum to the integral 1/2 x(0) ei o n d. converges to that integral. we conclude that

x[n] = 1/2 x(0) ei on do

where, x(0) = x[n] eion ……………………….(5)

the function x(0) is called the discrete time fourier transform of x[n] or the spectrum of x[n]. for any integer m,

x(0 + 2m) = x[n]e-i(0 + 2m)n

= x[n]e-jon e-i2mn

= x[n]e-ion = x(0)

so, x(0) is periodic of period 2 and we may choose as the domain of integration in eq. (5) any interval of length 2.

Formula (5) should look familiar. they are exactly the first fourier expansion that we saw but with the roles of the time and frequency domains exchanged. in theorem 1 of the notes fourier series, we saw that, if f(t) is continuous with continuous first derivative and is also periodic with period 2 then

f(t) = k = – 0 ckeikt, where ck = 1/2  f(t) e-ikt dt

if we make the substitutions t = 0 and k = – n, we get

f (0) = c-n e-ino

c-n = 1/2  f(0) eino do

which is exactly eq. (5) with x[n] = c-n and x(0) = f(0).

Example 1.  the discrete time signal

x[n] = an u[n] where u[n] = {1, if n > 0   0, if n < 0

is graphed in the figure

Sol.  it has the discrete fourier transform

x(0)  = x[n] e-i0n = an e-ion

= (ae -io)n = 1/ 1 – ae-io

we have now seen a number of different but closely related, fourier expansions. they are given in the following table.

observe that

. if in either domain, time or frequency, the function is periodic (not periodic) then the argument in the other domain is discrete (runs continuously).

. if in either domain, time or frequency, the function has a discrete argument (continuous argument), then the transformed function in the other domain is periodic (not periodic)

not surprisingly, all of four of these transforms have properties very similar to the linearity, time shifting, etc., properties of the fourier transform. the detailed versions of these properties are given in the following tables

Intro Exercise – 5

  1. The fourier series coefficient for the periodic signal shown below is

(a) 1

(b) cos (2 k)

(c) sin (2 k)

(d) 2

  1. The fourier series coefficient for the periodic signal shown below is

(a) 2a/jk sin ( 2 k)

(b) 2a/jk cos (2 k)

(c) 2a/k sin (2 k)

(d) 2a/k cos (2 k)

  1. The fourier series coefficient for the periodic signal shown below is

(a) a/2k (e-i (4k/3) – 1)

(b) j a/2k (e-j (4k/3) – 1)

(c) – j a/2k (e-j(4k/3) – 1)

(d) -a/2k (e -j (4k/3) – 1)

  1. The fourier series coefficient for the periodic signal shown below is

(a) a/k (1 – (-1) k)

(b) a/k (1 + (-1)k)

(c) a/jk (1 – (-1)k)

(d) a/jk (1 + (-1)k)

  1. The fourier series coefficient for the periodic signal x(t) = sin2 t is

(a) -1/4 [k – 1] + 1/2 [k] – 1/4 [k + 1]

(b) -1/4 [k – 2] + 1/2 [k] – 1/4 [k + 2)

(c) -1/2 [k – 1] + [k] – 1/2 [k + 1]

(d) -1/2 [k – 2] + [k] – 1/2 [k + 2]

  1. The fourier series coefficient of time domain signal x(t) is x[k] = j[k – 1] – j[k + 1] + [k + 3] + [k – 3].

the fundamental frequency of signal is = 2. the signal is

(a) 2(cos 3t – sin t)

(b) -2 (cos 3t – sin t)

(c) 2 (cos 6t – sin 2t)

(d) -2 (cos 6t – sin 2t)

  1. The fourier series coefficient of time domain signal x(t) is

x[k] = (-1/3)[k]

the fundamental frequency of signal is = 1. the signal is

(a) 4/5 + 3 sin t

(b) 5/4 + 3 sin t

(c) 5/4 + 3 cos t

(d) 4/5 + 3 cos t

  1. The fourier series coefficient of time domain signal x(t) is shown below.

the fundamental frequency of signal is the signal is

(a) 6 cos (2t + 4) – 3cos (3t – 4)

(b) 4 cos (4 t – 4) -2 cos (3 t + 4)

(c) 2 cos (2 t + 4) -2 cos (3 t – 4)

(d) 4 cos (4 t + 4) + 2 cos (3 t – 4)

  1. The fourier series coefficient of time-domain signal x(t) is shown below.

the fundamental frequency of signal is . the signal is

(a) 3 cos 3t + 2 cos 2t + cos t

(b) 3 sin 3 t + 2 sin 2t + sin t

(c) 6 sin 3 t + 4 sin 2 t + 2 sin t

(d) 6 cos 3 t + 4 cos 2t + 2 cos t

  1. The fourier series coefficient of time domain signal x(t) is shown below.

the fundamental frequency of signal is the signal x(t) is

(a) sin 9t/sin t

(b) sin 9 t/sin t

(c) sin 18 t/2 sin t

(d) none of these

  1. The fourier series coefficient of the signal y(t) = x(t – to) + x(t – to) is

(a) 2 cos (2/t kto) x[k]

(b) 2 sin (2/t kto) x[k]

(c) e-to x[k] + eto x[-k]

(d) 2 sin (2/t kto) x[k]

  1. The fourier series coefficient of the signal y(t) = ev {x{t}} is

(a) x[k] + x[-k]/2

(b) x[k] -x[-k]/2

(c) x[k] + x* [-k]/2

(d) x[k] + x* [-k]/2

  1. The fourier series coefficient of the signal y(t) = re {x(t)} is

(a) x[k] + x[-k]/2

(b) x[k] – x [-k]/2

(c) x[k] + x* [-k]/2

(d)x[k] + x* [-k]/2

  1. The fourier coefficient of the signal y(t) = d2 x(t)/dt2 is

(a) (2k/t)2 x[k]

(b) -(2k/t)2 x[k]

(c) j(2k/t)2 x[k]

(d) -j (2k/t)2 x[k]

  1. The fourier series coefficient of the signal y(t) = x(4t – 1) is

(a) 8/t x[k]

(b) 4/t x[k]

(c) e-jk (8/t) x[k]

(d) ejk (8/t) x[k]

  1. The discrete-time fourier coefficient cos (6/17 n + 3) is

(a) 1/2 e-j/3 [k – 3] + 1/2 ej3 [k + 3]

(b) 1/2 ej/3 [k – 3] + 1/2 e-i/3 [k + 3]

(c) 1/2 ej3 [k – 1] + 1/2 ei3 [k + 1]

(d) 1/2 ei3 [k + 1] + 1/2 e-j3 [k – 1]

  1. The discrete-time fourier coefficient of 2sin (4/19 n) + cos (4/19 n) + 1 is

(a) -j [k + 2] + j [k – 2] + 1/2 [k + 5] + 1/2 [k – 5] + [k]

(b) -j [k + 2] + j [k – 2] + 1/2 [k + 5] + 1/2 [k – 5] + 1

(c) -j [k – 2] + j [k + 2] + 1/2 [k – 5] + 1/2 [k + 5] + [k]

(d) none of the above

  1. The discrete-time fourier coefficient of cos2 (8 n) is

(a) 2 [k + 1] + 2[k] + [k – 1]

(b) 1/4j [k + 1] + 2 [k] + [k – 1])

(c) 1/4j [k + 1] + 2 [k] + [k – 1])

(d) none of the above

  1. The discrete time fourier coefficient x[n] = (……0,1,2,3,0, 1,2,3,0,1…..} is

(a) x[0] = 3/2, x[1] = x* [3] = 1/2 (1 – j) , x[2] = -1/2

(b) x[0] = 3/2 , x[1] = x* [3] = -1/2 (1-j), x[2] = -1/2

(c) x[0] = 1/2, x[1] = x* [3] = 1/2 (1 – j), x[2] = -3/2

(d) none of the above

  1. The discrete time fourier coefficients of [n – 4m] is

(a) -1/4, for all k

(b) 1/4, for all k

(c) 1/4 |k|, for all k

(d) e , for all k

Answers with Solutions

  1. (d)

x[k] = 1/t  a(t) e-jko t dt = a/t

a = 10, t = 5, x[k] = 2

  1. (c)

x[k] = 1/t     x(t)e-jko t = 1/t  2ae-jko t dt

= 2a/t [ejko t/jko] = 2a/k sin (k/2)

  1. (b)

t = 2, oo = 2/2 = 1

x(t) = {2a.  0 < t < 4/3   0,  4/3 < t < 2

x{k} = 1/2   x(t) e-jkt  dt = 1/2   2ae-jkt dt

= 2jA/2k [e-j(4k/3) – 1]

  1. (c)

t = 2, oo = 2/2

x(t) = {-a  – 1 < t > 0  a, 0 < t < 0

x[k] = 1/2   x(t) e-jkt dt

= 1/2 (1 – ejk/jk + e-jk – 1/-jk)

= A/jk (1 – (-1)k)

  1. (a)

sin2 t = (eit – e-jt/2j) = -1/4 (e2jt – 2 + e-2jt)

the fundamental period of sin2(t) is and = 2 = 2

x[k] = -1/4 [k – 1] + 1/2 [k] -1/4 [k + 1]

  1. (c)

x(t) = x[k]ej2kt

jej2t – je-j2t + ej6t + e-j6t

= – 2 sin 2t + 2cos 6t

  1. (d)

x(t) = x[k]ejkt

(-1/3) -k ejkt + (-1/3)k ejkt

= -1/3e-jt/1 + 1/3 e-jt + 1/1 + 1/3 ejt = 4/5 + 3 cos t

  1. (d)

x(t) = x[k]ejkt

= 2e-j/4 ej(-4) + ej/4ej(-3)t + ej/4ej(3) t + 2ej/4

= 2(e-j(4 t + 4) + ei(4 t + 4) ) + (e-j(3 t – 4) + ej(3 t – 4))

= 4 cos (4 t + 4) + 2 cos (3 t – 4)

  1. (d)

x[k] = |k|, – 3 < k < 3

x(t) = x[k] ejkt

= 3ej(-3)t + 2ej(-2)t + ej(-1) t + ej(1) t + 2ej(2) t + 3ej(3) t

= 6 cos 3t + 4 cos 2t + 2 cos t

  1. (a)

x[k] = e-j2k, – 4 < k < 4

x(t) = e-j2k ejkt

e-j2k(t – 1) = sin 9t/sin t

  1. (a)

x(t – to) is also periodic with t. the fourier series coefficient x1 [k] of x(t – to) is

x1 [k] = 1/t  x(t – to) e-jk t dt

= e-jk t/t  x e-jk t dt

= e-jk t/t x e-jk dt = e-jkt x[k]

similarly, the FS coefficient of x(t + to) is

x2[k] = ejk t x[k]

the FS coefficient of x(t – to) + x(t + to)

y[k] = x1[k] + x2[k]

= e-jkt x[k] + ejk t x[k]

= 2 cos (kt) x[k]

  1. (a)

ev {x(t)} = x(t) + x(-t)

the FS coefficient of x(t) is

x1[k] = 1/t x(-t) e-jk t dt

= 1/t x ejk  dt = x[-k]

therefore, the FS coefficient of ev {x(t)} is

y[k] = x[k] + x[-k]/2

  1. (c)

re {x(t)} = x(t) + x* (t)/2

the FS coefficient of x* (t) is

x1[k] = 1/t x* (t) e-jk t dt = x1* [-k]

x1*[k] = 1/t x(t) ejk t dt = x [-k]

x1[k] = x* [-k]

y[k] = x[k] + x* [-k]/2

  1. (b)

x(t) = x[k] ej2/t kt

dx(t)/dt = – j (2/t)k x[k]ej(2/t)kt

d2x(t)/dt2 = (2/t)2 k2 x[k] ej(2/t) kt

y[k] = – (2k/t)2 x[k]

  1. (c)

the period of x(4t) is a fourth of the period of x(t).the fourier series coefficient of x(4t) is still x[k] . using the analysis of q. 12, the fs coefficient of x(4t – 1) is ejk(8/t) x[k]

  1. (b)

N = 17, = 2/17

Cos (6/17 + 3) = 1/2 (e-j(6/17 + 3) + ej(6/17 + 3))

= 1/2 (ej/3ej(3) 2/17 + e-j 3 e-j(3)2/17)

x[k] = 1/2 ej3 [k – 3] + 1/2 e-j/3 [k + 3]

  1. (c)

x[n] = 2 sin (4/19 n) + cos (10/19 n) + 1

= 1/j (ej(4/19)n – e-j(4/19)n ) + 1/2 (ej(10/19)n – e-j(10/19)n) + 1

= – je j(2) 2/19 n + jej(-2) 2/19n + 1/2 ej(5) 2/19 n + 1/2 ej(5) 2/19 + 1/2 e-j(5) 2n /19 + 1/2 ej(0) 2/19 n

x[k] = – j[k – 2] + j[k + 2] + 1/2 [k – 5] + 1/2 [k + 5] + [k]

  1. (c)

n = 8, = 2/8 = 4

x[n] = cos2(8 n) = 1/4 (ej(8)n + e-j(8)n)2

= 1/4 (ej(2/8)n + 2 + e-j(2/8)n)2

x[k] = 1/4 [k – 1] + 1/4 [k + 1] + 1/2 [k]

  1. (b)

n = 4, = 2/4 = 2, e-j2 = – j, x[n] = n

x[k] = 1/4 x[n] (-j)kn = 1/4 n(-j)kn

= 1/4 [(-j)k + 2 (-j)2k + 3 (-j)3k]

x[0] = 3/2, x[1] = 1/2 (-1 + j), x[2] = -1/2

x[3] = 1/2 (-1 – j)

  1. (b)

n = 4, = 2/4 = 2

x[k] = 1/4  x[n] e-j (2)nk = 1/4 x [0] = 1/4, for all k