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Unit-3 Fourier, Laplace and Z transform Q1) Explain the Fourier series Representation of periodic signals?A1) A signal is said to be periodic if it satisfies the condition x (t) = x (t + T) or x (n) = x (n + N).Where T = fundamental time period,ω0= fundamental frequency = 2π/TThere are two basic periodic signals:x(t) = cos wot (sinusoidal)x(t) = e jwot (complex exponential) These two signals are periodic with period T=2π/ω0 A set of harmonically related complex exponentials can be represented as ɸk(t) ɸk(t) = { e jkwot} = { e jk(2π/T)t} where k=0,±1,±2,±3,………………(1) All these signals are periodic with period T. According to orthogonal signal space approximation of a function f(x) with n mutually orthogonal functions is given by x(t) =

e jkwot ……………………………………………….(2) =

k ejkwot where ak= Fourier coefficient = coefficient of approximation.This signal x(t) is also periodic with period T.Equation 2 represents Fourier series representation of periodic signal x(t).The term k = 0 is constant.The term k=±1 having fundamental frequency ω0, is called as 1st harmonics.The term k=±2 having fundamental frequency 2ω0, is called as 2nd harmonics, and so on...The term k=±n having fundamental frequency nω0, is called as nth harmonics. Q2) Write a short note on waveform symmetries?A2)

Even function symmetry

A function is defined to be even if and only if f(t) = f(-t) -------------------------------------------------------------(1)If the condition is satisfied then eq(1) is said to be even because polynomial functions with even components For any even periodic functions, the equations for the Fourier coefficients av = 2/T

dt ---------------------------------------------------(2) ak = 4/T

coskwot dt --------------------------------------------(3) bk=0 for all k --------------------------------------------------(4) In eq(4) all b co-effecients are zero if the function is even. The figure depicts the even periodic function.

Figure 1. Even symmetry

av = 1/T dt

= 1/T dt +

Odd function symmetry

A periodic function is defined to be odd if f(t) = -f(t) ---------------------------------(1)The function that satisfies eq(1) is said to be odd because polynomial funtions with odd exponents. The expression for Fourier co-effecients are: av=0 ak=0 for all k; bk = 4T

sinkwo dt ---------------------------------(2)

Figure 2. Odd function symmetry

Half Wave Symmetry

A function is said to have half-wave symmetry if it satisfies the following constraint:f(t) = -f(t - T/2) ------------------------------------ (1)Equation 1 expresses that a periodic function has a half-wave symmetry if, after it has been shifted by one-half of a period and inverted, it is said to be identical to the original periodic function.For instance, the periodic functions illustrated in Figures possess half-wave symmetry.

Figure 3. Half wave Symmetry

Quarter Wave Symmetry

If a function has half-wave symmetry and symmetry about the midpoint of the positive and negative half-cycles, the periodic function is said to have quarter--wave symmetry. This function is illustrated in Figure

Figure 4. Quarter Wave Symmetry Q3) Explain the calculation of Fourier co-efficient ?A3) We know that x(t) =

e jkwot ------------------------(1)Multiply e-jnwot on both sides we get x(t) e -jnwot =

e jkwot . e- jnwot Consider integral on both sides we get

e jkwot dt =

e jkwot . e- jnwot =

e j(k-n)wot dt =

e jkwot dt =

j(k-n)wot dt --------------------------(2)By Eulers formula

j(k-n)wot dt =

wo dt + j

wo dt

j(k-n)wot dt = { T k=n0 k

nHence in equation(2) the integral is zero for all values of k except at k=n. Put k=n is equation 2 =

j(k-n)wot dt = anT=an = 1/T

-jnwot Replace n by k we get = ak = 1/T

-jkwot dt x(t) =

e j(k-n) wot where ak = 1/T

-jkwot dt Q4) Derive Fourier Transform?A4) The (CT) Fourier transform (or spectrum) of x(t) isX(jw) =

e -jwt dt ---------------------------------------(1)x(t) can be reconstructed from its spectrum using the inverse Fourier transform x(t) = 1/ 2 π

e jwt dw ------------------------------------------(2) The above two equations are referred as Fourier transform pair with the first one being the analysis equation and the second being the synthesis equation. Notation X(jw) = F{x(t)}x(t) = F -1 {X(jw)}x(t) and X(jw) form a Fourier transform pair denoted by

x(t) F X(jw) Q5) Explain convolution?A5) Imagine we have a function f[t] whose Fourier transform is F[w] and another function g[t] whose transform is G[w]. Then the convolution is f[t] * g[t] =

g[t-u] du We write g[t-u] in terms of Inverse Fourier transform g[t-u] = 1/2π

E Iw(t-u) dw dwThus f[t] * g[t] =

1/2π

E Iw(t-u) dw dw du = 1/2π

E Iwt

E -Iwu du dw But the right hand integral above is the Fourier transform of f[u] so f[t] * g[t] = 1/2π

G[w] E -Iwu dw The convolution property states that: x(t) * h(t) > X(w) H(w)Let y(t) =

h(t-τ) dτ = x(t) * h(t)Y(w) =

e -jwt dt =

h(t-τ) dτ e -jwt dt If we switch the order of the two integrals we get Y(w) =

e -jwt dt =

h(t-τ) dτ e -jwt dt let u=t-τ=

h(u) e -jw(u+τ) du dτ =

e -jwτ dτ

h(u) e -jwu du = X(w) H(w)Therefore, y(t) = x(t) * h(t) <-> Y(w) = X(w) H(w)Fourier Transform of the convolution of two functions is simply the product of the Fourier Transforms of the functions. Q6) Explain the magnitude and phase response?A6) H(e jw) = H R ( e jw) + j H I ( e jw) = | H ( e jw) | e j

1 (w)

Thus, |H(ejω)| and angle H(ejω) are commonly referred to as the gain and the phase shift of the system, respectively. In magnitude and phase plots, as ω goes through a zero on the unit circle, the magnitude will go to zero and the phase will flip by π, as shown in the figure below.

Figure . Magnitude and Phase response Q7) Explain DTFT?A7) The discrete-time Fourier transform (DTFT) or the Fourier transform of a discrete–time sequence x[n] is a representation of the sequence in terms of the complex exponential sequence ejωn.The DTFT sequence x[n] is given byX(w) =

e-jwn ---------------(1)Here X(w) is a complex function of real frequency variable w and can be written as X(w) = Xre (w) + j X img(w)where Xre (w) , j X img(w) are real and Imaginary parts of X(w)Xre(w) = |X(w)| cos

(w)Ximg(w) = |X(w)| sin

(w)|X(w)| 2 = |Xre(w)| 2 + |Xim(w)| 2And | X(w)| can be represented asX(w) = |X(w)| e j

(w)Inverse Discrete Fourier Transform is given by x(n) = 1/2π

e jwn dw Q8) Find the DFT of the sequence f(n) ={ 8,4,8,0}A8) F[n] =

e -j π/2 nk =

(-j) nk

Figure. Magnitude of the four point sequence. Q9) Explain Parseval’s theorem?A9) Consider two periodic signals x1(t) and x2(t) with equal period T. If the Fourier series co-efficient of these two signals are cn and dn then

1/T x2(t) = 1/T n e j n Ωot [ m e jmΩot ] dt---------------------(1)

= 1/T n d *m e j(n-m)Ωot dt ------------------------------(2)

= 0 n≠ m

= n d *n n=m --------------------------------(3)

If x1(t) = x2(t) = x(t) then eq(3) becomes

1/T 2 = 2 --------------------------------(4)

The above equation can be written as

2 = c0 2 + 2

n≠0

= c0 2 + n c *n

n≠0

a0 2 + [Re(c 2 n ) + Im (cn ) 2]

= a0 2 + 2 n /2 + b 2 n /2

Q10) Find H(z), poles and zeros for the difference equation given by A10) y[n] – 3/8 y[n-1] – 7/16 y[n-2] = x[n] + x[n-2] Solution:Here ao=1,a1=-3/8,a2=-7/6,bo=1,b1=0,b2=1 H[z] = 1 + z -2 / 1 -3/8 z-1 -7/16 z -2 = (1+j z-1 ) (1 -jz-1) / (1-7/8 z-1) ( 1+1/2 z-1) Zeros : z=

j represented by o.Poles z= 7/8 , z=-1/2 represented by x.

Im(z)

Re(z)

Poles and Zeros

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