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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
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)
1 (w) = angle H(e jw) Example:h[n] = - δ[n] | H (ejw)| = 1 angle H(e jw) = πIn the magnitude or phase representation a real valued frequency response does not mean that the system is zero-phase. Using this representation,|Y (e jw) | = | H ( ejw)| | X (ejw)|
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
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.
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) 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.
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av = 1/T
= 1/T
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sinkwo dt ---------------------------------(2)
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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) 1 (w) = angle H(e jw) Example:h[n] = - δ[n] | H (ejw)| = 1 angle H(e jw) = πIn the magnitude or phase representation a real valued frequency response does not mean that the system is zero-phase. Using this representation,|Y (e jw) | = | H ( ejw)| | X (ejw)|
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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
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1/T
= 1/T
= 0 n≠ m
= If x1(t) = x2(t) = x(t) then eq(3) becomes 1/T
The above equation can be written as
n≠0
= c0 2 + n≠0 a0 2 +
= a0 2 +
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j represented by o.Poles z= 7/8 , z=-1/2 represented by x.
Poles and Zeros
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