Signals and Systems Test 4

Concept:

The Nyquist Rate is obtained by sampling the signal at twice the maximum frequency.

Let x(t) have a Fourier Transform X(ω)

ie. \(x\left( t \right)\overset{F.T}{\mathop{\leftrightarrow }}\,X\left( ω \right)\)

\(Y\left( ω \right)=X\left( ω \right)+{{e}^{-jω {{ω }_{0}}}}X\left( ω \right)\)

\(=X\left( ω \right)\left[ 1+{{e}^{-jω {{ω }_{0}}}} \right]\)

|Y(ω)| = |X(ω)|

The magnitude spectrum of both the signals are same.

∴ the maximum frequency of the signal is still limited by X(ω).

Hence sampling Frequency = ω_o

Concept:

For a signal x(t) sampled with a sampling interval T_s, the discrete sequence can be written as:

\(x\left( n \right) = x{\left. {\left( t \right)} \right|_{t = n{T_s}}}\) 

T_s = sampling interval

Application:

Given:

x(t) = 2 + cos (50 πt)

f_s = 40 Hz

\(\therefore {T_s} = \frac{1}{{40}} = 0.025\;sec\)

\(x\left( n \right) = x{\left. {\left( t \right)} \right|_{t = n{T_s}}}\)

x(n) = 2 + cos (50π nT_s)

Substituting the respective values, we get:

x(n) = 2 + cos (50 π n × 0.025)

x(n) = 2 + cos (1.25 πn)

= 2 + cos (1.25 πn – 2πn)

x(n) = 2 + cos (-0.75 πn)

\(=2 + \cos \left( {\frac{{3\pi }}{4}n} \right)\)

Concept:

For an N-point DFT as shown, the number of multiplication:

(M)_DFT = N(rows) × [N multiplication per Row]

M_(DFT) = N²

^{\(\left[ {\begin{array}{*{20}{c}} 1&2&3& \ldots &N\\ 1& \ldots & \ldots & \ldots & \ldots \\ 1& \ldots & \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ N& \ldots & \ldots & \ldots &N \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ \ldots \\ \ldots \\ \ldots \\ N \end{array}} \right]\)}

And for an N-point FFT, the number of multiplication equals the number of stages × Multiplications per stage, i.e.

\({\left( M \right)_{FFT}} = {\log _2}N \times \frac{N}{2}\)

Calculation:

(M)_DFT = N² = 256

\({\left( M \right)_{FFT}} = \frac{{16}}{2}{\log _2}\left( {16} \right)\)

\( = \frac{{16}}{2} \times 4 = 32\)

(M)_DFT – (M)_FFT = 256 – 32 = 224

\(\begin{array}{l} x\left( k \right) = \mathop \sum \limits_{n = 0}^7 x\left( n \right){e^{\frac{{ - j2\pi nk}}{8}}}\\ = \mathop \sum \limits_{n = 0}^3 x\left( n \right){e^{ - j\frac{{2\pi }}{8}nk\;}} + \mathop \sum \limits_{n = 4}^7 x\left( n \right){e^{ - j\frac{{2\pi }}{8}nk}} \end{array}\)

let n – 4 = l → n = 4 + l

Second summation becomes

\(\mathop \sum \limits_{l = 0}^3 x\left( {4 + l} \right){e^{ - j\frac{{2\pi }}{8}\left( {4 + l} \right)k}}\)

\(\mathop \sum \limits_{l = 0}^3 x\left( {4 + l} \right){e^{ - j\frac{{2\pi }}{8}lk}}{e^{ - j\pi k}}\)

Changing the variable

l → n, 2^nd summation becomes

\(\mathop \sum \limits_{n = 0}^3 x\left( {n + 4} \right){e^{ - j\frac{{2\pi }}{8}nk}}{e^{ - j\pi k}}\)      ______(1)

Symmetry property in Question

\(x\left( {n + \frac{N}{2}} \right) = - x\left( n \right)\)

x(n + 4) = -x(n)          _____(2)

Substitute in (1)

\(- \mathop \sum \limits_{n = 0}^3 x\left( n \right){e^{ - j\frac{{2\pi }}{8}nk}}{e^{ - j\pi k}}\)

e^-jπk = (-1)^k

The total summation is

\(X\left[ k \right] = \mathop \sum \limits_{n = 0}^3 x\left( n \right){e^{ - j\frac{{2\pi }}{8}nk}} - \mathop \sum \limits_{n = 0}^3 x\left( n \right){e^{ - j\frac{{2\pi }}{8}nk}}{\left( { - 1} \right)^k}\)

For k = even (0, 2, 4, 6)

X[0] = X[2] = X[4] = X[6] = 0

Concept:

If DFT of x(n) is X(k) i.e.

If \(x(n) \overset{DFT}{\longleftrightarrow} X(k)\)

Then \(x(n-n_0)\overset{DFT}{\longleftrightarrow} X(k)e^{-jk\omega n_0}\)

Calculation:

Given X(k) = [1, 1 – j, 1, 0, 1, 0, 1, 1+j] (N = 8 point DFT)

y(n) = x(n) ⊗ δ (n – 2) = x(n – 2) (Convolution with impulse property)

⇒ Y(k) = X(k)e^-jω2k 

\(=X\left( k \right).{{e}^{-j\frac{2\pi .2k}{8}}}\)

\(=X\left( k \right){{e}^{-\frac{jk\pi }{2}}}\)

So, Y(k) = [1.(-j)⁰, (1 – j).(-j)¹, 1.(-j)², 0.(-j)³, 1.(-j)⁴, 0.(-j)⁵, 1.(-j)⁶, (1 + j).(-j)⁷]

= [ 1, -1 – j, -1, 0, 1, 0, -1, -1 + j]

Concept:

N-point DFT of a sequence x(n) defined for n = 0, 1, 2 ... N-1 is given by:

\(X\left( k \right)=\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}\)

Also If \(x(n)\overset{DFT}{\longleftrightarrow} X(k)\)

The Conjugate symmetric property of DFT states:

X(k) = X*(N – k) 'or' X(N – k) = X*(k) 

Calculation:

Given, X(k) = {5, 1 - 3j, 0, 3 – 4j, 0, 3 + 4j, A, B]

Where A and B are the missing variable values.

Using the conjugate symmetric property of DFT: X(k) = X*(N – k)

We find X(6) = A = X*(8 – 6) = X*(2)

With X(2) = 0, X*(2) = 0

So, X(6) = A = 0

Similarly,

B = X(7) = X*(8 – 7) = X*(1)

With, X(1) = 1 – 3j

X(7) = X*(1) = 1+3j

So, A = 0 and B = 1 + 3j