Reference no: EM131724789

Lab - Frequency and Spectral Analysis

Theory:

Many sinusoidal signals encode information. These sinusoidal signals have a period or frequency. This is true of naturally occurring signals, as well as those that have been created by humans. Speech is a result of vibration of the human vocal cords; stars and planets change their brightness as they rotate on their axes and revolve around each other; ship's propellers generate periodic displacement of the water, and so on. The time-domain representation of these waveforms is often complicated and difficult to understand. Analysis of these signals is often easier to perform in the frequency domain where the frequency, phase and amplitude of the component sinusoids is easily obtained.

Prelab Assignment:

Lab:

1. Let x₁(n) = {1, 2, 2, 1} A new sequence x₂(n) is formed using

a. Express X₂(e^jω) in terms of X₁(e^jω) without explicitly computing X₁(e^jω)

b. Verify your results using MATLAB by computing and plotting magnitudes of the respective DTFTs.

2. Using the definition of the DTFT, determine the sequences corresponding to the following DTFTs.
a. X(e^jω) = 3 + 2cos(ω) + 4cos(2ω)
b. X(e^jω) = [1- 6cos(3ω) + 8cos(5ω)]e^-j3ω]
c. X(e^jω) = 2 + j4sin(2ω) - 5cos(4ω)
d. X(e^jω) = [ 1+ 2cos(ω) + 3cos(2ω)]cos(ω/2)e^-j5ω/2
e. X(e^jω) = j[3 + 2cos(ω) + 4cos(2ω)]sin(ω)e^-j3ω

3. Using the definition of the inverse DTFT, determine the sequences corresponding to the following DTFTs.

4. For each of the linear, shift-invariant systems described by the impulse response, determine the frequency response function H(e^jω). Plot the magnitude response |H(e^jω)| and the phase response, H(e^jω) over the interval [-Π,Π].

a. h(n) = (0.9)|n|
b. h(n) = sinc(0.2n)[u(n+20) - u(n -20)], where sinc 0 = 1
c. h(n) = sinc(0.2n)[u(n) - u(n -40)]
d. h(n) = [(0.5)ⁿ + (0.4)ⁿ]u(n)
e.h(n) = (0.5)^|n| cos(0.1Πn) = 1/2.0.5^|n|e^j.0.1Πn + 1/2.0.5^|n|e^-j0.1Πn

5. Determine H(e^jω)and plot its magnitude and phase for each of the following systems.
a. y(n) = 1/5 _m=0∑⁴ x(n-m)
b. y(n) = x(n) - x(n-2) + 0.95y(n -1) - 0.9025y(n-2)
c. y(n) = x(n) - x(n-1) + x(n - 2) + 0.95y(n-1) - 0.9025y(n-2)
d. y(n) = x(n) - 1.7678x(n-1) + 1.5625x(n - 2) + 1.1314y(n-1) - 0.64y(n-2)
e. y(n) = x(n) - _l=1∑⁵ (0.5)^ly(n-l)