Three discrete-time signals are defined as x₁ [n] = n(u[n] - u[n-5]), x2 [n] = 4(u[n - 5] - u[n — 11]) and x3[n] = −28[n — 8]. Let x[n] be the discrete-time signal defined as x[n] = x1[n] + x2[n] + x3[n]. In MATLAB do the following: 1. define the unit-step function u[n] as an anonymous function in MATLAB as u= @(n) (n >= 0) and the unit-impulse function as an anonymous function in MATLAB as delta = @(n)(n = 0) 2. Using subplot, plot x₁ [n], x2 [n], x3[n] and x[n] in a 2-by-2 plot window. Note that for discrete signals, stem() is used instead of plot(). Make sure to label your plots. 3. In a separate plot window, plot the odd component xo[n] = z[n]-x[-n]¸ 2 4. In a separate plot window, plot the even component [n] = a[n]+x[-n] 5. Use the MATLAB defined function sum() to calculate the total energy Ex(x) = x-x[n]|² of x[n], the total energy Ex(e) = x-xe[n]² of xe[n] and the total energy Ex(x) = -xo[n]² of xo[n]. 6. Compare the total energy of x[n] Ex(x) and Ex(xe) + Ex(xo).

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Three discrete-time signals are defined as x₁ [n] = n(u[n] - u[n − 5]), x₂[n] = 4(u[n- 5] - u[n-11])
and x3[n] = -28 [n-8]. Let x[n] be the discrete-time signal defined as x[n] = x₁ [n] + x₂ [n] + x3[n].
In MATLAB do the following:
1. define the unit-step function u[n] as an anonymous function in MATLAB as
@(n) (n >= 0)
u =
and the unit-impulse function as an anonymous function in MATLAB as
delta
@(n) (n = 0)
2. Using subplot, plot x₁ [n], x2 [n], x3[n] and x[n] in a 2-by-2 plot window. Note that for discrete
signals, stem() is used instead of plot(). Make sure to label your plots.
3. In a separate plot window, plot the odd component xo[n] = [n]-x[-n]
2
4. In a separate plot window, plot the even component re[n] = x[n]+x[-n]
2
2=-∞0
5. Use the MATLAB defined function sum() to calculate the total energy Ex(x) = x - x[n]|²
of x[n], the total energy Ex(x) = Σn-xe[n]|² of xe[n] and the total energy Ex(xo)
- xo[n]² of xo[n].
6. Compare the total energy of x[n] Ex(x) and Ex(xe) + Ex(xo).
Transcribed Image Text:Three discrete-time signals are defined as x₁ [n] = n(u[n] - u[n − 5]), x₂[n] = 4(u[n- 5] - u[n-11]) and x3[n] = -28 [n-8]. Let x[n] be the discrete-time signal defined as x[n] = x₁ [n] + x₂ [n] + x3[n]. In MATLAB do the following: 1. define the unit-step function u[n] as an anonymous function in MATLAB as @(n) (n >= 0) u = and the unit-impulse function as an anonymous function in MATLAB as delta @(n) (n = 0) 2. Using subplot, plot x₁ [n], x2 [n], x3[n] and x[n] in a 2-by-2 plot window. Note that for discrete signals, stem() is used instead of plot(). Make sure to label your plots. 3. In a separate plot window, plot the odd component xo[n] = [n]-x[-n] 2 4. In a separate plot window, plot the even component re[n] = x[n]+x[-n] 2 2=-∞0 5. Use the MATLAB defined function sum() to calculate the total energy Ex(x) = x - x[n]|² of x[n], the total energy Ex(x) = Σn-xe[n]|² of xe[n] and the total energy Ex(xo) - xo[n]² of xo[n]. 6. Compare the total energy of x[n] Ex(x) and Ex(xe) + Ex(xo).
Expert Solution
Step 1: Algorithm of the Source Code:

Algorithm to Analyze Discrete-Time Signals:

  1. Define the unit-step function u[n] and the unit-impulse function delta[n] as anonymous functions:

    • u(n) = (n >= 0)
    • delta(n) = (n == 0)
  2. Define the range of n over which you want to analyze the signals:

    • n = -10:20
  3. Define the three discrete-time signals:

    • x1[n] = n * (u[n] - u[n-5])
    • x2[n] = 4 * (u[n-5] - u[n-1])
    • x3[n] = -28 * (n - 8)
    • x[n] = delta[n] + x2[n] + x3[n]
  4. Create a 2-by-2 plot window and use the stem() function to plot the four signals x1[n], x2[n], x3[n], and x[n], labeling each plot.

  5. Calculate the odd component co[n] as the difference between x[n] and its time-reversed version x[-n]. Plot co[n].

  6. Calculate the even component ce[n] as the average of x[n] and x[-n]. Plot ce[n].

  7. Calculate the total energy for each signal:

    • Ex_x is the total energy of x[n]
    • Ex_e is the total energy of the even component ce[n]
    • Ex_o is the total energy of the odd component co[n]
  8. Compare the total energy of x[n] with the sum of energies of the even and odd components (Ex_e + Ex_o):

    • If |Ex_x - (Ex_e + Ex_o)| < 1e-6, then print "The total energy of x[n] is approximately equal to Ex_e + Ex_o."
    • Otherwise, print "The total energy of x[n] is not equal to Ex_e + Ex_o."
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