Show work. How they solve this problem 1241.0Ak0.0=am =bm=3. SIGNAL ANALYSIS IN THE FREQUENCY DOMAIN{ /0.5Time (sec)FIGURE 3.8 A half-triangle waveform used in Example 3.1.Find the first four magnitude and phase components (i.e., m = 1, 2, 3, 4). Constructplots of the magnitude components.Solution: Use Equations 3.8 and 3.9 to find the am and bm coefficients. Then convert to magnitudCm, and phase, 0m, using Equations 3.6 and 3.7. Plot Cm and 0m against the associated frequenoStart by evaluating either the sine or cosine coefficients; we begin with the sine coefficients usif=mfi.Equation 3.9:7 6²--7/am =x(1) =247²m²To find the cosine coefficients, use Equation 3.8:[x(t)cos(2amfit)dt*x(t)sin (2 mft) dt = ²²1247² m²247m² sin (2amt) - 2mmt cos(2mt)]0.50<1≤0.50.5 <1 ≤11.0[cos(TM)[sin(am) - nm cos(TM)]1-1 1 -1bm=2T 4T 6T 8T ;... = 0.159; -0.080; 0.053; -0.040;...=0.5220.52=47 m² cos (2πmt) - 2 mt sin(2πmt)10.5t sin(2+mt)dt-12n²m²-am sin(am) - 1] =t cos(2nmt)dt[cos(TM)]-127m² (cos(m) - 1]frequenc0;... = -0.101; 0; -0.0118; 0; ...

Given a periodic signal, mathematically defined over a period,To find:First four magnitude and phase…

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Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Show work. How they solve this problem

124
1.0
Ak
0.0
=
am =
bm=
3. SIGNAL ANALYSIS IN THE FREQUENCY DOMAIN
{ /
0.5
Time (sec)
FIGURE 3.8 A half-triangle waveform used in Example 3.1.
Find the first four magnitude and phase components (i.e., m = 1, 2, 3, 4). Construct
plots of the magnitude components.
Solution: Use Equations 3.8 and 3.9 to find the am and bm coefficients. Then convert to magnitud
Cm, and phase, 0m, using Equations 3.6 and 3.7. Plot Cm and 0m against the associated frequeno
Start by evaluating either the sine or cosine coefficients; we begin with the sine coefficients usi
f=mfi.
Equation 3.9:
7 6²
--7/
am =
x(1) =
2
47²m²
To find the cosine coefficients, use Equation 3.8:
[x(t)cos(2amfit)dt
*x(t)sin (2 mft) dt = ²²
1
2
47² m²
2
47m² sin (2amt) - 2mmt cos(2mt)]0.5
0<1≤0.5
0.5 <1 ≤1
1.0
[cos(TM)
[sin(am) - nm cos(TM)]
1
-1 1 -1
bm=
2T 4T 6T 8T ;... = 0.159; -0.080; 0.053; -0.040;...
=
0.5
2
2
0.5
2
=
47 m² cos (2πmt) - 2 mt sin(2πmt)10.5
t sin(2+mt)dt
-1
2n²m²
-am sin(am) - 1] =
t cos(2nmt)dt
[cos(TM)]
-1
27m² (cos(m) - 1]
frequenc
0;... = -0.101; 0; -0.0118; 0; ...

Expert Solution

Step 1: what is given and what to do:

Given a periodic signal, mathematically defined over a period

$x open parentheses t close parentheses equals open curly brackets table row t cell semicolon 0 less than t less or equal than 0.5 end cell row 0 cell semicolon 0.5 less than t less or equal than 1 end cell end table close$ ,

To find:

First four magnitude and phase components.

Note:

For magnitude and phase components, we need to compute Fourier series.

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