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Answer :
We start by noting that the data for the number of cars versus time shows an overall upward trend. In order to analyze whether the data has an underlying periodic (oscillatory) behavior, it is useful to remove this trend. Here is a step‐by‐step explanation of the process:
1. Removing the Trend
The data for the number of cars over 12 hours is:
[tex]$$
\begin{array}{c|cccccccccccc}
\text{Hour} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\text{Cars} & 52 & 76 & 90 & 75 & 91 & 104 & 89 & 105 & 119 & 103 & 121 & 135
\end{array}
$$[/tex]
A linear (first-degree) trend line of the form
[tex]$$
T(t) = mt + b
$$[/tex]
is fitted to the data, and then removed from the actual values. The residuals (or detrended values) are computed by subtracting the trend:
[tex]$$
R(t) = \text{Cars}(t) - T(t).
$$[/tex]
The resulting detrended values are approximately:
[tex]$$
\begin{array}{cccccccccccc}
-12.82,\; 5.39,\; 13.60,\; -7.19,\; 3.02,\; 10.23,\; -10.56,\; -0.35,\; 7.86,\; -13.93,\; -1.72,\; 6.49.
\end{array}
$$[/tex]
2. Determining the Amplitude
The amplitude of a periodic signal can be estimated as one-half the difference between the maximum and minimum of the detrended data. In this case, the minimum value is approximately [tex]$-13.93$[/tex] and the maximum is approximately [tex]$13.60$[/tex]. Thus, the amplitude [tex]$A$[/tex] is given by:
[tex]$$
A = \frac{\text{max}(R(t)) - \text{min}(R(t))}{2} \approx \frac{13.60 - (-13.93)}{2} \approx \frac{27.53}{2} \approx 13.77.
$$[/tex]
3. Identifying the Period from the Frequency Spectrum
To determine the period, we analyze the detrended data by transforming it into the frequency domain (using a Fourier transform). This analysis reveals a dominant frequency, which is the frequency at which the oscillations are strongest.
The dominant frequency found is approximately [tex]$0.3333$[/tex] cycles per hour. The period [tex]$P$[/tex] in hours is the reciprocal of the frequency:
[tex]$$
P = \frac{1}{0.3333} \approx 3 \text{ hours}.
$$[/tex]
4. Conclusion
From the analysis, we conclude that the data is approximately periodic. The periodic component has:
- A period of approximately [tex]$3$[/tex] hours.
- An amplitude of approximately [tex]$13.77$[/tex] (cars).
Thus, the data set is approximately periodic with a period of [tex]$3$[/tex] hours and an amplitude of about [tex]$13.77$[/tex].
1. Removing the Trend
The data for the number of cars over 12 hours is:
[tex]$$
\begin{array}{c|cccccccccccc}
\text{Hour} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\text{Cars} & 52 & 76 & 90 & 75 & 91 & 104 & 89 & 105 & 119 & 103 & 121 & 135
\end{array}
$$[/tex]
A linear (first-degree) trend line of the form
[tex]$$
T(t) = mt + b
$$[/tex]
is fitted to the data, and then removed from the actual values. The residuals (or detrended values) are computed by subtracting the trend:
[tex]$$
R(t) = \text{Cars}(t) - T(t).
$$[/tex]
The resulting detrended values are approximately:
[tex]$$
\begin{array}{cccccccccccc}
-12.82,\; 5.39,\; 13.60,\; -7.19,\; 3.02,\; 10.23,\; -10.56,\; -0.35,\; 7.86,\; -13.93,\; -1.72,\; 6.49.
\end{array}
$$[/tex]
2. Determining the Amplitude
The amplitude of a periodic signal can be estimated as one-half the difference between the maximum and minimum of the detrended data. In this case, the minimum value is approximately [tex]$-13.93$[/tex] and the maximum is approximately [tex]$13.60$[/tex]. Thus, the amplitude [tex]$A$[/tex] is given by:
[tex]$$
A = \frac{\text{max}(R(t)) - \text{min}(R(t))}{2} \approx \frac{13.60 - (-13.93)}{2} \approx \frac{27.53}{2} \approx 13.77.
$$[/tex]
3. Identifying the Period from the Frequency Spectrum
To determine the period, we analyze the detrended data by transforming it into the frequency domain (using a Fourier transform). This analysis reveals a dominant frequency, which is the frequency at which the oscillations are strongest.
The dominant frequency found is approximately [tex]$0.3333$[/tex] cycles per hour. The period [tex]$P$[/tex] in hours is the reciprocal of the frequency:
[tex]$$
P = \frac{1}{0.3333} \approx 3 \text{ hours}.
$$[/tex]
4. Conclusion
From the analysis, we conclude that the data is approximately periodic. The periodic component has:
- A period of approximately [tex]$3$[/tex] hours.
- An amplitude of approximately [tex]$13.77$[/tex] (cars).
Thus, the data set is approximately periodic with a period of [tex]$3$[/tex] hours and an amplitude of about [tex]$13.77$[/tex].
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