Amplitude, Period, Phase Shift, Vertical Shift, and Frequency - Problem 1

Given the equation y = 2 sin (4x − 0.5) + 3.

Find the following:

• Amplitude 
• Period 
• Phase Shift 
• Vertical Shift 
• Frequency 

Solution:

\begin{equation} A = 2 \end{equation}

Answer: \begin{equation} A = 2 \end{equation}

\begin{equation} P = \frac{2 \pi}{B} \end{equation}

\begin{equation} P = \frac{2 \pi}{4} \end{equation}

\begin{equation} P = \frac{\pi}{2} \end{equation}

Answer: \begin{equation} P = \frac{\pi}{2} \end{equation}

\begin{equation} C = -0.5 \end{equation}

Answer: \begin{equation} C = -0.5 \end{equation} or 0.5 unit shifted to the right

\begin{equation} D = 3 \end{equation}

Answer: \begin{equation} D = 3 \end{equation} or 3 units shifted upward

\begin{equation} F = \frac{1}{P} \end{equation}

\begin{equation} F = \frac{1}{\frac{\pi}{2}} \end{equation}

\begin{equation} F \approx 0.637 \end{equation}

Answer: \begin{equation} F = 0.637 \end{equation}

Amplitude, Period, Phase Shift, Vertical Shift, and Frequency

Some functions (like sine and cosine) repeat forever and are called Periodic Functions.

Amplitude

• It is the height from the center line to the peak (or to the trough). Or we can measure the height from highest to lowest points and divide that by 2.

Period

• It goes from one peak to the next (or from any point to the next matching point).

Phase Shift

• It is how far the function is shifted horizontally from the usual position.

Vertical Shift

• It is how far the function is shifted vertically from the usual position.

Frequency

• It is how often something happens per unit of time (per "1").

General Equation:

\begin{equation} y = A sin ( B x \pm C ) \pm D \end{equation}

\begin{equation} P = \frac{2\pi}{B} \end{equation}

\begin{equation} F = \frac{1}{P} \end{equation}

Where:

A = Amplitude

C = Phase Shift (when positive, it is shifted to the left; when negative, it is shifted to the right)

D = Vertical Shift (when positive, it is shifted upward; when negative, it is shifted downward)

P = Period

F = Frequency

Electronic Distance Measuring Equipments - Problem 5

A 750m. baseline distance is measured with an electronic total station. If the vendor's accuracy specification of the instrument is ±(3mm. + 2ppm). Determine the precision of the measured distance to the nearest thousand.

Solution:

\begin{equation} RA = \frac{e}{b} \end{equation}

\begin{equation} e = \pm A + (B \times C) \end{equation}

\begin{equation} e = \pm 3mm. + [(750mm. \times \frac{1000mm.}{1m.}) \times (2 \times 10^{-6})] \end{equation}

\begin{equation} e = \pm 4.5mm. \end{equation}

\begin{equation} RA = \frac{4.5mm.}{750m. \times \frac{1000mm.}{1m.}} \end{equation}

\begin{equation} RA = \frac{1}{166666.667} \end{equation}

\begin{equation} RA \approx \pm \frac{1}{166000} \end{equation}

Answer: \begin{equation} RA = \pm \frac{1}{166000} \end{equation}

Electronic Distance Measuring Equipments - Problem 4

If the error in EDM measurements is ±0.016ft in 10ft. distance. What is the relative accuracy of the EDM?

Solution:

\begin{equation} RA = \frac{e}{B} \end{equation}

\begin{equation} RA = \frac{0.016ft.}{10ft.} \end{equation}

\begin{equation} RA = \pm \frac{1}{625} \end{equation}

Answer: \begin{equation} RA = \pm \frac{1}{625} \end{equation}

Electronic Distance Measuring Equipments - Problem 3

If the error in EDM measurements is ±0.021ft in 1000ft. distance. What is the relative accuracy of the EDM?

Solution:

\begin{equation} RA = \frac{e}{B} \end{equation}

\begin{equation} RA = \frac{0.021ft.}{1000ft.} \end{equation}

\begin{equation} RA = \pm \frac{1}{47619} \end{equation}

Answer: \begin{equation} RA = \pm \frac{1}{47619} \end{equation}