2.1: Features and Fundamental Equations

1. Classification
   1. Deterministic: written functional form of independent variable \(t\) \[ x(t) = \alpha t^{n}\mathbf{e}^{-t/\beta}\]
   2. Adjective Periodic: repeats at some fundamental frequency
      1. Simple: implies only a single frequency
      2. Complex: multiple frequencies are possible including through summation: \[ y(t) = D_{0} + \sum_{i}C_{i}\cos(2\pi f_{i} t+\phi_{i}) \]
2. Combining periodics
   1. Parameters of general simple period equation \[ y(t) = D_{0}+C\sin(\omega t + \phi) \]
      1. \(D_{0}\) is the mean offset from zero
      2. \(C\) is the amplitude of oscillation
      3. \(\omega\) circular frequency [rad/s]
      4. \(f\) cyclic frequency [rev/time] (\(f = \frac{\omega}{2\pi})\)
      5. \(T\) period is the inverse of cyclic frequency [time] (\(T = 1/f\))
      6. \(\phi\) phase offset (left-right shift)
      7. Link to unscripted video with student interpretation and subsequent instruction ...
   2. Combining sines and cosines into one \[ A\cos(\omega t) + B\sin(\omega t) + D_{0}= C\sin(\omega t + \phi^{*}) +D_{0} = C\cos(\omega t -\phi) + D_{0} \]
      1. \(C\) is the norm of the amplitudes: \(C = \sqrt{A^2+B^2}\)
      2. Phase is a 4-quadrant inverse tangent \[\phi = \tan^{-1}\left(\frac{B}{A}\right)\] \[ \phi^{*} = \tan^{-1}\left(\frac{A}{B}\right) \] *sign of numerator and denominator are crucial to get values beyond \(\frac{-\pi}{2} \leq \phi \leq \frac{\pi}{2}\)
      3. \(D_{0}\) is the mean offset (up or down depending on positive or negative sign)
3. Statistical features
   1. Mean: average value over specific time or data range
      1. Continuous Form starting at \(t_{1}\) and finishing at \(t_{2}\) \[ \bar{x} = \frac{\int_{t_{1}}^{t_{2}}x(t)dt}{\int_{t_{1}}^{t_{2}}dt} \]
      2. Discrete is the standard version from list of \(x_{i}\) values \[ \bar{x} = \frac{1}{N}\sum_{i=1}^{N} x_{i} \]
      3. Be wary of the time range on dynamic signal (major part of some laboratory assignments)
   2. Root mean square (RMS): indicates variation of the mean
      1. Continuous Signal
         1. The square \(x(t)^{2}\)
         2. Of which a mean is calculated \(\frac{\int_{t_{1}}^{t_{2}}x(t)^{2}dt}{\int_{t_{1}}^{t_{2}}dt}\)
         3. Then rooted \(\sqrt{\frac{\int_{t_{1}}^{t_{2}}x(t)^{2}dt}{\int_{t_{1}}^{t_{2}}dt}}\)
         4. In the simplified form \[ x_{rms} = \sqrt{\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}x(t)^{2}dt} \]
      2. Similar in discrete form \[ x_{rms} = \sqrt{\frac{1}{N}\sum_{i=1}^{N}x_{i}^{2}} \]
      3. RMS is NOT the standard deviation when \(\bar{x}\neq 0\)
         1. Standard deviation is relative to the mean \(\bar{x}\) \[ x_{std} = \sqrt{\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}\left[x(t)-\bar{x}\right]^{2}dt} = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_{i}-\bar{x})^{2}} \]
4. Some Example Problems

   1. Exercise \(\PageIndex{1}\)

      Convert the following into a cosine only and a sine only functions with the appropriate phase: \[ x(t) = -5\sin(10\pi t) + 8\cos(10\pi t) \nonumber \]

      Produce a plot of at least three periods using all three functions at different timesteps or markers to prove each are equivalent.

      Answer

      The amplitude will be the norm of the two amplitudes \(C = \sqrt{(-5)^{2} + 8^{2}} = 9.43\).

      The phase will depend on whether it is cosine or sine. 

      \(\phi = \tan^{-1}(\frac{-5}{8}) = -0.559\) and \(\phi^{*} = \tan^{-1}(\frac{8}{-5}) = 2.13\)

      which results in \[ x(t) = 9.43 \cos (10\pi t + 0.559) = 9.43 \sin(10\pi t + 2.13) \nonumber \]

      With cyclic frequency of 5~Hz, the time range for the plot would need to be\(0 \leq t \leq \frac{3}{5}\)

      Exercise \(\PageIndex{2}\)

      Using the function\(x(t)\)in the prior Problem, calculate the mean for analog (integrate) and discrete (summation) methods for the following time intervals where\(T\)is the period of the original function:

      • \(0 < t < 3T\)~sec
      • \(1.5T < t < 4T\)~sec
      • \(0 < t < 1.1T\)~sec

      Interpret how the mean value changes as a function of the portion of the period included in the integration/summation range.

      Answer

      • \(0 < t < 3T\) sec: The period\(T = \frac{2\pi}{\omega} = \frac{2\pi}{10\pi} = 0.2\). The continuous integral will be \[ \bar{x} = \frac{1}{0.6-0} \int_{0}^{0.6}-5\sin(10\pi t) + 8\cos(10\pi t)dt \] \[\bar{x} = \frac{5}{0.6(10\pi)} \cos\left(10\pi t\right|_{0}^{0.6} + \frac{8}{0.6(10\pi)} \sin\left(10\pi t\right|_{0}^{0.6}\] \[ \bar{x} = \frac{5}{2\pi} (1-1) \] \[ \bar{x} = 0 \nonumber \] The discrete mean is straightforward, but will depend on whether you include the final timestep at\(t_{max}=0.2\). If using the functional values to the end and\(\delta t = 0.001\)~sec up yields a mean value\(\bar{x} = 0.0398\)whereas using only up to\(t_{max}=0.199\)~s makes\(\bar{x} = 9.23\times 10^{-16}\)(which is machine precision \ldots). The zero value is because the range was over a full period of the original signal.
      • \(1.5T < t < 4T\) sec: In continuous form \[ \bar{x} = \frac{1}{0.8-0.3} \int_{0.3}^{0.8}-5\sin(10\pi t) + 8\cos(10\pi t)dt = 0.637 \nonumber \] and in discrete form\(\bar{x} = 0.621\). Note that the fewer periods means that the partial period will shift the mean slightly.
      • \(0 < t < 1.1T\) sec: In continuous form \[ \bar{x} = \frac{1}{0.22-0} \int_{0}^{0.22}-5\sin(10\pi t) + 8\cos(10\pi t)dt = 0.542 \nonumber \] and in discrete form\(\bar{x} = 0.552\). The very narrow window and discrete result creates the numerical difference. If\(\Delta t\)were decreased it would more closely match the continuous form.