Sinusoidal Signals

BME253L - Fall 2025

Dr. Mark Palmeri, M.D., Ph.D.

Duke University

September 29, 2025

Learning Objectives

• Review sinusoidal signals and their properties.

• Root mean square (RMS) values of sinusoidal signals.

• Complex numbers and Euler’s formula.

• Phasor representation of sinusoidal signals.

Sinusoidal Source Signal

• A sinusoidal voltage source can be represented as:

\[ v(t) = V_{A} \sin(2 \pi f t + \phi) = V_{A} \sin(\omega t + \phi) = V_{A} \sin(2 \pi (t/T) + \phi) \]

• Where:
  • \(V_{A}\) is the amplitude (peak value) of the signal.
  • \(f\) is the frequency in Hertz (Hz).
  • \(\omega = 2 \pi f\) is the angular frequency in radians per second (rad/s).
  • \(T = 1/f\) is the period of the signal in seconds (s).
  • \(\phi\) is the phase angle in radians.
• Lets draw this…

Complex Plane Representation

• We can represent a fixed frequency sinusoidal signal as a rotating vector in the complex plane.
  • The horizontal axis represents the real part (cosine component).
  • The vertical axis represents the imaginary part (sine component).
  • The length of the vector represents the amplitude \(V_{A}\).
  • The angle of the vector with respect to the horizontal axis represents the phase angle \(\phi\).
  • As time progresses, the vector rotates counterclockwise at an angular velocity \(\omega\).
  • One full rotation corresponds to one period \(T\) of the sinusoidal signal.
• Lets draw this…

Euler’s Formula and Complex Exponentials

• Euler’s formula:

\[ e^{j\theta} = \cos(\theta) + j \sin(\theta) \]

• We can express the sinusoidal signal as the imaginary part of a complex exponential:

\[ v(t) = \text{Im} \{ V_{A} e^{j(\omega t + \phi)} \} \]

• We can also express the (cos)sinusoidal signal as the real part of a complex exponential:

\[ v(t) = \text{Re} \{ V_{A} e^{j(\omega t + \phi - \pi/2)} \} \]

• What does this look like on the complex plane?

Power in Sinusoidal Signals

• Why not just use average value?

  • The average value of a sinusoidal signal over one period is zero, regardless of frequency or amplitude.

  • This is because the positive and negative halves cancel out.

  • Therefore, average value is not a good measure of the signal’s power.

• Does this make sense?

• Instead, we use the root mean square (RMS) value.

Calculating RMS Value

\[ V_{RMS} = \sqrt{\frac{1}{T} \int_0^T [V_A \sin(\omega t + \phi)]^2 dt} \]

\[ P_R(t) = i_R v_R = \frac{v_R^2}{R} = \frac{V_A^2 \sin^2(\omega t + \phi)}{R} \]

Let’s calculate the average power over one period:

\[ P_{avg} = \frac{1}{T} \int_0^T P_R(t) dt = \frac{1}{T} \int_0^T \frac{V_A^2 \sin^2(\omega t + \phi)}{R} dt = \frac{V_A^2}{R} \cdot \frac{1}{T} \int_0^T \sin^2(\omega t + \phi) dt \]

Since the integral of \(\sin^2\) over one period is \(T/2\), we have:

\[ P_{avg} = \frac{V_A^2}{R} \cdot \frac{1}{T} \cdot \frac{T}{2} = \frac{V_A^2}{2R} \]

RMS Value and Power

\[P_{avg} = \frac{V_{RMS}^2}{R}\]

Important

\(V_{RMS}\) is the equivalent DC voltage that would deliver the same average power to a resistive load as the AC voltage.

How does \(V_{RMS}\) relate to \(V_A\)?

\[ \begin{gather} V(t) = V_A \cos(\omega t + \phi) \\ V_{RMS} = \sqrt{\frac{1}{T} \int_0^T V_A^2 \cos^2(\omega t + \phi) dt} \\ = V_A \sqrt{\frac{1}{T} \int_0^T \cos^2(\omega t + \phi) dt} \\ \textrm{Remember that:} \cos^2(x) = \frac{1 + \cos(2x)}{2} \\ = V_A \sqrt{\frac{1}{T} \int_0^T \frac{1 + \cos(2(\omega t + \phi))}{2} dt} \\ = V_A \sqrt{\frac{1}{T} \cdot \frac{T}{2}} \\ = \frac{V_A}{\sqrt{2}} \end{gather} \]

\[ V_{RMS} = \frac{V_A}{\sqrt{2}} \approx 0.707 V_A \]

What does this “look like”?