Sinusoidal Signals
BME253L - Fall 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
\[P_{avg} = \frac{V_{RMS}^2}{R}\]
\(V_{RMS}\) 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”?