Chapter 8 Practice Problems — AC Signals and Sinusoidal Waveforms
These problems are meant to build your problem-solving intuition. A hint is provided for each — try the problem on your own first before reading it.
Problem 1 — Sinusoidal Parameters
A voltage signal is given by \(v(t) = 8\sin(500\pi t - 30°)\) volts.
(a) Identify the peak amplitude \(V_m\), angular frequency \(\omega\), frequency \(f\), period \(T\), and phase angle \(\phi\).
(b) Calculate the RMS voltage.
(c) Write the equivalent cosine expression for this signal. (Recall: \(\sin\theta = \cos(\theta - 90°)\).)
Hint
(a) Match the signal to the general form \(v(t) = V_m \sin(\omega t + \phi)\): the coefficient of t inside the sine gives \(\omega\). Then \(f = \omega/(2\pi)\) and \(T = 1/f\). The phase is the constant term in the argument.
(b) For any sinusoid: \(V_{rms} = V_m / \sqrt{2}\). This factor of \(\sqrt{2} \approx 1.414\) appears because the sinusoid's squared average equals half its peak squared.
(c) Use the identity \(\sin(\theta) = \cos(\theta - 90°)\). Add −90° to the existing phase angle to convert the sine to a cosine form.
Problem 2 — RMS Values and Power
A sinusoidal current \(i(t) = 5\cos(120\pi t + 45°)\) A flows through a 10 Ω resistor.
(a) Find the RMS current \(I_{rms}\).
(b) Calculate the average power dissipated in the resistor using \(P = I_{rms}^2 R\).
(c) What peak current would a DC source need to dissipate the same average power in the same resistor? Compare this to the AC peak current and explain the factor you observe.
Hint
(a) The cosine form has the same RMS formula as sine: \(I_{rms} = I_m / \sqrt{2}\) where \(I_m = 5\) A. Phase has no effect on RMS.
(b) Use \(P = I_{rms}^2 R\). This is the average (not instantaneous) power and is what a wattmeter would read.
(c) For DC, \(P = I_{DC}^2 R\), so \(I_{DC} = \sqrt{P/R} = I_{rms}\). The DC equivalent current equals the RMS current — that is precisely the definition of RMS: the equivalent DC that delivers the same power.
Problem 3 — Phase Relationships
Two sinusoidal voltage signals share the same frequency \(f = 1\) kHz:
[v_1(t) = 10\sin(2\pi \cdot 1000 \cdot t)\text{ V}] [v_2(t) = 10\sin(2\pi \cdot 1000 \cdot t + 60°)\text{ V}]
(a) Which signal leads and by how many degrees? By how many milliseconds?
(b) Represent each signal as a phasor in polar form using its RMS amplitude and phase angle.
(c) Using the phasor representations, add the two phasors. Convert to rectangular form, add, and express the result in polar form. Then write the corresponding time-domain expression.
Hint
(a) The signal with the larger (more positive) phase angle leads. The time delay between them is \(\Delta t = \Delta\phi / (360° \cdot f)\).
(b) The phasor magnitude is the RMS value: \(V_{rms} = V_m/\sqrt{2} \approx 7.07\) V. The phase angle is taken directly from the time-domain expression. Phasors: \(\mathbf{V_1} = 7.07\angle 0°\) V and \(\mathbf{V_2} = 7.07\angle 60°\) V.
(c) Convert each phasor to rectangular form (\(a + jb\)), add real and imaginary parts separately, then convert back to polar (\(r\angle\theta\)). The time-domain sum uses the peak amplitude \(V_m = r\sqrt{2}\) and the phase from the polar result.
Problem 4 — Frequency Spectrum of Non-Sinusoidal Waveforms
A square wave with peak amplitude 4 V and fundamental frequency \(f_1 = 1\) kHz can be approximated by its Fourier series:
(a) Identify the frequencies, amplitudes, and periods of the first three harmonic components.
(b) Calculate the RMS value of the first harmonic (fundamental) alone.
(c) At what frequency does the 7th harmonic occur, and what is its amplitude? Explain in one sentence why higher harmonics have smaller amplitudes.
Hint
(a) The fundamental is at \(f_1\), the third harmonic at \(3f_1\), and the fifth harmonic at \(5f_1\). The amplitude of the nth harmonic is \(\frac{16}{\pi n}\) (for odd n only — the square wave contains only odd harmonics). The period of each harmonic is \(T = 1/f\).
(b) RMS of the fundamental: \(V_{rms,1} = A_1/\sqrt{2}\) where \(A_1 = 16/\pi\) V. This is the same formula as for a pure sinusoid.
(c) The 7th harmonic is at \(7f_1 = 7\) kHz with amplitude \(16/(\pi \cdot 7)\) V. Higher harmonics have smaller amplitudes because a square wave's sharp transitions contain mostly low-frequency energy; the high-frequency components are responsible only for the steepness of the edges and contribute progressively less to the overall waveform shape.