Question 2 of 7: Half-wave rectifier, a.c. coupling and integration of a cosine (20 marks)

Question 2 — Half-wave rectifier, a.c. coupling and integration of a cosine (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Input $x(t) = A\cos(2\pi f_0 t)$ of period $T_0 = 1/f_0$ and radian frequency $\omega_0 = 2\pi f_0$. Cascade: ideal half-wave rectifier (passes the positive half cycles, blocks the negative ones), then a d.c. block, then an ideal unity-gain integrator, and in part (b) an ideal low-pass filter of bandwidth $2.5 f_0$. Steady-state operation is to be assumed.

Find. (a) the steady-state integrator output $z(t)$, plotted with every parameter labelled; (b) the filter output in terms of $A$ and $f_0$.

Approach. Integrating a periodic waveform only yields a bounded, periodic result if that waveform has zero mean — which is precisely what the d.c. block guarantees. So the natural route is to compute the rectifier’s Fourier series, remove its d.c. term, integrate term by term for the spectral answer of part (b), and integrate the waveform directly in closed form for the plot of part (a).

1. Write the Fourier series of the half-wave rectified cosine. Taking the cosine to conduct on $|t| \le T_0/4$ (mod $T_0$), the standard result is $$g(t) = \frac{A}{\pi} + \frac{A}{2}\cos\omega_0 t + \frac{2A}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{4k^{2}-1}\cos 2k\omega_0 t .$$ The spectrum therefore contains a d.c. term, the fundamental at half the input amplitude, and even harmonics only. Numerically integrating the rectified waveform confirms $A/\pi = 0.3183A$ for the mean, $A/2$ for the fundamental, $2A/3\pi = 0.2122A$ for the second harmonic and $-2A/15\pi = -0.0424A$ for the fourth.
2. Remove the d.c. term. The a.c.-coupling stage subtracts the mean, leaving $$v(t) = g(t) - \frac{A}{\pi} = \frac{A}{2}\cos\omega_0 t + \frac{2A}{\pi}\sum_{k\ge1}\frac{(-1)^{k+1}}{4k^{2}-1}\cos 2k\omega_0 t .$$ Because $v$ now has zero average, its integral does not accumulate a ramp, and a genuine steady state exists — the whole reason the d.c. block is in the chain.
3. Integrate in closed form over the conducting interval. On $|t| \le T_0/4$ the rectifier is passing, so $v = A\cos\omega_0 t - A/\pi$ and $$z(t) = \int v\,dt = \frac{A}{\omega_0}\sin\omega_0 t - \frac{A}{\pi}t .$$ The constant of integration is zero because $z$ inherits the odd symmetry of the integral of an even, zero-mean waveform. Evaluating at the end of the conducting interval, $$z\!\left(\frac{T_0}{4}\right) = \frac{A}{\omega_0} - \frac{A}{\pi}\cdot\frac{T_0}{4} = \frac{A}{\omega_0} - \frac{A}{2\omega_0} = \frac{A}{2\omega_0}.$$
4. Integrate over the non-conducting interval. On $T_0/4 \le t \le 3T_0/4$ the diode is off, so $v = -A/\pi$ is constant and $z$ falls along a straight line of slope $-A/\pi$: $$z(t) = \frac{A}{2\omega_0} - \frac{A}{\pi}\left(t - \frac{T_0}{4}\right), \qquad z\!\left(\frac{3T_0}{4}\right) = \frac{A}{2\omega_0} - \frac{A}{\pi}\cdot\frac{T_0}{2} = -\frac{A}{2\omega_0},$$ which is exactly the negative of the value a half period earlier. The waveform therefore closes on itself with period $T_0$ — the formal confirmation of steady state.
5. Locate and evaluate the peak. The maximum occurs where $dz/dt = v = 0$, i.e. where $\cos\omega_0 t = 1/\pi$: $$\omega_0 t^{*} = \arccos\!\left(\frac{1}{\pi}\right) = 1.2468\ \text{rad} \qquad\Longrightarrow\qquad t^{*} = 0.1984\,T_0 .$$ Substituting back, $$z_{\max} = \frac{A}{\omega_0}\left[\sqrt{1 - \frac{1}{\pi^{2}}} - \frac{\arccos(1/\pi)}{\pi}\right] = 0.5512\,\frac{A}{\omega_0} \qquad\Longrightarrow\qquad \boxed{\;z_{\max} = 0.0877\,\frac{A}{f_0},\quad z_{\min} = -0.0877\,\frac{A}{f_0}\;}$$ so the peak-to-peak swing is $0.1754\,A/f_0$ and the output period is $T_0 = 1/f_0$. A numerical integration of the whole cascade reproduces the same extrema to four figures.

The plot shows a smooth sinusoidal-plus-ramp arch on each conducting quarter cycle joined by a straight discharge ramp of slope $-A f_0/\pi$ — not the pure sine wave that a careless term-by-term sketch would suggest. Note also that the peak does not sit at $t = T_0/4$: the d.c. subtraction moves the zero-crossing of $v$ inwards to $0.1984\,T_0$.

6. Integrate the series term by term for the filtered answer. An ideal integrator maps $\cos n\omega_0 t$ to $\sin(n\omega_0 t)/(n\omega_0)$, so $$z(t) = \frac{A}{2\omega_0}\sin\omega_0 t + \frac{2A}{\pi\omega_0}\sum_{k\ge1}\frac{(-1)^{k+1}}{2k(4k^{2}-1)}\sin 2k\omega_0 t .$$ Each harmonic is attenuated by an extra factor $1/n$, which is why the integrated waveform looks so much smoother than the rectifier output.
7. Apply the $2.5f_0$ low-pass filter. The filter passes d.c., the line at $f_0$ and the line at $2f_0$, and rejects everything from $4f_0$ upwards (there is no line at $3f_0$; a half-wave rectified cosine has no odd harmonics beyond the fundamental). Keeping the two surviving terms with $\omega_0 = 2\pi f_0$: $$\boxed{\;y_{\mathrm{LPF}}(t) = \frac{A}{4\pi f_0}\sin(2\pi f_0 t) + \frac{A}{6\pi^{2}f_0}\sin(4\pi f_0 t)\;}$$ that is $0.07958\,(A/f_0)\sin\omega_0 t + 0.01689\,(A/f_0)\sin 2\omega_0 t$. Numerically extracting these two Fourier coefficients from the simulated waveform gives the same values, confirming both the series and the integration.

The second harmonic is only 21 % of the fundamental in amplitude, so the filtered output is a mildly distorted sine wave of frequency $f_0$ — the integrator plus low-pass filter has effectively recovered a smoothed version of the original tone from the rectified signal.