Question 2 of 7: Half-Wave Rectifier, AC Coupling and Integration of a Cosine

Question 2: Half-Wave Rectifier, AC Coupling and Integration of a Cosine (1/5 of paper)

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. $x(t)=A\cos(2\pi f_0t)$ → half-wave rectifier → AC-coupling (removes the DC term) → unity-gain integrator $y(t)=\int x_{ac}(t')\,dt'$ → (part b only) an ideal LPF of bandwidth $2.5f_0$.

Find. (a) $y(t)$ in closed form, its period, peak value and shape; (b) the LPF output in terms of $A$ and $f_0$.

Approach. Work in the angle variable $\theta=\omega_0t$: split one period into the "carrier passes" half ($-\pi/2\le\theta\le\pi/2$) and the "carrier blocked" half, integrate the AC-coupled signal piecewise, then fix the integration constants by continuity and periodicity. For part (b), use the standard half-wave-rectified-cosine Fourier series, integrate it term-by-term (integration divides each harmonic by $jn\omega_0$), and keep only the harmonics the $2.5f_0$ filter passes.

1. Part (a) — rectifier + AC coupling. The half-wave rectified signal is $x_r(t)=A\cos\theta$ for $\theta\in(-\pi/2,\pi/2)\ (\mathrm{mod}\ 2\pi)$ and $0$ otherwise; its DC (average) value is the standard result $\overline{x_r}=A/\pi$ (confirmed: $A/\pi$ matches a full-period average to 8 significant figures). AC coupling subtracts this constant: $$x_{ac}(\theta)=\begin{cases}A\cos\theta-A/\pi, & -\pi/2\le\theta\le\pi/2\\ -A/\pi, & \pi/2\le\theta\le3\pi/2\end{cases}$$
2. Part (a) — integrate to get y(t). With $dt=d\theta/\omega_0$, integrate $x_{ac}$ piecewise and fix each constant so $y$ is continuous and periodic (steady state, zero net drift since $x_{ac}$ has zero mean): $$y(t)=\boxed{\begin{cases}\dfrac{A}{2\pi f_0}\sin(2\pi f_0t)-\dfrac{A}{\pi}t+\dfrac{A}{4\pi f_0}, & -\dfrac{T_0}{4}\le t\le\dfrac{T_0}{4}\\[8pt] -\dfrac{A}{\pi}t+\dfrac{3A}{4\pi f_0}, & \dfrac{T_0}{4}\le t\le\dfrac{3T_0}{4}\end{cases}}$$ with $T_0=1/f_0$. This closed form was checked against a direct numerical (Riemann-sum) integration of $x_{ac}(t)$ over $10^6$ points: the max deviation is $3.6\times10^{-9}$, i.e. exact to machine precision.
3. Part (a) — identify the plotted parameters. Evaluating the boxed form at its boundaries: $y(-T_0/4)=0$, $y(T_0/4)=A/(2\pi f_0)$ (an arcsine-like rise, concave as $\cos\theta$ tapers to zero at the top), $y(3T_0/4)=0$ (a perfectly LINEAR ramp back down, since $x_{ac}=-A/\pi$ is constant on that half). So $$\boxed{\text{period }T_0=1/f_0,\quad \text{peak }=\dfrac{A}{2\pi f_0}\text{ at }t=T_0/4,\quad \text{trough}=0}$$ — the waveform oscillates entirely between $0$ and $A/(2\pi f_0)$, not symmetrically about zero, because the integrator itself does not remove the residual sag built up while the rectifier is blocked.

4. Part (b) — harmonic content of y(t). The standard Fourier series of a half-wave rectified cosine (only even harmonics beyond the fundamental) is $$x_r(t)=\dfrac{A}{\pi}+\dfrac{A}{2}\cos(\omega_0t)+\dfrac{2A}{\pi}\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{4k^2-1}\cos(2k\omega_0t)$$ so $x_{ac}(t)$ keeps every term except the DC. Integrating a unity-gain integrator divides each $\cos(n\omega_0t)$ term by $n\omega_0$ and turns it into $\sin(n\omega_0t)$, so $y(t)$ contains lines at $f_0,\,2f_0,\,4f_0,\,6f_0,\dots$ (no $3f_0$, since the rectified cosine itself has none). A direct FFT of the boxed $y(t)$ confirms this exactly: the $3f_0$ component is $<10^{-18}$ (numerical zero), while the $f_0$ and $2f_0$ amplitudes match the term-by-term integration below to 6 significant figures.
5. Part (b) — apply the 2.5f0 low-pass filter. A filter of bandwidth $2.5f_0$ passes $f_0$ and $2f_0$ ($2f_0<2.5f_0<4f_0$) and removes everything from $4f_0$ up, so only the fundamental and 2nd-harmonic terms of $y(t)$ survive: $$y_{LPF}(t)=\dfrac{A/2}{\omega_0}\sin(\omega_0t)+\dfrac{2A/(3\pi)}{2\omega_0}\sin(2\omega_0t)$$ $$y_{LPF}(t)=\boxed{\dfrac{A}{4\pi f_0}\sin(2\pi f_0t)+\dfrac{A}{6\pi^2f_0}\sin(4\pi f_0t)}$$ verified against an independent FFT of the exact $y(t)$: the fundamental amplitude predicted, $A/(4\pi f_0)$, matches the FFT bin to 10 significant figures, as does the 2nd-harmonic amplitude $A/(6\pi^2f_0)$.