22-Elec-A3 Signals and Communications · December 2017

Question 1 of 5: Frequency Doubler — Full-Wave Rectifier and Ideal Band-Pass Filter

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Frequency Doubler — Full-Wave Rectifier and Ideal Band-Pass Filter (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.

Find. The Fourier series of the rectified waveform; the centre frequency and bandwidth of the output filter; the exact time-domain output of the doubler; and the insertion loss in dB.

Approach. Expand $|x(t)|$ in a Fourier series (the rectifier halves the period, so only even multiples of $f_0$ survive), size the band-pass filter to admit the $2f_0$ line while rejecting DC and $4f_0$, then compare the normalised power of the surviving line with the power of the input sinusoid.

1. Recognise what full-wave rectification does to the period. Writing $\theta = 2\pi f_0 t + \pi/4$, the rectifier output is $$y(t) = |x(t)| = 2\,\bigl|\cos\theta\bigr| .$$ Taking the absolute value folds the negative half-cycles up, so $y$ repeats every half period of $x$: the fundamental period becomes $T_0/2 = 1/(2f_0) = 50$ ns and the fundamental frequency is $2f_0 = 20\ \text{MHz}$. This is exactly why the arrangement doubles frequency — the rectifier manufactures energy at $2f_0$ that was not present at the input.
2. Write the Fourier series of the rectified cosine. The standard expansion of a full-wave-rectified cosine of unit amplitude is $$|\cos\theta| = \frac{2}{\pi} + \frac{4}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{4n^{2}-1}\,\cos(2n\theta).$$ Multiplying by the amplitude $A = 2$ and substituting $\theta = 2\pi f_0 t + \pi/4$ gives the required series, $$\boxed{\,y(t) = \frac{4}{\pi} + \frac{8}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{4n^{2}-1}\,\cos\!\bigl(2\pi (2nf_0) t + n\tfrac{\pi}{2}\bigr)\,}$$ Note carefully that the $n$-th term sits at frequency $2nf_0$ and carries phase $n\pi/2$, because the argument $2n\theta$ doubles both the frequency and the phase offset $n$ times over.
3. Evaluate the first few coefficients numerically. With $A = 2$ V the one-sided amplitudes are $c_0 = 4/\pi = 1.2732$ V (DC), $c_1 = 8/(3\pi) = 0.8488$ V at 20 MHz, $c_2 = -8/(15\pi) = -0.1698$ V at 40 MHz and $c_3 = 8/(35\pi) = 0.0728$ V at 60 MHz. The $1/n^{2}$ decay means the wanted line at $2f_0$ dominates everything above DC by a factor of five.
Figure 1.2 — One-sided line spectrum of the rectified signal. Only the 2f₀ = 20 MHz line falls inside the shaded passband.
4. (b) Specify the band-pass filter. The doubler must deliver the $2f_0$ component alone, so the filter is centred on that line: $$f_c = 2f_0 = 20\ \text{MHz}, \qquad B = \frac{f_c}{Q} = \frac{20\ \text{MHz}}{10} = \boxed{2\ \text{MHz}}$$ The ideal passband therefore runs from $f_c - B/2 = 19$ MHz to $f_c + B/2 = 21$ MHz, with unity gain in band and zero gain outside. Checking the rejection requirement: the nearest unwanted components are DC (0 MHz) and $4f_0 = 40$ MHz, both far outside a 19–21 MHz window, so a selectivity of 10 is generous — even a $Q$ of 2 would have separated the lines. The specification also implies a group delay flat across the 2 MHz band if waveform fidelity matters.
5. (c) Time-domain output. An ideal filter passes its in-band input unchanged, so only the $n = 1$ term of the series survives: $$z(t) = \frac{8}{3\pi}\cos\!\Bigl(2\pi(2f_0)t + \frac{\pi}{2}\Bigr)$$ that is, $$\boxed{\,z(t) = 0.8488\,\cos\!\bigl(4\pi\times 10^{7}\,t + \tfrac{\pi}{2}\bigr) = -0.8488\,\sin\!\bigl(4\pi\times 10^{7}\,t\bigr)\ \text{V}\,}$$ The output is a clean 20 MHz sinusoid of amplitude 0.8488 V whose phase is twice the input phase, as it must be for any true frequency-doubling operation.
6. (d) Insertion loss. Insertion loss compares the useful signal power delivered at the output with the power supplied at the input. Using normalised (1 Ω) powers, a sinusoid of amplitude $A$ carries $P = A^{2}/2$: $$P_{\text{in}} = \frac{A^{2}}{2} = \frac{2^{2}}{2} = 2.000\ \text{W}, \qquad P_{\text{out}} = \frac{(8/3\pi)^{2}}{2} = \frac{0.8488^{2}}{2} = 0.3603\ \text{W}$$ Hence $$\text{IL} = 10\log_{10}\frac{P_{\text{in}}}{P_{\text{out}}} = 10\log_{10}\frac{2.000}{0.3603} = 10\log_{10}(5.5516) = \boxed{7.44\ \text{dB}}$$ Equivalently, $\text{IL} = 20\log_{10}(2 \div 0.8488) = 7.44$ dB on an amplitude basis. Roughly 82 % of the input power is discarded — it ends up in the DC term and the higher even harmonics that the filter throws away. That is the intrinsic price of this passive doubling scheme.