22-Elec-A3 Signals and Communications · May 2015

Question 1 of 7: Periodic Ramp Train — Fourier Series, Power Spectral Density and Ideal Filtering

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

Question 1: Periodic Ramp Train — Fourier Series, Power Spectral Density and Ideal Filtering (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. A single ramp pulse \(p(t) = t/a\) supported on \(|t| \le a\) (total width \(2a\)), with \(a = 0.5\text{ ms}\); the signal \(x(t)\) is the sum of that pulse repeated at every multiple of \(a\). The filter of part (d) has impulse response \(h(t) = \sin(2\pi t/a)/(2\pi t/a)\).

Find. The trigonometric and exponential Fourier series of \(x(t)\), its power spectral density, and the filter output \(y(t) = x(t) * h(t)\).

Approach. Because the pulse is twice as wide as the repetition interval, first collapse the overlapping copies into a single closed-form periodic waveform; then compute the Fourier coefficients of that waveform, read the line spectrum from them, and finally treat \(h(t)\) as an ideal low-pass filter that keeps only the lines inside its passband.

1. Collapse the overlapping copies into one periodic waveform. A copy \(p(t-ka)\) is non-zero when \(|t-ka| \le a\), i.e. when \(ka\) lies in the interval \([t-a,\ t+a]\). That interval has width \(2a\) while the shifts are spaced \(a\) apart, so exactly two copies are active at any instant. For \(0 \le t \lt a\) those are \(k=0\) and \(k=1\): $$x(t) = \frac{t}{a} + \frac{t-a}{a} = \frac{2t-a}{a}, \qquad 0 \le t \lt a$$ and the pattern repeats every \(a\). Hence \(x(t)\) is a sawtooth of period \(T_0 = a = 0.5\text{ ms}\) sweeping linearly from \(-1\) to \(+1\), with fundamental frequency $$\boxed{\,f_0 = \frac{1}{a} = 2\ \text{kHz}\,}$$
2. Sketch the resulting waveform. The overlap has turned a one-sided ramp into a symmetric, zero-mean sawtooth — the key structural insight of the question.
   Figure Q1.1 — the two overlapping ramp copies collapse to a sawtooth of period a = 0.5 ms sweeping from -1 to +1.
3. Real (trigonometric) Fourier series. The waveform is odd about the mid-point of each period, so the DC term and all cosine terms vanish, \(a_0 = a_n = 0\). The sine coefficients follow from $$b_n = \frac{2}{a}\int_{0}^{a}\frac{2t-a}{a}\,\sin\!\left(\frac{2\pi n t}{a}\right)dt$$ Integrating by parts, the \(2t/a\) term contributes \(-2/(n\pi)\) and the constant term contributes nothing, giving $$\boxed{\,b_n = -\frac{2}{n\pi},\qquad x(t) = -\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin\!\left(\frac{2\pi n t}{a}\right)\,}$$ The first three harmonics are therefore \(b_1 = -0.6366\), \(b_2 = -0.3183\), \(b_3 = -0.2122\) — the classic \(1/n\) sawtooth roll-off, present at every harmonic (unlike a square or triangle wave, which drop the even orders).
4. Complex (exponential) Fourier series. Writing \(\sin\theta = (e^{j\theta}-e^{-j\theta})/2j\) converts the sine series directly: $$c_n = \frac{b_n}{2j} = \frac{-2/(n\pi)}{2j} \;\Longrightarrow\; \boxed{\,c_n = \frac{j}{n\pi}\ (n \ne 0),\qquad c_0 = 0\,}$$ $$x(t) = \sum_{n \ne 0}\frac{j}{n\pi}\,e^{\,j 2\pi n t/a}$$ Direct integration of \(c_n = \frac{1}{a}\int_0^a x(t)e^{-j2\pi n t/a}dt\) reproduces the same result. Note \(c_{-n} = c_n^{*}\) as required for a real signal, and that the coefficients are purely imaginary because \(x(t)\) is real and odd.
5. Power spectral density. For a periodic signal the PSD is a line spectrum with a weight \(|c_n|^2\) at each harmonic: $$\boxed{\,S_x(f) = \sum_{n \ne 0}\frac{1}{n^2\pi^2}\,\delta\!\left(f - \frac{n}{a}\right)\,}$$ so the lines sit at \(\pm 2,\ \pm 4,\ \pm 6\ \text{kHz}\dots\) with weights \(0.1013,\ 0.0253,\ 0.0113\dots\) The total power follows from Parseval, $$P = \sum_{n \ne 0}|c_n|^2 = \frac{2}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{2}{\pi^2}\cdot\frac{\pi^2}{6} = \frac{1}{3}$$ which agrees exactly with the time-domain average \(\frac{1}{a}\int_0^a\left(\frac{2t-a}{a}\right)^2dt = 1/3\) — a useful self-check.
   Figure Q1.2 — power spectral density of x(t): lines of weight 1/(nπ)² at every multiple of 2 kHz; total power 1/3.
6. Identify the filter. Writing \(W = 1/a\), the impulse response is \(h(t) = \dfrac{\sin(2\pi W t)}{2\pi W t} = \dfrac{1}{2W}\cdot\dfrac{\sin(2\pi W t)}{\pi t}\), whose transform is an ideal brick-wall low-pass characteristic $$H(f) = \frac{1}{2W}\,\mathrm{rect}\!\left(\frac{f}{2W}\right) = \frac{a}{2}\quad \text{for } |f| \le \frac{1}{a} = 2\ \text{kHz},\ \ 0 \text{ elsewhere}$$ The cut-off therefore lands exactly on the fundamental: every harmonic \(n \ge 2\) is removed and only the \(n = \pm 1\) pair survives.
7. Filter output. Scaling the surviving fundamental by the passband gain \(a/2\), $$y(t) = \frac{a}{2}\,b_1 \sin\!\left(\frac{2\pi t}{a}\right) = \boxed{\,-\frac{a}{\pi}\sin\!\left(\frac{2\pi t}{a}\right) = -1.59\times 10^{-4}\,\sin\!\left(2\pi\,(2000)\,t\right)\,}$$ The rich sawtooth has been reduced to a single 2 kHz sinusoid of amplitude \(159\ \mu\text{V}\) (per unit input), inverted with respect to the reference sine.
   Figure Q1.3 — filter output: only the 2 kHz fundamental survives, inverted and scaled by the passband gain a/2.