22-Elec-A3 Signals and Communications · December 2015

Question 2 of 6: Fourier Series, Power Spectral Density and Band-Pass Filtering

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

Question 2: Fourier Series, Power Spectral Density and Band-Pass 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. From the figure, $x(t)$ is a two-level periodic waveform: it takes the value $+2$ over a window of width $\tau$ centred on each multiple of $3\tau$, and $-1$ everywhere else. The pulses are drawn centred at $t = 0$ and $t = 3\tau$, so the repetition period is $T_{0} = 3\tau$.

Find. (a) the trigonometric (real) Fourier series of $x(t)$; (b) its power spectral density; (c) the time-domain output of an ideal band-pass filter of centre frequency 10 kHz and bandwidth 5 kHz.

[Figure not reproduced: Figure 2.1 — The periodic waveform read from the examination figure: a $+2$ V pulse of width $\tau$ once every $3\tau$, sitting on a $-1$ V floor. The duty cycle is exactly $1/3$, which makes the signal zero-mean. See the official exam paper.]

Approach. Decompose $x(t)$ into a constant plus a scaled unit pulse train, so the coefficients follow from the standard rectangular-pulse-train result without evaluating a single integral; then square the coefficients for the power spectral density and test which harmonics survive the filter passband.

1. Split the waveform into a DC term and a pulse train. Let $p(t)$ be the unit-height train of $\tau$-wide pulses repeating every $T_{0} = 3\tau$. Then the waveform, which sits at $-1$ and jumps up by $3$ during each pulse, is $$x(t) = -1 + 3\,p(t)$$
2. Check the mean value. Over one period the signal spends $\tau$ at $+2$ and $2\tau$ at $-1$, so $$\langle x \rangle = \frac{2\tau + (-1)(2\tau)}{3\tau} = 0$$ The waveform is zero-mean, so the DC coefficient must vanish — a useful check on the algebra that follows.
3. Quote the exponential coefficients of the pulse train. For a unit-height train of width $\tau$ and period $T_{0}$, the standard result is $c_{n}^{(p)} = (\tau/T_{0})\operatorname{sinc}(n\tau/T_{0})$ with $\operatorname{sinc}(x) = \sin(\pi x)/(\pi x)$. Here $\tau/T_{0} = 1/3$, so scaling by 3 and subtracting the DC term gives $$c_{n} = -\delta_{n0} + 3 \cdot \tfrac{1}{3}\operatorname{sinc}(n/3) = \operatorname{sinc}(n/3), \quad n \neq 0; \qquad c_{0} = -1 + 1 = 0$$ confirming the zero mean found in Step 2.
4. Convert to the real (cosine) series. The waveform is even about $t = 0$, so all sine terms vanish and $a_{n} = 2c_{n}$: $$\boxed{\,x(t) = \sum_{n=1}^{\infty} a_{n}\cos(2\pi n f_{0}t), \qquad a_{n} = 2\operatorname{sinc}\!\left(\frac{n}{3}\right) = \frac{6\sin(n\pi/3)}{n\pi},\qquad f_{0} = 3333.33\ \text{Hz}\,}$$
5. Evaluate the leading coefficients. Because $\sin(n\pi/3)$ vanishes whenever $n$ is a multiple of 3, every third harmonic is absent: $$a_{1} = 1.6540,\quad a_{2} = 0.8270,\quad a_{3} = 0,\quad a_{4} = -0.4135,\quad a_{5} = -0.3308,\quad a_{6} = 0,\quad a_{7} = 0.2363$$ Physically the nulls sit at multiples of $1/\tau = 10$ kHz, which is where the sinc envelope of a $\tau$-wide pulse crosses zero — and $3f_{0} = 10$ kHz exactly.

Part (b) follows immediately, because a periodic power signal has a purely discrete power spectral density: each spectral line carries the mean-square power of its own complex exponential.

6. Write the power spectral density. With two-sided coefficients $c_{n} = \operatorname{sinc}(n/3)$, $$\boxed{\,S_{x}(f) = \sum_{n=-\infty}^{\infty} |c_{n}|^{2}\,\delta(f - nf_{0}) = \sum_{n \neq 0,\,n \neq 3k} \operatorname{sinc}^{2}\!\left(\frac{n}{3}\right)\delta(f - nf_{0})\,}$$ with line weights $|c_{1}|^{2} = |c_{-1}|^{2} = 0.6839$, $|c_{2}|^{2} = 0.1710$, $|c_{4}|^{2} = 0.0427$, and zeros at every multiple of 10 kHz.
7. Confirm by Parseval. The mean square computed in the time domain is $\left[2^{2}\tau + (-1)^{2}(2\tau)\right]/3\tau = 2\ \text{V}^{2}$, and summing $|c_{n}|^{2}$ over all $n$ converges to the same 2 V². The PSD therefore accounts for all the signal power.
8. Identify which harmonics reach the filter passband. An ideal band-pass filter centred at 10 kHz with bandwidth 5 kHz passes $7.5\ \text{kHz} \le f \le 12.5\ \text{kHz}$. The harmonic frequencies are $nf_{0} = 3.333n$ kHz, so $n = 2$ lands at 6.667 kHz (below the band), $n = 3$ lands at 10.000 kHz (inside), and $n = 4$ lands at 13.333 kHz (above the band). Exactly one harmonic is inside the passband.
9. Evaluate that harmonic. The only in-band line is $n = 3$, and from Step 5 its amplitude is $a_{3} = 6\sin(\pi)/(3\pi) = 0$. The filter therefore passes nothing at all: $$\boxed{\,y(t) = 0 \quad \text{for all } t\,}$$

This is the point of the question rather than an accident of the numbers. The duty cycle $\tau/T_{0} = 1/3$ places the sinc envelope's first null at $1/\tau = 10$ kHz, which coincides exactly with the filter's centre frequency, and the harmonic spacing $f_{0} = 10/3$ kHz is coarse enough that the nearest surviving lines fall 2.5 kHz and 0.83 kHz outside the band edges. A candidate who computes the passband but forgets to check the coefficient will confidently report a 10 kHz sinusoid that does not exist.