22-Elec-A3 Signals and Communications · May 2016

Question 1 of 6: Fourier Transform, Energy and Ideal Low-Pass Filtering

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

Question 1: Fourier Transform, Energy and Ideal Low-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. A triangular pulse read from the figure on page 2 of the paper:

Find. (a) the Fourier transform $X(f)$ in closed form, (b) the total signal energy $E$, and (c) the spectrum at the output of an ideal low-pass filter of bandwidth 1 Hz, drawn to scale.

Approach. Recognise the pulse as a shifted triangle, use the standard triangle-transform pair together with the time-shift property, then compute the energy directly in the time domain (and confirm it with Parseval's theorem) before truncating the spectrum to the filter passband.

1. Write the pulse in standard triangular form. The unit triangle of half-width $T$ is $\Lambda(t/T)=1-|t|/T$ for $|t|\le T$ and zero elsewhere. The figure shows a triangle of peak $A=2$ and half-width $T=2$ s centred at $t_0=12$ s, so $$x(t)=A\,\Lambda\!\left(\frac{t-t_0}{T}\right)=2\left(1-\frac{|t-12|}{2}\right),\qquad 10\le t\le 14,$$ and $x(t)=0$ outside that interval. Checking two points confirms the reading: $x(12)=2$ and $x(11)=x(13)=1$.
2. Transform the centred triangle. The triangle is the convolution of a rectangle with itself, so its transform is the square of a sinc: $$\mathcal{F}\left\{A\,\Lambda(t/T)\right\}=A\,T\,\operatorname{sinc}^2(fT), \qquad \operatorname{sinc}(u)\equiv\frac{\sin \pi u}{\pi u}.$$ With $A=2$ and $T=2$ the leading constant is $AT=4$.
3. Apply the time-shift property. A delay of $t_0$ multiplies the transform by $e^{-j2\pi f t_0}$ and leaves the magnitude untouched. Substituting $t_0=12$ s, $$\boxed{\;X(f)=4\,\operatorname{sinc}^{2}(2f)\;e^{-j24\pi f} =4\left[\frac{\sin 2\pi f}{2\pi f}\right]^{2}e^{-j24\pi f}\;}$$ so $|X(f)|=4\operatorname{sinc}^2(2f)$ and $\angle X(f)=-24\pi f$ (a pure linear phase, as it must be for a symmetric pulse that has only been delayed). A useful sanity check is the value at the origin: $X(0)=4$, which equals the area of the triangle, $\tfrac{1}{2}\times 4\ \text{s}\times 2=4$.
4. Compute the energy in the time domain. Energy is the integral of the squared signal, and by symmetry it is twice the integral over the rising half: $$E=\int_{-\infty}^{\infty}|x(t)|^{2}\,dt =2\int_{0}^{T}A^{2}\left(1-\frac{u}{T}\right)^{2}du=\frac{2A^{2}T}{3}.$$ Substituting $A=2$ and $T=2\ \text{s}$ gives $$\boxed{\;E=\frac{2(2)^{2}(2)}{3}=\frac{16}{3}=5.333\ \text{J}\;}$$ (joules per ohm, if $x(t)$ is taken as a voltage across a 1 Ω load).
5. Truncate the spectrum to the filter passband. An ideal low-pass filter of bandwidth $B=1$ Hz has $H(f)=1$ for $|f|\le 1$ Hz and $H(f)=0$ elsewhere, so the output spectrum is simply $$Y(f)=X(f)H(f)=\begin{cases}4\operatorname{sinc}^{2}(2f)\,e^{-j24\pi f}, & |f|\le 1\ \text{Hz}\\[2pt] 0,&|f|\gt 1\ \text{Hz}.\end{cases}$$ Because $\operatorname{sinc}(2f)$ has nulls wherever $2f$ is a non-zero integer, the passband contains the whole main lobe (nulls at $f=\pm 0.5$ Hz) plus the first pair of side lobes, and the cut-off at $f=\pm 1$ Hz falls exactly on a null — so the truncated spectrum is continuous, with no step at the band edge.
6. Quantify how little is lost. Integrating $|X(f)|^{2}$ over the passband and comparing with the total gives $$\frac{E_{\text{out}}}{E}=\frac{5.331\ \text{J}}{5.333\ \text{J}} \;\Rightarrow\;\boxed{\;E_{\text{out}}=5.331\ \text{J}\ \ (99.96\%\ \text{of }E)\;}$$ The triangular pulse is smooth (continuous, with only its derivative discontinuous), so its spectrum decays as $1/f^{2}$ and essentially all of the energy already lies inside 1 Hz. The time-domain output is therefore a very slightly rounded version of the input triangle.