Scaling, Shifting and Modulation of a Sinc Pulse: 22-Elec-A3 Signals and Communications, May 2014, Q3

Question 3: Scaling, Shifting and Modulation of a Sinc Pulse (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. $x(t)=2\operatorname{sinc}(2\pi t)$, with the paper's convention $\operatorname{sinc}(x)=\sin x/x$ and the table entry $2B\operatorname{sinc}(2\pi Bt)\leftrightarrow\operatorname{rect}(f/2B)$.

Find. $X(f)$; then the magnitude and phase spectra of the compressed-and-delayed signal $y_1$, and of the modulated signal $y_2$.

Approach. Get $X(f)$ from the table, then apply the three transform properties in the order the signal is built: time scaling, time shifting, and finally frequency shifting (modulation). Each property changes exactly one feature of the picture.

Part (a): read the pair off the table. Matching $2B\operatorname{sinc}(2\pi Bt)$ against $2\operatorname{sinc}(2\pi t)$ gives $B=1$, and the amplitude $2B = 2$ then matches exactly, so $$\boxed{X(f)=\operatorname{rect}\!\left(\frac{f}{2}\right)}$$ a unit-height rectangle occupying $|f| \lt 1$ Hz, with zero phase everywhere (the signal is real and even).

X(f) = rect(f/2): unit height over |f| < 1 Hz, zero elsewhere.

Part (b) applies two properties in sequence. It is worth doing them in the right order: the argument $2(t-0.5)$ means "compress by 2, then delay by 0.5 s", and writing it as $x(2t-1)$ tempts one into using a delay of 1 s, which is wrong.

Apply time scaling. With $g(t)=x(2t)$, the table gives $g(t)=x(at)\leftrightarrow \frac{1}{|a|}X(f/a)$ with $a=2$: $$G(f)=\tfrac12 X\!\left(\tfrac{f}{2}\right)=\tfrac12\operatorname{rect}\!\left(\frac{f}{4}\right)$$ Compressing the signal in time by 2 halves the spectral height and doubles the width, conserving the area.
Apply the time shift. A delay of $t_0 = 0.5$ s multiplies the transform by $e^{-j2\pi f t_0}$ and leaves the magnitude alone: $$\boxed{Y_1(f)=\tfrac12\operatorname{rect}\!\left(\frac{f}{4}\right)e^{-j\pi f}}$$ so $|Y_1(f)| = 1/2$ for $|f| \lt 2$ Hz and zero outside, while $\arg Y_1(f) = -\pi f$ rad is a straight line through the origin of slope $-\pi$ rad/Hz, running from $+2\pi$ at $f=-2$ Hz to $-2\pi$ at $f=+2$ Hz.

y1(t) = x(2(t − 0.5)): time compression by 2 halves the height and doubles the width (1/2 over |f| < 2 Hz); the 0.5 s delay adds the linear phase −πf, reaching ∓2π at the band edges.

Check: phase is plotted unwrapped. The phase above is drawn as the continuous line $-\pi f$ over the band. A plotting routine that wraps to $(-\pi,\pi]$ would show the same information as a sawtooth with jumps of $2\pi$ at $f = \pm 1$ Hz; both are correct, and the unwrapped form is preferred here because it displays the constant group delay $\tau_g = -\tfrac{1}{2\pi}\,d(\arg Y_1)/df = 0.5$ s directly as the slope.

Part (c): identify the modulation frequency. Writing $\cos 10\pi t = \cos(2\pi f_0 t)$ gives $f_0 = 5$ Hz. Since $y_2(t)=y_1(t)\cos 2\pi f_0 t$, the frequency-convolution (modulation) property applies: $$\boxed{Y_2(f)=\tfrac12\big[Y_1(f-5)+Y_1(f+5)\big]}$$
Translate and halve. Each copy of $Y_1$ keeps its shape but is halved in height and re-centred on $\pm 5$ Hz. Since $Y_1$ occupied $|f| \lt 2$ Hz, the copies occupy $3 \lt |f| \lt 7$ Hz with magnitude $\tfrac12\times\tfrac12 = 1/4$. The two bands do not overlap (the carrier at 5 Hz exceeds the 2 Hz baseband bandwidth), so no distortion occurs.
Carry the phase across with the shift. The linear phase travels with each copy and is re-referenced to its own centre: on the upper band $\arg Y_2(f) = -\pi(f-5)$ and on the lower band $\arg Y_2(f) = -\pi(f+5)$, each again sweeping from $+2\pi$ to $-2\pi$ across its 4 Hz width. The slope, and therefore the group delay of 0.5 s, is unchanged by modulation — which is exactly why double-sideband modulation transmits a pulse without smearing it.

y2(t) = y1(t) cos 10πt: the spectrum of y1 is halved in height and translated to ±5 Hz, giving bands of height 1/4 over 3 < |f| < 7 Hz. Each band keeps y1's own phase slope, re-centred on its carrier.

Question 3 — final results
Quantity	Result
(a) $X(f)$	$\operatorname{rect}(f/2)$: height 1 over $\|f\| \lt 1$ Hz
(b) $\|Y_1(f)\|$	$1/2$ over $\|f\| \lt 2$ Hz
(b) $\arg Y_1(f)$	$-\pi f$ rad (slope $-\pi$ rad/Hz; $\mp 2\pi$ at $f=\pm 2$ Hz)
Group delay	0.5 s, constant
(c) Carrier frequency	5 Hz
(c) $\|Y_2(f)\|$	$1/4$ over $3 \lt \|f\| \lt 7$ Hz
(c) $\arg Y_2(f)$	$-\pi(f\mp 5)$ on the upper/lower band

Topic. The scaling, shifting and modulation properties of the Fourier transform, and how each one alters a magnitude/phase pair.

Key relations. $x(at)\leftrightarrow \frac{1}{|a|}X(f/a)$; $x(t-t_0)\leftrightarrow X(f)e^{-j2\pi ft_0}$; $x(t)\cos 2\pi f_0t \leftrightarrow \tfrac12[X(f-f_0)+X(f+f_0)]$; group delay $\tau_g = -\frac{1}{2\pi}\frac{d\theta}{df}$.

Why this works. Each property isolates one degree of freedom. Scaling trades time width against bandwidth at constant area, so it touches the magnitude only. Shifting multiplies by a unit-modulus exponential, so it touches the phase only, and because that exponential is linear in $f$ the added phase is a straight line whose slope is the delay. Modulation convolves with two impulses, which translates the whole complex spectrum without deforming it. Applying them in the order in which the signal is built keeps the bookkeeping trivial.

Common pitfall. Misreading $x(2(t-0.5))$ as a delay of 1 s: the shift must be extracted after factoring the scale out of the argument, and using $t_0=1$ would double the phase slope. The other frequent error is forgetting the extra factor $\tfrac12$ that modulation introduces, giving bands of height $1/2$ instead of $1/4$.

Source. Lathi, Linear Systems and Signals, 2nd ed., Ch. 7 (Continuous-Time Fourier Transform), §7.3 properties table.

Question 3 of 10: Scaling, Shifting and Modulation of a Sinc Pulse

Question 3: Scaling, Shifting and Modulation of a Sinc Pulse (20 marks)