Question 3 of 10: FT Properties — Time-Scaling, Shift and Modulation of a sinc Pulse

Question 3: FT Properties — Time-Scaling, Shift 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. $x(t)=2\,\text{sinc}(2\pi t)$; $y_1(t)=x(2(t-0.5))$; $y_2(t)=y_1(t)\cos(10\pi t)$.

Find. $X(f)$, and the magnitude/phase spectra of $y_1(t)$ and $y_2(t)$.

Approach. Identify $X(f)$ from the table pair $2B\,\text{sinc}(2\pi Bt)\leftrightarrow\text{rect}(f/2B)$, then apply the time-scaling property to get $x(2t)$'s spectrum, the time-shift property to add the $0.5$ s delay, and the modulation (frequency-shift) property for the $\cos10\pi t=\cos(2\pi\cdot5t)$ carrier.

1. Part (a) — X(f). Matching $x(t)=2\,\text{sinc}(2\pi\cdot1\cdot t)$ to $2B\,\text{sinc}(2\pi Bt)\leftrightarrow\text{rect}(f/2B)$ with $B=1$, $$X(f)=\boxed{\text{rect}(f/2)}$$ unit height for $|f|<1$ Hz, zero beyond — $x(t)$ is bandlimited to 1 Hz.

2. Part (b) — Y_1(f). Write $y_1(t)=x_1(t-0.5)$ with $x_1(t)=x(2t)$. Time-scaling ($g(at)\leftrightarrow\frac1{|a|}G(f/a)$, $a=2$) gives $X_1(f)=\tfrac12\,\text{rect}(f/4)$ (band widens to $\pm2$ Hz, amplitude halves). Time-shifting by $0.5$ s ($g(t-t_0)\leftrightarrow G(f)e^{-j2\pi ft_0}$) then gives $$Y_1(f)=X_1(f)e^{-j\pi f}=\tfrac12\,\text{rect}(f/4)\,e^{-j\pi f}.$$ So $$|Y_1(f)|=\boxed{\tfrac12\,\text{rect}(f/4)}\quad\text{(height 1/2, band } |f|<2\text{ Hz)},$$ $$\angle Y_1(f)=\boxed{-\pi f\ \text{rad}}\ \text{ for } |f|<2\text{ Hz (linear, the phase of a 0.5 s delay); undefined (zero magnitude) outside the band.}$$

3. Part (c) — Y_2(f). $\cos10\pi t=\cos(2\pi\cdot5t)$ is a carrier at $f_0=5$ Hz. The modulation property $g(t)\cos(2\pi f_0t)\leftrightarrow\tfrac12[G(f-f_0)+G(f+f_0)]$ gives $$Y_2(f)=\tfrac12\big[Y_1(f-5)+Y_1(f+5)\big]=\tfrac14\Big[\text{rect}\big(\tfrac{f-5}4\big)e^{-j\pi(f-5)}+\text{rect}\big(\tfrac{f+5}4\big)e^{-j\pi(f+5)}\Big].$$ Since $Y_1(f)$ occupies only $|f|<2$ Hz, the two shifted copies land in $3\le f\le7$ Hz and $-7\le f\le-3$ Hz — disjoint bands, no overlap: $$|Y_2(f)|=\boxed{\tfrac14\ \text{ on } 3\le|f|\le7\text{ Hz, zero elsewhere}},$$ $$\angle Y_2(f)=-\pi(f\mp5)\ \text{rad on each band (linear about its own centre } f=\pm5\text{ Hz),}$$ i.e. $-\pi(f-5)$ on the upper band and $-\pi(f+5)$ on the lower one. Both carry the same negative slope $-\pi$ (the 0.5 s delay is unchanged by modulation) and each passes through zero phase at its own band centre, so $\angle Y_2(f)$ is an odd function of $f$, as it must be for the real signal $y_2(t)$.