Practice CTFT computation rect and sinc ECE438F11 - Rhea

Question

Compute the Continuous-time Fourier transform of the two following functions:

$ x(t)= \text{rect}(t) = \left\{ \begin{array}{ll} 1, & \text{ if } |t|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right. $

$ y(t)= \frac{ \sin ( \pi t )}{\pi t} $

Justify your answer.

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Answer 1

Fourier Transform of rect(t):

$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx =\frac{e^{-j2\pi ft}}{-j2\pi f} $ from t=-1/2 to t=1/2

$ =\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} =\frac{sin(\pi f)}{\pi f} $

Instructor's comments: Technically, you should look at the case f=0 separately, because your solution involves a division by f. -pm

Fourier Transform of $ \frac{sin(\pi t)}{\pi t} $:

Guess: $ X(f)=rect(t) $

Proof:

$ x(t)=\int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{j2\pi ft} df =\frac{e^{j2\pi ft}}{j2\pi t} $ from f=-1/2 to f=1/2

$ =\frac{e^{j\pi t}-e^{j\pi t}}{j2\pi t} =\frac{sin(\pi t)}{\pi t} $

Instructor's comments: Guessing the answer and proving it using the inverse Fourier transform is a good trick. One could also obtain this Fourier transform using the duality property and your previous answer. -pm

Answer 2

$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt =\int_{-\infty}^{\infty} rect(t)e^{-j2\pi ft} dt =\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $

$ = -\frac{e^{-j2\pi ft}}{-j2\pi f} $ integrating from -0.5 to +0.5. $ = \frac{e^{-j\pi f} - e^{j\pi f}}{-j2\pi f} = \frac{sin(\pi f)}{\pi f} $

For y(t), we know that

$ y(t) = \int_{-\infty}^{\infty} Y(f)e^{j2\pi ft} df $

$ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{-j\pi t} - e^{j\pi t}}{\pi t} $

it should be $ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} $

For the above equation to be true,

$ Y(f) = \frac{\delta(f - \frac{1}{2})}{\pi t} - \frac{\delta(f + \frac{1}{2})}{\pi t} $

TA's comments: The first part is correct. For the second part, the denominator j2pi*t is not taken into account, which leads to the wrong answer. The CTFT should only have variable f. See answer 9 for reference.

Answer 3

$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt = \int_{-1/2}^{1/2} e^{-j2\pi ft} dt = \frac{e^-j2\pi ft}{j2 \pi f},where x from-1/2 to 1/2 =sinc(f) $ $ use duality, Y(f)=rect(x) $

Answer 4

Using Euler's equation, we know

$ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} $

Do Fourier transform of rect(t),

$ \begin{align}\mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt \\ = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt \\ = \frac{e^{-j2\pi ft}}{-j2\pi f}\vert \end{align} $ evaluate from -1/2 to 1/2,

$ \begin{align} Y(f)=\frac{e^{j\pi f}-e^{-j\pi f}}{j2\pi f} =\frac{sin(\pi f)}{\pi f} \end{align} $

use duality, $ Y(f)=rect(x) $

TA's comments: From duality, you get Y(f)=rect(-t). Since rect(t) is an even function, rect(t)=rect(-t)=Y(f).

Answer 5

Since the function is periodic we can integrate over one period:

$ F[rect(t)] = \frac{1}{1} \int_{-\frac{1}{2}}^{\frac{1}{2}}\ (1) e^{j2\pi (1)t} df $

$ = \frac{1}{-j2\pi} e^{-j2\pi t} \Bigg|_{-\frac{1}{2}}^{\frac{1}{2}} = \frac{-1}{j2\pi k } (e^{-j2\pi f\frac{1}{2}} - e^{j2\pi f\frac{1}{2}}) $

$ = \frac{sin(\pi f)}{\pi f} $

To find the FT of the sinc function, simply use duality from the first solution:

$ F[sinc(t)] = rect(-f) $

and since $ rect $ is even,

$ rect(-f)= \text{rect}(f) = \left\{ \begin{array}{ll} 1, & \text{ if } |f|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right. $

TA's comments: It doesn't matter whether it's periodic or not. Just integrate over the real line, which is equivalent to integrating from -0.5 to 0.5 in this problem.

Answer 6

$ \begin{align} X(f)&=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx \\ &=\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx \\ &=\frac{e^{-j2\pi ft}}{-j2\pi f} \big| _{-1/2}^{1/2} \\ &=\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} \\ &=\frac{sin(\pi f)}{\pi f} \end{align} $

duality:

$ f(t) \Leftrightarrow F(-\omega) $

$ F(t) \Leftrightarrow 2\pi f(-\omega) $

so

$ rect(t) \Leftrightarrow sinc(\omega/2) $

$ 2 \pi sinc(t) \Leftrightarrow 2 \pi rect(-\omega/2\pi) = rect(f) $

TA's comments: You can use the duality for continuous time signal directly.

Answer 7

Start with x(t) = rect(t)

$ X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt $

$ = \int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $

$ = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg| _{-1/2}^{1/2} $

$ = \frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} $

$ = \frac{ \sin (\pi f)}{\pi f} $

Now the FT of the sinc function is much easier. Use Duality to show:

$ F{(sinc(t))} = rect(-f) = rect(f) $

Answer 8

$ X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt $

Since x(t) only exists from -1/2<t<1/2, $ = \int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $

$ = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg| _{-1/2}^{1/2} $

$ = \frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} $

$ = \frac{ \sin (\pi f)}{\pi f} $

TA's comments: You should consider the case where f=0 separately to make it more rigorous.

Answer 9

$ X(f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt $

$ = \int_{-\infty}^{\infty} rect(t)e^{-j2\pi ft} dt $

$ = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $

$ = \frac{e^{-j2\pi ft}}{-j2\pi f} \Bigg|_{-\frac{1}{2}}^{\frac{1}{2}} $

$ = \frac{1}{\pi f} \frac{e^{-j\pi f} -e^{j\pi f}}{2j} $

$ = \frac{\sin\pi f}{\pi f} = sinc(\pi f) $

By duality:

$ F[y(t)]\rightarrow rect(-f) $

Since rect() is an even function,

$ rect(-f) = rect(f) $

Instructor's comments: This answer is almost completely correct. The only issue is that if f=0, then your integration is incorrect. You should really consider the case f=0 separately. -pm

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