Fourier Transform: 14 Practice Problems with Solutions

$$F(\omega)=\int_{-a}^{a}e^{-i\omega x}\,dx=\left[\frac{e^{-i\omega x}}{-i\omega}\right]_{-a}^{a}=\frac{e^{ia\omega}-e^{-ia\omega}}{i\omega}=\frac{2\sin a\omega}{\omega}.$$

At $\omega=0$ the value is the limit $2a$, the area under the pulse.

Answer: $\displaystyle F(\omega)=\frac{2\sin a\omega}{\omega}$

$$F(\omega)=\int_{-\infty}^{0}e^{(a-i\omega)x}\,dx+\int_{0}^{\infty}e^{-(a+i\omega)x}\,dx=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}=\frac{2a}{a^2+\omega^2}.$$

The transform is real and even, as it must be for a real even function.

Answer: $\displaystyle F(\omega)=\frac{2a}{a^2+\omega^2}$

$$F(\omega)=\int_{0}^{\infty}e^{-(a+i\omega)x}\,dx=\left[\frac{e^{-(a+i\omega)x}}{-(a+i\omega)}\right]_0^\infty=\frac{1}{a+i\omega}.$$

In real and imaginary parts: $F(\omega)=\dfrac{a-i\omega}{a^2+\omega^2}$. The function is not even, so the transform is complex.

Answer: $\displaystyle F(\omega)=\frac{1}{a+i\omega}$

From Problem 2, $f(x)=e^{-|x|}$ has $F(\omega)=\frac{2}{1+\omega^2}$. Shifting by $3$ multiplies the transform by $e^{-3i\omega}$:

$$G(\omega)=\frac{2\,e^{-3i\omega}}{1+\omega^2}.$$

A shift changes only the phase: $|G(\omega)|=|F(\omega)|$.

Answer: $\displaystyle G(\omega)=\frac{2e^{-3i\omega}}{1+\omega^2}$

Completing the square in the exponent:

$$F(\omega)=e^{-\omega^2/(4a)}\int_{-\infty}^{\infty}e^{-a\left(x+\frac{i\omega}{2a}\right)^2}dx=\sqrt{\frac{\pi}{a}}\;e^{-\omega^2/(4a)}.$$

The shift of the integration line by $\frac{i\omega}{2a}$ does not change the integral (Cauchy’s theorem, or differentiate $F$ in $\omega$ and solve $F'=-\frac{\omega}{2a}F$).

A Gaussian transforms into a Gaussian: narrow in $x$ (large $a$) means wide in $\omega$.

Answer: $\displaystyle F(\omega)=\sqrt{\frac{\pi}{a}}\,e^{-\omega^2/(4a)}$

Because $f$ is even, the sine part cancels:

$$F(\omega)=2\int_0^1(1-x)\cos\omega x\,dx=2\left(\left[\frac{(1-x)\sin\omega x}{\omega}\right]_0^1+\frac{1}{\omega}\int_0^1\sin\omega x\,dx\right)=\frac{2(1-\cos\omega)}{\omega^2}.$$

With $1-\cos\omega=2\sin^2\frac{\omega}{2}$, this is $F(\omega)=\dfrac{4\sin^2(\omega/2)}{\omega^2}$, which is never negative. Problem 13 explains why.

Answer: $\displaystyle F(\omega)=\frac{2(1-\cos\omega)}{\omega^2}=\frac{4\sin^2(\omega/2)}{\omega^2}$

Problem 2 with $a=1$ gives $e^{-|x|}\to\dfrac{2}{1+\omega^2}$. Dividing by $2$ (the transform is linear):

$$\frac12e^{-|x|}\ \to\ \frac{1}{1+\omega^2}.$$

Check at $x=0$: $f(0)=\frac{1}{2\pi}\int\frac{d\omega}{1+\omega^2}=\frac{1}{2\pi}\cdot\pi=\frac12$. It matches.

Answer: $f(x)=\frac12e^{-|x|}$

From Problem 5, $F(\omega)=\sqrt{\pi}\,e^{-\omega^2/4}$. Differentiating under the integral, $F'(\omega)=\int(-ix)f(x)e^{-i\omega x}dx$, so the transform of $xf(x)$ is $iF'(\omega)$:

$$G(\omega)=i\cdot\sqrt{\pi}\left(-\frac{\omega}{2}\right)e^{-\omega^2/4}=-\frac{i\sqrt{\pi}\,\omega}{2}\,e^{-\omega^2/4}.$$

$g$ is real and odd, so its transform is purely imaginary and odd, as found.

Answer: $\displaystyle G(\omega)=-\frac{i\sqrt{\pi}}{2}\,\omega\,e^{-\omega^2/4}$

Multiplying by $e^{i\omega_0x}$ turns $F(\omega)$ into $F(\omega-\omega_0)$. With $F(\omega)=\frac{2}{1+\omega^2}$:

$$G(\omega)=\frac12\left[F(\omega-2)+F(\omega+2)\right]=\frac{1}{1+(\omega-2)^2}+\frac{1}{1+(\omega+2)^2}.$$

The peak at $\omega=0$ splits into two peaks at $\omega=\pm2$, the frequency of the cosine.

Answer: $\displaystyle G(\omega)=\frac{1}{1+(\omega-2)^2}+\frac{1}{1+(\omega+2)^2}$

With $a=1$, $F(\omega)=\frac{2\sin\omega}{\omega}$. The inversion formula at $x=0$, a point of continuity, gives

$$1=f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sin\omega}{\omega}\,d\omega\quad\Longrightarrow\quad\int_{-\infty}^{\infty}\frac{\sin\omega}{\omega}\,d\omega=\pi.$$

The integrand is even, so half of that comes from $(0,\infty)$.

Answer: $\displaystyle\int_0^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$

For the pulse with $a=1$, $\int_{-\infty}^{\infty}|f|^2dx=\int_{-1}^{1}1\,dx=2$. Parseval’s theorem gives

$$2=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4\sin^2\omega}{\omega^2}\,d\omega\quad\Longrightarrow\quad\int_{-\infty}^{\infty}\frac{\sin^2\omega}{\omega^2}\,d\omega=\pi.$$

Answer: $\pi$

The left side of Parseval is $\int_{-\infty}^{\infty}e^{-2|x|}dx=2\cdot\frac12=1$. The right side is $\frac{1}{2\pi}\int\frac{4}{(1+\omega^2)^2}d\omega$. So

$$1=\frac{2}{\pi}\int_{-\infty}^{\infty}\frac{d\omega}{(1+\omega^2)^2}\quad\Longrightarrow\quad\int_{-\infty}^{\infty}\frac{d\omega}{(1+\omega^2)^2}=\frac{\pi}{2}.$$

The same value comes from the substitution $x=\tan\theta$, which turns the integral into $\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=\frac{\pi}{2}$.

Answer: $\frac{\pi}{2}$

(a) $(p*p)(x)=\int p(s)p(x-s)\,ds$ counts the overlap of two unit intervals whose centers are $x$ apart. The overlap is $1-|x|$ when $|x|<1$ and $0$ otherwise: the triangle function of Problem 6.

(b) Problem 1 with $a=\frac12$ gives $P(\omega)=\frac{2\sin(\omega/2)}{\omega}$. The convolution theorem says

$$\mathcal{F}\{p*p\}=P(\omega)^2=\frac{4\sin^2(\omega/2)}{\omega^2},$$

which is exactly the transform found in Problem 6. This also explains why that transform is never negative: it is a square.

Answer: (a) $(p*p)(x)=1-|x|$ for $|x|<1$, $0$ otherwise  (b) $P(\omega)^2=\frac{4\sin^2(\omega/2)}{\omega^2}$, matching Problem 6

Let $\hat u(\omega,t)$ be the transform of $u$ in $x$. The equation becomes $\hat u_t=-\omega^2\hat u$, so $\hat u(\omega,t)=\hat u(\omega,0)\,e^{-\omega^2t}$. By Problem 5 with $a=1$:

$$\hat u(\omega,t)=\sqrt{\pi}\,e^{-\omega^2/4}\,e^{-\omega^2t}=\sqrt{\pi}\,e^{-\omega^2(1+4t)/4}.$$

Problem 5 with $a$ replaced by $b$ says $e^{-bx^2}\to\sqrt{\pi/b}\,e^{-\omega^2/(4b)}$. Here $b=\frac{1}{1+4t}$, so $e^{-x^2/(1+4t)}\to\sqrt{\pi(1+4t)}\,e^{-\omega^2(1+4t)/4}$. Dividing by $\sqrt{1+4t}$:

$$u(x,t)=\frac{1}{\sqrt{1+4t}}\,e^{-x^2/(1+4t)}.$$

The bump spreads out and its height falls like $1/\sqrt{1+4t}$, while its area $\sqrt{\pi}$ stays constant.

Answer: $\displaystyle u(x,t)=\frac{e^{-x^2/(1+4t)}}{\sqrt{1+4t}}$