Properties of the Fourier transform

I run every operation on this page on one signal, the unit pulse: 1 for $\lvert t\rvert<\tfrac12$ and 0 elsewhere. First I move it 1 s later. Does every frequency turn by the same angle? Watch the size of the spectrum, then the three dials under it.

One signal, five operations

Every operation you can do to a signal in time has a matching change in its spectrum. You can delay a signal, squeeze it, multiply it by another, convolve it, or differentiate it. This page takes the operations one at a time and gives the matching rule for each.

I run all of them on the unit pulse, written $\mathrm{rect}(t)$. On the page From series to transform (8.1) you saw that its spectrum is $X(j\omega)=\dfrac{\sin(\omega/2)}{\omega/2}$, with $\omega$ in rad/s. That spectrum is 1 at $\omega=0$ and first reaches zero at $\omega=2\pi$ rad/s. The width of the pulse, 1 s, times that first zero, $2\pi$ rad/s, is $2\pi$.

The instruments on this page use one fixed frame: time on top, spectrum under it, with a faint copy of the untouched pulse so you can see what moved. The spectrum panel shows one quantity throughout, the size $\lvert X(j\omega)\rvert$. A delay makes $X$ complex, and a complex number cannot be drawn as one signed curve, so I draw its size.

One rule needs no instrument. The transform is an integral, and an integral of a sum is the sum of the integrals, so the transform of $ax(t)+by(t)$ is $aX(j\omega)+bY(j\omega)$. This is linearity. It is why a spectrum can be built, and taken apart, one arrow at a time.

A delay turns every frequency by a different angle

The picture at the top of the page moved the pulse 1 s later. Here is a way to guess what it showed. The same song played a second later has the same notes at the same loudness. Only the timing changed. So I expect the size of the spectrum to stay put, and the angles to carry the delay.

Delay the signal by $t_0$ seconds and its transform becomes

To see why, substitute $u=t-t_0$ inside the integral $\int x(t-t_0)e^{-j\omega t}\,dt$. Then $t=u+t_0$, and the factor $e^{-j\omega t_0}$ comes out of the integral, leaving $X(j\omega)$. The factor $e^{-j\omega t_0}$ has size 1, so $\lvert X\rvert$ is untouched. It turns frequency $\omega$ by $-\omega t_0$ radians, so higher frequencies turn further. This is the rule from Fourier series coefficients (7.2): there, line $k$ turned by $-k\omega_0t_0$, and now every $\omega$ does.

Now look at the picture again. The size curve does not move, and the faint unit pulse’s spectrum stays exactly under the bold curve, while the dials turn. At $t_0=0.5$ s the dials read $-45°$, $-90°$ and $-135°$: the dial at $\pi$ rad/s has turned twice as far as the one at $\pi/2$. On to $t_0=1$ s, each dial keeps turning in proportion to $\omega$, to $-90°$, $-180°$ and $-270°$. Check one with the rule: $-\omega t_0=-\pi/2$ rad $=-90°$ for $\omega=\pi/2$ and $t_0=1$ s.

I chose those three frequencies because the spectrum is positive and nonzero there: $0.900$, $0.637$ and $0.300$. So a dial never sits on a zero or on a sign flip. The readout counts the whole turn, so a turn past $-180°$ is never a jump.

Plotted against $\omega$, the turn $-\omega t_0$ is a straight line with slope $-t_0$. I come back to that line in Frequency response and Bode plots (8.4).

A squeeze in time stretches the spectrum

Squeeze the pulse to half its length and keep its height at 1. Its spectrum has to change, and the rule says how. For $x(at)$ with $a>0$, substitute $u=at$ in the integral. Then $dt=du/a$, and the exponent becomes $e^{-j(\omega/a)u}$:

For $a<0$ the same steps hold with the absolute value in front. With $a=-1$ this gives the reversal rule: $x(-t)\leftrightarrow X(-j\omega)$.

Watch the spectrum as $a$ eases from 1 to 2, and keep an eye on the product of the width and the first zero.

Shorter in time means wider in frequency, and lower. Halfway, at $a=1.5$, the width is $0.667$ s, the spectrum at $\omega=0$ is $0.667$ and the first zero is $3\pi=9.425$ rad/s. At the end the pulse is half as long, the spectrum is twice as wide and half as tall: width $0.5$ s, height $0.5$, first zero $4\pi=12.566$ rad/s. Notice that “width × first zero” reads $2\pi$ at every frame.

That steady product is the time–bandwidth trade: a shorter signal needs a wider band of frequencies. It differs slightly from the trade on page 8.1. There I kept the pulse’s area at 1, so the centre of the spectrum stayed at 1. Here I keep the pulse’s height at 1, so its area halves, and the centre of the spectrum halves with it.

You may meet the general form of this trade, with widths measured by spread, under the name uncertainty principle of signals. It is a fact about the shapes of a signal and its spectrum. It is not about measuring badly. The general form is on the reference card at the end.

A sound shows the same thing. A tone burst played at double speed is half as long and an octave higher.

Multiply by a cosine and the spectrum splits in two

First, multiply by one arrow. If you multiply $x(t)$ by $e^{j\omega_1t}$, the integral becomes $\int x(t)e^{-j(\omega-\omega_1)t}\,dt$, which is $X$ evaluated at $\omega-\omega_1$. So the whole spectrum slides up by $\omega_1$:

I call this the frequency shift. It is the mirror of the delay: a shift in time multiplies the spectrum by an arrow, and a shift in frequency multiplies the signal by one. Here $\omega_1$ is the shift frequency. It is not the fundamental $\omega_0$ of the series pages.

A cosine is two arrows of half size, one at $+\omega_1$ and one at $-\omega_1$, as on the page Complex exponentials & phasors (3.4). So multiplying by $\cos\omega_1t$ gives two copies of the spectrum, each at half height:

This is modulation. Watch the two half copies of the spectrum as $\omega_1$ eases from 0 to $4\pi$ rad/s, which is two turns per second.

At the start $\omega_1=0$, and since $\cos0=1$ the two half copies sit on top of each other. Then they move apart. At $\omega_1=2\pi$ rad/s the copies are at $\pm2\pi$ and the sum reads $0.5$ there. At the end the copies sit at $\pm4\pi$ rad/s. The sum reads $0.5$ at $\pm4\pi$ and $0$ at $\omega=0$. Each copy has the pulse’s shape at half the height.

Where the side lobes of the two copies overlap with opposite signs, they partly cancel. So the solid size of the sum can sit below a dashed copy there. At $\omega_1=4\pi$ the tails cancel exactly at the centres, because $X(j8\pi)=0$, which is why the readout is exactly $0.5$.

A voice multiplied by a carrier cosine is moved up to the carrier’s frequency. That is how a radio signal is made.

Convolution becomes multiplication

On the page Continuous-time convolution (5.3) you slid two unit pulses across each other and got a triangle. That page put its pulses on 0 to 1, so its triangle peaked at $t=1$. Centred pulses give the same triangle centred at 0. I call it $\mathrm{tri}(t)=\mathrm{rect}*\mathrm{rect}$. It has its base from $-1$ to $1$, a peak of 1, and an area of 1.

Why should the spectrum turn this into a product? The pulse $x$ is a sum of arrows: that is the inverse transform of 8.1. Each arrow goes through a system with impulse response $h$ and comes out multiplied by one number, $H(j\omega)$, as on the page Fourier series and LTI systems (7.4). So the output is the same sum of arrows, each scaled by $H(j\omega)$:

This is the convolution rule. Two smoothing filters in a row multiply their responses, point by point. For the triangle, $h$ is also the pulse, so its spectrum is the pulse’s spectrum times itself.

The curves in the next instrument are different from the earlier ones. It draws the signed $X(j\omega)$, which can be negative, as on page 8.1. It does not draw the size. The sign matters here, because the product of two negative values is positive.

A marker moves up in $\omega$, and at each frequency the two values multiply. Watch the product trace a faint curve, the triangle’s own spectrum, which I computed by integrating $\mathrm{tri}(t)$ and not from the product.

At $\omega=0$ the product is $1\times1=1$, the area of the triangle. At $\pi$ rad/s it is $0.637\times0.637=0.405$, and the faint curve says the same. At $3\pi$, $X=-0.212$ and the product is $(-0.212)\times(-0.212)=0.045$. Point by point, the product traces the faint curve, and it never goes negative. Slide-and-add in time is multiply-point-by-point in frequency.

Differentiation multiplies by jω

Each arrow $e^{j\omega t}$ differentiates to $j\omega\,e^{j\omega t}$, as on page 7.4. So the derivative $\dot x$ has the spectrum $j\omega X(j\omega)$:

Differentiating multiplies frequency $\omega$ by $j\omega$. That turns up the high frequencies: a fast wiggle changes quickly, so its derivative is large.

Check it on the pulse. The derivative of $\mathrm{rect}(t)$ is two spikes: $\delta(t+\tfrac12)-\delta(t-\tfrac12)$. A spike $\delta(t-t_0)$ is an arrow labelled with its area, as on the page Impulse, step and ramp (3.1), and it transforms to $e^{-j\omega t_0}$. So the two spikes give

which is $j\omega X(j\omega)$. At $\omega=\pi$ the size is $2=\pi\times0.637$.

The reverse operation, integration, needs a spike at $\omega=0$. It belongs to the next page, Transforms of common signals (8.3).

A real signal has a mirrored spectrum

For a real signal, the spectrum at $-\omega$ is the conjugate of the spectrum at $\omega$:

Here $z^*$ is the conjugate, the mirror image across the real axis, as on the page Complex numbers for signals (3.3). So the size at $-\omega$ equals the size at $\omega$, and the angle is the opposite. That is why the instruments draw only positive-frequency dials. A delay turns frequency $-\omega$ the other way by the same amount.

Duality: the roles of time and frequency swap

The pair for the pulse and the pair for its spectrum are the same two shapes with the roles swapped, times $2\pi$. In symbols, if $x(t)\leftrightarrow X(j\omega)$, then $X(jt)\leftrightarrow2\pi\,x(-\omega)$. This is duality. It comes from the symmetry between the transform and its inverse, which differ by the sign in the exponent and the factor $1/2\pi$.

Scaling by $a=2\pi$, with the rule from the squeeze section, turns the bottom pair into a friendlier one: $\dfrac{\sin\pi t}{\pi t}\leftrightarrow\mathrm{rect}\!\left(\dfrac{\omega}{2\pi}\right)$, a spectrum of height 1 for $\lvert\omega\rvert<\pi$. This is the brick wall of 8.4. Its time zeros at the nonzero integers match its band edge $\pi$.

Parseval: energy counted in time or in frequency

The energy of a signal is $\int\lvert x\rvert^2dt$, from How big is a signal (1.3). The unit pulse has energy 1. Counted in frequency, $\tfrac1{2\pi}\int\lvert X\rvert^2d\omega$ is 1 as well. That is Parseval’s relation for the transform.

The spectrum has a main lobe, $\lvert\omega\rvert<2\pi$, and side lobes outside it. The main lobe alone holds $0.903$ of the energy. So a band of frequencies only $4\pi$ rad/s wide carries most of the pulse.

Worked example

1. Delay. Delay the unit pulse by $t_0=0.5$ s. Then $X=\dfrac{\sin(\omega/2)}{\omega/2}e^{-j0.5\omega}$. At $\omega=\pi$ this is $0.6366\angle-90°$, which is $-0.6366j$. At $t_0=1$ the dials at $\pi/2$, $\pi$ and $3\pi/2$ rad/s turn by $-90°$, $-180°$ and $-270°$, where $X$ is $0.9003$, $0.6366$ and $0.3001$.
2. Squeeze. With $a=2$ the spectrum is $\tfrac12X(j\omega/2)$. Its height is $0.5$, its first zero is $4\pi=12.566$ rad/s, and the width is $0.5$ s. The product is $0.5\times12.566=2\pi$.
3. Modulation. $\mathrm{rect}(t)\cos4\pi t$ has the spectrum $\tfrac12[X(j(\omega-4\pi))+X(j(\omega+4\pi))]$. At $\omega=4\pi$ this is $\tfrac12[1+X(j8\pi)]=0.5$. At $\omega=0$ it is $X(j4\pi)=0$. With $\omega_1=2\pi$, at $\omega=2\pi$ it is $\tfrac12[1+X(j4\pi)]=0.5$.
4. Convolution. $\mathrm{rect}*\mathrm{rect}=\mathrm{tri}$ has the transform $\left[\dfrac{\sin(\omega/2)}{\omega/2}\right]^2$. It is $1$ at $0$, $0.6366^2=0.4053=4/\pi^2$ at $\pi$, and $0.2122^2=0.0450$ at $3\pi$.
5. Duality. The signal $x(t)=\dfrac{\sin(\omega_ct)}{\pi t}$ has $X=1$ for $\lvert\omega\rvert<\omega_c$ and $0$ elsewhere, and $x(0)=\omega_c/\pi$. With $\omega_c=\pi$ this is $\sin(\pi t)/(\pi t)$, with zeros at the nonzero integers and $x(0)=1$.
6. Differentiation. $\tfrac{d}{dt}\mathrm{rect}(t)=\delta(t+\tfrac12)-\delta(t-\tfrac12)\leftrightarrow2j\sin(\omega/2)=j\omega\,\dfrac{\sin(\omega/2)}{\omega/2}$. At $\omega=\pi$ the size is $2=\pi\times0.6366$.
7. Parseval. The unit pulse has energy 1 in time, and $\tfrac1{2\pi}\int\bigl[\tfrac{\sin(\omega/2)}{\omega/2}\bigr]^2d\omega=1$ in frequency. The share in $\lvert\omega\rvert<2\pi$ is $0.9028$. For $e^{-t}u(t)$ the energy is $0.5$ either way, and the share in $\lvert\omega\rvert<W$ is $\arctan(W)/\pi$: $0.25$ for $W=1$ and $0.4683$ for $W=10$.

Where you’ll meet this

The delay rule is why a recording stays recognisable when it starts a second late, and why a filter that delays each frequency by a different amount changes a sound’s shape. The convolution rule is how filters are designed: you choose the response $H(j\omega)$ and let multiplication do the work. The modulation rule is how a radio puts a voice on a carrier.

The next page, Transforms of common signals (8.3), builds a table of pairs and uses these rules to extend it. After it comes Frequency response and Bode plots (8.4), where a cascade of systems multiplies its responses. Cutting a signal off, called windowing, multiplies it in time and smears its spectrum. The same rules return for sampled signals in the chapter on the DFT.

Reference card