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Signals and Systems · 144 h · Topic 24 of 34

Properties of the transform

Each property says what happens to the spectrum when you touch the signal. With them, problems are solved without integrating again, and out comes the rise-time rule used every day in the lab.

Properties Shifting Scaling Modulation Duality

01The table you need to know how to read

The properties of the transform are not curiosities to memorize: each one says what happens to the spectrum when you modify the signal, and vice versa. They are what lets you solve problems without ever integrating again.

In timeIn frequencyWhat it means
\( a\,x(t) + b\,y(t) \)\( a\,X(j\omega) + b\,Y(j\omega) \) Linearity: everything can be decomposed
\( x(t - t_0) \)\( X(j\omega)\,e^{-j\omega t_0} \) The magnitude does not change; only linear phase is added
\( x(t)\,e^{j\omega_0 t} \)\( X\big(j(\omega - \omega_0)\big) \) Modulation: the spectrum is shifted. That is radio
\( x(at) \)\( \dfrac{1}{|a|}X\!\left(\dfrac{j\omega}{a}\right) \) Compressing in time widens in frequency
\( \dfrac{dx}{dt} \)\( j\omega\,X(j\omega) \) Differentiating boosts the sharp parts, with +90° of phase
\( \displaystyle\int_{-\infty}^{t} x\,d\tau \) \( \dfrac{X(j\omega)}{j\omega} + \pi X(0)\delta(\omega) \) Integrating attenuates the sharp parts
\( x(t) * h(t) \)\( X(j\omega)\,H(j\omega) \) Convolution becomes a product
\( x(t)\,h(t) \)\( \dfrac{1}{2\pi}X * H \) Multiplying in time convolves the spectra
\( x^{*}(t) \)\( X^{*}(-j\omega) \) This is where the Hermitian symmetry of real signals comes from

The two rows in bold are the ones used most. The time-shift row explains why a delay cannot be heard: the ear perceives the magnitude of the spectrum, not the phase. And the convolution row is the one that turns all circuit analysis into multiplying numbers.

02Seeing both sides at once

The way to make these properties stop being a table is to look at both domains at the same time while you modify the signal. The lab does exactly that: on the left \( x(t) \), on the right its magnitude and phase, and the original signal in gray for comparison.

Lab · one operation, two domains

Choose the operation and move its parameter. Watch which of the two plots on the right changes and which one stays the same.

03The trade-off you cannot avoid

The scaling property has a consequence that shows up everywhere. If \( x(at) \) has spectrum \( X(j\omega/a)/|a| \), then shortening a signal makes it wider in frequency, in the same proportion. The product of duration and bandwidth is a constant:

\[ \Delta t \cdot \Delta f \ge \text{constant} \] There is no way to have a signal that is both short and narrow. It is the same principle that in physics is known as the uncertainty relation.
SituationConsequence
A 1 μs pulseOccupies on the order of 1 MHz of bandwidth
A narrow radio channelCannot carry fast pulses
Doubling the speed of a linkRequires twice the bandwidth
A 100 MHz oscilloscopeCannot display an edge faster than 3.5 ns

The last row is the most widely used rule of thumb in the lab: the rise time of an instrument and its bandwidth satisfy \( t_r \approx 0.35/B \). If the edge you want to measure is faster than that, what you see on the screen is the oscilloscope, not the signal.

04Modulation: shifting the spectrum

Multiplying by a complex exponential shifts the whole spectrum. With a real cosine, which is two exponentials, the spectrum splits into two copies:

\[ x(t)\cos(\omega_0 t) \;\longleftrightarrow\; \tfrac12 X\big(j(\omega - \omega_0)\big) + \tfrac12 X\big(j(\omega + \omega_0)\big) \] Two replicas at half amplitude, one on each side of the carrier. That is exactly amplitude modulation.

This is where the sidebands come from: if the modulating signal extends up to 5 kHz, the modulated signal occupies 10 kHz around the carrier. And it explains something that was already anticipated when discussing LTI systems: to shift a signal in frequency you need a multiplier, because an LTI system cannot do it.

Duality: everything can be read backwards

The transform and its inverse are so similar that each property has its counterpart. If a rectangular pulse in time gives a sinc in frequency, then a sinc in time gives a rectangular pulse in frequency —the ideal filter—. And if shifting in time adds linear phase, shifting in frequency multiplies by an exponential in time. Once you have learned one property, you get the other for free.

05In the lab

Activity 1 · A delay does not change the magnitude

Capture a pulse and the same pulse delayed by 1 ms, and compare the two transforms using the oscilloscope’s FFT. Verify that the magnitude is identical and that the phase differs by a straight line with slope \( -t_0 \).

Activity 2 · Measure the rise time of the instrument

Inject an edge much faster than the bandwidth of the oscilloscope and measure the rise time it displays. Compare with \( 0.35/B \) and conclude which of the two is being measured.

Activity 3 · The sidebands

Modulate a 1 MHz carrier with a 10 kHz tone and observe the spectrum. Identify the carrier and the two sidebands, and verify that they are at half amplitude.

06Common mistakes

  • Believing that a delay changes the spectrum. It changes the phase, not the magnitude.
  • Mixing up the direction of scaling. Compressing in time widens in frequency.
  • Forgetting the \( 1/|a| \) in scaling: the area of the spectrum has to be conserved.
  • Confusing convolution with multiplication. Convolution in one domain is a product in the other.
  • Asking for a signal that is both short and narrowband. It does not exist; the product is bounded.

07Self-assessment

What does a delay do to the spectrum?

It multiplies by \( e^{-j\omega t_0} \): it adds linear phase and leaves the magnitude intact.

If a signal is compressed to half its duration, what happens to its bandwidth?

It doubles, and the amplitude of the spectrum is halved.

How much bandwidth does a 5 kHz signal modulated in amplitude occupy?

10 kHz: two sidebands of 5 kHz each, around the carrier.

What is the minimum rise time of a 100 MHz oscilloscope?

\( 0.35/100\text{ MHz} = 3.5 \) ns. A faster edge is displayed distorted.

What does convolution become when moving to frequency?

A product. That is why cascading filters means multiplying responses.

What does duality say?

That each property has its counterpart when the domains are swapped: a pulse gives a sinc, and a sinc gives a pulse.

08Further reading

  • Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1997. Chapter 4, with the complete table of properties and their proofs.
  • Ronald N. Bracewell. The Fourier Transform and Its Applications. 3rd ed., McGraw-Hill, 1999. Every property illustrated with the corresponding pair of plots. Unbeatable for building intuition.
  • Leon W. Couch. Digital and Analog Communication Systems. 7th ed. (Spanish edition, “Sistemas de comunicación digitales y analógicos”), Pearson Prentice Hall, 2007. Modulation as a direct application of the frequency shift.
Development of the topic “Properties of the transform” of Signals and Systems (Level 2), based on the curriculum of the UTN Electronic Engineering program, 2023 curriculum — Ordinance No. 1849 of the UTN Higher Council. Back to the Topic Map · catto.ar