EE 351 · Communication Systems

Week 4

Amplitude Modulation: From DSB to Hilbert-Transform SSB

Understanding sidebands through time, frequency, and quadrature signals.

By the end of this lesson you should be able to: explain DSB-SC and SSB in time and frequency; describe the Hilbert transform as a frequency-dependent 90° phase shift; derive the phase-shift method for USB and LSB; decide which sideband survives from the sign in the SSB equation; calculate SSB bandwidth; compare DSB and SSB spectra; and explain coherent SSB demodulation.

Week 3 stopped at DSB-TC. This week answers the question it left open: if both sidebands carry the same information, can we transmit only one?

01 — Starting point

A message sits at baseband. It has to move.

A speech or audio message occupies a band from near zero up to some highest frequency \(B\). It cannot be radiated efficiently from an antenna of practical size, and every such message would occupy the same band.

The question of this week: having moved the message to a carrier, how much spectrum does the result actually need?

A real message has both positive and negative frequency content.

02 — Generation

Multiply the message by a carrier

Suppressed-carrier AM is a single multiplication. No carrier line is produced.

Week 3 called this mixing or heterodyning. Nothing new yet — but the spectrum it produces sets up the whole of Week 4.

03 — Frequency domain

Multiplication translates the spectrum

\[m(t)\cos(\omega_c t)\;\longleftrightarrow\;\tfrac12 M(f-f_c)+\tfrac12 M(f+f_c).\]

Around each carrier location the message appears twice: mirrored below (LSB) and upright above (USB).

04 — Single tone

One message tone becomes two lines

\[\cos(\omega_m t)\cos(\omega_c t)=\tfrac12\cos[(\omega_c-\omega_m)t]+\tfrac12\cos[(\omega_c+\omega_m)t].\]

Open AM Visualizer — DSB-SC

05 — Design question

Both sidebands carry the same information

For a real message the upper and lower sidebands are mirror images of one another. Either one is enough to reconstruct \(m(t)\) completely. Yet DSB transmits both:

\[B_{\mathrm{DSB}}=(f_c+B)-(f_c-B)=2B.\]

06 — The goal

Keep one sideband. But how?

The obvious idea: filter it out

Generate DSB-SC, then pass it through a sharp bandpass filter that keeps one sideband.

This works when the message has little low-frequency content. When the message extends close to DC, the two sidebands meet at \(f_c\) with no gap between them, and the filter would need an impossibly sharp transition.

The better question

Rather than building a signal with a redundant half and then removing it, can we generate only one sideband in the first place?

Filter design itself is not the subject of this week.

07 — Building intuition

What is exactly 90° away from a cosine?

Start from something familiar. Shift \(\cos(\omega t)\) by 90° and you obtain \(\sin(\omega t)\); the two are said to be in quadrature.

08 — Key distinction

A fixed time delay cannot do this job

Delaying a signal by \(\tau\) shifts a sinusoid of frequency \(\omega\) in phase by

\[\theta(\omega)=-\omega\tau .\]

The shift is proportional to frequency: a delay giving \(-90^\circ\) at 1 kHz gives \(-180^\circ\) at 2 kHz. A message carries many frequencies at once, so one fixed delay cannot put all of them into quadrature. The ideal Hilbert transformer instead holds the shift flat at \(-90^\circ\) for every positive frequency.

One response is a ramp, the other is a constant. That is the whole reason a delay line cannot replace a Hilbert transformer.

09 — What we need

Unit gain, and ±90° depending on the sign of \(f\)

\[H_H(f)=-j\,\operatorname{sgn}(f)=\begin{cases}-j, & f>0\\[2pt] +j, & f<0\end{cases}\]

10 — Notation

The transform changes phase, never amplitude

Write the Hilbert transform of \(m(t)\) as \(\hat{m}(t)=\mathcal{H}\{m(t)\}\). Because \(|H_H(f)|=1\) for all \(f\neq0\),

\[|\hat{M}(f)|=|M(f)|,\qquad \angle\hat{M}(f)=\angle M(f)\mp 90^\circ .\]

Same amplitude spectrum. Different phase. That is the whole content of the operation.

Mathematical form of the ideal Hilbert transformer

The impulse response corresponding to \(H_H(f)=-j\operatorname{sgn}(f)\) is \(h_H(t)=\dfrac{1}{\pi t}\), so \(\hat{m}(t)\) is the convolution \(m(t)*\frac{1}{\pi t}\), interpreted as a Cauchy principal value. You are not asked to derive or evaluate this integral in this course.

11 — Worked case

Under our convention, cosine becomes sine

\[\cos(\omega_m t)=\tfrac12 e^{j\omega_m t}+\tfrac12 e^{-j\omega_m t}.\]

Apply \(-j\) to the positive-frequency term and \(+j\) to the negative-frequency term:

\[\mathcal{H}\{\cos(\omega_m t)\}=-\tfrac{j}{2}e^{j\omega_m t}+\tfrac{j}{2}e^{-j\omega_m t}=\sin(\omega_m t).\]

12 — Structure

Two DSB-SC branches in quadrature

The phase-shift (Hartley) SSB modulator. Both branches are ordinary DSB-SC signals; only their combination is special.

13 — The equation

Upper sideband

\[s_{\mathrm{USB}}(t)=m(t)\cos(\omega_c t)-\hat{m}(t)\sin(\omega_c t)\]

Do not memorise this yet. On the next slide we prove what it does.

14 — Single-tone proof

Only \(f_c+f_m\) remains

Take \(m(t)=\cos(\omega_m t)\), so \(\hat{m}(t)=\sin(\omega_m t)\):

\[s_{\mathrm{USB}}(t)=\cos(\omega_m t)\cos(\omega_c t)-\sin(\omega_m t)\sin(\omega_c t).\]

Recognise the cosine addition identity \(\cos A\cos B-\sin A\sin B=\cos(A+B)\):

\[s_{\mathrm{USB}}(t)=\cos[(\omega_c+\omega_m)t]\]

One line, at \(f_c+f_m\). The lower sideband is gone — not filtered away, but never created.

15 — Change one sign

Adding the branches gives the LSB

\[s_{\mathrm{LSB}}(t)=m(t)\cos(\omega_c t)+\hat{m}(t)\sin(\omega_c t).\]

With the same single tone, \(m=\cos(\omega_m t)\) and \(\hat{m}=\sin(\omega_m t)\):

\[s_{\mathrm{LSB}}(t)=\cos(\omega_m t)\cos(\omega_c t)+\sin(\omega_m t)\sin(\omega_c t).\]

Now the identity \(\cos A\cos B+\sin A\sin B=\cos(A-B)\) applies:

\[s_{\mathrm{LSB}}(t)=\cos[(\omega_c-\omega_m)t]\]

One line, at \(f_c-f_m\). The sign in front of the quadrature branch chooses the sideband.

16 — The mechanism

Each branch is DSB. The signs do the work.

Branch A produces \(\tfrac12\) at both sidebands. Branch B produces \(\tfrac12\) at the lower and \(-\tfrac12\) at the upper. Switch the branches on and off, and change the combining sign:

17 — General message

Does this work for any message?

Yes. The single tone was only a convenient way to see the cancellation. For an arbitrary real message, define the analytic signal

\[m_a(t)=m(t)+j\hat{m}(t).\]

Every frequency component of \(m(t)\) is handled independently by the Hilbert transformer, so the cancellation that removed one line removes the whole sideband.

18 — Frequency-domain reason

Adding \(j\hat{m}\) deletes the negative frequencies

\(M_a(f)=2M(f)\) for \(f>0\) and \(0\) for \(f<0\). One-sidedness in the message is what becomes one-sidedness about the carrier.

19 — Compact engineering form

The same result in one line

\[s_{\mathrm{USB}}(t)=\operatorname{Re}\!\left\{\left[m(t)+j\hat{m}(t)\right]e^{j\omega_c t}\right\}.\]

Expanding the product and taking the real part returns exactly the two-branch expression:

\[s_{\mathrm{USB}}(t)=m(t)\cos(\omega_c t)-\hat{m}(t)\sin(\omega_c t).\]

Shifting a one-sided spectrum up to \(f_c\) is precisely what "single sideband" means. Use this form once the mechanism is already clear.

20 — Comparison

The sign decides the sideband

\[s_{\mathrm{USB}}=m\cos\omega_c t-\hat{m}\sin\omega_c t\]

\[s_{\mathrm{LSB}}=m\cos\omega_c t+\hat{m}\sin\omega_c t\]

21 — Suppressed carrier

Half the spectrum, and no carrier

Removing the redundant sideband halves the transmission bandwidth: \(B_{\mathrm{DSB}}=2B\) becomes \(B_{\mathrm{SSB}}=B\).

Only one sideband is transmitted and no power is spent on a carrier. Because there is no carrier line to lock to, recovery requires a coherent local oscillator.

Visualizer: USB Visualizer: LSB

22 — Transmitted carrier

Sending a reference back with the signal

\[s_{\mathrm{SSB\text{-}TC}}(t)=A_c\cos(\omega_c t)+s_{\mathrm{SSB\text{-}SC}}(t).\]

Adding a carrier — or a reduced-level pilot — gives the receiver a reference from which to regenerate its local oscillator, at the cost of power spent on a component that carries no message.

Open AM Visualizer — compare SSB-TC with SSB-SC

23 — Recovery

Multiply again, then low-pass

The product returns a baseband copy of the message plus components near \(2f_c\); the low-pass filter keeps the former.

The local oscillator must match the transmitter carrier in frequency and phase. How that lock is achieved is a later topic.

24 — Engineering insight

What a small frequency error does to SSB

Suppose the local oscillator runs at \(\omega_c+\Delta\omega\) instead of \(\omega_c\). Every recovered message component is then shifted by the same \(\Delta\omega\):

\[f_m\;\longrightarrow\;f_m+\Delta f .\]

This is a frequency translation, not a scaling, so the harmonic relationships inside speech are broken. A few tens of hertz of error already makes voice sound unnatural — the characteristic reason SSB receivers need a fine tuning control.

25 — Synthesis

What each form costs and buys

	DSB-SC	SSB-SC
Sidebands transmitted	Both	One
Transmission bandwidth	\(2B\)	\(B\)
Carrier	Suppressed	Suppressed
Generation	One multiplier	Two multipliers and a Hilbert transformer
Detection	Coherent	Coherent, and sensitive to frequency error

DSB-SC in the visualizer SSB-SC in the visualizer

26 — Check your understanding

Five questions to answer without notes

Why does DSB-SC contain two sidebands?
What exactly does the Hilbert transform do in the frequency domain?
Why is a fixed time delay not equivalent to an ideal Hilbert transform?
Why does one sideband cancel when the two branches are combined?
Under our convention, which sign produces USB and which produces LSB?

EE 351 · Week 4

Assessment readiness

Three short ungraded checks. Nothing is scored or stored — work them on paper, then confirm with the visualizer.

\(f_m=2\) kHz and \(f_c=100\) kHz. Where are the two DSB sidebands?
If only the upper sideband is transmitted, which single frequency remains?
For a message bandwidth \(B=5\) kHz, what are the DSB and SSB transmission bandwidths?

B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 5th ed., Oxford University Press, 2018 — Chapter 4, single-sideband modulation and the Hilbert transform.

R. W. Stewart, K. W. Barlee, D. S. W. Atkinson, and L. H. Crockett, Software Defined Radio using MATLAB & Simulink and the RTL-SDR, Chapter 6.