MIT EECS: 6.003 Signal Processing lecture notes (Fall 2019)

6.003: Signal Processing

Communications Systems

5 December 2019

Communications Systems

Some of largest and fastest growing applications of signal processing.

Examples:

• cellular communications

• wifi

• broadband

• cable

• bluetooth

• GPS (the Global Positioning System)

• private networks: fire departments, police

• radar and navigation systems

• IOT (the Internet of Things)

− smart house / smart appliances

− smart car

− medical devices

• many more

Telephone

Popular thirst for communications has been evident since the early

days of telephony.

mic amp telephone wire amp speaker

Patented by Alexander Graham Bell (1876) this technology flour-

ished first as a network of copper wires and later as optical fibers

(“long-distance” network) connecting virtually every household in

the US by the 1980’s.

Bell Labs became a premier research facility, developing information

theory and a host of wired and wireless communications technologies

that built on that theory, as well has hadware inovations such as the

transistor and the laser.

Cellular Communication

First demonstrated by Motorola in 1973, cellular communications

quickly revolutionized the field.

There are now more cell phones than people in the world.

sound in

sound out

cellphone

tower towercell

phonesound

in

E/M optic

fiber

E/M soundout

Much of the popularity and convenience of cellular communications

is that the communication is wireless (at least to the local tower).

Wireless Communication

Wireless signals are transmitted via electromagnetic (E/M) waves.

mic amp E/M wave amp speaker

For energy-efficient transmission and reception, the length of the

antenna should be on the order of the wavelength.

Telephone-quality speech contains frequencies from 200 to 3000 Hz.

How long should the antenna be?

Check Yourself

For energy-efficient transmission and reception, the length of

the antenna should be on the order of the wavelength.

Telephone-quality speech contains frequencies between 200 Hz

and 3000 Hz.

How long should the antenna be?

1. < 1 mm

2. ∼ cm

3. ∼m

4. ∼ km

5. > 100 km

Check Yourself

Wavelength is λ = c/f .

The lowest frequencies (200 Hz) produce the longest wavelengths

λ = c

f= 3× 108 m/s

200 Hz= 1.5× 106 m = 1500 km .

and highest frequencies (3000 Hz) produce the shortest wavelengths

λ = c

f= 3× 108 m/s

3000 Hz= 105 m = 100 km .

The size of the antenna should be on the order of 900 miles!

Check Yourself

For energy-efficient transmission and reception, the length of

the antenna should be on the order of the wavelength.

Telephone-quality speech contains frequencies between 200 Hz

and 3000 Hz.

How long should the antenna be? 5

1. < 1 mm

2. ∼ cm

3. ∼m

4. ∼ km

5. > 100 km

Check Yourself

What frequency E/M wave is well matched to an antenna

with a length of 10 cm (about 4 inches)?

1. < 100 kHz

2. 1 MHz

3. 10 MHz

4. 100 MHz

5. > 1 GHz

Check Yourself

A wavelength of 10 cm corresponds to a frequency of

f = c

λ∼ 3× 108 m/s

10 cm≈ 3 GHz .

Modern cell phones use frequencies near 2 GHz.

Check Yourself

What frequency E/M wave is well matched to an antenna

with a length of 10 cm (about 4 inches)? 5

1. < 100 kHz

2. 1 MHz

3. 10 MHz

4. 100 MHz

5. > 1 GHz

Wireless Communication

Speech is not well matched to the wireless medium.

Matching the message to the medium is important in all communi-

cations systems.

Example media:

• radio (E/M) waves

• cable (coaxial wires)

• fiber optics

Today we will introduce simple matching strategies based on

modulation, which underlie virtually all communication schemes.

Check Yourself

Construct a signal Y that codes the audio frequency information

in X using frequency components near 2 GHz.

ω

|X(ω)|

ωωc

|Y (ω)|

Determine an expression for y(·) in terms of x(·).

1. y(t) = x(t) e jωct 2. y(t) = x(t) ∗ e jωct

3. y(t) = x(t) cos(ωct) 4. y(t) = x(t) ∗ cos(ωct)5. none of the above

Check Yourself

Construct a signal Y that codes the audio frequency information in

X using frequency components near 2 GHz.

ω

|X(ω)|

ωωc

|Y (ω)|

Y (ω) = X(ω − ωc)

y(t) = 12π

∫Y (ω) ejωtdω = 1

2π

∫X(ω − ωc) ejωtdω

= 12π

∫X(λ) ej(λ+ωc)tdλ where λ = ω − ωc

= ejωct(

12π

∫X(λ) ejλtdλ

)= ejωctx(t)

Check Yourself

Construct a signal Y that codes the audio frequency information

in X using frequency components near 2 GHz.

ω

|X(ω)|

ωωc

|Y (ω)|

Determine an expression for Y in terms of X. 1

1. y(t) = x(t) e jωct 2. y(t) = x(t) ∗ e jωct

3. y(t) = x(t) cos(ωct) 4. y(t) = x(t) ∗ cos(ωct)5. none of the above

Amplitude Modulation (Time Domain)

Multiplying a signal by a sinusoidal carrier signal is called amplitude

modulation. The signal “modulates” the amplitude of the carrier.

×x(t) y(t)

cos(ωct)

tx(t) cos(ωct)

tx(t)

tcos(ωct)

Check Yourself!

Multiplying a signal by a sinusoidal carrier signal is called amplitude

modulation. The signal “modulates” the amplitude of the carrier.

×x(t) y(t)

cos(ωct)

ω

|X(ω)|

What does the resulting signal’s CTFT Y (·) look like?

Amplitude Modulation (Frequency Domain)

Multiplying a signal by a sinusoidal carrier signal is called amplitude

modulation (AM). AM shifts the frequency components of X by ±ωc.

×x(t) y(t)

cos(ωct)

ω

|X(ω)|

ωωc−ωc

ωωc−ωc

|Y (ω)|

Amplitude Modulation

How could you recover x(t) from y(t)?

×x(t) y(t)

cos(ωct)

Synchronous Demodulation

x(·) can be recovered by multiplying by the carrier and then low-pass

filtering. This process is called synchronous demodulation.

y(t) = x(t) cos(ωct)

z(t) = y(t) cos(ωct) = x(t)× cos(ωct)× cos(ωct) = x(t)(

12

+ 12

cos(2ωct))

Synchronous Demodulation

Synchronous demodulation: convolution in frequency.

ωωc−ωc

|Y (ω)|

ωωc−ωc

ω2ωc−2ωc

|Z(ω)|

Synchronous Demodulation

We can recover X by low-pass filtering.

ωωc−ωc

|Y (ω)|

ωωc−ωc

ω2ωc−2ωc

|Z(ω)|2

Frequency-Division Multiplexing

Multiple transmitters can co-exist, as long as the frequencies that

they transmit do not overlap.

x1(t)

x2(t)

x3(t)

z1(t)

z2(t) z(t)y(t)

z3(t)

cos w1t

cos w2t cos wct

cos w3t

LPF

Frequency-Division Multiplexing

Multiple transmitters simply sum (to first order).

x1(t)

x2(t)

x3(t)

z1(t)

z2(t) z(t)y(t)

z3(t)

cos w1t

cos w2t cos wct

cos w3t

LPF

Frequency-Division Multiplexing

Multiple transmitters can co-exist, as long as the frequencies that

they transmit do not overlap.

ω

X1(ω)

ω

X2(ω)

ω

X3(ω)

Frequency-Division Multiplexing

Multiple transmitters can co-exist, as long as the frequencies that

they transmit do not overlap.

ω

Z1(ω)

ω1

ω

Z2(ω)

ω2

ω

Z3(ω)

ω3

Frequency-Division Multiplexing

Multiple transmitters can co-exist, as long as the frequencies that

they transmit do not overlap.

ω

Z1(ω)

ω1

ω

Z2(ω)

ω2

ω

Z3(ω)

ω3

ω

Z(ω)

ω1 ω2 ω3

Broadcast Radio

“Broadcast” radio was championed by David Sarnoff, who previously

worked at Marconi Wireless Telegraphy Company (point-to-point).

• envisioned “radio music boxes”

• analogous to newspaper, but at speed of light

• receiver must be cheap (as with newsprint)

• transmitter can be expensive (as with printing press)

Sarnoff (left) and Marconi (right)

Modernity: What About Sending Digital Data?

Consider input stream x[n] of 1’s and 0’s.

Transmitter:

• Construct new stream by replicating each bit some number of

times.

• Modulate with a cosine.

• Transmit.

Receiver:

• Multiply received stream with a cosine (demodulate).

• Low-pass filter.

• Threshold result to find 1’s and 0’s, and decimate.

Simple Receiver

The problem with making the simple strategy we’ve discussed is that

you must know the carrier signal exactly! Moreover, delay between

the sender and receiver generally introduce a phase difference.

z[n]x[n] y[n]

cos(Ωcn) cos(Ωcn+ φ)

LPF

Check Yourself

The problem with making the simple strategy we’ve discussed

is that you must know the carrier signal exactly!

z[n]x[n] y[n]

cos(Ωcn) cos(Ωcn+ φ)

LPF

What happens if there is a phase shift φ between the signal

used to modulate and that used to demodulate?

Check Yourself

y[n] = x[n]× cos(Ωcn)× cos(Ωcn+ φ)

= x[n]×(

12

cos(φ) + 12

cos(2Ωcn+ φ))

Passing y[n] through a low pass filter yields 12x[n] cos(φ).

If φ = π/2, the output is zero!

If φ changes with time, then the signal “fades.”

Fixing Phase Problems: Quadrature Demodulation

Phase errors (and channel delay) result in scaling the output ampli-

tude, where the magnitude of the scaling can’t be determined when

we design the system:

• channel delay varies on mobile devices

• phase difference between transmitter and receiver is arbitrary

One strategy to mitigate this effect is quadrature demodulation:

• Multiply by cosine to find I[n] = x[n−D]× cos(φ)• Multiply by sine to find Q[n] = x[n−D]× sin(φ)

Then, if we let w[n] = I[n] + jQ[n], we have:

|w[n]| =√I[n]2 +Q[n]2

=√

(x[n−D] cos(φ))2 + (x[n−D] sin(φ))2

= |x[n−D]|√

cos2(φ) + sin2(φ) = |x[n−D]|

Example: Sending Data via Audio

Example: audio-based communication link

Phase-shift Keying

We can do better (in terms of mitigating noise) if we encode our

bitsream as +1 and −1 instead of as 1 and 0. Called binary phase-

shift keying because the carrier changes phase when x[n] switches

from 0 to 1 (and vice versa).

However, this introduces an ambiguity between 0’s and 1’s! However,

we can fix this multiple ways:

• Send an agreed-on preamble (known bit sequence) at the start

of each message, or

• Use a different encoding (for example, a 1 corresponds to step-

ping the phase, and a 0 corresponds to keep the phase the same.

WiFi

Brief overview of WiFi.

Additional challenges! AM radio is single source→ multiple receivers.

WiFi needs to be:

• multiple source (computers) → single receiver (access point)

• single sender (access point) → single receiver (computers)

How is this accomplished (simple version)?

• Frequency division multiplexing

• Medium access control protocols for effectively sharing a channel