Digital Band-pass Modulation

PROF. MICHAEL TSAI

2011/11/10

Band-pass Signal Representation

• General form:

𝒈 𝒕 = 𝒂 𝒕 𝒄𝒐𝒔 𝟐𝝅𝒇𝒄𝒕 + 𝝓 𝒕

• Envelope is always non-negative, or we can switch the phase by 180 degree

• This is called the canonical representation of a band-pass signal

Envelope Phase

𝑎 𝑡

2𝜋𝑓𝑐𝑡 + 𝜙 𝑡

𝑔 𝑡

Band-pass Signal Representation

• 𝒈 𝒕 = 𝒂 𝒕 𝒄𝒐𝒔 𝟐𝝅𝒇𝒄𝒕 + 𝝓 𝒕 can be re-arranged into

• 𝒈 𝒕 = 𝒈𝑰 𝒕 𝒄𝒐𝒔 𝟐𝝅𝒇𝒄𝒕 − 𝒈𝑸 𝒕 𝒔𝒊𝒏 𝟐𝝅𝒇𝒄𝒕

• 𝒈𝑰 𝒕 = 𝒂 𝒕 𝒄𝒐𝒔 𝝓 𝒕 and 𝒈𝑸 𝒕 = 𝒂 𝒕 𝒔𝒊𝒏 𝝓 𝒕

• 𝒈𝑰 𝒕 and 𝒈𝑸 𝒕 are called inphase and quadrature

components of the signal g(t), respectively

• Then 𝒂 𝒕 = 𝒈𝑰𝟐 𝒕 + 𝒈𝑸

𝟐 𝒕 and 𝝓 𝒕 = 𝒕𝒂𝒏−𝟏𝒈𝑸 𝒕

𝒈𝑰 𝒕

Band-pass Signal Representation

• We can also represent g(t) as

𝒈 𝒕 = 𝑹𝒆 𝒈 𝒕 𝒆𝒙𝒑 𝒋𝟐𝝅𝒇𝒄𝒕

• 𝒈 𝒕 = 𝒈𝑰 𝒕 + 𝒋𝒈𝑸 𝒕

• 𝒈 𝒕 is called the complex envelope of the band-pass signal.

• This is to remove the annoying 𝒆𝒙𝒑 𝒋𝟐𝝅𝒇𝒄𝒕 in the analysis.

𝑎 𝑡

𝜙 𝑡

𝑔 𝑡

Sinusoidal Functions’ Fourier Transform • Complex exponential function

• 𝐹 exp 𝑗2𝜋𝑓𝑐𝑡 = 𝛿(𝑓 − 𝑓𝑐).

• Sinusoidal functions:

• cos 2𝜋𝑓𝑐𝑡 =1

2exp 𝑗2𝜋𝑓𝑐𝑡 + exp −𝑗2𝜋𝑓𝑐𝑡

• 𝐹 cos 2𝜋𝑓𝑐𝑡 =1

2𝛿 𝑓 − 𝑓𝑐 + 𝛿 𝑓 + 𝑓𝑐

• sin 2𝜋𝑓𝑐𝑡 =1

2exp 𝑗2𝜋𝑓𝑐𝑡 − exp −𝑗2𝜋𝑓𝑐𝑡

• 𝐹 sin 2𝜋𝑓𝑐𝑡 =1

2𝛿 𝑓 − 𝑓𝑐 − 𝛿 𝑓 + 𝑓𝑐

5

f

𝐺(𝑓)

𝑓𝑐

f

𝐺(𝑓)

𝑓𝑐 −𝑓𝑐 −𝑓𝑐

f

𝐺(𝑓)

𝑓𝑐

Band-pass Signal Transmitter

Signal Encoder

90 degree

shift

×

×

∼

Message Source Σ

Band-pass Signal g(t)

cos(2𝜋𝑓𝑐𝑡)

sin(2𝜋𝑓𝑐𝑡)

𝑔𝐼(𝑡)

𝑔𝑄(𝑡)

𝑔 𝑡 = 𝑔𝐼 𝑡 cos 2𝜋𝑓𝑐𝑡 + 𝑔𝑄 𝑡 sin 2𝜋𝑓𝑐𝑡

+

+

Maps each bit into 𝑔𝐼 𝑡 and 𝑔𝑄 𝑡

Assumption

• The channel is linear: flat-fading channel.

• 𝐵𝑐 > 𝐵𝑠

• Negligible distortion to 𝑔(𝑡)

• The received signal s(t) is perturbed by AWGN

• noise w(t) ~𝑁 0,𝑁0

2

•𝑁0

2 is the PSD of the noise and also its variance (since it’s white)

AWGN Channel

Channel 𝐴𝑐

Band-pass Signal g(t)

Σ +

+

White Gaussian Noise 𝑤 𝑡

Received Signal plus Noise 𝑠 𝑡 + 𝑤(𝑡)

Channel path loss or attenuation Add “additive”

𝑥 𝑡 = 𝑠 𝑡 + 𝑤 𝑡 = 𝐴𝑐𝑔 𝑡 + 𝑤(𝑡)

Band-pass Signal Receiver

Band-pass Filter

90 degree

shift

×

×

∼

Message Sink

Received Signal plus Noise

cos(2𝜋𝑓𝑐𝑡)

sin(2𝜋𝑓𝑐𝑡)

𝑔𝑄(𝑡)

Filters out out-of-band signals and noises

𝑥 𝑡 = 𝑠 𝑡 + 𝑛(𝑡)

Low-pass Filter

Signal Detector

Low-pass Filter

1

2𝐴𝑐𝑔𝐼 𝑡 + 𝑛𝐼 𝑡

1

2𝐴𝑐𝑔𝑄 𝑡 + 𝑛𝑄 𝑡

Mixer

Band-pass Filter • The band-pass filter at the frontend filters out out-of-band

signals and noises

1. Signal s(t) is within the band not affected

2. White noise w(t) becomes narrowband noise n(t)

• Much smaller since now we only include noises within the band

• Still “white over the bandwidth of the signal”

3. Other signal (out-of-band) is filtered out

Band-pass Filter

𝑓 𝑓𝑐

𝑓𝑐 −𝐵

2 𝑓𝑐 +

𝐵

2

Other signals

Noise

s(t)

Up-conversion (TX)

×

∼ cos(2𝜋𝑓𝑐𝑡)

𝑓 𝑓

𝐴 𝑡 exp 𝑗𝜃 𝑡 Accos 2𝜋𝑓𝑐𝑡 ×

In time domain

In frequency domain

∗

Convolution

−𝑓𝑐 𝑓𝑐 𝑓

−𝑓𝑐 𝑓𝑐

Down-conversion (RX)

×

∼ cos(2𝜋𝑓𝑐𝑡)

s t Ac′ cos(2𝜋𝑓𝑐𝑡 + 𝜙) Accos 2𝜋𝑓𝑐𝑡 ×

In time domain

In frequency domain

∗

Convolution

Low-pass Filter

𝑓 −𝑓𝑐 𝑓𝑐

𝑓 −𝑓𝑐 𝑓𝑐

𝑓 −2𝑓𝑐 2𝑓𝑐

Low-pass Filter

Signal Detector

• The signal detector:

• Observes complex representation of the received signal, 𝒈𝑰 𝒕 + 𝒏𝑰 𝒕 + 𝒋[𝒈𝑸 𝒕 + 𝒏𝑸 𝒕 ],

• For a duration of T seconds (symbol/bit period)

• And the make its best estimate of the corresponding transmitted signal 𝒈𝑰 𝒕 + 𝒋𝒈𝑸 𝒕

• 𝒈𝑰 𝒕 + 𝒋𝒈𝑸 𝒕 bit stream

Signal Detector

Time synchronization

• To simplify, we assume we have time synchronization between the TX and the RX

• Symbol boundary needs to be same for TX and RX

• In practice, a timing recovery circuit is required

𝑡

Where does each symbol start and end?

Coherent & non-coherent

• Sometimes, the receiver is phase-locked to the transmitter

• That means, the in TX and in RX generate 𝒄𝒐𝒔(𝟐𝝅𝒇𝒄𝒕) with no phase difference.

• RX looks at the received signal to lock onto TX’s carrier

• When that happens, we say

• The receiver is a coherent receiver, carrying out coherent detection

• Otherwise, we say

• The receiver is a non-coherent receiver, carrying out non-coherent detection

∼

Basic forms of digital modulation

Amplitude Shift Keying

Frequency Shift Keying

Phase Shift Keying

Keying == Switching

(Binary) Amplitude Shift Keying (BASK)

• Fixed Amplitude/fixed frequency for a duration of 𝑻𝒃 to represent “1”

• No transmission to represent “0”

• Or, more formally,

• 𝑠1 𝑡 = 𝐴𝑐 cos(2𝜋𝑓𝑐𝑡)

• 𝑠0 𝑡 = 0

for a duration of 𝑇𝑏

(Binary) Phase Shift Keying (BPSK)

• Same amplitude, same frequency

• Send the original carrier to represent “1”

• Send an inverted carrier (phase difference 180 degrees) to represent “0”

• Or, more formally,

• 𝑠1 𝑡 = 𝐴𝑐 cos(2𝜋𝑓𝑐𝑡)

• 𝑠0 𝑡 = 𝐴𝑐 cos 2𝜋𝑓𝑐𝑡 + 𝜋 = −𝐴𝑐 cos(2𝜋𝑓0𝑡)

(Binary) Phase Shift Keying (BPSK)

(Binary) Frequency Shift Keying (BFSK)

𝑇𝑏

• Same amplitude

• Send a carrier at 𝒇𝟏to represent “1”

• Send a carrier at 𝒇𝟎to represent “0”

• Or, more formally,

• 𝑠1 𝑡 = 𝐴𝑐 cos(2𝜋𝑓1𝑡)

• 𝑠0 𝑡 = 𝐴𝑐 cos(2𝜋𝑓0𝑡)

for a duration of 𝑇𝑏

(Binary) Frequency Shift Keying (BFSK)

• Usually we have 𝒇𝟏 = 𝒇𝒄 +𝚫𝐟, 𝐟𝟎 = 𝐟𝐜 − 𝚫𝐟

• 𝑠1 𝑡 = 𝐴𝑐 cos[2𝜋 𝑓𝑐 + Δ𝑓 𝑡]

• 𝑠0 𝑡 = 𝐴𝑐 cos[2𝜋 𝑓𝑐 − Δ𝑓 𝑡]

• Then,

• 𝑠1 𝑡 = 𝑅𝑒 𝐴𝑐 exp 𝑗2𝜋 𝑓𝑐 + Δf 𝑡 = 𝑔 𝑡 exp 𝑗2𝜋𝑓𝑐𝑡

• 𝑠0 𝑡 = 𝑅𝑒 𝐴𝑐 exp 𝑗2𝜋 𝑓𝑐 − Δf 𝑡 = 𝑔 𝑡 exp 𝑗2𝜋𝑓𝑐𝑡

• So,

• For “1”, 𝑔 𝑡 = 𝑔𝐼 𝑡 + 𝑗𝑔𝑄 𝑡 = 𝐴𝑐exp[−𝑗2𝜋Δ𝑓𝑡]

• For “0”,𝑔 𝑡 = 𝑔𝐼 𝑡 + 𝑗𝑔𝑄 𝑡 = 𝐴𝑐exp[+𝑗2𝜋Δ𝑓𝑡]

I

Q

Coherent Detection of FSK and PSK signals

• Since 𝒇𝒄 is large compared to 𝟏

𝑻𝒃 (symbol rate, or bit

rate), we can say that the same signal energy 𝑬𝒃 is transmitted in a bit interval 𝑻𝒃:

𝐸𝑏 = 𝑠02 𝑡 𝑑𝑡

𝑇𝑏

0

= 𝑠12 𝑡 𝑑𝑡

𝑇𝑏

0

=𝐴𝑐2𝑇𝑏2

Two-path correlation receiver (general case)

𝑑𝑡𝑇𝑏

0

𝑑𝑡𝑇𝑏

0

×

×

Σ +

−

Detection device

Choose 1 if 𝑙 > 0

Otherwise, choose 0

𝑥(𝑡)

x(t): received signal

𝑠1(𝑡)

𝑠0(𝑡)

Correlator: see how similar 𝑥 𝑡 and 𝑠1(𝑡) are

Correlator: see how similar 𝑥 𝑡 and 𝑠0(𝑡) are

𝑙

Coherent Detection

• 𝒘(𝒕): AWGN, 𝑵 𝟎,𝑵𝟎

𝟐

• 𝑯𝟎: 𝒙 𝒕 = 𝒔𝟎 𝒕 + 𝒘(𝒕)

• 𝑯𝟏: 𝒙 𝒕 = 𝒔𝟏 𝒕 + 𝒘(𝒕)

• Receiver output:

• Decision level: 0

• If 𝑙 is larger than 1, than 𝑥(𝑡) is “more similar” to 𝑠1(𝑡)

• If 𝑙 is smaller than 1, than 𝑥(𝑡) is “more similar” to 𝑠0(𝑡)

𝑙 = 𝑥 𝑡 𝑠1 𝑡 − 𝑠0 𝑡 𝑑𝑡𝑇𝑏

0

Coherent Detection

• 𝑯𝟏:

• Since the noise w(t) is zero-mean,

• 𝝆: the correlation coefficient of the signals 𝒔𝟎(𝒕) and 𝒔𝟏 𝒕

𝑙 = 𝑠1 𝑡 𝑠1 𝑡 − 𝑠0 𝑡 𝑑𝑡𝑇𝑏

0

− 𝑤 𝑡 𝑠1 𝑡 − 𝑠0 𝑡 𝑑𝑡𝑇𝑏

0

𝑙 = 𝑥 𝑡 𝑠1 𝑡 − 𝑠0 𝑡 𝑑𝑡𝑇𝑏

0

𝐸 𝐿 𝐻1 = 𝑠1 𝑡 𝑠1 𝑡 − 𝑠0 𝑡 𝑑𝑡𝑇𝑏

0

= 𝐸𝑏(1 − 𝜌)

𝜌 = 𝑠0 𝑡 𝑠1 𝑡 𝑑𝑡𝑇𝑏0

𝑠02 𝑡 𝑑𝑡

𝑇𝑏0

𝑠12 𝑡 𝑑𝑡

𝑇𝑏0

12

=1

𝐸𝑏 𝑠0 𝑡 𝑠1 𝑡 𝑑𝑡𝑇𝑏

0

L: the random variable whose value is 𝑙

0 ≤ 𝜌 ≤ 1

Coherent Detection

• Similarly,

• L’s variance is the same for 𝑯𝟏 and 𝑯𝟎. Since 𝒔𝟏(𝒕) and 𝒔𝟎(𝒕) is deterministic given the transmitted bit, we have

𝐸 𝐿 𝐻0 = −𝐸𝑏 1 − 𝜌

𝑉𝑎𝑟 𝐿 = E L − E L 2

= E 𝑤 𝑡 𝑤 𝑢 𝑠1 𝑡 − 𝑠0 𝑡 𝑠1 𝑢 − 𝑠0 𝑢 𝑑𝑡𝑇𝑏

0

𝑑𝑢𝑇𝑏

0

= 𝑬 𝒘 𝒕 𝒘 𝒖 𝑠1 𝑡 − 𝑠0 𝑡 𝑠1 𝑢 − 𝑠0 𝑢 𝑑𝑡𝑇𝑏

0

𝑑𝑢𝑇𝑏

0

= 𝛿(𝑡 − 𝑢) 𝑠1 𝑡 − 𝑠0 𝑡 𝑠1 𝑢 − 𝑠0 𝑢 𝑑𝑡𝑇𝑏

0

𝑑𝑢𝑇𝑏

0

= 𝛿(𝑡 − 𝑢) 𝑠1 𝑡 − 𝑠0 𝑡 𝑠1 𝑢 − 𝑠0 𝑢 𝑑𝑡𝑇𝑏

0

𝑑𝑢𝑇𝑏

0

= 𝑁02

𝑠1 𝑡 − 𝑠0 𝑡 2𝑑𝑡𝑇𝑏

0

= 𝑁0𝐸𝑏(1 − 𝜌)

• Therefore, we know that L conditioned on 𝑯𝟎 is a Gaussian distributed random variable: 𝑵 𝑬𝒃 𝟏 − 𝝆 ,𝑵𝟎𝑬𝒃 𝟏 − 𝝆

Q Function

• Q function is defined over the CDF of Gaussian distribution 𝑵(𝟎, 𝟏)

𝑄 𝑥 =1

2𝜋 exp −

𝑢2

2𝑑𝑢

∞

𝑥

= 1 −Φ(𝑥)

CDF of 𝑵(𝟎, 𝟏)

N(0,1)’s PDF

u

f(u)

Integration (Area under the curve) x

Bit Error Rate 𝑳|𝑯𝟏~𝑵 𝑬𝒃 𝟏 − 𝝆 ,𝑵𝟎𝑬𝒃 𝟏 − 𝝆

𝑬𝒃 𝟏 − 𝝆

𝝇𝟐~𝑵𝟎𝑬𝒃 𝟏 − 𝝆

Integration (Area under the curve)

0

How to express this area with Q function?

−𝑬𝒃 𝟏 − 𝝆 0

Shift left by 𝑬𝒃 𝟏 − 𝝆

−𝑬𝒃 𝟏 − 𝝆

𝑵𝟎𝑬𝒃 𝟏 − 𝝆

0

Divide u by 𝑁0𝐸𝑏 1 − 𝜌

𝑵 𝟎,𝑵𝟎𝑬𝒃 𝟏 − 𝝆 𝑵(𝟎, 𝟏)

Bit Error Rate

−𝑬𝒃 𝟏 − 𝝆

𝑵𝟎𝑬𝒃 𝟏 − 𝝆

0

𝑵(𝟎, 𝟏)

𝑬𝒃 𝟏 − 𝝆

𝑵𝟎𝑬𝒃 𝟏− 𝝆

0

𝑵(𝟎, 𝟏)

𝑃𝑒 = 𝑄𝐸𝑏 1 − 𝜌

𝑁0

𝑃𝑒 = 𝑄2𝐸𝑏𝑁0

𝑃𝑒 = 𝑄𝐸𝑏𝑁0

For BPSK, 𝜌 = −1 For BFSK, 𝜌 = 0

Signal Space - BPSK

Inphase

Quadrature

sin(2𝜋𝑓𝑐𝑡)

cos(2𝜋𝑓𝑐𝑡) −𝐴𝑐 𝐴𝑐

Bit 0 Bit 1

𝑠1 𝑡 = 𝐴𝑐cos(2𝜋𝑓𝑐𝑡)

𝑠0 𝑡 = −𝐴𝑐cos(2𝜋𝑓𝑐𝑡)

Noise

Energy

Signal Space - QPSK

Inphase

Quadrature

sin(2𝜋𝑓𝑐𝑡)

cos(2𝜋𝑓𝑐𝑡)

Bit 01 Bit 11

𝑠11 𝑡 = 𝐴𝑐 cos 2𝜋𝑓𝑐𝑡 +𝜋

4

𝑠00 𝑡 = 𝐴𝑐 cos 2𝜋𝑓𝑐𝑡 +5𝜋

4

Bit 10 Bit 00

𝐴𝑐

2

−𝐴𝑐

2

−𝐴𝑐

2

𝐴𝑐

2

𝐴𝑐

𝑠01 𝑡 = 𝐴𝑐 cos 2𝜋𝑓𝑐𝑡 +3𝜋

4

𝑠10 𝑡 = 𝐴𝑐 cos 2𝜋𝑓𝑐𝑡 +7𝜋

4

𝜙

M-ary Modulation

Inphase

Quadrature

𝐴𝑐

𝜙

8-PSK

Inphase

Quadrature

4-PSK

Increasing M would increase the data rate (given the same signal bandwidth)

M-ary Modulation

Inphase

Quadrature 16-QAM

M-PAM BER versus SNR

𝐵𝐸𝑅 ≤ 𝑃𝑒 ≤ 2 × 1 −1

𝑀× 𝑄

3 × 𝑆𝑁𝑅

𝑀2 − 1

𝑆𝑁𝑅 =𝐸𝑏

𝜍𝑛2

M-QAM BER versus SNR

𝐵𝐸𝑅 ≤ 𝑃𝑒 ≤ 4 × 1 −1

𝑀× 𝑄

3 × 𝑆𝑁𝑅

𝑀 − 1

𝐵𝐸𝑅 ≤ 𝑃𝑒 ≤ 4 × 1 −1

2𝑀× 𝑄

3 × 𝑆𝑁𝑅

31 ×𝑀32 − 1

M-PSK BER versus SNR

𝐵𝐸𝑅 ≤ 𝑃𝑒 ≤ 2 × 𝑄 2 × 𝑆𝑁𝑅 × sin𝜋

𝑀