Chapter 9. Amplitude Modulation
1
Goals
Modulate and demodulate the double sideband-suppressed carrier (DSB-SC) amplitude modulation (AM) signals.
Investigate the effects of frequency and phase errors on the demodulation performance.
Modulate and demodulate AM signals by using the sampling and band pass filter (BPF) technique.
2
Double side-band suppressed carrier(DSB-SC)
3
𝑓𝑓 𝑡𝑡
cos 𝜔𝜔
𝑐𝑐𝑡𝑡
∅(𝑡𝑡)=𝑓𝑓 𝑡𝑡 × cos(𝜔𝜔
𝑐𝑐𝑡𝑡)
Amplitude modulations(AM)
DSB-SC, DSB-LC, SSB, VSB, etc.
DSB-SC
Modulated signal ∅(𝑡𝑡)=𝑓𝑓 𝑡𝑡 cos(𝜔𝜔
𝑐𝑐𝑡𝑡)
• 𝑓𝑓 𝑡𝑡 : information signal
• cos 𝜔𝜔
𝑐𝑐𝑡𝑡 : carrier signal, 𝜔𝜔
𝑐𝑐= carrier frequency.
Transmitter(modulator) structure
Spectrum of DSB-SC
F ∅ 𝑡𝑡 = F ∅ 𝑡𝑡 = F 𝑓𝑓 𝑡𝑡 cos(𝜔𝜔
𝑐𝑐𝑡𝑡) = 1
2 𝐹𝐹 𝜔𝜔 + 𝜔𝜔
𝑐𝑐+ 1
2 𝐹𝐹 𝜔𝜔 − 𝜔𝜔
𝑐𝑐• Recall that multiplying sinusoid = frequency shift
Double side-band suppressed carrier(DSB-SC)
DSB-SC
Receiver (demodulator) structure
[CH19.2B9] Proof:
𝑔𝑔 𝑡𝑡 = ∅ 𝑡𝑡 cos 𝜔𝜔
𝑐𝑐𝑡𝑡 = 𝑓𝑓 𝑡𝑡 cos(𝜔𝜔
𝑐𝑐𝑡𝑡)
2= 𝑓𝑓(𝑡𝑡)
1+cos(2𝜔𝜔2 𝑐𝑐𝑡𝑡)=
12
𝑓𝑓 𝑡𝑡 +
12𝑓𝑓 𝑡𝑡 cos(2𝜔𝜔
𝑐𝑐𝑡𝑡)
𝐺𝐺(𝜔𝜔)=
12𝐹𝐹 𝜔𝜔 +
14𝐹𝐹 𝜔𝜔 − 2𝜔𝜔
𝑐𝑐+
14𝐹𝐹 𝜔𝜔 + 2𝜔𝜔
𝑐𝑐• Here, let us define LPF[x]: LPF[x] ≜ Ideal(Perfectly flat magnitude response within BW, no delay) LPF output of x.
Then,
LPF[ 𝐺𝐺(𝜔𝜔)]=
12𝐹𝐹 𝜔𝜔 +
14𝐹𝐹 𝜔𝜔 − 2𝜔𝜔
𝑐𝑐+
14𝐹𝐹 𝜔𝜔 + 2𝜔𝜔
𝑐𝑐=
12
𝐹𝐹 𝜔𝜔
LPF 𝑔𝑔 𝑡𝑡 = LPF
12𝑓𝑓 𝑡𝑡 +
12𝑓𝑓 𝑡𝑡 cos(2𝜔𝜔
𝑐𝑐𝑡𝑡) =
12𝑓𝑓(𝑡𝑡)
4∅(𝑡𝑡)=𝑓𝑓 𝑡𝑡 cos(𝜔𝜔
𝑐𝑐𝑡𝑡)
LPF 12𝑓𝑓 𝑡𝑡 : demodulated signal
cos 𝜔𝜔
𝑐𝑐𝑡𝑡 :local carrier signal
𝑔𝑔(𝑡𝑡)
Waveforms and Spectra of DSB-SC
[CH9.1F4~1F6]If the BW of LPF is larger than W, noise power in the demodulated signal increases.
[CH9.1F4~1F6]If the BW of LPF is smaller than W, the demodulated information is distorted. 5
LPF
cos 𝜔𝜔𝑐𝑐𝑡𝑡
Effect of phase error
What if there exists a phase error between the carrier and the local carrier?
Say, local carrier = cos 𝜔𝜔
𝑐𝑐𝑡𝑡 +
𝜃𝜃not cos 𝜔𝜔
𝑐𝑐𝑡𝑡 . Then,
𝑔𝑔 𝑡𝑡 = ∅ 𝑡𝑡 cos 𝜔𝜔
𝑐𝑐𝑡𝑡 +
𝜃𝜃= 𝑓𝑓 𝑡𝑡 cos(𝜔𝜔
𝑐𝑐𝑡𝑡)cos 𝜔𝜔
𝑐𝑐𝑡𝑡 +
𝜃𝜃= 𝑓𝑓(𝑡𝑡)
12cos 2𝜔𝜔
𝑐𝑐𝑡𝑡 +
𝜃𝜃+ cos −𝜃𝜃
=
12𝑓𝑓 𝑡𝑡 cos 2𝜔𝜔
𝑐𝑐𝑡𝑡 +
𝜃𝜃+
12𝑓𝑓 𝑡𝑡 cos
𝜃𝜃 [CH9.2A4] the demodulated signal(LPF output in the receiver) is given as:
LPF 𝑔𝑔 𝑡𝑡 = LPF 1
2 𝑓𝑓(𝑡𝑡) cos 2𝜔𝜔
𝑐𝑐𝑡𝑡 +
𝜃𝜃+ 1
2 𝑓𝑓(𝑡𝑡)cos
𝜃𝜃=LPF
12𝑓𝑓(𝑡𝑡) cos 2𝜔𝜔
𝑐𝑐𝑡𝑡 +
𝜃𝜃+LPF
12𝑓𝑓(𝑡𝑡)cos
𝜃𝜃=
12𝑓𝑓 𝑡𝑡 cos(𝜃𝜃)
Phase error results in decreased amplitude of the demodulated signal by the factor of cos(𝜃𝜃).
Special case: if
𝜃𝜃=
𝜋𝜋2
, then the signal disappears.
6
Effect of phase error
[CH9.2B]Consider the background noise 𝑛𝑛 𝑡𝑡 is added to the received signal 𝑓𝑓 𝑡𝑡 cos(𝜔𝜔𝑐𝑐𝑡𝑡):
Then, the multiplier output in the receiver 𝑔𝑔 𝑡𝑡 = 𝑓𝑓 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡 + 𝑛𝑛(𝑡𝑡) cos 𝜔𝜔𝑐𝑐𝑡𝑡 +𝜃𝜃
= 𝑓𝑓 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡 +𝜃𝜃 + 𝑛𝑛(𝑡𝑡) cos 𝜔𝜔𝑐𝑐𝑡𝑡 +𝜃𝜃
𝑛𝑛 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡 + 𝜃𝜃 = the noise component in 𝑔𝑔 𝑡𝑡 . We denote 𝑛𝑛𝜃𝜃 𝑡𝑡 = 𝑛𝑛 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡 + 𝜃𝜃
.
[CH9.2B10a] ACF of 𝑛𝑛𝜃𝜃 𝑡𝑡
IF 𝜏𝜏=0, 𝑅𝑅 𝑛𝑛𝜃𝜃 𝜏𝜏 = 0 = lim𝑇𝑇→∞∫− ⁄𝑇𝑇 2
⁄
𝑇𝑇 2 𝑛𝑛 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡+𝜃𝜃 ∗𝑛𝑛 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡+𝜃𝜃 𝑑𝑑𝑡𝑡
𝑇𝑇 =
𝑇𝑇→∞lim
∫− ⁄𝑇𝑇 2𝑇𝑇 2⁄ 𝑛𝑛 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡+𝜃𝜃 2𝑑𝑑𝑡𝑡
𝑇𝑇 =Power of 𝑛𝑛𝜃𝜃 𝑡𝑡
• Two sample cases of 𝑛𝑛𝜃𝜃 𝑡𝑡
• These samples illustrate that Power of 𝑛𝑛𝜃𝜃 𝑡𝑡 = Power of 𝑛𝑛 𝑡𝑡 × Power of cos(𝜔𝜔𝑡𝑡+𝜃𝜃), where power of cos(𝜔𝜔𝑡𝑡+𝜃𝜃) =1/2 irrespective of 𝜃𝜃 .
𝑅𝑅𝑛𝑛𝜃𝜃 𝜏𝜏 = 0 =0.5 𝑅𝑅𝑛𝑛 𝜏𝜏 = 0 7
𝜃𝜃 = 0 𝜃𝜃 = 𝜋𝜋/3
𝑛𝑛 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡 + 𝜃𝜃 𝑛𝑛𝜃𝜃 𝑡𝑡 = 𝑛𝑛 𝑡𝑡 cos 𝜔𝜔𝑐𝑐𝑡𝑡 + 𝜃𝜃
Effect of phase error
[CH9.2B10a] (continued) ACF of 𝑛𝑛
𝜃𝜃𝑡𝑡
𝑅𝑅
𝑛𝑛𝜃𝜃𝜏𝜏 = Correlation between 𝑛𝑛 𝑡𝑡 cos 𝜔𝜔
𝑐𝑐𝑡𝑡 + 𝜃𝜃 and 𝑛𝑛 𝑡𝑡 + 𝜏𝜏 cos(𝜔𝜔
𝑐𝑐(𝑡𝑡 +
8
Effect of phase error
Demodulated signal (LPF output)
= signal term + noise term
=
12
𝑓𝑓 𝑡𝑡 cos(𝜃𝜃) + LPF[𝑛𝑛
𝜃𝜃𝑡𝑡 ]
Signal amplitude decreases by the factor of cos 𝜃𝜃 Signal power decreases by the factor of cos
2𝜃𝜃
On the other hand, noise power is constant irrespective of 𝜃𝜃
[CH9.2B4] Signal to Noise power Ratio (SNR) SNR (𝜃𝜃) =
Signal powerNoise power
=
Power of1
2𝑓𝑓 𝑡𝑡 cos(𝜃𝜃)
Power of LPF[𝑛𝑛𝜃𝜃 𝑡𝑡 ]
=
cos2 𝜃𝜃 ×Power of1 2𝑓𝑓 𝑡𝑡 Power of LPF[𝑛𝑛𝜃𝜃 𝑡𝑡 ]
= cos
2𝜃𝜃 SNR(𝜃𝜃 = 0)
Phase error results in SNR decrease by the factor of cos
2𝜃𝜃
[CH9.2B5] SNR loss = cos
2𝜃𝜃 = 10𝑙𝑙𝑙𝑙𝑔𝑔
10cos
2𝜃𝜃 dB
9
Effect of frequency error
Now we consider the case when
local carrier =cos 𝜔𝜔𝑐𝑐 + ∆𝜔𝜔 𝑡𝑡 not cos 𝜔𝜔𝑐𝑐𝑡𝑡 .
We can rewrite
local carrier = cos 𝜔𝜔𝑐𝑐𝑡𝑡 + ∆𝜔𝜔𝑡𝑡 =cos 𝜔𝜔𝑐𝑐𝑡𝑡 + 𝜃𝜃 where 𝜃𝜃
=
∆𝜔𝜔𝑡𝑡.
• This implies that the frequency error results in linearly increasing phase error.
• [CH9.2C1] So, we only need to change 𝜃𝜃 into ∆𝜔𝜔𝑡𝑡 in all the derivations for the phase error case.
– For example : demodulated signal=1
2𝑓𝑓 𝑡𝑡 cos(∆𝜔𝜔𝑡𝑡)
So, Signal to Noise power Ratio (SNR)
SNR(∆𝜔𝜔𝑡𝑡) =cos2 ∆𝜔𝜔𝑡𝑡 SNR ∆𝜔𝜔 = 0
• SNR is time varying according to cos2 ∆𝜔𝜔𝑡𝑡 .
• [CH9.2C2,C3,C6,C7] Even with a very tiny frequency error, cos2 ∆𝜔𝜔𝑡𝑡 periodically changes from 0 to 1.
• [CH9.2C2,C3,C6,C7] Whenever cos2 ∆𝜔𝜔𝑡𝑡 approaches 0, SNR instantaneously becomes 0. Critical problem!
10
Generating AM signal without Oscillator
To generate carrier signal cos(𝜔𝜔𝑐𝑐𝑡𝑡), we need an oscillator which is expensive.
[CH9.3B] DSB-SC is equal to frequency shifting to center frequencies = ±𝜔𝜔𝑐𝑐
Recall, in chapter 7, frequency shifting is possible by sampling and BPF
technique if we properly set the sampling frequency and properly design BPF.
• BPF output spectrum=𝑃𝑃−𝑛𝑛𝑋𝑋 𝜔𝜔 + 𝑛𝑛𝜔𝜔0 + 𝑃𝑃𝑛𝑛𝑋𝑋 𝜔𝜔 − 𝑛𝑛𝜔𝜔0
• BPF output signal= 𝑓𝑓(𝑡𝑡)2 𝑃𝑃𝑛𝑛 cos 𝑛𝑛𝜔𝜔0𝑡𝑡 + 𝜃𝜃𝑃𝑃𝑛𝑛
= DSB-SC modulated signal with a carrier signal =2 𝑃𝑃𝑛𝑛 cos 𝑛𝑛𝜔𝜔0𝑡𝑡 + 𝜃𝜃𝑃𝑃𝑛𝑛
[CH9.3B4] Parameter setting
Condition 1: 𝑛𝑛𝜔𝜔0=carrier frequency 𝜔𝜔𝑐𝑐, n=positive integer
Condition 2: The sampling frequency 𝜔𝜔0 ≥ 2 ×Bandwidth of 𝑓𝑓 𝑡𝑡 .
• Possible sampling frequency 𝜔𝜔0 ∶ 𝑆𝑆𝑆𝑆𝑙𝑙𝑆𝑆𝑆𝑆𝑡𝑡 𝑙𝑙𝑛𝑛𝑆𝑆 𝑙𝑙𝑓𝑓 𝜔𝜔𝑐𝑐,𝜔𝜔2𝑐𝑐,𝜔𝜔3𝑐𝑐,𝜔𝜔4𝑐𝑐, … under
the condition 2 11
P0
P1
P0
P1 Pn −nω0 −ω0 ω0
nω0
P−1
P−n
P0
P1 Pn
nω0
−
Sampling of 𝑓𝑓 𝑡𝑡
by a periodic signal 𝑝𝑝(𝑡𝑡)
BPF BPF output
=Spectrum of DSB-SC