EE 102 spring 2001-2002 Handout #25

Lecture 12 Modulation and Sampling

• The Fourier transform of the product of two signals

• Modulation of a signal with a sinusoid

• Sampling with an impulse train

• The sampling theorem

12–1 Convolution and the Fourier transform suppose f(t), g(t) have Fourier transforms F (ω), G(ω) the convolution y = f ∗ g of f and g is given by

∞ y(t)= f(τ)g(t − τ) dτ −∞

(we integrate from −∞ to ∞ because f(t) and g(t) are not necessarily zero for negative t) from the table of Fourier transform properties:

Y (ω)=F (ω)G(ω) i.e.,convolution in the time domain corresponds to multiplication in the frequency domain

Modulation and Sampling 12–2 Multiplication and the Fourier transform the Fourier transform of the product

y(t)=f(t)g(t) is given by 1 ∞ Y (ω)= F (λ)G(ω − λ)dλ 2π −∞ 1 Y = (F ∗ G) 2π i.e.,multiplication in the time domain corresponds to convolution in the frequency domain

Modulation and Sampling 12–3 example: 2 f(t)=e−|t|,F(ω)= 1+ω2 g(t)=cos20t, G(ω)=πδ(ω − 20) + πδ(ω + 20) the Fourier transform of y(t)=e−|t| cos 20t is given by 1 ∞ Y (ω)= F (λ)G(ω − λ) dλ 2π −∞ 1 ∞ 1 ∞ = F (λ)δ(ω − λ − 20) dλ + F (λ)δ(ω − λ + 20) dλ 2 −∞ 2 −∞ 1 1 = ( − 20) + ( + 20) 2F ω 2F ω 1 1 = + 1+(ω − 20)2 1+(ω + 20)2

Modulation and Sampling 12–4 2 1

0.8 1.5 ) ) 0.6 t ω ( ( 1 f 0.4 F

0.5 0.2

0 0 −3 −2 −1 t0 1 2 3 −30 −20 −10 ω0 10 20 30

1 1

0.8 0.5

) 0.6 ) ω t

0 ( (

y 0.4 Y

−0.5 0.2

0 −1 −3 −2 −1 t0 1 2 3 −30 −20 −10 ω0 10 20 30

Modulation and Sampling 12–5 Sinusoidal amplitude modulation (AM)

u(t) y(t)=u(t)cosω0t

cos ω0t Fourier transform of y 1 Y (ω)= U(ω) ∗ (πδ(ω − ω0)+πδ(ω + ω0)) 2π 1 1 = ( − )+ ( + ) 2U ω ω0 2U ω ω0

• cos ω0t is the carrier signal • y(t) is the modulated signal • the Fourier transform of the modulated signal is the Fourier transform of the input signal, shifted by ±ω0

Modulation and Sampling 12–6 Sinusoidal amplitude modulation

Baseband Signal U(ω)

0 ω

Carrier * πδ(ω+ω0) πδ(ω−ω0)

0 ω = Modulated (1/2)U(ω+ω0) (1/2)U(ω−ω0) Signal

0 ω

Modulation and Sampling 12–7 example: u(t)=2+cost, ω0 =20

3 U(ω)=4πδ(ω)+πδ(ω−1)+πδ(ω+1) 2.5 4π ) t

( 2 u

1.5 π π 1 ω 0 2 4 t 6 8 10 −10 1

3 1 1 2 Y (ω)= U(ω − 20) + U(ω + 20) 2 2 1 2π 2π ) t

( 0 y −1

−2 π/2 π/2 π/2 π/2 −3 0 2 4 6 8 10 ω t −21 −19 0 19 21

Modulation and Sampling 12–8 demodulation

y(t)=u(t)cosω0tuz(t) lowpass (t) filter

cos ω0t Fourier transform of y and z: 1 1 ( )= ( − )+ ( + ) Y ω 2U ω ω0 2U ω ω0 1 Z(ω)= Y (ω) ∗ (πδ(ω − ω0)+πδ(ω + ω0)) 2π 1 1 = ( − )+ ( + ) 2Y ω ω0 2Y ω ω0 1 1 1 = ( − 2 )+ ( )+ ( +2 ) 4U ω ω0 2U ω 4U ω ω0 if U is bandlimited, we can eliminate the 1st and 3rd term by lowpass filtering

Modulation and Sampling 12–9 Sinusoidal amplitude demodulation

Modulated (1/2)U(ω+ω0) (1/2)U(ω−ω0) Signal

0 ω

Demodulation * πδ(ω+ω ) πδ(ω−ω ) Cosine 0 0

0 ω = Demodulated Signal Lowpass (1/2)U(ω) Filter (1/4)U(ω−2ω0) (1/4)U(ω+2ω0) 0 ω

Modulation and Sampling 12–10 Application

Suppose for example that u(t) is an audio signal (frequency range 10Hz − 20kHz) We rather nottransmitu directly using electromagnetic waves:

• the wavelength is several 100 km, so we’d need very large antennas

• we’d be able to transmit only one signal at a time

• the Navy communicates with submerged submarines in this band

Modulating the signal with a carrier signal with frequency 500 kHz to 5 GHz:

• allows us to transmit and receive the signal

• allows us to transmit many signals simultaneously (frequency division multiplexing)

Modulation and Sampling 12–11 Sampling with an impulse train

Multiply a signal x(t) with a unit impulse train with period T ∞ p(t)= δ(t − kT) k=−∞

x(t) y(t)

∞ ∞ Sampled signal: y(t)=x(t) δ(t − kT)= x(kT)δ(t − kT) k=−∞ k=−∞ x(t) y(t)

t T 2T t

(a train of impulses with magnitude . . . , x(−T ), x(0), x(T ), x(2T ),...)

Modulation and Sampling 12–12 The Fourier transform of an impulse train

∞ train of unit impulses with period T : p(t)= δ(t − kT) k=−∞ 1

t −T 0 T 2T 3T 4T 5T 6T

∞ 2π 2πk Fourier transform (from table): P (ω)= δ(ω − ) T k=−∞ T 2π/T

−2π 2π 4π 6π 8π 10π 20π ω T 0 T T T T T T

Modulation and Sampling 12–13 Consequences of Sampling

ω << 2π/T

3Τ 4Τ 0 Τ 2Τ t ω = 2π/T

03Τ 2Τ Τ 4Τ t

ω = 4π/T

03Τ 2Τ Τ 4Τ t

• Frequencies well below the sampling rate (ω<<2π/T)are“sampled” in the sense we expect. • Frequencies at multiples of the sampling rate (ω =2πn/T)looklike they are constant. We can’t tell them from DC. These frequencies “alias” as DC.

Modulation and Sampling 12–14 Frequency domain interpretation of sampling

The Fourier transform of the sampled signal is 1 Y = (X ∗ P ), 2π 2π ∞ 2πk i.e.,theconvolution of X with P (ω)= T k=−∞ δ(ω − T ) 1 ∞ Y (ω)= X(λ)P (ω − λ) dλ 2π −∞   ∞ 1 ∞ 2πk = X(λ)  δ(ω − λ − ) dλ T −∞ k=−∞ T ∞ 1 ∞ 2πk = X(λ)δ(ω − λ − ) dλ T k=−∞ −∞ T ∞ 1 2πk = X(ω − ) T k=−∞ T

Modulation and Sampling 12–15 example: sample x(t)=e−|t| at different rates

2 1

0.8 1.5 ) ) 0.6 t ω ( ( 1 x

0.4 X

0.5 0.2

0 0 −3 −2 −1 t0 1 2 3 −30 −20 −10 ω0 10 20 30 x sampled with T =1(2π/T =6.3)

1 2

0.8 1.5 ) ) 0.6 t ω ( ( 1 y

0.4 Y

0.5 0.2

0 0 −3 −2 −1 t0 1 2 3 −30 −20 −10 ω0 10 20 30

Modulation and Sampling 12–16 x sampled with T =0.5 (2π/T =12.6)

4 1 3.5

0.8 3 )

) 2.5 0.6 t ω ( ( 2 y

0.4 Y 1.5

1 0.2 0.5

0 0 −3 −2 −1 t0 1 2 3 −30 −20 −10 ω0 10 20 30 x sampled with T =0.2 (2π/T =31.4)

1 10

0.8 8 ) ) 0.6 6 t ω ( ( y

0.4 Y 4

0.2 2

0 0 −3 −2 −1 t0 1 2 3 −50 ω0 50

Modulation and Sampling 12–17 The sampling theorem can we recover the original signal x from the sampled signal y ? example: a band-limited signal x (with bandwidth W ) X(ω)

1

ω −W W

Fourier transform of y(t)= k x(kT)δ(t − kT): Y (ω)

1/T

ω −W W 2π/T 4π/T

Modulation and Sampling 12–18 suppose wefiltery through an ideal lowpass filter with cutoff frequency ωc, i.e.,wemultiply Y (ω) with 1 −ωc ≤ ω ≤ ωc H(ω)= 0 |ω|≥ωc if W ≤ ωc ≤ 2π/T − W ,thenthe result is H(ω)Y (ω)=X(ω)/T , i.e.,we recover X exactly Y (ω) H(ω) 1/T

ω −W W 2π/T 4π/T Y (ω)H(ω)

1/T

−W W

Modulation and Sampling 12–19 same signal, sampled with T = π/W Y (ω)

ω 2π/T 4π/T we can still recover X(ω) perfectly by lowpass filtering with ωc = W sample with T>π/W Y (ω)

ω 2π/T 4π/T

X(ω) cannot be recovered from Y (ω) by lowpass filtering

Modulation and Sampling 12–20 the sampling theorem suppose x is a band-limited signal with bandwidth W , i.e.,

X(ω)=0for |ω| >W and we sample at a rate 1/T

∞ y(t)= x(kT)δ(t − kT) k=−∞ then we can recover x from y if T ≤ π/W

• the sampling ratemustbeatleast1/T = W/π samples per second (W/π is called the Nyquist rate)

• the distortion introduced by sampling below the Nyquist rate is called aliasing

Modulation and Sampling 12–21