EE 102 spring 2001-2002 Handout #25

Lecture 12Modulation 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 necessarilyzero 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 thefrequency 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

Y (ω) =12π

∫ ∞

−∞F (λ)G(ω − λ)dλ

Y =12π

(F ∗ G)

i.e., multiplication in the time domain corresponds to convolution in thefrequency domain

Modulation and Sampling 12–3

example:

f(t) = e−|t|, F (ω) =2

1 + ω2

g(t) = cos 20t, G(ω) = πδ(ω − 20) + πδ(ω + 20)

the Fourier transform of y(t) = e−|t| cos 20t is given by

Y (ω) =12π

∫ ∞

−∞F (λ)G(ω − λ) dλ

=12

∫ ∞

−∞F (λ)δ(ω − λ − 20) dλ +

12

∫ ∞

−∞F (λ)δ(ω − λ + 20) dλ

=12F (ω − 20) +

12F (ω + 20)

=1

1 + (ω − 20)2+

11 + (ω + 20)2

Modulation and Sampling 12–4

−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

t

f(t

)

−30 −20 −10 0 10 20 300

0.5

1

1.5

2

ω

F(ω

)

−3 −2 −1 0 1 2 3

−1

−0.5

0

0.5

1

t

y(t

)

−30 −20 −10 0 10 20 30

0

0.2

0.4

0.6

0.8

1

ω

Y(ω

)

Modulation and Sampling 12–5

Sinusoidal amplitude modulation (AM)

cos ω0t

u(t) y(t) = u(t) cos ω0t

Fourier transform of y

Y (ω) =12π

U(ω) ∗ (πδ(ω − ω0) + πδ(ω + ω0))

=12U(ω − ω0) +

12U(ω + ω0)

• cosω0t is the carrier signal

• y(t) is the modulated signal

• the Fourier transform of the modulated signal is the Fourier transformof the input signal, shifted by ±ω0

Modulation and Sampling 12–6

Sinusoidal amplitude modulation

ω

U(ω)

0

ω0

ω

(1/2)U(ω−ω0)

0

πδ(ω−ω0)πδ(ω+ω0)

(1/2)U(ω+ω0)

*

=

Baseband Signal

Carrier

ModulatedSignal

Modulation and Sampling 12–7

example: u(t) = 2 + cos t, ω0 = 20

0 2 4 6 8 10

1

1.5

2

2.5

3

t

u(t

)

U(ω) = 4πδ(ω)+πδ(ω−1)+πδ(ω+1)

−1 0 1

π

4π

π

ω

0 2 4 6 8 10−3

−2

−1

0

1

2

3

t

y(t

)

Y (ω) =1

2U(ω − 20) +

1

2U(ω + 20)

−21 −19 0 19 21

π/2

2π

π/2 π/2

2π

π/2

ω

Modulation and Sampling 12–8

demodulation

cos ω0t

z(t) lowpass

filter

y(t) = u(t) cos ω0t u(t)

Fourier transform of y and z:

Y (ω) =12U(ω − ω0) +

12U(ω + ω0)

Z(ω) =12π

Y (ω) ∗ (πδ(ω − ω0) + πδ(ω + ω0))

=12Y (ω − ω0) +

12Y (ω + ω0)

=14U(ω − 2ω0) +

12U(ω) +

14U(ω + 2ω0)

if U is bandlimited, we can eliminate the 1st and 3rd term by lowpassfiltering

Modulation and Sampling 12–9

Sinusoidal amplitude demodulation

ω0

πδ(ω−ω0)πδ(ω+ω0)*

=

ω

(1/2)U(ω)

0

ω

(1/2)U(ω−ω0)

0

(1/2)U(ω+ω0)

(1/4)U(ω−2ω0)(1/4)U(ω+2ω0)

LowpassFilter

ModulatedSignal

DemodulationCosine

Demodulated Signal

Modulation and Sampling 12–10

Application

Suppose for example that u(t) is an audio signal (frequency range10Hz − 20kHz)

We rather not transmit u 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 5GHz:

• allows us to transmit and receive the signal

• allows us to transmit many signals simultaneously (frequency divisionmultiplexing)

Modulation and Sampling 12–11

Sampling with an impulse train

Multiply a signal x(t) with a unit impulse train with period T

x(t) y(t)

p(t) =

∞�

k=−∞δ(t − kT )

Sampled signal: y(t) = x(t)∞∑

k=−∞δ(t − kT ) =

∞∑k=−∞

x(kT )δ(t − kT )

t

x(t)

t

y(t)

T 2T

(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) =∞∑

k=−∞δ(t − kT )

t0 T 2T 3T 4T 5T 6T−T

1

Fourier transform (from table): P (ω) =2π

T

∞∑k=−∞

δ(ω − 2πk

T)

ω0 2π

T4πT

6πT

8πT

10πT

20πT

−2πT

2π/T

Modulation and Sampling 12–13

Consequences of Sampling

t

t

t

ω << 2π/T

ω = 2π/T

ω = 4π/T

3Τ 4Τ

Τ 2Τ0

Τ 2Τ0 3Τ 4Τ

Τ 2Τ0 3Τ 4Τ

• 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 ) look likethey 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

Y =12π

(X ∗ P ),

i.e., the convolution of X with P (ω) = 2πT

∑∞k=−∞ δ(ω − 2πk

T )

Y (ω) =12π

∫ ∞

−∞X(λ)P (ω − λ) dλ

=1T

∫ ∞

−∞X(λ)

∞∑

k=−∞δ(ω − λ − 2πk

T)

dλ

=1T

∞∑k=−∞

∫ ∞

−∞X(λ)δ(ω − λ − 2πk

T) dλ

=1T

∞∑k=−∞

X(ω − 2πk

T)

Modulation and Sampling 12–15

example: sample x(t) = e−|t| at different rates

−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

t

x(t

)

−30 −20 −10 0 10 20 300

0.5

1

1.5

2

ω

X(ω

)

x sampled with T = 1 (2π/T = 6.3)

−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

t

y(t

)

−30 −20 −10 0 10 20 300

0.5

1

1.5

2

ω

Y(ω

)

Modulation and Sampling 12–16

x sampled with T = 0.5 (2π/T = 12.6)

−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

t

y(t

)

−30 −20 −10 0 10 20 300

0.5

1

1.5

2

2.5

3

3.5

4

ω

Y(ω

)

x sampled with T = 0.2 (2π/T = 31.4)

−3 −2 −1 0 1 2 30

0.2

0.4

0.6

0.8

1

t

y(t

)

−50 0 500

2

4

6

8

10

ω

Y(ω

)

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 )

ω

X(ω)

1

W−W

Fourier transform of y(t) =∑

k x(kT )δ(t − kT ):

ω

Y (ω)

1/T

2π/T 4π/TW−W

Modulation and Sampling 12–18

suppose we filter y through an ideal lowpass filter with cutoff frequency ωc,i.e., we multiply Y (ω) with

H(ω) ={

1 −ωc ≤ ω ≤ ωc

0 |ω| ≥ ωc

if W ≤ ωc ≤ 2π/T −W , then the result is H(ω)Y (ω) = X(ω)/T , i.e., werecover X exactly

ω

Y (ω)H(ω)

1/T

2π/T 4π/TW−W

Y (ω)H(ω)

1/T

W−W

Modulation and Sampling 12–19

same signal, sampled with T = π/W

ω

Y (ω)

2π/T 4π/T

we can still recover X(ω) perfectly by lowpass filtering with ωc = W

sample with T > π/W

ω

Y (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) = 0 for |ω| > W

and we sample at a rate 1/T

y(t) =∞∑

k=−∞x(kT )δ(t − kT )

then we can recover x from y if T ≤ π/W

• the sampling rate must be at least 1/T = W/π samples per second(W/π is called the Nyquist rate)

• the distortion introduced by sampling below the Nyquist rate is calledaliasing

Modulation and Sampling 12–21