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6.02 Spring 2012 Lecture 15, Slide #1

6.02 Fall 2011

Lecture #15

• More on signal spectra

• Modulation & demodulation

6.02 Spring 2012 Lecture 15, Slide #2

Again, by inspection: since the cos

and sin are at different frequencies,

we can analyze them separately.

Again, P is odd here (=11), so the

end points of the frequency scale

are at ±(π– (π/P)), not ±π.

A0 = average value = 1

A±3 = 2(1/2) = 1 [from cos term]

A-5 = –3(j/2) = –1.5j [from sin term]

A5 = –3(–j/2) = 1.5j

Ak = 0 otherwise

6.02 Spring 2012 Lecture 15, Slide #3

The DTFS is also good for

finite-duration signals!

Claim: Over any contiguous interval of length P that we may be

interested in --- say n=0,1,…,P–1 for concreteness --- an arbitrary

DT signal x[n] can be written in the form

What’s going on here? If we know we will only be interested in

the interval [0,P–1], then it doesn’t matter that our

representation above will create periodically repeating

extensions outside the interval of interest.

6.02 Spring 2012 Lecture 15, Slide #4

Spectrum of Digital Transmissions

|Ak|

Ak

n

k

Time Sequence

Frequency Spectrum P=196

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6.02 Spring 2012 Lecture 15, Slide #5

Application

h[.] x[n] y[n]

Suppose x[n] is nonzero only over the time interval [0 , nx],

and h[n] is nonzero only over the time interval [0 , nh] .

In what time interval can the non-zero values of y[n] be

guaranteed to lie? The interval [0 , nx + nh] .

Since all the action we are interested in is confined to this

interval, choose P – 1 ≥ nx + nh , then use the DTFS to

represent x[n] and y[n] over this interval.

This is actually the much more common use of the DTFS!

6.02 Spring 2012 Lecture 15, Slide #6

Spot Quiz

Determine non-zero spectral components:

1. Ak=? , when P=5, k= -2 … 2

2. Ak=?, when P=10, k= -5 … 4

)5

2cos(][ nnx

For P=5, W1 = 2/5, and

so A-1 = 0.5 and A1 = 0.5 and Ak =0 for other k.

For P=10, W1 = 2/10, and

so A-2 = 0.5 and A2 = 0.5 and Ak =0 for other k.

njnjeenx 11 5.05.0][

WW

njnjnjnjeeeenx 2211 5.05.05.05.0][

22 WWWW

6.02 Spring 2012 Lecture 15, Slide #7

From Baseband to Modulated Signal, and

Back

codeword

bits in

codeword

bits out

101110101 DAC

ADC

NOISY & DISTORTING ANALOG CHANNEL

modulate

101110101 demodulate

& filter

generate

digitized

symbols

sample &

threshold

x[n]

y[n]

6.02 Spring 2012 Lecture 15, Slide #8

Using Some Piece of the Spectrum

• You have: a band-limited signal x[n] at baseband

(i.e., centered around 0 frequency).

Re(ak)

Im(ak)

+kx −kx

Re(ak)

Im(ak)

+kc −kc

Signal centered at 0 Signal centered at kc

modulation

demodulation

Spectrum of

baseband

signal

Spectrum of

transmitted

signal

• You want: the same signal, but centered around

some specific frequency kcW1.

• Modulation: convert from baseband up to kcW1.

• Demodulation: convert from kcW1 down

to baseband

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6.02 Spring 2012 Lecture 15, Slide #9

Modulation

× x[n]

cos(kcW1n)

t[n]

WW

W

njknjk

k

kk

njk

kcc

x

x

eeeAnt 111

2

1

2

1][

Re(ak)

Im(ak)

+kc −kc

A/2

A/2

For band-limited signal

Ak are nonzero only for

small range of ±kx i.e., just replicate baseband

signal at ±kc, and scale

by ½.

W

W

x

x

c

x

x

c

k

kk

nkkj

k

k

kk

nkkj

k eAeA 11

2

1

2

1

6.02 Spring 2012 Lecture 15, Slide #10

Effect of Band-limiting a Transmission

6.02 Spring 2012 Lecture 15, Slide #11

How Low Can We Go?

7 samples/bit → 14 samples/period → k=(N/14)=(196/14)=14

6.02 Spring 2012 Lecture 15, Slide #12

Example: Modulation (time) Shaped pulses!

Chosen to band-limit

the transmit signal Baseband input x[n]

Carrier signal

Transmitted signal t[n]

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6.02 Spring 2012 Lecture 15, Slide #13

Example: Modulation

(freq domain picture) Band-limited x[n] cos(35W1n) t[n]

6.02 Spring 2012 Lecture 15, Slide #14

Demodulation

× t[n]

cos(kcW1n)

z[n]

What we want

Hmm. So z[n] has what

we want at baseband,

but has signal we don’t

want at ±2kcW1

Assuming no

distortion or

noise on

channel, so

what was

transmitted

is received

6.02 Spring 2012 Lecture 15, Slide #15

Demodulation Frequency Diagram

Re(ak)

Im(ak)

+kc −kc

A/2

A/2

Re(ak)

Im(ak)

−2kc

A/2

A/2

+2kc

t[n]

z[n]

Re(ak)

Im(ak)

+kc −kc

1/2

cos(kcW1n)

Re(ak)

Im(ak)

−2kc

A/2

A/2

+2kc

+

6.02 Spring 2012 Lecture 15, Slide #16

Demodulation Frequency Diagram

Re(ak)

Im(ak)

+kc −kc

A/2

A/2

Re(ak)

Im(ak)

−2kc

A/2

A/2

+2kc

t[n]

z[n]

What we want

Note combining of signals around 0

results in doubling of amplitude

Re(ak)

Im(ak)

+kc −kc

1/2

cos(kcW1n)

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6.02 Spring 2012 Lecture 15, Slide #17

Example: Demodulation (freq)

x[n] t[n] z[n]

Only want these frequencies…

6.02 Spring 2012 Lecture 15, Slide #18

Demodulation + LPF

× t[n] z[n] LPF y[n]

Cutoff @ ±kx

Gain = 2 cos(kcW1n)

6.02 Spring 2012 Lecture 15, Slide #19

Demodulation

× y[n]

cos(kcW1n)

z[n]

W

W

W

WW

W

W

WW

WW

WW

WW

x

x

x

x

c

x

x

c

cc

x

x

c

x

x

c

cc

k

kk

njk

kcchcch

k

kk

nkkj

kcch

k

kk

nkkj

kcch

njknjkk

kk

nkkj

kcch

k

kk

nkkj

kcch

njknjk

eAkkHkkH

eAkkHeAkkH

eeeAkkHeAkkH

eenynz

1

11

1111

11

))()((4

1

)(4

1)(

4

1

2

1

2

1)(

2

1)(

2

1

2

1

2

1][][

11

2

1

2

1

11

What we want

Hmm. So z[n] has what

we want at baseband,

but has signal we don’t

want at ±2kcW1

t[n] Channel

Hch(W)