6.02 Fall 2014 Lecture 11, Slide #1

6.02 Fall 2014�Lecture #11�

•  Convolu'on  

6.02 Fall 2014 Lecture 11, Slide #2

Unit Step

A simple but useful discrete-time signal is the unit step signal or function, u[n], defined as

u[n]= 0, n < 01, n ≥ 0

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6.02 Fall 2014 Lecture 11, Slide #3

Unit Sample

Another simple but useful discrete-time signal is the unit sample signal or function, δ[n], defined as

δ[n]= u[n]−u[n−1]= 0, n ≠ 01, n = 0

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6.02 Fall 2014 Lecture 11, Slide #4

Unit Sample and Unit Step Responses

S δ[n] h[n]

Unit sample Unit sample response

The unit sample response of a system S is the response of the system to the unit sample input. We will always denote the unit sample response as h[n]. For a causal linear system, h[n] = 0 for n < 0.

S u[n] s[n]

Unit step Unit step response

Similarly, the unit step response s[n]:

6.02 Fall 2014 Lecture 11, Slide #5

δ[n]= u[n]−u[n−1] h[n]= s[n]− s[n−1]

Relating h[n] and s[n] of an LTI System

S u[n] s[n]

Unit step signal Unit step response

S δ[n] h[n]

Unit sample signal Unit sample response

from which it follows that (assuming , e.g., a causal LTI system; more generally, assuming a “right-sided” unit sample response)

s[n]= h[k]k=−∞

n

∑s[−∞]= 0

6.02 Fall 2014 Lecture 11, Slide #6

h[n] s[n]

6.02 Fall 2014 Lecture 11, Slide #7

h[n] s[n]

6.02 Fall 2014 Lecture 11, Slide #8

h[n] s[n]

6.02 Fall 2014 Lecture 11, Slide #9

Unit Step Decomposition

“Rectangular-wave” digital signaling waveforms, of the sort we have been considering, are easily decomposed into time-shifted, scaled unit steps --- each transition corresponds to another shifted, scaled unit step. e.g., if x[n] is the transmission of 1001110 using 4 samples/bit:

x[n]

= u[n]

−u[n− 4]

+u[n−12]

−u[n− 24]

6.02 Fall 2014 Lecture 11, Slide #10

… so the corresponding response is

y[n]

= s[n]

− s[n− 4]

+ s[n−12]

− s[n− 24]

x[n]

= u[n]

−u[n− 4]

+u[n−12]

−u[n− 24]

Note how we have invoked linearity and time invariance!

6.02 Fall 2014 Lecture 11, Slide #11

Example

6.02 Fall 2014 Lecture 11, Slide #12

Transmission Over a Channel

6.02 Fall 2014 Lecture 11, Slide #13

Receiving the Response

Digitization threshold = 0.5V

6.02 Fall 2014 Lecture 11, Slide #14

Faster Transmission

Noise margin? 0.5 − y[28]

6.02 Fall 2014 Lecture 11, Slide #15

Step response of audiocom channel�

6.02 Fall 2014 Lecture 11, Slide #16

Unit Sample Decomposition�

A  discrete-‐'me  signal  can  be  decomposed  into  a  sum  of  'me-‐shi9ed,  scaled  unit  samples.    Example:  in  the  figure,  x[n]  is  the  sum  of    x[-‐2]δ[n+2]  +  x[-‐1]δ[n+1]  +  …  +  x[2]δ[n-‐2].    In  general:  

x[n]= x[k]δ[n− k]k=−∞

∞

∑

For  any  par'cular  index,  only  one  term  of  this  sum  is  non-‐zero  

6.02 Fall 2014 Lecture 11, Slide #17

If  system  S  is  both  linear  and  'me-‐invariant  (LTI),  then  we  can  use  the  unit  sample  response  to  predict  the  response  to  any  input  waveform  x[n]:                  Indeed,  the  unit  sample  response  h[n]  completely  characterizes  the  LTI  system  S,  so  you  o9en  see  

S x[n]= x[k]δ[n− k]k=−∞

∞

∑ y[n]= x[k]h[n− k]k=−∞

∞

∑

Sum of shifted, scaled unit samples Sum of shifted, scaled responses

Modeling LTI Systems�

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

CONVOLUTION SUM

6.02 Fall 2014 Lecture 11, Slide #18

Evalua'ng  the  convolu'on  sum        for  all  n  defines  the  output  signal  y  in  terms  of  the  input  x  and  unit-‐sample  response  h.  Some  constraints  are  needed  to  ensure  this  infinite  sum  is  well  behaved,  i.e.,  doesn’t  “blow  up”  -‐-‐-‐  we’ll  discuss  this  later.    We  use          to  denote  convolu'on,  and  write  y=x      h.  We  can  then  write  the  value  of  y  at  'me  n,  which  is  given  by  the  above  sum,  as                                                                                                                                                                                                                                                                                                                  or                                                                                                                                              

y[n]= x[k]h[n− k]k=−∞

∞

∑

Convolution�

∗ ∗

y[n]= (x∗h)[n] y[n]= x∗h[n]

Instead  you’ll  find  people  wri'ng                                                                                ,  where  the  poor  index  n  is  doing  double  or  triple  duty.  This  is  lousy  nota'on,  but  a  super-‐majority  of  engineering  professors  (including  at  MIT)  will  inflict  it  on  their  students.  

y[n]= x[n]∗h[n]

6.02 Fall 2014 Lecture 11, Slide #19

Properties of Convolution�

(x∗h)[n]≡ x[k]h[n− k]k=−∞

∞

∑ = h[m]x[n−m]m=−∞

∞

∑

The  second  equality  above  establishes  that  convolu'on  is  commuta've:        Convolu'on  is  associa've:        Convolu'on  is  distribu've:        (Some  condi'ons  needed,  typically  guaranteeing  that  the  individual  convolu'ons  are  well-‐behaved.)  

x∗h = h∗ x

x∗ (h1 ∗h2 ) = x∗h1( )∗h2

x∗ h1 + h2( ) = (x∗h1)+ (x∗h2 )

6.02 Fall 2014 Lecture 11, Slide #20

Series Interconnection of LTI Systems�

h1[.]  x[n]   h2[.]  

y[n]  

y = h2 ∗w = h2 ∗ h1 ∗ x( ) = h2 ∗h1( )∗ x

(h2∗h1)[.]  x[n]   y[n]  

w[n]  

(h1∗h2)[.]  x[n]   y[n]  

h2[.]  x[n]   h1[.]  

y[n]  

6.02 Fall 2014 Lecture 11, Slide #21

Parallel Interconnection of LTI Systems�

h1[.]  

x[n]  

y1[n]  

h2[.]  

+

y2[n]  

y[n]  

(h1+h2)[.]  x[n]   y[n]  

y = y1 + y2 = (h1 ∗ x)+ (h2 ∗ x) = h1 + h2( )∗ x

6.02 Fall 2014 Lecture 11, Slide #22

Unit Sample Response of a �Scale-&-Delay System�

S  x[n]   y[n]=Ax[n-‐D]  

If  S  is  a  system  that  scales  the  input  by  A  and  delays  it  by  D  'me  steps  (nega've  ‘delay’  D  =  advance),  is  the  system    

             'me-‐invariant?      

             linear?  

                                                                                                 Yes!                                                                                  Yes!    Unit  sample  response  is  h[n]=Aδ[n-‐D]  

General  unit  sample  response                h[n]=…  +  h[-‐1]δ[n+1]  +  h[0]δ[n]  +  h[1]δ[n-‐1]+…      for  an  LTI  system  can  be  thought  of  as  resul'ng  from    many  scale-‐&-‐delays  in  parallel  

6.02 Fall 2014 Lecture 11, Slide #23

y[n]= h[m]x[n−m]m=−∞

∞

∑

h[.]  x[n]= x[k]δ[n− k]k=−∞

∞

∑ y[n]= x[k]h[n− k]k=−∞

∞

∑

A Complementary View of Convolution�

h[.]=…+h[-‐1]δ[n+1]+h[0]δ[n]+h[1]δ[n-‐1]+…            x[n]           y[n]  

So  instead  of  the  picture:  

we  can  consider  the  picture:  

from  which  we  get    

6.02 Fall 2014 Lecture 11, Slide #24

(side by side)�y[n]=

(x∗h)[n]= x[k]h[n− k]k=−∞

∞

∑ = h[m]x[n−m]= (h∗ x)[n ]m=−∞

∞

∑

Input  term  x[0]  at  'me  0  launches    scaled  unit  sample  response  x[0]h[n]  at    output    Input  term  x[k]  at  'me  k  launches    scaled  shi9ed  unit    sample  response    x[k]h[n-‐k]  at  output    

Unit  sample  response  term  h[0]  at  'me  0    contributes  scaled  input  h[0]x[n]  to  output      Unit  sample  response    term  h[m]  at  'me  m    contributes  scaled  shi9ed  input  h[m]x[n-‐m]    to  output    

6.02 Fall 2014 Lecture 11, Slide #25

To Convolve (but not to “Convolute”!)

A  simple  graphical  implementa'on:    Plot  x[.]  and  h[.]  as  a  func'on  of  the  dummy  index    (k  or  m  above)    Flip  (i.e.,  reverse)  one  signal  in  'me,    slide  it  right  by  n  (slide  le9  if  n  is  –ve),  take  the  dot.product  with  the  other.    This  yields  the  value  of  the  convolu'on  at    the  single  'me  n.      ‘flip  one    &  slide  by  n  ….  dot.product  with  the  other’      

x[k]h[n− k]k=−∞

∞

∑ = h[m]x[n−m]m=−∞

∞

∑

6.02 Fall 2014 Lecture 11, Slide #26

Example �•  From  the  unit  sample  response  h[n]  to  the  unit  step  response    

                                                                           s[n]  =  (h∗u)[n]    •  Flip  u[k]  to  get  u[-‐k]  •  Slide  u[-‐k]  n  steps  to  right  (i.e.,  delay  u[-‐k])  to  get  u[n-‐k]),            place  over  h[k]  •  Dot  product  of  h[k]  and  u[n-‐k]  wrt  k:  

s[n]= h[k]k=−∞

n

∑

6.02 Fall 2014 Lecture 11, Slide #27

Channels as LTI Systems�

Many  transmission  channels  can  be  effec'vely  modeled  as  LTI  systems.    When  modeling  transmissions,  there  area    few  simplifica'ons  we  can  make:    

y[n]= x[k]h[n− k]k=−∞

∞

∑ = x[k]h[n− k]k=0

∞

∑ = x[k]h[n− k]k=0

n

∑ = x[n− j]h[ j]j=0

n

∑

These  two  observa'ons  allow  us  to  rework  the  convolu'on  sum  when  it’s  used  to  describe  transmission  channels:  

 •  We’ll  call  the  'me  transmissions  start  t=0;  the  signal  before  the  start  is  

0.    So  x[m]  =  0  for  m  <  0.      

•  Real-‐word  channels  are  causal:  the  output  at  any  'me  depends  on  values  of  the  input  at  only  the  present  and  past  'mes.    So  h[m]  =  0  for  m  <  0.  

j=n-k start at t=0 causal

6.02 Fall 2014 Lecture 11, Slide #28

“Deconvolving”  Output  of  Echo  Channel  

Channel,        h1[.]  

Receiver  filter,    h2[.]  

x[n]   y[n]   z[n]  

Suppose  channel  is  LTI  with      

 h1[n]=δ[n]+0.8δ[n-‐1]      Find  h2[n]  such  that  z[n]=x[n]  

Good  exercise  in  applying    Flip/Slide/Dot.Product  

(h2*h1)[n]=δ[n]  

6.02 Fall 2014 Lecture 11, Slide #29

“Deconvolving” Output of�Channel with Echo �

Channel, h1[.]

Receiver filter, h2[.]

x[n] y[n] + w[n]

z[n]+v[n]

Even  if  channel  was  well  modeled  as  LTI  and  h1[n]    was  known,  noise  on  the  channel  can  greatly  degrade  the  result,  so  this  is  usually  not  prac'cal.