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16 Oct'09 Comp30291 Section 3 1
University of Manchester
School of Computer Science
Comp30291: Digital Media Processing 2009-10
Section 3 :
Discrete-time LTI systems
16 Oct'09 Comp30291 Section 3 2
3.1 Introduction
• Consider a DSP system as implemented by a digital processor.
• Takes discrete time input signal { x[ n ] },& produces an output signal { y[ n ] }.
SYSTEM input {x[n]}
output {y[n]}
16 Oct'09 Comp30291 Section 3 3
• {x[n]} is sequence whose value at t=nT is x[n].
• Similarly for {y[n]}.
• T is sampling interval in seconds (T = 1/FS).
• {x[n-N]} is sequence whose value at t=nT is x[n-N].
{x[n-N]} is {x[n]} with every sample delayed by N
sampling intervals.
Sequences representing digital signals
16 Oct'09 Comp30291 Section 3 4
(i) Discrete time ‘amplifier’:
y[n] = A . x[n].
• Described by ‘difference equation’: y[n] = A x[n].
• Represented in diagram form by a ‘signal flow graph’:
x[n] y[n] A
Examples of discrete time systems
16 Oct'09 Comp30291 Section 3 5
(ii) Non-recursive digital filter
• Output at t=nT obtained by weighting & summing present & previous input samples: e.g.
y[n] = A0 x[n] + A1 x[n-1] + A2 x[n-2] + A3 x[n-3] + A4x[n-4]
• This is a ‘non-recursive difference equation’ • Represented by signal flow graph below.• Boxes marked ‘ z -1 ‘ produce a delay of one sampling interval.
z-1 z-1 z-1 z-1 x[n]
A0 A1 A2 A3 A4
y[n]
16 Oct'09 Comp30291 Section 3 6
(iii) Recursive digital filter
• Output at t= nT calculated from a recursive difference equation:
e.g. y[n] = A 1 x[n] - B 2 y[n-1]
• Represented by signal flow graph below.
z-1
x[n] A1
B2
y[n]
• Recursive means that previous values of y[n] as well as present & previous values of x[n] are used to calculate y[n].
16 Oct'09 Comp30291 Section 3 7
(iv) A non-linear system
• Output at t=nT calculated from some non-linear equation, e.g. y[n] = (x[n]) 2
• Represented below:
x[n]
y[n]
16 Oct'09 Comp30291 Section 3 8
(I) A DSP system is linear if: • Given any two discrete time signals {x 1 [n]} & {x 2 [n]},
if {x 1 [n]} {y 1 [n]} & {x 2 [n]} {y 2 [n]} then response to k 1{x 1[n]} + k 2{x 2[n]} must be k 1{y1[n]} + k 2 {y 2 [n]} for any values of k 1 and k 2 ,
•To multiply a sequence by k, multiply each element by k, k{x[n]} = {k x[n]}. •To add two sequences together, add corresponding samples, {x[n]} + {y[n]} = {x[n] + y[n]}.)
3.2. Linearity & time-invariance
16 Oct'09 Comp30291 Section 3 9
(II) A DSP system is ‘time-invariant’ if:
• Given any discrete time signal {x[n]}, if response to {x[n]} is {y[n]}, response to {x[n-N]} must be {y[n-N]} for any N.
• Delaying input by N samples only delays output by N samples.
• An LTI system is both linear & time-invariant
• Examples (i), (ii) and (iii) are LTI whereas (iv) is not LTI .
16 Oct'09 Comp30291 Section 3 10
3.3. Discrete time unit impulse
• Useful to consider response of LTI systems to a discrete time unit impulse , or in short an impulse denoted by {d [n] } with:
0 n : 0
0 n : 1d[n]
... ...
1
-3 -2 -1 1 2 3 4 5
n
0
d[n-4]
... ...
1
-3 -2 -1 1 2 3 4 5
d[n]
n
0
{d[n-N]} is delayed impulse where the only non-zero sample occurs at n=N rather than at n=0.
16 Oct'09 Comp30291 Section 3 11
• When input is {d[n]}, output is impulse-response {h[n]}.
• If impulse-response of an LTI system is known, its response to any other input signal may be obtained.
Impulse-response
16 Oct'09 Comp30291 Section 3 12
3.4. Implementing signal-flow-graphs
• Consider the non-recursive signal flow graph below with A1, A2, A3, A4, A5 set to specific constants. • Notice the labels X1, X2, etc.
z-1 z-1 z-1 z-1 x[n]
A1 A2 A3 A4 A5
y[n]
X1
Y
X5X4X2 X3
• Realised by MATLAB program & flow-diagram on next slides.
16 Oct'09 Comp30291 Section 3 13
Set X1, X2, X3, X4, X5 to zero
Set values of A1,A2, A3, A4, A5
INPUT X1
Y = A1*X1 + A2*X2 + A3*X3 +A3*X4 + A5*X5
OUTPUT Y
X5 = X4; X4 = X3; X3 = X2; X2 =X1
Flow-diagram for non-recursive signal-flow-graph
16 Oct'09 Comp30291 Section 3 14
clear all;
A1=1; A2=2; A3=3; A4=-4; A5=5;
X1=0; X2=0; X3=0; X4=0; X5=0;
while 1
X1 = input( 'X1 = ');
Y= A1*X1 + A2*X2 + A3*X3 + A4*X4 + A5*X5 ;
disp([' Y = ' num2str(Y)]);
X5 = X4 ; X4 = X3; X3 = X2 ; X2 =X1;
end;
MATLAB program for non-recursive signal flow graph
16 Oct'09 Comp30291 Section 3 15
More efficient version
A = [1 2 3 -4 5 ]’ ;x = [0 0 0 0 0 ]’ ; while 1 x(1) = input( ‘x(1) = '); Y=0; for k = 1 : 5 Y = Y + A(k)*x(k); end; disp([' Y = ' num2str(Y)]); for k=5:-1:2 x(k) = x(k-1); end;end;
16 Oct'09 Comp30291 Section 3 16
Even more efficient version
A = [1 2 3 -4 5 ]' ;x = [0 0 0 0 0 ]' ; while 1
x(1) = input( 'x(1) = ');Y = A(1)*x(1);for k = 5 : -1: 2
Y = Y + A(k)*x(k);x(k) = x(k-1);
end;disp(['Y = ' num2str(Y)]);
end;
16 Oct'09 Comp30291 Section 3 17
• The ‘while 1’ statement initiates an infinite loop.
• Program runs for ever or until interrupted by ‘CONTROL+C’
• Either of the following prints out value of Y:
disp(['Y = ' num2str(Y)]);
disp(sprintf(‘Y=%d’,Y));
• A = [1 2 3 -4 5]’ makes A a column vector (not a row). Note ’.
Comments
• Ready to be converted to DSP assembler.
Y = Y + A(k)*x(k); x(k) = x(k-1);
One DSP instruction:Mult-acc-shift - MACS
16 Oct'09 Comp30291 Section 3 18
clear all;x=[0 1 0 0 0 0 0 0 0 0]';a = [1 2 3 -4 5]';y=filter(a,1,x);y
Use of ‘filter’ to implement non-rec signal-flow-graph- & thus to produce its impulse-response
• Just writing ‘y’ without ‘;’ prints out the array.
16 Oct'09 Comp30291 Section 3 19
Impulse-response for non-rec signal-flow-graph• Use any of previous versions & enter values for X1 :
0, 0, 0, 1, 0, 0, 0, 0, .... • Sequence of output samples printed out will be :
0, 0, 0, A1, A2, A3, A4, A5, 0, 0, .... • Impulse-response can also be obtained by tabulation (later). • Output must be zero until input becomes equal to 1 at n=0• Impulse response is:
{..., 0, ..., 0, A1, A2, A3, A4, A5, 0, 0, ... ,0, ...}
where the sample at n=0 is underlined. • Only five non-zero output samples are observed.• This is a ‘ finite impulse-response ‘ (FIR).
16 Oct'09 Comp30291 Section 3 20
(i) y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4]
(ii) y[n] = 4 x[n] - 0.5 y[n-1]
• Difference eqn (i) will produce a finite impulse-response.• Difference eqn (ii) produces infinite response whose samples
gradually reduce in amplitude but never quite become zero.
Exercise 3.1
•Calculate impulse-responses, by tabulation, for:
16 Oct'09 Comp30291 Section 3 21
n x[n] x[n-1] x[n-2] x[n-3] x[n-4] y[n]-1 0 0 0 0 0 00 1 0 0 0 0 11 0 1 0 0 0 22 0 0 1 0 0 33 0 0 0 1 0 -44 0 0 0 0 1 55 0 0 0 0 0 0: : : : : : :
Impulse-response for example (i) by tabulation
Impulse response is: {.. 0, .., 0, 1, 2, 3, -4, 5, 0, .., 0, ...}
y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4]
16 Oct'09 Comp30291 Section 3 22
n x[n] y[n-1] y[n]0 1 0 41 0 4 -22 0 -2 13 0 1 -0.54 0 -0.5 0.255 0 0.25 -0.125: : : :
Impulse-response by tabulation for Example (ii)
Impulse response is: {.., 0, .., 0, 4, -2, 1, -0.5, 0.25, -0.125, ...)
y[n] = 4 x[n] - 0.5 y[n-1]
16 Oct'09 Comp30291 Section 3 23
n x[n] y[n-1] y[n]0 1 0 41 0 4 -82 0 -8 163 0 16 -324 0 -32 645 0 64 128: : : :
Further example: Impulse-response by tabulation
• Impulse response is: {.., 0, .., 0, 8, 16, -32, 64, 128, ...)
• This IIR filter is ‘unstable’ (see later)
y[n] = 4 x[n] - 2 y[n-1]
16 Oct'09 Comp30291 Section 3 24
• Difference equation (i) is "non-recursive" & produces a finite impulse response (FIR). • Difference equation (ii) is " recursive " . • Impulse-response of a recursive difference equation can have an infinite number of non-zero terms. In this case it is an infinite impulse-response (IIR).
3.5. FIR & IIR digital filters
• A ‘digital filter’ is a digitally implemented LTI discrete time system governed by a difference equation of finite order; e.g. :
(i) y[n] = x[n] + 2 x[n-1] + 3 x[n-2] - 4 x[n-3] + 5 x[n-4]
(ii) y[n] = 4 x[n] - 0.5 y[n-1]
16 Oct'09 Comp30291 Section 3 25
clear all;
Y2=0;
while 1
X1 = input( 'X1 = ');
Y1= 4*X1 - 0.5*Y2 ;
Y2 = Y1; % for next time round
disp([' Y1 = ' num2str(Y)]);
end;
MATLAB program for recursive signal flow graph
- example (ii)
16 Oct'09 Comp30291 Section 3 26
y = filter(A, B, x) filters signal in array x to create array y.
For FIR example (i), A = [ 1 2 3 -4 5 ] & B = [1].
For IIR example (ii), A = [4], B = [1 0.5]
Consider a third IIR example:
y[n] = 2x[n] + 3x[n-1] + 4x[n-2] -0.5 y[n-1] - 0.25 y[n-2]
In this case set A = [2 3 4]
and B = [1 0.5 0.25].
Why are A & B defined in this way?
Use of ‘filter’ for FIR & IIR digital filters
16 Oct'09 Comp30291 Section 3 27
• A digital filter has a ‘system function’ which is
MM
NN
zazbzbb
zazazaazH
...
...)(
22
110
22
110
with b0 = 1
for the difference equation:
][...]2[]1[
][...]2[]1[][][
21
210
Mnybnybnyb
Mnxanxanxanxany
M
N
• A contains [a0 a1 ... aN] & B contains [b0, b1, ..., bM] .
• Reasons for this & more details will be given later in course.
• ‘filter’ can be used without knowing why H(z) is defined in this way.
Definition of A & B in ‘y = filter(A,B,x);’
16 Oct'09 Comp30291 Section 3 28
If impulse-response of an LTI system is {h[n]} its response to any input {x[n]} is an output {y[n] } whose samples are given by the following ‘convolution’ formulae:
m
m] x[nh[m] y[n]
k
k]h[n x[k]y[n]or
3.6. Discrete time convolution
• Formulae are equivalent.
• Clearly, if we know impulse-response {h[n]} we can produce the response to any other input sequence, from either of these formulae.
•Proof of convolution in Appendix 3A.
16 Oct'09 Comp30291 Section 3 29
Calculate response of a system with impulse response: {h[n]} = { ..., 0,..., 0, 1, 2, 3, -4, 5, 0, .....0, .... } to {x[n]} = { ... 0, ... , 0 , 1, 2, 3, 0, ..., 0, ....}Solution: By first discrete time convolution formula,
4]5x[n3]4x[n2]3x[n1]2x[n x[n]
zero are msr eother t all since m] x[nh[m]
m] x[nh[m] y[n]
4
0m
-m
Example
This is difference equation for an LTI system with impulse response {.., 0, .., 0, 1, 2, 3, -4, 5, 0, .., 0, ..}
16 Oct'09 Comp30291 Section 3 30
Completing the Example
• Program discussed earlier implements this difference equation,
• We could run it, and enter {0 1 2 3 0 0 0 0...},
• Output sequence produced is what we want.
• Alternatively, we could use tabulation as follows:
16 Oct'09 Comp30291 Section 3 31
Response of: y[n] = x[n] +2x[n-1]+3x[n-2]-4x[n-3] +5x[n-4]to input sequence {...,0,1,2,3,0,...} by tabulation
n x[n] x[n-1] x[n-2] x[n-3] x[n-4] y[n]: : : : : : :-1 0 0 0 0 0 00 1 0 0 0 0 11 2 1 0 0 0 42 3 2 1 0 0 103 0 3 2 1 0 84 0 0 3 2 1 65 0 0 0 3 2 -26 0 0 0 0 3 157 0 0 0 0 0 0: : : : : : :{y[n]} = { .... 0, ....., 0, 1, 4, 10, 8, 6, -2, 15, 0, ...., 0, ....}
16 Oct'09 Comp30291 Section 3 32
3.7. Stability An LTI system is stable if its impulse-response {h[n]} satisfies:
finite is ][
n
nh
This means that {h[n]} must be either an FIR or an IIR whose samples decay towards zero as n .
16 Oct'09 Comp30291 Section 3 33
3.8. Causality
An LTI system operating in real time must be ‘causal’ which
means that its impulse-response {h[n]} must satisfy:
h[n] = 0 for n < 0.
Non-causal system would need “crystal ball ” to predict future.
16 Oct'09 Comp30291 Section 3 34
Illustration of stability & causality
Looks stable
but is not causal.
n
h[n] h[n]
n
Causal, but not stable
h[n]
Causal & looks stable.
n
16 Oct'09 Comp30291 Section 3 35
• = T is ‘relative frequency’ of sampled sinusoid. • Units of are 'radians per sample'.
3.9. Relative Frequency
• Study effect of digital filters on sinusoids of different frequencies.• Discrete time sinusoid obtained by sampling A cos(t + ). • If sampling frequency is Fs Hz, and T=1/Fs, we obtain:
x[n] = A cos(nT + ) = A cos(n + )
16 Oct'09 Comp30291 Section 3 36
Radians per sample
• To convert back to true frequency (radians/s ) multiply by Fs.
• radians / sample samples / second = radians / second
• Analogue signals in range 0 to FS/2 Hz =1/(2T)
• Restricts to the range 0 to .
16 Oct'09 Comp30291 Section 3 37
T
3T-T
-3T 4T-4T
cos( t )
1
3-1
-3 4-4
cos( T n )
t
n
= 2 / 8
=2 / (8T)
16 Oct'09 Comp30291 Section 3 38
Values of & corresponding true frequencies
Relative frequency True frequency(radians/sample) (radians/s) (Hz)
0 0 0/6 fS/6 fS/12/4 fS/4 fs/8/3 fS/3 fs/6/2 fS/2 fs/42/3 2fS/3 fs/3 fS fS/2
‘radians / sample’ ‘samples / second’ = ‘radians / second’
16 Oct'09 Comp30291 Section 3 39
3.10. Relative frequency response
It us useful to analyse the response to a sampled sinusoid:
x[n] = Acos(n + )
To begin with, set A=1, =0 and remember de Moivre’s Theorem:
)sin()cos( njne nj
Easier to calculate response to the complex signal x[n] = ejn
than to cos(n) directly.
16 Oct'09 Comp30291 Section 3 40
Relative frequency response (cont)
If x[n] = ejn is applied to a system with impulse-response {h[n]}, output would be, by convolution :
(DTFT) ][ ][)( where
)(
][
][
][][ )(
n
nj
m
mjj
jnj
m
mjnj
m
mjnj
m
mnj
enhemheH
eHe
emhe
eemh
emhny
16 Oct'09 Comp30291 Section 3 41
• H( e j ) is the DTFT of { h[n] }.
• It is called the ‘relative frequency-response’
• Complex number for any value of .
• Note similarity to the analogue Fourier transform.
(FT) )()(
(DTFT) ][)(
a dtethjH
enheH
tja
n
njj
Discrete time Fourier Transform (DTFT)
16 Oct'09 Comp30291 Section 3 42
• If input is x[n]=e j n, output is same sequence with each
element multiplied by H( e j ).
n
njj enheH ][)(
e j n
{h[n]}
e j n H(e j ) LTI
Recap
16 Oct'09 Comp30291 Section 3 43
3.11. Gain & phase responses
• G() = |H( e j )| is ‘gain’ • () = arg ( H( e j ) ) is ‘phase lead’. • Both vary with . • Can express: H( e j ) = G() e j ()
• If input is {A cos(n)}, output is: { G()A cos(n + ()) }
When input is sampled sinusoid of relative frequency , output is sinusoid of same frequency , but with amplitude scaled by G() & phase increased by ().
16 Oct'09 Comp30291 Section 3 44
Gain & phase response graphs again
/4 /2 3/40
20log10[G()] dB-()
-()
G() in dB
16 Oct'09 Comp30291 Section 3 45
Graphs of G() & () against .
• G() often converted to dBs by calculating 20 log10( G() ).
• Restrict to lie in range 0 to
• Adopt a linear horizontal frequency scale.
16 Oct'09 Comp30291 Section 3 46
Example Derive frequency-response of FIR digital filter below.
z-1 z-1 z-1 z-1 x[n]
1 2 3 -4 5
y[n]
• Impulse-response is: {.., 0, .., 0, 1, 2, 3, -4, 5 ,0, .., 0,..}. • By the formula established above, • H( e j ) = 1 + 2 e - j + 3 e -2 j - 4 e -3 j + 5 e - 4 j
16 Oct'09 Comp30291 Section 3 47
Example: Plot gain & phase responses of this FIR filter
• To do this ‘by hand’, we would need to take modulus & phase of expression for H( e j ).• Fortunately we almost never need to do this ‘by hand’ (except for very simple expressions).• Best done by MATLAB.
• To calculate gain & phase responses for H( e j ) = 1 + 2 e - j + 3 e -2 j - 4 e -3 j + 5 e - 4 j
start MATLAB and type: freqz( [1 2 3 -4 5] );
16 Oct'09 Comp30291 Section 3 48
16 Oct'09 Comp30291 Section 3 49
‘freqz’ graphs of gain & phase responses
• Frequency scale normalised to fS/2 & labelled 0 to 1 instead of 0 to .
• To plot freq-response for an IIR filter equally straightforward using ‘freqz’ if you know how to define the system function.
•See earlier.
•More about this in a later section.
16 Oct'09 Comp30291 Section 3 50
• In next section we see how FIR filter coefficients can be chosen to achieve a particular type of gain & phase response.
• Tells us how the MATLAB function ‘fir1’ works in principle.
• In later sections, principles of IIR digital filter design will be considered also
FIR & IIR digital filter design
16 Oct'09 Comp30291 Section 3 51
Phase delay
• Assume input is {A cos(n)},
• Output is: { G() A cos ( n + () ) }
= { G() A cos ( [ n + () / ] ) }
= { G() A cos ( [ n - (-() / ) ] ) }
• Phase lead () delays sine-wave by -()/ sampling intervals
• This is ‘phase-delay’.
16 Oct'09 Comp30291 Section 3 52
3.12. Linear phase response
• If -()/ remains constant for all values of ,
i.e. if -() = k for constant k,
system has ‘linear phase’ response.
• Graph of -() against on linear scale would then be
straight line with slope k where k is "phase delay"
i.e. the delay measured in sampling intervals.
• This need not be an integer.
16 Oct'09 Comp30291 Section 3 53
-()/
/4 /2 3/40
-()
G() in dB
-()/
Gain, phase & ‘phase-delay’ response graphs
Not ‘linear-phase’.
16 Oct'09 Comp30291 Section 3 54
-()/
/4 /2 3/40
-()
G() in dB
-()/
Gain, phase & ‘phase-delay’ response graphs again
This is ‘linear-phase’ as -()/ is constant.
k
16 Oct'09 Comp30291 Section 3 55
Why is linear phase a good property?
• All sinusoids delayed by same number of sampling intervals.
• Input signal expressed as Fourier series will have all its
sinusoidal components delayed by same amount of time.
•No ‘phase distortion’ due to different delays at different
frequencies.
• LTI systems are not necessarily linear phase.
• A certain class of digital filter can be made linear phase.
16 Oct'09 Comp30291 Section 3 56
3.13. Inverse DTFT
• Frequency-response H( e j ) is DTFT of {h[n]}.
• Inverse DTFT formula allows {h[n]} to be deduced from H(e j ) :
<<-: )(2
1][ ndeeHnh njj
• This formula requires H(ej) for in range - to .• Negative frequencies?? • Not a problem because when { h[n] } is real, H( e -j ) = H*( e j ) where * denotes complex conj.
16 Oct'09 Comp30291 Section 3 57
Similarities with inverse analogue Fourier Transform
<<-: )(2
1][ ndeeHnh njj
Notice: (i) (1/2) factor, (ii) sign of jn (n replaces t).
(iii) variable of integration (d)
<<-: )(
2
1)( tdejHth tj
aa
Inverse DTFT
Inverse FT
16 Oct'09 Comp30291 Section 3 58
Also note that, with inverse DTFT• Range of integration is now - to rather than - to .
• DTFT is a summation & inverse DTFT is an integral.
• This is because {h[n]} is a sequence
whereas H(e j ) is function of the continuous variable .
16 Oct'09 Comp30291 Section 3 59
3.14 Problems1. Why is y[n]=(x[n]) 2 not LTI.2. If {h[n]}= { .. 0, .. , 0, 1, -1, 0, .. 0, .. }, calculate response to {x[n]} = { .... 0, .., 0, 1, 2, 3, 4, 0, .., 0, ... }.3.. Produce a signal flow graph for each of the following difference equations: (i) y[n] = x[n] + x[n-1] (ii) y[n] = x[n] - y[n-1] (iii) y[n] = 2x[n] +3x[n-1] +2x[n-2] - 0.5 y[n-1] -0.25 y[n-2]4. For difference equations (i) & (ii) in question 3, determine the impulse- response, & deduce whether the system it represents is stable and/or causal.5. Calculate, by tabulation, the output from difference equation (ii) in question 3 when the input is the impulse-response of difference equation (i).6. If fS =8000 Hz, what true frequency corresponds to /5 radians/sample?7. Sketch the gain & phase responses of the system referred to in question 2. (You don't need a computer program.) Is it a linear phase system?8. Calculate the impulse-response for y[n] = 4x[n] + 2y[n-1]. Is it stable?9. Show that input {cos(n)} produces an output {G()cos(n+())}.10. For y[n]=x[n]+2x[n-1]+3x[n-2]+2x[n-3]+x[n-4], sketch phase-response and comment on its significance. Show that () = -2.
16 Oct'09 Comp30291 Section 3 60
Appendix 3A: Proof of discrete time convolution formulae
Only use linearity, time-invariance & impulse-response. Since d[n-m] is non-zero only at n = m, given any sequence {x[n]},
m
m]x[m]d[nx[n]
m]}[m]{d[nx{x[n]}
m
{x[n]} is sum of infinite number of delayed impulses{d[n-m]} each multiplied by a single element, x[m]. Response to {d[n-m]} is {h[n-m]} for any value of m. response to {x[n]} is :
n allfor m][m]h[nxy[n] i.e. m]}[m]{h[nx{y[n]}
mm
16 Oct'09 Comp30291 Section 3 61
• Replacing n-m by k gives the alternative formula. • Study the graphical explanation of this proof in Section 3.17.
n
2
3
x[n]
2
h[n]
n
{h[n]} = { ..., 0, ..., 0, 1, 2, 1, 2, 0, ..., 0...}
{x[n]} = { ..., 0, ..., 0, 1, 2, 3, 0, 0, ..., 0, ...}
16 Oct'09 Comp30291 Section 3 62
Discrete time convolution graphically
n
23
x[n]
d[n]
n
2
2d[n-1]
n
3
3d[n-2]
n
2
h[n]
n
2
42h[n-1]
n
3
63h[n-2]
n
{h[n]} = { ..., 0, ..., 0, 1, 2, 1, 2, 0, ..., 0,...}
{x[n]} = { ..., 0, ..., 0, 1, 2, 3, 0, 0, ..., 0, ...}
16 Oct'09 Comp30291 Section 3 63
2
4
6
8
10
y[n]
n
Express:
{x[n]} = {d[n]} + 2 {d[n-1]} + 3 {d[n-2]}
Response is:
{y[n]} = {h[n]} + 2 {h[n-1} + 3 {h[n-2]}
16 Oct'09 Comp30291 Section 3 64
2
4
6
8
10y[n]
n
n
2
3
x[n]
{ d[n] } {h[n]}
+ 2{ d[n-1 ]} + 2 { h[n-1]}
+ 3 { d[n-2] } + 3 { h[n-2] }
{h[n]} = { ..., 0, ..., 0, 1, 2, 1, 2, 0, ..., 0,...}
{x[n]} = { ..., 0, ..., 0, 1, 2, 3, 0, 0, ..., 0, ...}
16 Oct'09 Comp30291 Section 3 65
d[n]
n
2
2d[n-1]
n
3
3d[n-2]
n
2
4
2h[n-1]
n
3
63h[n-2]
n
2
h[n]
n