Discrete Signal Processing
Lecture 2
Signals and systems review Part 2
DSP background material
The Unit Impulse (Unit Sample)
ƒ Review of properties of the unit impulse
ƒ Definition
ƒ Shifting
ƒ Sifting
ƒ Connection to the unit step
ƒ System Properties
ƒ Inpulse response
ƒ Convolution
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Unit impulse (unit sample)
ƒ Defined as:
δ [n] = 0, n ≠ 0
= 1, n = 0
Shifted (Delayed)
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2
1.8
1.8
δ [n]
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1.4
1.6
δ [n − 3]
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1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
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-6
-4
-2
0
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Unit impulse and Unit step
ƒ Close relationship
ƒ Unit impulse is first difference of unit step
δ [n] = u[n] − u[n − 1]
ƒ Remember that CT impulse was the first derivative of the
CT step
ƒ Alternatively:
u[n] =
∞
∑ δ [ m]
m = −∞
Called a running sum.
Analogous to integration
Only non zero value is at m = 0, so that summation
agrees with definition of the unit step.
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Sifting Property
ƒ Multiplying any sequence
signal
by a unit impulse selects or
‘sifts’ out the value of the sample
signal at the instant of the
impulse:
x[n] ⋅ δ [n] = x[0]
x[n] ⋅ δ [n − k ] = x[k ]
The well known sifting property of impulses
ƒ We can ‘build’ any sequence using this property
Called Sampling
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DT sequence from Impulses
Does this
look
familiar?
A convolution
type summation
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Summarising
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Summarising
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Systems
ƒ A discrete-time system transforms an input sequence
into an output sequence according to some
input/output rule.
ƒ Can be done sample by sample
ƒ Thus:
H
[ x0 , x1 , x2 , L xn ] ⎯⎯→
[ y0 , y1 , y2 ,L yn ]
Called sample by sample processing
ƒ Real-time systems would use this technique
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Systems
OR
⎡ x0 ⎤
⎡ y0 ⎤
⎢y ⎥ = y
H
x = ⎢⎢ x1 ⎥⎥ ⎯⎯→
⎢ 1⎥
⎢⎣ xn ⎥⎦
⎢⎣ yn ⎥⎦
Called block processing
For
systems
where data
is stored
⎡2 0 0⎤ ⎡ x0 ⎤
y = Hx = ⎢⎢0 3 0⎥⎥ ⎢⎢ x1 ⎥⎥
⎢⎣0 0 4⎥⎦ ⎢⎣ x3 ⎥⎦
ƒ The technique used depends on the particular DSP
application
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Required System Properties
ƒ Linearity
ƒ If input consists of a number of weighted signals
ƒ Overall response is a linear superposition of
responses to each individual signal.
Thus:
F [a1 x1 (n) + L + an xn (n)] = a1F [ x1 (n)] + L + an F [ xn (n)]
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Properties of DSP systems
ƒ Time-invariance
ƒ Relationship between input and output does not
change with time
If
y[n ] = F [ x[n ]] then
y[n − k ] = F [ x[n − k ]] and
y[ n ] = y[ n − k ]
ƒ Causality
ƒ System is causal if the output of system only
exists for n ≥ 0
ƒ Most commonly encountered type of systems
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Causal Signal
ƒ Discrete-time signal
ƒ Made up of a set of weighted unit samples
(impulses) with appropriate delay.
ƒ Thus
Will discuss this in
detail when dealing
∞
with sampling
x ( n) =
x(k )δ (n − k )
∑
k =0
No signal before
n=0 (causal)
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Impulse Response
ƒ Linear time invariant (LTI) systems are uniquely
characterised by their impulse response
ƒ Recall this from earlier
on
So that each
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x[n] =
∞
∑ x[k ]δ [n − k ]
k = −∞
δ [n − k ] → h[n − k ]
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Impulse Response
ƒ Therefore y[ n] =
∞
∑ x(k )h[(n − k ]
k = −∞
•This is also expressed as:
n = 0, 1, 2, K
y[n] = x[n] ∗ h[n]
The convolution sum
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Impulse Response
ƒ If h[n] does not converge
ƒ Infinite Impulse Response system
ƒ IIR
ƒ If h[n] converges
ƒ Finite Impulse Response system
ƒ FIR
To be discussed
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The Convolution Equation
ƒ Let us assume a causal sequence
∞
y[n] = ∑ x(k )h[(n − k ] n = 0, 1, 2, K
k =0
∞
y[0] = ∑ x(k )h[(0 − k )]
k =0
What is h[-k]?
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That is simply ©h[k]
reversed (‘flipped’)
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Continuing
y[0] = x(0)h(0) + x(1)h( −1) + x(2)h(−2) + K +
∞
y[1] = ∑ x(k )h[(1 − k )]
k =0
h[-k] shifted right
once! That is a slide
y[1] = x(0)h(1 − 0) + x(1)h(1 − 1) + x(2)h(1 − 2) + K +
y[1] = x(0)h(1) + x(1)h(0) + x(2)h(−1) + K +
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Eventually
y[2] = x(0)h(2) + x(1)h(1) + x(2) h(0) + K +
ƒ Let us apply this to an example and observe
the pattern
ƒ x[n]=[1 3 4]
ƒ h[n]=[5 4 2]
ƒ Find y[n]=x[n]*h[n]
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Start
n
x[k]
h(-(k))
-2
-1
2
4
0
1
5
1
3
Reversed
sequence
2
4
3
4
5
6
7
Overlap
begins here
y[0] = x(0)h(0) + x(1)h( −1) + x(2)h(−2) + K +
y[0] = (1)(5) = 5
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Shift once (‘slide’)
n
x[k]
h(-(k))
h(1-k)
-2
-1
0
1
1
3
2
4
5
2
4
3
4
5
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7
Overlap zone
y[1] = x(0)h(1) + x(1)h(0) + x(2)h(−1) + K +
y[1] = (1)(4) + (3)(5) = 19
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Slide again
n
x[k]
h(-(k))
h(1-k)
h(2-k)
-2
-1
0
1
1
3
2
4
2
4
5
3
4
5
6
7
Overlap zone
y[2] = x(0)h(2) + x(1)h(1) + x(2) h(0) + K +
y[2] = (1)(2) + (3)(4) + (4)(5) = 34
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Finish off
n
x[k]
h(-(k))
h(1-k)
h(2-k)
h(3-k)
h(4-k)
h(5-k)
-2
-1
2
4
2
0
1
5
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2
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3
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2
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y(n)
y(0)=5
y(1)=19
y(2)=34
y(3)=22
y(4)=8
y(5)=0
5
ƒ x[n]=[1 3 4]
ƒ h[n]=[5 4 2]
ƒ Find
y[n]=x[n]*h[n]
ƒ Answer
ƒ [5 19 34 ©22
8]
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Overlap ends
here
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Another approach
∞
y[n] = ∑ x(k )h[(n − k )]
k =0
y[0] = x(0)h(0) + x(1)h( −1) + x(2)h(−2) + K +
y[1] = x(0)h(1) + x(1)h(0) + x(2)h(−1) + K +
y[2] = x(0)h(2) + x(1)h(1) + x(2) h(0) + K +
.
.
.
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If we are dealing with causal systems
then h(-1), h(-2) etc are zero
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Another approach
∞
y[n] = ∑ x(k )h[(n − k )]
k =0
y[0] = x(0)h(0)
y[1] = x(0)h(1) + x(1) h(0)
y[2] = x(0)h(2) + x(1)h(1) + x(2) h(0)
.
.
.
Let’s enter this into a table
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Using 4 terms for x[n] and 3 for h[n]
x[0]
h[0]
h[0]x[0]
h[1]
h[1]x[0]
x[1]
x[2]
x[3]
x[4]
h[0]x[4]
h[2]
h[3]
h[3]x[0]
h[3]x[4]
Called a Convolution Table
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Enter the values from our example
Sum across the anti-diagonals
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3
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20
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2
2
6
8
y[0] = x(0)h(0) = 5
y[1] = x(0)h(1) + x(1)h(0) = 19
y[2] = x(0)h( 2) + x(1)h(1) + x(2)h(0) = 34 y[3] = 22
y[4] = 6
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Sequence Xn
Sequence Hn
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0
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2.5
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0
1
1.5
2
2.5
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Sequence Yn
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ƒ Graphical
representation of
results
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0
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Example
Find the output of a finite
response system h[nT ] = nT to
the input signal x[nT ] = e − nT ,
where T = 0.1s and n = 0 to 3
ƒ Step 1
ƒ Find the sequences
ƒ h[nT]=?
ƒ x[nT]=?
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Using MATLAB conv function
%
%
%
Use of the conv function
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n=0:3;
Xn=exp(-0.1*n);
Hn=0.1*n;
Yn=conv(Xn,Hn);
Subplot(2,2,1),stem(Xn,'fill');Title('Sequence Xn');
Subplot(2,2,2),stem(Hn,'fill');Title('Sequence Hn');
Subplot(2,2,3),stem(Yn,'fill');Title('Sequence Yn');
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Using MATLAB conv function
Sequence Xn
Sequence Hn
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0.6
0.2
0.4
0.1
0.2
0
1
2
3
4
0
1
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Sequence Yn
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The results agree with
those obtained before
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