Discrete-Time Impulse Signal
Let δ[k] be a discrete-time impulse function,
a.k.a. the Kronecker delta function:
δ [k ] =
1 k =0
0 k≠0
Impulse response h[k]: response of a
discrete-time LTI system to a discrete
impulse function
Finite impulse response filter
Non-zero extent of impulse response is finite
Can be in continuous time or discrete time
Also called a tapped delay line
1
Discrete-time Convolution
By linear and time-invariant properties, linear
convolution
For each value of k, compute a different
(possibly) infinite summation for y[k]
y[k ] = x[k ]∗ h[k ] =
∞
x[m] h[k − m] =
m = −∞
∞
h[m] x[k − m]
m = −∞
h[k]
Averaging filter
impulse response
1
2
y[k] = h[0] x[k] + h[1] x[k-1]
k
= ( x[k] + x[k-1] ) / 2
2
1
Comparison to Continuous Time
Continuous-time convolution of x(t) and h(t)
y(t ) = x(t ) ∗ h(t ) =
∞
−∞
x(λ ) h(t − λ ) dλ =
∞
−∞
h(λ ) x(t − λ ) dλ
For each value of t, we compute a different
(possibly) infinite integral.
Discrete-time definition is the continuous-time
definition with the integral replaced by summation
LTI system
If we know the impulse response and the input,
we can determine the output
Impulse response uniquely characterizes an LTI
system
3
Convolution Demos
Johns Hopkins University Demonstrations
http://www.jhu.edu/~signals
Convolution applet to animate convolution of
simple signals and hand-sketched signals
Convolving two rectangular pulses of same width
gives a triangle whose width is twice the width of
the rectangular pulses
4
2
Linear Time-Invariant Systems
The Fundamental Theorem of Linear Systems
If one inputs a complex sinusoid into an LTI system, then
the output will be a complex sinusoid of the same
frequency that has been scaled by the frequency response
of the LTI system at that frequency
Scaling may attenuate the signal and shift it in phase
Example in discrete time. Let x[k] = e j Ω k,
y[ k ] =
∞
e
j Ω (k −m )
h[m] = e
jΩk
m = −∞
∞
h[m] e − j Ω m = e j Ω k H (Ω )
m = −∞
H (Ω )
H(Ω) is the discrete-time Fourier transform of h[k] and is
also called the frequency response
5
Frequency Response
For continuous-time systems, response to
complex sinusoid is
frequency response
e j Ω t → H ( jΩ ) e j Ω t
cos(Ω t ) → H ( jΩ ) cos(Ω t + ∠H ( jΩ ))
For discrete-time systems, response to
complex sinusoid is
( )
H (e ) cos(ω k + ∠H (e ))
e jω k → H e jω e jω k
frequency response
cos(ω k ) →
jω
jω
6
3
Example: Ideal Delay
Continuous Time
Discrete Time
Delay by T seconds
Delay by 1 sample
x(t)
y(t)
T
x[k]
y (t ) = x(t − T )
z −1
y[k]
y[ k ] = x[k − 1]
Impulse response
Impulse response
h(t ) = δ (t − T )
h[ k ] = δ [ k − 1]
Frequency response
Frequency response
H (Ω ) = e
| H (Ω ) | = 1
∠H (Ω ) = − Ω T
H (ω ) = e − j ω
| H (ω ) | = 1
∠H (ω ) = − ω
− jΩT
7
Example: Lowpass Filter
System response to complex exponential e j Ωt
for all possible frequencies Ω where Ω=2π
πf :
∠|H(Ω)|
|H(Ω)|
stopband
Linear
phase
stopband
−Ωs −Ωp
Ωp Ωs
passband
Ω
Ω
delay (Ω ) = −
d
∠H (Ω) = T
dΩ
Passes low frequencies, a.k.a. lowpass filter
FIR filters are only realizable LTI filters that can
have linear phase over all frequencies
8
4
Mandrill Demo (DSP First)
Five-tap averaging FIR filter with input x[k] and
output y[k]
y[k ] =
1
1
1
1
1
x[k ] + x[ k − 1] + x[ k − 2] + x[ k − 3] + x[ k − 4]
5
5
5
5
5
Lowpass filter (smooth/blur input signal)
Impulse response is {1/5, 1/5, 1/5, 1/5, 1/5}
First-order difference FIR filter
Highpass filter (sharpens input h[k]
signal)
1
Impulse response is {1, -1}
y[k ] = x[k ] − x[ k − 1]
First-order difference
impulse response
k
−1
9
Mandrill Demo (DSP First)
From lowpass filter to highpass filter
original → blurry → sharpened
From highpass to lowpass filter
original → sharpened → blurry
Frequencies that are zeroed out (e.g. DC)
can never be recovered
Order of two LTI systems in cascade can be
switched under the assumption that the
computations are performed in exact
precision
10
5
Finite Impulse Response Filters
Duration of impulse response h[k] is finite,
i.e. non-zero for k =∞ 0, 1, … N-1: N −1
y[k ] = x[k ]∗ h[k ] =
h[m ] x[k − m] = h[m] x[k − m ]
m = −∞
m =0
Output depends on current input and previous N-1 inputs
Summation to compute y[k] reduces to a vector dot product
between N input samples in the vector
{ x[k ], x[k − 1], ..., x[k − ( N − 1)] }
and N nonzero values of the impulse response in vector
{ h[0], h[1], ..., h[ N − 1] }
What instruction set/architecture features would you
add to accelerate FIR filtering?
11
Discrete-time Tapped Delay Line
Assuming that h[k] has finite duration from
k =0,…,N-1
y[k ] =
N −1
h[ m] x[ k − m]
m= 0
Block diagram of an implementation (finite
impulse response digital filter)
x[k]
h[0]
z-1
z-1
…
h[1]
h[2]
… h[N-1]
Σ
z-1
y[k]
12
6
FIR Implementation: Circular Buffer
Shifting the elements in the entire array is
inefficient
Time index
n-1
n
0
1
1
2
3
2
3
address index
4
Better approach is to use circular buffers and
updating address index
Time index
0 3
1
2
n-1
n
address index
4
0
1
4
3
1
2
3
1
2
2
3
13
DSP Hardware for Circular Buffer
DSPs have hardware support to implement circular
buffers even more efficiently
Let N be number of coefficients and then, N is
circular buffer block size
BK is a block size register
Beginning address of the circular buffer array is a
multiple of the smallest power of 2 strictly greater
than the filter length N.
Data memory
31
Beginning address
Value of register, BK
AR
K-1
XXXX
0…………0
XXXX
L………….L
XXXX
LSBs of BK
0
0
LAST+1
14
7