3. Discrete-time Processing of Continuous-time Signals
A major application of discrete-time systems is in the processing of
continuous-time signals.
The overall system is equivalent to a continuous-time system, since it
transforms the continuous-time input signal 𝑥 𝑠 𝑡 into the continuous
time signal 𝑦𝑟 𝑡 .
3
4. Continuous to discrete (Ideal C/D) converter
The impulse train (sampling Signal) is
Where
The sampled signal in Time-domain (Time domain multiplication) is
4
n
cs nTttxtstxtx )()()(
n
nTtts
n
snj
T
jS
2
continuous
F. T.
Ts /2
5. Continuous to discrete (Ideal C/D) converter (Cont.)
The sampled signal in Frequency-domain is
Hence, the continuous Fourier transforms of 𝑥 𝑠(𝑡) consists of
periodically repeated copies of the Fourier transform of 𝑥 𝑐(𝑡).
Review of Nyquist sampling theorem:
Aliasing effect: If 𝛺 𝑠 < 2𝛺ℎ, the copies of 𝑋𝑐(𝑗𝛺) overlap, where 𝛺ℎ is the
highest nonzero frequency component of 𝑋𝑐(𝑗𝛺). 𝛺ℎ is referred to as the
Nyquist frequency.
5
)()(
2
1
)( jSjXjX cs
n
scs njX
T
jX
1
7. Continuous to discrete (Ideal C/D) converter (Cont.)
In the above, we characterize the relationship of 𝑥 𝑠(𝑡) and 𝑥 𝑐(𝑡) in the
continuous F.T. domain.
From another point of view, 𝑋𝑠(𝑗𝛺) can be represented as the linear
combination of a serious of complex exponentials:
If 𝑥(𝑛𝑇) ≡ 𝑥[𝑛], its DTFT is
The relation between digital frequency and analog frequency is 𝝎 = 𝛀𝐓 7
n
cs nTtnTxtx since
n
Tnj
cs enTxjX
n
nj
n
njj
enTxenxeX
)(][)(
8. Continuous to discrete (Ideal C/D) converter (Cont.)
Combining these properties, we have the relationship between the
continuous F.T. and DTFT of the sampled signal:
where
: represent continuous F.T.
: represent DTFT
Thus, we have the input-output relationship of C/D converter
8
n
s
c
n
sc
jw
T
n
T
jX
T
njX
T
eX
11
jw
eX
jXc
Tj
T
j
s eXeXjX
10. Ideal reconstruction filter
The ideal reconstruction filter
is a continuous-time filter,
with the frequency response
being H 𝑟(𝑗𝛺) and impulse
response ℎ 𝑟(𝑡).
10
)/(
)/sin(
)(
Tt
Tt
thr
11. Combine C/D, discrete-time system, and D/C
Consider again the discrete-time processing of continuous signals
Let 𝐻(𝑒 𝑗𝜔) be the frequency response of the discrete-time system in
the above diagram. Since Y 𝑒 𝑗𝜔 = 𝐻 𝑒 𝑗𝜔 𝑋(𝑒 𝑗𝜔)
11
12. D/C converter revisited
An ideal low-pass filter 𝐻(𝑒 𝑗𝜔) that has a cut-off frequency Ω 𝑐 = Ω 𝑠/2
= 𝜋/𝑇 and gain 𝐴 is used for reconstructing the continuous signal.
Frequency domain of D/C converter: [𝐻(𝑒 𝑗𝜔) is its frequency response]
Remember that the corresponding impulse response is a 𝑠𝑖𝑛𝑐 function, and
the reconstructed signal is
12
n
r
TnTt
TnTt
nyty
/
/sin
otherwise
TeYA
eYjHjY
Tj
Tj
rr
,0
/,
14. Practical Reconstruction
We cannot build an ideal Low-Pass-Filter.
Practical systems could use a zero-order hold block
This distorts signal spectrum, and compensation is needed
14
15. Practical Reconstruction (Zero-Order Hold)
Rectangular pulse used to analyze zero-order hold reconstruction
15
s
s
Ttt
Tt
th
,0,0
0,1
)(0
n
s
n
s
nTthnx
nTtnxth
thtxtx
)(][
)(][)(
)()()(
0
0
00
)2/sin(
2)(
)()()(
2/
0
00
sTj T
ejH
jXjHjX
s
16. Practical Reconstruction
(Zero-Order Hold)
Effect of the zero-order hold in the
frequency domain.
a) Spectrum of original continuous-time
signal.
b) spectrum of sampled signal.
c) Magnitude of Ho(j).
d) Phase of Ho(j).
e) Magnitude spectrum of signal
reconstructed using zero-order hold.
16
17. Practical Reconstruction (Anti-imaging Filter)
Frequency response of a compensation filter used to eliminate some of
the distortion introduced by the zero-order hold.
17
ms
m
s
s
c T
T
jH
||,0
||,
)2/sin(2)(
Anti-imaging filter.
19. Analog Processing using DSP
Block diagram for discrete-time processing of continuous-time signals.
19
(b) Equivalent continuous-time system.
(a) A basic system.
20. Idea: find the CT system
0th-order S/H:
20
)()()(such that)()( jXjGjYjGtg FT
)2/sin(
2)( 2/
0
sTj T
ejH s
)()()( jXjHjX aa
n
ssa
Tj
c
s
n
ssa
Tj
s
ss
n
ssa
s
n
sa
s
kjXkjHeHjHjH
T
jY
kjXkjHeH
T
jY
TkjXkjH
T
kjX
T
jX
s
s
))(())(()()(
1
)(
))(())((
1
)(
/2,))(())((
1
))((
1
)(
0
21. Idea: find the CT system (Cont.)
If no aliasing, the anti-imaging filter Hc(jw) eliminates frequency
components above s/2, leaving only k=0 terms
If anti-aliasing and anti-imaging filters are chosen to compensate the
effects of sampling and reconstruction, then
21
)()()(
1
)(
)()()()(
1
)(
0
0
jHeHjHjH
T
jG
jXjHeHjHjH
T
jY
a
Tj
c
s
a
Tj
c
s
s
s
sTj
acs
eHjG
jHjHjHT
)(Thus,
1)()()()/1( 0
23. Digital Signal Processing
Provided that :
𝐻 𝑎 𝑗Ω =
1 Ω ≤
Ω 𝑠
2
0 Ω >
Ω 𝑠
2
𝐺 𝑒𝑓𝑓 𝑗Ω =
𝐻 𝑎 𝑗Ω 𝐺 𝑒 𝑗Ω𝑇
𝐻 𝑜 𝑗Ω 𝐻𝑟 𝑗Ω Ω ≤
Ω 𝑠
2
0 Ω >
Ω 𝑠
2 23
Digital Signal
Processor
C/D
Sample
nT
𝑥(𝑡)
𝑋(𝑗Ω)
Antialiasing
Analog
Filter
𝑯 𝒂(𝒋𝛀)
𝑥 𝑎(𝑡)
𝑋 𝑎(𝑗Ω)
𝑥[𝑛]
𝑋(𝑒 𝑗𝜔
)
Reconstruction
Analog
Filter
𝑯 𝒓(𝒋𝛀)
𝑮(𝒆𝒋𝝎)
𝑦[𝑛]
𝑌(𝑒 𝑗𝜔
)
𝑦𝑎(𝑡)
𝑌𝑎(𝑗Ω)𝑯 𝒐(𝒋𝛀)
D/C
Zero order
hold
𝑦(𝑡)
𝑌(𝑗Ω)
𝑮 𝒆𝒇𝒇(𝒋𝛀)
𝑦(𝑡)
𝑌(𝑗Ω)𝑋(𝑗Ω)
𝑥(𝑡)
24. Introduction
In single-rate DSP systems, all data is sampled at the same rate no
change of rate within the system.
In Multirate DSP systems, sample rates are changed (or are different)
within the system
Multirate can offer several advantages
reduced computational complexity
reduced transmission data rate.
24
25. Example: Audio sample rate conversion
Recording studios use 192 kHz
CD uses 44.1 kHz
wideband speech coding using 16 kHz
Master from studio must be rate-converted by a factor 44.1/192
25
26. Example: Oversampling ADC
Consider a Nyquist rate ADC in which the signal is sampled at the
desired precision and at a rate such that Nyquist’s sampling criterion is
just satisfied.
Bandwidth for audio is 20 Hz < f < 20 kHz
The required Antialiasing filter has very demanding specification
Requires high order analogue filter such as elliptic filters that have very
nonlinear phase characteristics
• hard to design, expensive and bad for audio quality.
26
27. Nyquist Rate Conversion Anti-aliasing Filter.
Building a narrow band filters 𝐻 𝑎(𝑗Ω) and 𝐻 𝑜(𝑗Ω) is difficult &
expensive.
27
28. Consider oversampling the signal at, say, 64 times the Nyquist rate but
with lower precision. Then use multirate techniques to convert sample
rate back to 44.1 kHz with full precision.
New (over-sampled) sampling rate is 44.1 × 64 kHz.
Requires simple antialiasing filter
Could be implemented by simple filter (eg. RC network)
Recover desired sampling rate by downsampling process.
28
30. Multirate Digital Signal Processing
The solution is changing sampling rate using downsampling and
upsampling.
Low cos:
Analog filter (eg. RC circuit)
Storage / Computation
Price:
Extra Computation for ↓ 𝑴 𝟏 and ↑ 𝑴 𝟐
Faster ADC and DAC
30
DSPC/D
Sample
nT
𝑥(𝑡)
𝑋(𝑗Ω)
Antialiasing
Analog
Filter
𝑯 𝒂(𝒋𝛀)
𝑥 𝑎(𝑡) 𝑥[𝑛]
Reconstruction
Analog
Filter
𝑯 𝒓(𝒋𝛀)
𝑮(𝒆𝒋𝝎
)
𝑦[𝑛] 𝑦𝑎(𝑡)
𝑯 𝒐(𝒋𝛀)
D/C
Zero order
hold 𝑦(𝑡)
𝑌(𝑗Ω)
↓ 𝑴 𝟏
Downsampler
𝑥1[𝑛]
↑ 𝑴 𝟐
Upsampler
𝑦2[𝑛]
31. Example (1)
Given a 2nd order analog filter
Evaluate 𝐻 𝑎 𝑗Ω for
Anti-aliasing filter
Zero-Order-Hold
Anti-imaging filter
At sampling frequency:
Sampling frequency = 300Hz.
Sampling frequency = 2400Hz. (8x Oversampling)
31
𝐻 𝑎 𝑗Ω =
(200𝜋)2
(𝑗Ω + 200𝜋)2
, 𝑠𝑖𝑔𝑛𝑎𝑙 𝐵𝑊 − 50 < 𝑓 < 50
32. Example (1): Ideal Anti-aliasing Filter
The required specification is
𝐻 𝑎 𝑗Ω =
1 Ω ≤ 100𝜋
0 Ω > 100𝜋
In case of Ω 𝑠 = 600𝜋
In case of Ω 𝑠 = 4800𝜋
Passband constraint same for both cases
𝐻 𝑎 𝑗100𝜋 =
(200𝜋)2
200𝜋(𝑗
1
2
+1)
2 =
4
5
= 0.8
Stopband actual
𝐻 𝑎 𝑗500𝜋 =
(200𝜋)2
200𝜋(𝑗
5
2
+1)
2 =
4
29
≅ 0.14
𝐻 𝑎 𝑗4700𝜋 =
(200𝜋)2
200𝜋(𝑗
47
2
+1)
2 =
4
472+4
≅ 0.002
32
76𝑋 𝑎𝑡𝑡𝑒𝑛𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 1
33. Example (1): Ideal Anti-imaging Filter
The required specifications are shown in the figure
In case of Ω 𝑠 = 600𝜋,
Desired at Ω = 100𝜋
• 𝐻𝑟 𝑗100𝜋 =
𝜋/6
𝑠𝑖𝑛 𝜋/6
≅ 1.06
Actual filter at
• Passband: 𝐻 𝑎 𝑗100𝜋 =
(200𝜋)2
200𝜋(𝑗
1
2
+1)
2 =
4
5
= 0.8
• Stopband: 𝐻 𝑎 𝑗500𝜋 =
(200𝜋)2
200𝜋(𝑗
5
2
+1)
2 =
4
29
≅ 0.14
In case of Ω 𝑠 = 4800𝜋
Desired at Ω = 100𝜋
• 𝐻𝑟 𝑗100𝜋 =
𝜋/48
𝑠𝑖𝑛 𝜋/48
≅ 1.007
Actual filter at
• Passband: 𝐻 𝑎 𝑗100𝜋 =
(200𝜋)2
200𝜋(𝑗
1
2
+1)
2 =
4
5
= 0.8
• Stopband: 𝐻 𝑎 𝑗500𝜋 =
(200𝜋)2
200𝜋(𝑗
5
2
+1)
2 =
4
29
≅ 0.002
33
)/sin(
/
)2/sin(2 s
s
s
s
r
T
T
jH
34. Downsampling (Decimation ):
To relax design of anti-aliasing filter and anti-imaging filters, we wish to
use high sampling rates.
High-sampling rates lead to expensive digital processor
Wish to have:
• High rate for sampling/reconstruction
• Low rate for discrete-time processing
This can be achieved using downsampling/upsampling
34
35. Downsampling
Let 𝑥1 𝑛 and 𝑥2 𝑛 be obtained by sample 𝑥(𝑡) with sampling interval 𝑇 and MT,
respectively.
35
𝑥 𝑎(𝑡)
𝑋 𝑎(𝑗Ω) 𝑋1(𝑒 𝑗𝜔)
𝑥1[n]A/D
Sample @T
Ω 𝑠
𝑥 𝑎(𝑡)
𝑋 𝑎(𝑗Ω) 𝑋2(𝑒 𝑗𝜔
)
𝑥2[n]A/D
Sample @MT
Ω 𝑠/𝑀
T
MT
𝟐𝝅 𝟒𝝅𝟐𝝅𝟒𝝅 T𝛀 𝒐T𝛀 𝒐
𝟐𝝅 𝟒𝝅𝟐𝝅𝟒𝝅
36. Downsampling Block
𝑥[n]
𝑋(𝑒 𝑗𝜔
)
Discrete Time
Low-Pass Filter
𝐻 𝑑(𝑒 𝑗𝜔
)
𝑤[n]
𝑊(𝑒 𝑗𝜔
) 𝑌(𝑒 𝑗𝜔
)
𝑦 n = 𝑤[Mn]Discard Values
𝑛 ≠ 𝑀𝑙
𝝅
𝑴
𝝅
𝝅
𝑴
(d) Spectrum after Downsampling.
(c) Spectrum of filter output.
(b) Filter frequency response.
(a) Spectrum of oversampled input signal.
Noise is depicted as the shaded portions
of the spectrum.
𝑌(𝑒 𝑗𝜔)
𝑦 nDownsampler
↓ 𝑀
𝑥[n]
𝑋(𝑒 𝑗𝜔)
37. Downsampling Block
Note: if 𝑦 n = 𝑤[Mn], then
𝑌 𝑒 𝑗𝜔 =
1
𝑀
𝑚=0
𝑀−1
𝑋 𝑒 𝑗(𝜔−2𝜋𝑚)/𝑀
What does 𝑌 𝑒 𝑗𝜔
=
1
𝑀 𝑚=0
𝑀−1
𝑋 𝑒 𝑗(𝜔−2𝜋𝑚)/𝑀
represent?
Stretching of 𝑋(𝑒 𝑗𝜔) to 𝑋(𝑒 𝑗𝜔/𝑀)
Creating M − 1 copies of the stretched versions
Shifting each copy by successive multiples of 2𝜋 and superimposing (adding)
all the shifted copies
Dividing the result by M
37
Discrete Time
Low-Pass Filter
𝐻 𝑑(𝑒 𝑗𝜔
)
𝑤[n]
𝑊(𝑒 𝑗𝜔) 𝑌(𝑒 𝑗𝜔
)
𝑦 n = 𝑤[Mn]Discard Values
𝑛 ≠ 𝑀𝑙
38. Upsampling
Let 𝑥1 𝑛 and 𝑥2 𝑛 be obtained by sample 𝑥(𝑡) with sampling interval 𝑇 and T/M,
respectively.
38
𝑥 𝑎(𝑡)
𝑋 𝑎(𝑗Ω) 𝑋1(𝑒 𝑗𝜔)
𝑥1[n]A/D
Sample @T
Ω 𝑠
𝑥 𝑎(𝑡)
𝑋 𝑎(𝑗Ω) 𝑋2(𝑒 𝑗𝜔
)
𝑥2[n]A/D
Sample @T/M
𝑀 Ω 𝑠
M
T
T
𝟐𝝅 𝟒𝝅𝟐𝝅𝟒𝝅
𝟐𝝅 𝟒𝝅𝟐𝝅𝟒𝝅
39. Upsampling (zero padding) Block Diagram
39
Insert (M-1) zeros
between each two
successive samples
𝑥2[n]
𝑋2(𝑒 𝑗𝜔
)
Discrete Time
Low-Pass Filter
𝐻𝑖(𝑒 𝑗𝜔
)
𝑋𝑖(𝑒 𝑗𝜔
)
𝑥𝑖 nUpsampler
↑ 𝑀
𝑥[n]
𝑋(𝑒 𝑗𝜔
)
41. Upsampling (zero padding) Block Diagram
41
𝑥[n]
𝑋(𝑒 𝑗𝜔
)
Insert (M-1) zeros
between each two
successive samples
𝑥2[n]
𝑋2(𝑒 𝑗𝜔
) 𝑋𝑖(𝑒 𝑗𝜔
)
𝑥𝑖 nDiscrete Time
Low-Pass Filter
𝐻𝑖(𝑒 𝑗𝜔
) 𝑋𝑖(𝑒 𝑗𝜔)
𝑥𝑖 nUpsampler
↑ 𝑀
𝑥[n]
𝑋(𝑒 𝑗𝜔
)
(a) Spectrum of original sequence.
(b) Spectrum after inserting (M – 1) zeros in
between every value of the original sequence.
(c) Frequency response of a filter for removing
undesired replicates located at 2/M, 4/M, …,
(M – 1)2/M.
(d) Spectrum of interpolated sequence.
𝝅
42. Continuous Signal Processing Using DSP
Block diagram of a system for discrete-time processing of continuous-
time signals including decimation and interpolation.
42
43. Assignment
For the shown Digital Signal Processing system, find the DTFT and FT
representations for 𝑥 𝑛 , 𝑥 𝑑 𝑛 , 𝑦 𝑛 , 𝑎𝑛𝑑 𝑦𝑢 𝑛 .
If the input is a real signal with its Fourier transform shown in Fig.2 (a). Where
𝑯 𝒑𝒇(𝛀) is a Low Pass Filter and 𝑯 𝑩𝑷𝑭(𝒆𝒋𝝎) is an ideal digital Band Pass Filter
with passband from 𝜔1
𝜋
4
to 𝜔2
3𝜋
4
, and their frequency response is given as
sown in Fig.2 (b).
43
Fig. (b) Fig. (a)