[PPT] - H ( z ) x ( t ) CT DT DT CT y ( t ) X ( j ) X (e j ) X i ( j ) T PowerPoint Presentation

SLIDE 1

Overview of DT Processing of CT Signals H(z)

x[n] y[n] x(t) y(t) Ts Ts CT ⇒ DT DT ⇒ CT

Many systems (1) sample a signal, (2) process it in discrete-time,

and (3) convert it back to a continuous-time signal

Called discrete-time processing of continuous-time signals
Most modern digital signal processing (DSP) uses this architecture
The first step is called sampling
The last step is called interpolation
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Overview of Sampling Topics

(Shannon) sampling theorem
Impulse-train sampling
Interpolation (continuous-time signal reconstruction)
Aliasing
Relationship of CTFT to DTFT
DT processing of CT signals
DT sampling
Decimation & interpolation
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Signals x(t) → x[n] → xi(t) X(jω) → X(ejω) → Xi(jω)

For now, just consider the two conversions

– Sampling (CT → DT) – Interpolation (DT → CT)

Suppose we want xi(t) to be as close to x(t) as possible
Need to consider relationships in both time- and

frequency-domains

Three signals, three transforms
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Amplitude versus Time

In this class we are working only with continuous-valued signals
Signals with discrete values that have been quantized are called

digital signals

Analog-to-Digital converters (ADC) convert continuous-valued

signals to discrete-valued signals. These often also convert from CT to DT.

Digital-to-analog converters (DAC) convert discrete-valued

signals to continuous-valued signals. These often also convert from DT to CT.

Signals that are both continuous-valued and continuous-time are

usually called analog signals

Signals that are both discrete-valued and discrete-time are usually

called digital

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SLIDE 2

Use of the Bridge Signal

Define the bridge signal by relating Xδ(jω) to X(ejΩ)
Determine how x(t) is related to xδ(t) in the time domain
Determine how X(jω) is related to Xδ(jω)
Use the relationships of X(ejΩ) and X(jω) to Xδ(jω) to

determine their relationship to one another

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Guiding Questions and Objectives x[n] = x(t)|t=nTs

The operation of sampling (CT → DT conversion) is trivial
The challenge is to understand the limits and tradeoffs
The remainder of these slides is dedicated to answering the

following three questions

1. How is the CTFT of x(t), X(jω) related to the DTFT of

x[n] = x(nTs), X(ejΩ)?

2. Under what conditions can we synthesize x(t) from x[n]:

x[n] → x(t)?

3. How do we perform this DT → CT conversion?
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Defining the Bridge Signal

The bridge signal xδ(t) is a CT representation of a DT signal x[n]
It is defined as having the same transform, within a scale factor,

as the DT signal

Suppose x[n] = Aej(Ωn+φ)
What is the CT equivalent?
Let us pick x(t) = Aej(ωt+φ)
How do we relate the DT frequency Ω to the CT frequency ω?

– ω has units of radians/second – Ω has units of radians/sample

Let us use the conversion factor of Ts = f −1

s

seconds/sample Ω (radians/sample) = ω (radians/second) × Ts (seconds/sample) ω (radians/second) = Ω (radians/sample) × fs (samples/second)

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Need for a Bridge Signal Consider three signals Type Time Domain Frequency Domain Original Signal CT x(t) X(jω) Sampled Signal DT x[n] X(ejΩ) Bridge Signal CT xδ(t) Xδ(jω)

Goals: to determine the relationship of x(t) to x[n] = x(nTs) in

the time and frequency domains

In the time domain, the relationship x[n] = x(nTs) is clear
But what is the relationship of X(jω) to X(ejΩ)?
The transforms differ in character

– X(ejΩ) is periodic – X(ejΩ) has units of radians per sample

A bridge signal xδ(t) is the only way (that I know of) to determine

how X(jω) is related to X(ejΩ)

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SLIDE 3

Why Isn’t the Inverse Transform Similar to x(t)? DTFT x[n] = 1 2π

2π

X(ejω) ejΩn dΩ CTFT x(t) = 1 2π +∞

−∞

X(jω) ejωt dω

We generated xδ(t) from x[n] by equating a CTFT to the DTFT
f x[n]
This seems reasonable at first
But consider the range of the CTFT and DTFT synthesis

equations

The DTFT synthesizes x[n] out of a finite range of frequencies
The CTFT synthesizes x(t) out of all frequencies
This is why, if x[n] = x(nTs), that

xδ(t) = F−1 {Xδ(jω)} = F−1 X(ejωTs)

= x(t)
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Defining the Bridge Signal Xδ(jω) = X(ejΩ)

Ω=ωTs = X(ejωTs)
The bridge signal is a CT representation of a DT signal
We define it by equating the CTFT and DTFT with an

appropriate scaling factor for frequency

Note that this is a highly unusual CT signal

– The CTFT is periodic – What does this tell us about xδ(t)?

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Impulse Sampling

Ts 2Ts 3Ts 4Ts 5Ts

Ts
2Ts
3Ts
4Ts

p(t)

1

p(t) =

∞

n=−∞

δ(t − nTs) xδ(t) =

∞

n=−∞

x(nTs) δ(t − Tsn) = x(t) p(t)

xδ(t) can also be formed from a CT signal
This is called impulse sampling
We can model sampling by use of the periodic impulse train
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Solving for the Bridge Signal X(ejΩ) =

∞

n=−∞

x[n] e−jΩn Xδ(jω) = X(ejΩ)

Ω=ωTs =

∞

n=−∞

x[n] e−jωTsn xδ(t)

FT

⇐ ⇒ Xδ(jω) δ(t − to)

FT

⇐ ⇒ e−jωto δ(t − Tsn)

FT

⇐ ⇒ e−jωTsn

∞

n=−∞

x[n] δ(t − Tsn)

FT

⇐ ⇒

∞

n=−∞

x[n] e−jωTsn xδ(t) =

∞

n=−∞

x[n] δ(t − Tsn)

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SLIDE 4

Impulse Sampling Terminology H(z)

x[n] y[n] x(t) y(t) Ts Ts CT ⇒ DT DT ⇒ CT

xδ(t) = x(t) p(t) =

∞

n=−∞

x(nTs)δ(t − nTs)

The impulse train p(t) is called the sampling function
p(t) is periodic with fundamental period Ts
Ts, the fundamental period of p(t), is called the sampling period
fs ≡

1 Ts and ωs = 2π Ts are called the sampling frequency

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Rectangular Window and Impulse Train Notation

Ts 2Ts 3Ts 4Ts 5Ts

Ts
2Ts
3Ts
4Ts

p(t)

1

Note that a similar symbol was used for rectangular windows

pT (t) =

1

|t| < T Otherwise but p(t) = pT (t)

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Fourier Transforms of Periodic Signals Overview xδ(t) =

∞

n=−∞

x(nTs) δ(t − Tsn) = x(t) p(t) Xδ(jω) = 1 2π X(jω) ∗ P(jω)

To determine how Xδ(jω) is related to X(jω), we need to

calculate P(jω)

p(t) is a periodic function with infinite energy
The CTFT clearly doesn’t converge
Is easier to

– Calculate the Fourier series coefficients P[k] for p(t) – Solve for P(jω) from P[k] using the general relationship between the CTFS and the CTFT

Since periodic signals have infinite energy, the CTFT of these

signals consists of impulses

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Impulse Sampling Conceptual Example

Ts 2Ts 3Ts 4Ts 5Ts

Ts
2Ts
3Ts
4Ts

1 Ts 2Ts 3Ts 4Ts 5Ts

Ts
2Ts
3Ts
4Ts

1 Ts 2Ts 3Ts 4Ts 5Ts

Ts
2Ts
3Ts
4Ts

1

x(t) p(t) x(t)p(t)

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SLIDE 5

Example 1: Workspace

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Fourier Transforms of Periodic Signals Recall the Fourier series representations of periodic signals x(t) =

∞

k=−∞

X[k] ejkωt ejωot

FT

⇐ ⇒ 2π δ(ω − ωo) x(t) =

∞

k=−∞

X[k] ejkωot

FT

⇐ ⇒ 2π

∞

k=−∞

X[k] δ(ω − kωo)

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The Relationship of X(jω) to Xδ(jω) x(t) p(t)

FT

⇐ ⇒ 1 2π X(jω) ∗ P(jω) P(jω)

FT

⇐ ⇒ 2π Ts

∞

k=−∞

δ

ω − k 2π

Ts

ωs
2π

Ts x(t) p(t)

FT

⇐ ⇒ 1 2π X(jω) ∗ 2π Ts

∞

k=−∞

δ(ω − kωs) X(jω) ∗ δ(ω − ωo) = X (j(ω − ωo)) x(t) p(t)

FT

⇐ ⇒ 1 Ts

∞

k=−∞

X (j(ω − kωs))

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Example 1: CT Fourier Transform of an Impulse Train Solve for the Fourier transform of the impulse train p(t) =

∞

n=−∞

δ(t − nTs) Hint: the impulse train is periodic. X[k] = 1 T

T

x(t)e−jkωot dt X(jω) = +∞

−∞

x(t) e−jωt dt

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SLIDE 6

Observations x(t) p(t)

FT

⇐ ⇒ 1 Ts

∞

k=−∞

X (j(ω − kωs)) X(ejωTs) = 1 Ts

∞

k=−∞

X (j(ω − kωs))

What is Ωs?

Ωs = ωsTs = 2π Ts Ts = 2π

Thus ωs in CT corresponds to 2π radians/sample in DT
Recall that the fastest DT signal is ejπn = (−1)n and oscillates at

π radians/sample

This is the fastest signal we can observe due to ejΩn = ej(Ω±ℓ2π)n
What is this frequency in the CT domain?
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Summary Sampling: x[n] = x(nTs) Definition: Xδ(jω) = X(ejω)

Ω=ωTs

Inverse CTFT: xδ(t) =

∞

n=−∞

x[n]δ(t − nTs) Impulse Sampling: xδ(t) =

∞

n=−∞

x(nTs)δ(t − nTs) CTFT: P(jω) = 2π Ts

∞

k=−∞

δ (t − kωs) Multiplication Property: Xδ(jω) = 1 2π X(jω) ∗ P(jω) = 1 Ts

∞

k=−∞

X (j(ω − kωs))

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Expressing DT Transform in Terms of CT Transform X(ejωTs) = 1 Ts

∞

k=−∞

X (j(ω − kωs)) Ω = ωTs ω = Ω Ts X(ejΩ) = 1 Ts

∞

k=−∞

X

j( Ω

Ts − kωs)

ωs = Ωs

Ts = 2π Ts X(ejΩ) = 1 Ts

∞

k=−∞

X

j

Ω − k2π Ts

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Conceptual Diagram of the Relationship

1

2π Ts 1 Ts 1 Ts

ωs-ωx
ωs+ωx

ωs-ωx ωs+ωx ωx ωx ωx

ωx
ωx
ωx

ωs ωs ωs

ωs
ωs
ωs
Ωs-Ωx
Ωs+Ωx

Ωs-Ωx Ωs+Ωx Ωx

Ωx

Ωs

Ωs

X(jω) P (jω) Xδ(jω) X(ejω)

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SLIDE 7

Signal Relationships Summary x[n] = x(nTs) X(ejωTs) = 1 Ts

∞

k=−∞

X (j(ω − kωs)) xδ(t) =

∞

n=−∞

x(nTs) δ(t − Tsn) Xδ(jω) = 1 Ts

∞

k=−∞

X (j(ω − kωs)) xδ(t) =

∞

n=−∞

x[n] δ(t − Tsn) Xδ(jω) = X

ejωTs
We now know the relationships between all of the three signals
Note that xδ(t) was just a means to determining the relationship
f X(ejω) to X(jω)
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What Does this Mean? x(t) p(t)

FT

⇐ ⇒ 1 Ts

∞

k=−∞

X (j(ω − kωs))

Sampling with an impulse train results in a signal with a CTFT of

Xδ(jω) that is a periodic function of ω

There are replicas of X(jω) at each multiple of ωs
Note that the replicas will not overlap if ωx < ωs

2

In this case, X(jω) could be recovered from Xδ(jω) by applying a

lowpass filter with gain Ts and a cutoff frequency ωc such that ωx < ωc < ωs − ωx

This is a surprising result
This is the sampling theorem
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Guiding Questions and Objectives Revisited x[n] = x(nTs) X(ejωTs) = 1 Ts

∞

k=−∞

X (j(ω − kωs))

Recall our guiding questions

– How is the CTFT of x(t) related to the DTFT of x[n] = x(nTs)? – Under what conditions can we synthesize x(t) from x[n]? – How do we do this DT ⇒ CT conversion?

Only the last remains
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The Sampling Theorem Let x(t) be a bandlimited signal with X(jω) = 0 for |ω| > ωx. Then x(t) is uniquely determined by its samples x(nT), n = 0, ±1, ±2, . . . , if ωs > 2 ωx where ωs = 2π

Ts .

Thus, we can reconstruct any bandlimited signal x(t) exactly by

creating a scaled impulse train and lowpass filtering

This theorem is sometimes called the Shannon sampling

theorem

min ωs = 2ωx is called the Nyquist rate
max ωx = ωs

2 is called the Nyquist frequency

In other words, we must obtain at least two samples per a cycle of

the fastest sinusoidal component

This should sound familiar
Recall that the fastest perceivable frequency in discrete-time

signals is π radians per sample (i.e. 0.5 cycles per sample)

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SLIDE 8

Example 2: Equivalent Sinusoids

−3 −2 −1 1 2 3 −1 1 cos(ω t+θ Equivalent DT Sinusoids −3 −2 −1 1 2 3 −1 1 cos((2π − ω)t − θ) −3 −2 −1 1 2 3 −1 1 cos((ω t+2π)t + θ) Time (s)

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Example 2: Equivalent Sinusoids Suppose x[n] is a DT sinusoidal signal, x[n] = cos(Ωn + θ) Solve for all of the CT sinusoids x(t) = cos(ωt + φ) that satisfy the relationship x(t)|t=nTs = x[n]. Which of these CT sinusoids satisfy the sampling theorem criterion?

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Example 2: MATLAB Code

function [] = EquivalentSinusoids(); close all; t = -3:6/1000:3; n = floor(min(t)):ceil(max(t)); ph = 2*pi*rand; w = 0.6*pi; figure; FigureSet(1,’LTX’); subplot(3,1,1); h = plot(t,cos(w*t + ph),n,cos(w*n + ph),’k.’); set(h,’MarkerSize’,12); box off; ylabel(’cos(\omega t+\theta’); title(’Equivalent DT Sinusoids’); ylim([-1.05 1.05]); AxisLines; subplot(3,1,2); h = plot(t,cos((2*pi-w)*t - ph),n,cos(w*n + ph),’k.’); set(h,’MarkerSize’,12); box off; ylabel(’cos((2\pi - \omega)t - \theta)’); ylim([-1.05 1.05]); AxisLines; subplot(3,1,3); h = plot(t,cos((w+2*pi)*t + ph),n,cos(w*n + ph),’k.’); set(h,’MarkerSize’,12); box off; ylabel(’cos((\omega t+2\pi)t + \theta)’); xlabel(’Time (s)’); ylim([-1.05 1.05]); AxisLines; AxisSet(8);

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Example 2: Workspace

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SLIDE 9

Interpolation Possibilities

−3 −2 −1 1 2 3 −0.2 0.2 0.4 0.6 0.8 1 Time (s) Signal (scaled)

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print -depsc EquivalentSinusoids;

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Bandlimited Interpolation

X

y(t) x(t) xp(t) p(t) =

∞

n=−∞ δ(t − nTs)

H(jω)

Intuitively there are ∞ signals that are equal to x(t) and evenly

spaced samples

However, if

– The samples were taken from a bandlimited signal: X(jω) = 0 for ω > ωx – The conditions of the sampling theorem are satisfied: ωs > 2 ωx then there is only one bandlimited signal that passes exactly through the samples!

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Interpolation Introduction

CT ⇒ DT conversion is trivial: x[n] = x(nTs)
We now know that we can reconstruct x(t) from x[n] exactly
This should be surprising
There are many signals x(t) that satisfy the constraint

x[n] = x(nTs)

Which one is the original x(t)?
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SLIDE 10

Interpolation Simplified xδ(t) =

∞

n=−∞

x(nTs) δ(t − nT) xr(t) = xδ(t) ∗ h(t) = ∞

−∞

xδ(t − τ) h(τ) dτ = +∞

−∞

∞
n=−∞

x(nTs) δ(t − τ − nTs)

h(τ) dτ

=

∞

n=−∞

x(nT) ∞

−∞

δ(t − τ − nTs) h(τ) dτ =

∞

n=−∞

x(nTs) h(t − nTs)

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DT Processing of CT Signals H(z)

x[n] y[n] x(t) y(t) Ts Ts CT ⇒ DT DT ⇒ CT

Since x[n] completely represents x(t), we can process x[n] in

discrete-time

However, once we process the discrete-time signal and generate a

discrete-time output, y[n] we need to create a continuous-time

utput y(t)
We assume that y[n] represent samples from a continuous-time

signal y(t) that is bandlimited the same as x(t)

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Interpolation with Ideal Filters x[n] = x(nTs) Xδ(jω) = 1 Ts

∞

k=−∞

X (j(ω − kωs))

If the terms of the sum do not overlap, an ideal lowpass filter can

extract X(jω) from Xδ(jω)

The passband gain must be Ts
Recall that the ideal lowpass filter with cutoff frequency ωc and

gain Ts has an impulse response given by h(t) = Tsωc π sin(ωct) ωct = Tsωc π sinc ωct π

Thus, for ideal interpolation

xr(t) = Tsωc π

∞

n=−∞

x(nTs) sinc ωc(t − nTs) π

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Interpolation

X

xr(t) x(t) xδ(t) p(t) =

∞

n=−∞ δ(t − nTs)

H(jω)

There are many forms of interpolation

– Piecewise constant – Linear (point to point) – Splines (piecewise cubic)

For bandlimited signals, a lowpass filter can be used to exactly

recover the signal

Called bandlimited interpolation
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SLIDE 11

Introduction to Aliasing x(t) · p(t)

FT

⇐ ⇒ 1 Ts

+∞

k=−∞

X (j(ω − kωs))

If the sampling theorem is not satisfied (ωx > ωs

2 ), the sum of

replicated spectral components X(j(ω − kωs)) will overlap

This is called aliasing
In this case xr(t) = x(t), though the two will be equal at t = nTs
When aliasing occurs with pure tones (sinusoidal signals), signals

with frequencies above ωs

2 are reflected to lower frequencies

You should have experienced this in the lab with audio
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Notes on Bandlimited Interpolation x[n] = x(nTs) xr(t) = Tsωc π

∞

n=−∞

x(nTs) sinc ωc(t − nTs) π

If the sampling theorem criterion is satisfied, xr(t) = x(t)
However, this assumes that the frequency components of x[n] are

between −π/2 and +π/2

If the sampling criterion is not satisfied, high-frequency

components will appear as low-frequency components

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Example 3:Aliasing of Pure Tones

1 2 3 4 5 6 7 8 9 10 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Time (ms) Signals (scaled) Aliasing of Pure Tones Example Tone Frequency:500 Hz Sample Rate:2000 Hz True Signal Reconstructed Signal

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Bandlimited Interpolation Example

−5 −4 −3 −2 −1 1 2 3 4 5 0.5 1 Time (s) xr(t) −5 −4 −3 −2 −1 1 2 3 4 5 0.5 1 x[n] −5 −4 −3 −2 −1 1 2 3 4 5 0.5 1 x(t) Example of Band−limited Signal Reconstruction (Interpolation)

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SLIDE 12

Example 3:Aliasing of Pure Tones

1 2 3 4 5 6 7 8 9 10 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Time (ms) Signals (scaled) Aliasing of Pure Tones Example Tone Frequency:1500 Hz Sample Rate:2000 Hz True Signal Reconstructed Signal

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Example 3:Aliasing of Pure Tones

1 2 3 4 5 6 7 8 9 10 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Time (ms) Signals (scaled) Aliasing of Pure Tones Example Tone Frequency:900 Hz Sample Rate:2000 Hz True Signal Reconstructed Signal

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Example 3:Aliasing of Pure Tones

1 2 3 4 5 6 7 8 9 10 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Time (ms) Signals (scaled) Aliasing of Pure Tones Example Tone Frequency:1800 Hz Sample Rate:2000 Hz True Signal Reconstructed Signal

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Example 3:Aliasing of Pure Tones

1 2 3 4 5 6 7 8 9 10 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Time (ms) Signals (scaled) Aliasing of Pure Tones Example Tone Frequency:1100 Hz Sample Rate:2000 Hz True Signal Reconstructed Signal

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SLIDE 13

AxisSet(8); legend(’True Signal’,’Reconstructed Signal’); eval(sprintf(’print -depsc AliasingTones%04d;’,fx(c1))); end; return; n = -15:15; t = -5.5:0.01:5.5; xp = rand(size(n)); % Sampled signal wc = pi; T = 1; xr = zeros(size(t)); % Reconstructed signal subplot(3,1,3); for cnt = 1:length(n), st = (wc*T/pi)*xp(cnt)*sinc(wc*(t-n(cnt)*T)/pi); plot(t,st,’g’); hold on; xr = xr + st; end; h = plot(t,xr,’b’,n,xp,’ro’); set(h(2),’MarkerFaceColor’,’r’); set(h(2),’MarkerSize’,2); hold off; xlim([min(t) max(t)]); ylim([-0.2 1.2]); xlabel(’Time (s)’); ylabel(’Signal’); box off; subplot(3,1,2); plot([min(t) max(t)],[0 0],’k:’); hold on; h = stem(n,xp,’r’); set(h(1),’MarkerFaceColor’,’r’); set(h(1),’MarkerSize’,2);

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Example 3:Aliasing of Pure Tones

1 2 3 4 5 6 7 8 9 10 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 Time (ms) Signals (scaled) Aliasing of Pure Tones Example Tone Frequency:2200 Hz Sample Rate:2000 Hz True Signal Reconstructed Signal

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hold off; ylabel(’Sampled Signal’); subplot(3,1,1); h = plot(t,xr,’b’,n,xp,’ro’); set(h(2),’MarkerFaceColor’,’r’); set(h(2),’MarkerSize’,2); ylabel(’Reconstructed Signal’); xlim([min(t) max(t)]); ylim([-0.2 1.2]); box off; title(’Example of Band-limited Siganl Reconstruction (Interpolation)’);

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Example 3: MATLAB Code

function [] = AliasingTones(); close all; fs = 2e3; % 2 kHz sample rate fx = [500 900 1.1e3 1.5e3 1.8e3 2.2e3]; t = 0:10e-3/500:10e-3; % Span of 5 ms ts = -15e-3:1/fs:25e-3; % Sampled times T = 1/fs; ws = 2*pi*fs; wc = ws/2; for c1 = 1:length(fx), xt = cos(2*pi*fx(c1)*t); % True signal xs = cos(2*pi*fx(c1)*ts); % True signal sampled xr = zeros(size(t)); % Reconstructed signal memory allocation for c2 = 1:length(ts), st = (wc*T/pi)*xs(c2)*sinc(wc*(t-ts(c2))/pi); % Band-limited interpolation xr = xr + st; end; figure; FigureSet(1,’LTX’); h = plot(t*1e3,xt,’b’,t*1e3,xr,’g’,ts*1e3,xs,’ko’); set(h(1),’LineWidth’,1.5); set(h(2),’LineWidth’,0.8); set(h(3),’MarkerFaceColor’,’k’); set(h(3),’MarkerSize’,4); xlim([min(t*1e3) max(t*1e3)]); ylim([-1.1 1.1]); box off; xlabel(’Time (ms)’); ylabel(’Signals (scaled)’); title(sprintf(’Aliasing of Pure Tones Example Tone Frequency:%d Hz Sample Rate:%d Hz’,fx(c1),fs));

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SLIDE 14

Example 4:Aliasing of Speech

30 32 34 36 38 40 42 44 46 48 50 −0.5 −0.4 −0.3 −0.2 −0.1 0.1 0.2 0.3 0.4 0.5 Time (ms) Signals (scaled) Regis "Your" Sample Rate:8820.0 Hz True Signal Reconstructed Signal

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Example 4:Aliasing of Speech The following slides show a segment of speech for Regis Philbin stating part of the word “Your”. The signal was sampled at various rates and then interpolated. The difference between the true signal and the reconstructed signal demonstrates the hazard of under

sampling. The original signal was sampled at 44.1 kHz.

Why isn’t the aliasing apparent at 22 kHz?

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Example 4:Aliasing of Speech

30 32 34 36 38 40 42 44 46 48 50 −0.5 −0.4 −0.3 −0.2 −0.1 0.1 0.2 0.3 0.4 0.5 Time (ms) Signals (scaled) Regis "Your" Sample Rate:4410.0 Hz True Signal Reconstructed Signal

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Example 4:Aliasing of Speech

30 32 34 36 38 40 42 44 46 48 50 −0.5 −0.4 −0.3 −0.2 −0.1 0.1 0.2 0.3 0.4 0.5 Time (ms) Signals (scaled) Regis "Your" Sample Rate:22050.0 Hz True Signal Reconstructed Signal

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SLIDE 15

Aliasing Comments

In discrete-time, one of the key concepts of this class is that

complex exponentials with frequencies that differ by a multiple of 2π are indistinguishable: ej(ω+ℓ2π)n = ejωn

When we sample a signal, we convert a continuous-time signal to

a discrete-time signal

Sinusoidal components that are above ωs

2 are reflected down to

lower frequencies

You have experienced aliasing before

– Spoked wheels in street lights that strobe at 120 Hz – Spoked wagon wheels or helicopter blades in movies and television – Strobe lights applied to rapidly oscillating objects

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Example 4:Aliasing of Speech

30 32 34 36 38 40 42 44 46 48 50 −0.5 −0.4 −0.3 −0.2 −0.1 0.1 0.2 0.3 0.4 0.5 Time (ms) Signals (scaled) Regis "Your" Sample Rate:2205.0 Hz True Signal Reconstructed Signal

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Preventing Aliasing

x[n] C/D Conversion T

H(s)

x(t) Anti-Aliasing Filter

In practice, most data acquisition systems that convert CT signals

to DT signals take two measures to prevent aliasing – Apply an analog (continuous-time) lowpass filter prior to sampling to ensure the signal is bandlimited – Use a higher sampling rate than required by the sampling theorem (e.g. ωs = 2.2 · ωx)

These lowpass filters are often called anti-aliasing filters
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Effect of Aliasing in the Frequency Domain

1 1 1 1 2

2ws

2ws

ω ω ω ω ωx ωx ωx −ωx −ωx −ωx ωs ωs ωs ωs −ωs −ωs −ωs −ωs Xδ(jω) Xδ(jω) Xδ(jω) Xδ(jω)

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SLIDE 16

Fourier Example of CTFT & DTFT Relationship

1 1

ω ω Ω

1 T

−2π 2π ωs ωs −ωs −ωs X(jω) Xδ(jω) X(ejΩ)

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Anti-aliasing Filter Tradeoffs

x[n] C/D Conversion T

H(s)

x(t) Anti-Aliasing Filter

Over-sampling also has a number of benefits for the anti-aliasing filter (AAF).

The transition band can be wider
If the filter order is fixed, the passband and stopband attenuation

can be improved

If the passband and stopband attenuation are fixed, a lower order

filter can be used (easier and less expensive to design and build)

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CT to DT Conversion Comments X(ejΩ) = 1 Ts

∞

k=−∞

X

j

Ω − k2π Ts

If the sampling theorem is satisfied, the DTFT is a scaled version
f the CTFT

– The amplitude of the DTFT is scaled by

1 Ts

– The frequency ωs is mapped to Ωs = 2π

This means we can theoretically calculate the CTFT of a

band-limited signal exactly using the DTFT of the sampled signal x(nT)

For spectral estimation, the signal must first be windowed and

then the FFT is used to calculate the DTFT

This is exactly how the sampling oscilloscopes use the FFT to

estimate the Fourier transform of a CT signal

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Discrete-Time Processing of Continuous-Time Signals H(z)

x[n] y[n] x(t) y(t) Ts Ts CT ⇒ DT DT ⇒ CT

X(ejΩ) = 1 Ts

∞

k=−∞

X

j

Ω − k2π Ts

Most electronic “systems” that process signals do so in

discrete-time

There are many advantages to this approach

– Digital technology is cheap to manufacture and very advanced – Digital systems do not drift over time – Parameter tuning (expensive) is not necessary; repeatable – More flexible

The sampling theorem is the theoretical basis for DT processing of

CT signals

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