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Signals and Systems

Lecture 13 Wednesday 6

th

December 2017

DR TANIA STATHAKI

READER (ASSOCIATE PROFFESOR) IN SIGNAL PROCESSING

IMPERIAL COLLEGE LONDON

• Continuous time systems.

▪ Good for analogue signals and general understanding of signals and systems.

▪ Appropriate mostly to analogue electronic systems.

• Electronic devices are increasingly digital.

▪ E.g. mobile phones are all digital, TV broadcast will be 100 % digital in UK.

▪ We use digital ASIC chips, FPGAs and microprocessors to implement systems

and to process signals.

▪ Continuous signals are converted to numbers (discrete signals), they are

processed and then converted back to continuous signals.

𝑥 𝑡 𝑥[𝑛] 𝑦 𝑛 𝑦(𝑡)

Continuous time versus discrete time

Electronic devices

Digital

computers

A-to-D D-to-A

• Sampling theorem is the bridge between continuous-time and discrete-

time signals.

• It states how often we must sample in order not to loose any information.

𝑠

𝑠

𝑥[𝑛]

Sampling theorem

A continuous-time lowpass signal 𝑥(𝑡) with frequencies no higher than

𝑚𝑎𝑥

𝐻𝑧 can be perfectly reconstructed from samples taken every 𝑇 𝑠

units

of time, 𝑥 𝑛 = 𝑥(𝑛𝑇 𝑠

), if the samples are taken at a rate 𝑓 𝑠

𝑠

that is

greater than 2 𝑓 𝑚𝑎𝑥

Sampling theorem

• If a lowpass signal has a spectrum bandlimited to 𝐵𝐻𝑧, i.e., 𝑋 𝜔 = 0 for

𝜔 > 2 𝜋𝐵, it can be reconstructed from its samples without error if these

samples are taken uniformly at a rate 𝑓 𝑠

2 𝐵 samples per second.

• The minimum sampling rate 𝑓 𝑠

= 2 𝐵 required to reconstruct 𝑥(𝑡) from its

samples is called the Nyquist rate for 𝑥(𝑡) and the corresponding

sampling interval 𝑇 𝑠

1

2 𝐵

is called the Nyquist interval. Samples of a

continuous signal taken at its Nyquist rate are the Nyquist samples of that

signal.

• In other words the minimum sampling frequency is 𝑓 𝑠

• A bandpass signals whose spectrum exists over a frequency band

𝑐

𝐵

2

𝑐

𝐵

2

also has a bandwidth of 𝐵𝐻𝑧. Such a signal is still

uniquely determined by 2 𝐵 samples per second but the sampling scheme

is a bit more complex compared to the case of a lowpass signal.

Sampling theorem cont.

• The previous analysis is depicted below. Consider a signal, bandlimited

to 𝐵𝐻𝑧, with Fourier transform 𝑋(𝜔) (depicted real for convenience).

• The sampled signal has the following spectrum.

Depiction of previous analysis

• We graphically illustrate below the collection of the above mentioned

processes.

𝑠

𝑠

𝑥[𝑛]

Depiction of previous analysis

• The gap between two adjacent spectral repetitions is (𝑓 𝑠

• In order to reconstruct the original signal 𝑥(𝑡) we can use an ideal

lowpass filter on the sampled spectrum which has a bandwidth of any

value between 𝐵 and (𝑓 𝑠

• Reconstruction process is possible only if the shaded parts do not

overlap. This means that 𝑓 𝑠

must be greater that twice 𝐵.

• We can also visually verify the sampling theorem in the above figure.

Reconstruction of the original signal

• Frequency domain

The signal 𝑥 𝑛𝑇 𝑠

𝑠

) > has a spectrum 𝑋 𝑠

(𝜔) which is

multiplied with a rectangular pulse in frequency domain.

Reconstruction generic example

• Consider the signal 𝑥 𝑡 = sinc

2

5 𝜋𝑡 whose spectrum is:

𝑋 𝜔 = 0. 2 Δ(𝜔/ 20 𝜋) (Look at Table 7. 1 , page 702 , Property 20 by Lathi).

• The bandwidth is 𝐵 = 5 𝐻𝑧 ( 10 𝜋𝑟𝑎𝑑/𝑠). Consequently, the Nyquist

sampling rate is 𝑓 𝑠

= 10 𝐻𝑧; we require at least 10 samples per second.

The Nyquist interval is 𝑇 𝑠

1

10

Example

1. 2

− 10 𝜋 10 𝜋

• In that case we use the Nyquist sampling rate of 10 𝐻𝑧.
• The spectrum

𝑋 𝜔 consists of back-to-back, non-overlapping

repetitions of

1

𝑇 𝑠

𝑋 𝜔 repeating every 10 𝐻𝑧.

• In order to recover 𝑋(𝜔) from

𝑋 𝜔 we must use an ideal lowpass filter

of bandwidth 5 𝐻𝑧. This is shown in the right figure below with the dotted

line.

Example cont.

Nyquist sampling: Just about the correct sampling rate

• Sampling at lower than the Nyquist rate (in this case 5 𝐻𝑧) makes

reconstruction impossible.

• The spectrum

𝑋 𝜔 consists of overlapping repetitions of

1

𝑇 𝑠

repeating every 5 𝐻𝑧.

• 𝑋(𝜔) is not recoverable from

• Sampling below the Nyquist rate corrupts the signal. This type of

distortion is called aliasing.

Example cont.

Undersampling: What happens if we sample too slowly?

• Consider what happens when a 1𝐻𝑧 and a 6𝐻𝑧 sinewaves are sampled

at a rate of 5Hz.

• The 1𝐻𝑧 and 6𝐻𝑧 sinewaves are indistinguishable after sampling. The

two discrete signals produced are identical.

Aliasing

• Consider making a video of a clock face.
• The second hand makes one revolution per minute (𝑓 𝑚𝑎𝑥

1

60

• Critical Sampling: 𝑓 𝑠

1

30

𝑠

< 30 sec), anything below that sampling

frequency will create problems.

For example:

▪ When 𝑇 𝑠

= 60 sec (𝑓 𝑠

~ 0. 167 𝐻𝑧), the second hand will not move.

▪ When 𝑇 𝑠

= 59 sec ( 𝑓 𝑠

~ 0. 169 𝐻𝑧 ), the second hand will move

backwards.

• Watch the videos (optional)

https://www.youtube.com/watch?v=VNftf5qLpiA

https://www.youtube.com/watch?v=QOwzkND_ooU

Aliasing and the wagon wheel effect

Anti-aliasing filter

• To avoid distortion of a signal after sampling, one must ensure that the

signal being sampled at 𝑓 𝑠

is bandlimited to a frequency 𝐵, where 𝐵 <

𝑓 𝑠

2

• If the signal does not obey the above restriction, we may apply a lowpass

filter with cut-off frequency

𝑓 𝑠

2

before sampling.

Sampling Reconstruction

• If the original signal 𝑥(𝑡) is not bandlimited to

𝑓 𝑠

2

, perfect reconstruction is

not possible when sampling at 𝑓 𝑠

. However, the reconstructed signal 𝑥ො(𝑡) is

the best bandlimited approximation to 𝑥(𝑡) in the least-square sense.