6 Filtering

A complete description of a LTI system is shown below:

x(t) h(t) y(t) = h(t) ∗ x(t)

X(ω) H(ω) Y (ω) = H(ω)X(ω)

On the top is a time-domain description, in terms of time-domain signals, the system impulse
response h(t), and the convolution operation for finding the system output. On the bottom is
the frequency-domain description, using frequency-domain descriptions of signals, the frequency
response or transfer function H(ω), and multiplication for finding the system output. Given
the impulse response we can find the frequency response, and vice versa: the two descriptions
are entirely equivalent. However, since multiplication is a simpler operation to understand than
convolution, the second description is more intuitive and we prefer it.
A main applications of LTI systems is to perform filtering. A system takes an input signal.
It produces an output signal that is related to the input, according to the system properties. In
practice we can define systems that have desirable properties, and these systems take input signals
and produce output signals that are useful for our purposes.
Most of the time we specify the properties we require from a system in the frequency domain.
This involves specifying H(ω). In effect, we design a system that has a frequency response that
approximates the requirements for our application.

6.1 Ideal lowpass filter

Applying the duality property to the Fourier pair
τω 
F
pτ (t) ←→ τ sinc
2π
gives the new pair  
τt F
τ sinc ←→ 2πpτ (−ω) = 2πpτ (ω),
2π
since pτ (ω) = pτ (−ω) is an even function. By linearity,
 
τ τt F
sinc ←→ pτ (ω).
2π 2π

A sinc in the time domain transforms to a rectangular pulse in frequency.

Now consider the system

x(t) y(t)
H(ω)
X(ω) Y (ω)

with H(ω) = p2 (ω). The frequency response is the real-valued function below:

1
H(ω)

ω
−1 0 1

6-1
What is the output of the system when the input is x(t) = ejω0 t for some specific value of ω0 ?
The definition of the frequency response provides the most direct way of obtaining the solution.
Recall that for any LTI system the following is a valid input-output pair:

ejω0 t −→ H(ω0 )ejω0 t .

The (possibly complex) value H(ω0 ) completely determines the action of the system on a signal
at frequency ω0 , or in other words on the signal ejω0 t . Thus the output will be y(t) = H(ω0 )ejω0 t .
Looking at the transfer function H(ω) above we see that the output will depend on the value of
ω0 : (
ejω0 t −1 ≤ ω0 ≤ 1
y(t) =
0 otherwise.
If the frequency of the input complex exponential is in the passband of the filter, that is if |ω| ≤ 1,
then it propagates through without modification. The system passes low frequencies. However, if
the input is a complex exponential with a frequency |ω0 | > 1, then the output is zero: the system
completely blocks signals at high frequencies. The system is therefore a lowpass filter. It passes
low frequency signals without modification, and eliminates high frequencies. Also, the system is
ideal because the transition between the passband |ω| ≤ 1 and the stopband |ω| > 1 is infinitely
sharp.
Another way of looking at this scenario is to simply use frequency-domain reasoning. The input
signal x(t) = ejω0 t has a frequency representation X(ω) = 2πδ(ω − ω0 ), which consists of exactly
one impulse at ω = ω0 . The output of the filter is the product Y (ω) = H(ω)X(ω). If the impulse
lies in the passband |ω| ≤ 1 of the filter (that is, if |ω0 | ≤ 1), then the output is Y (ω) = 2πδ(ω−ω0 ).
The time-domain output in this case is just y(t) = ejω0 t , the same as the input. However, if |ω0 | > 1
then Y (ω) = 0, so y(t) = 0.
Since the ideal lowpass filter is LTI (otherwise the frequency response H(ω) would not exist), it
has an equivalent description in terms of an impulse response. The impulse response in this case
is  
2 2t
h(t) = sinc ,
2π 2π
which has the form of a sin(x)/x function in time. Since the system is governed by the covolution
y(t) = h(t) ∗ x(t), we see that an ideal lowpass filter convolves the time-domain input x(t) with a
sinc function to obtain the output. The output for x(t) = ejω0 t will be
 
1 t
y(t) = sinc ∗ ejω0 t .
π π

The result of this convolution is ejω0 t if |ω| ≤ 1 and zero if |ω0 | > 1. This is obvious from
frequency-domain reasoning, but completely non-intuitive in the time domain.

6.2 RC lowpass filter

An ideal lowpass filter is conceptually useful, but we can’t build one in practice. Because the
impulse response is infinitely long (there is no point in time beyond which it is always equal to
zero) we need to know the input x(t) for all time to calculate the output y(t) at any single instant
— and we can never measure a signal for all time. Also, note that the sin(x)/x impulse response
is not zero for negative time (that is, it is not right-sided), so the ideal lowpass filter is not causal.
This can be a problem if we want to filter a signal online in real time, which is often required.
Nonetheless, we often try to design systems that at least approximate the ideal lowpass filter
frequency response.
The simplest electrical system that we can construct to approximate a lowpass filter is the RC
circuit shown below:

6-2
 vR (t)

i(t)
x(t) = v(t) vC (t) = y(t)
C

The input is the voltage generated by the voltage source on the left, and the output is the voltage
across the capacitor on the right.
We know the time-domain models that govern the behaviour of resistors and capacitors, so we can
find a differential equation relating the input x(t) to the output y(t). The fundamental relationship
defining a resistor is V /I = R. However, this is only for constant V and I — the whole point
of signal analysis is to reason about quantities that vary with time. If R is constant but the
voltage and current can vary, then the usual equation generalises to v(t)/i(t) = R, or v(t) = Ri(t).
This relationship must hold for all t in the model of a resistor: the voltage across it must always
equal R times the current through it at any instant in time. In the diagram above the voltage
across the resistor is denoted by vR (t), so the governing equation in terms of the signals shown is
vr (t) = Ri(t).
Similarly we consider the voltage across the capacitor to be a function vC (t) of time. Using the
principles of electrostatics one finds the time-domain relationship linking the voltage vC (t) across
a capacitor and the current i(t) through it to be i(t) = C dvdt C (t)
, where C is defined to be the
capacitance. Unlike the resistor the equations governing a capacitor have to be specified in terms
of functions of time, since it is a dynamic device.
The last equation that is required to specify the input-output relation involves potentials. If we
consider the bottom wire of the circuit to be ground, then the potential above the voltage source
is v(t). However, this potential must be equal to the sum of the voltages across the resistor and
the capacitor (properly directed as shown), so v(t) = vC (t) + vR (t). This must also hold for all
instants in time.
The three equations that determine the behaviour of the circuit are therefore
dvC (t)
vR (t) = Ri(t), i(t) = C , and v(t) = vC (t) + vR (t).
dt
Eliminating vR (t) from the first and third equations gives
dvC (t)
i(t) = C and v(t) = vC (t) + Ri(t),
dt
and now eliminating i(t) gives
dvC (t)
v(t) = vC (t) + RC .
dt
Finally, noting that the input is defined to be x(t) = v(t) and the output vC (t) = y(t) gives the
overall input-output relationship:
dy(t)
x(t) = y(t) + RC .
dt
For any given input x(t) one can in principle solve this differential equation for y(t), so we now
have an input-output relation for the system.
Resistors and capacitors (and inductors) are linear and time-invariant devices, so any system
constructed from these passive devices is LTI. (Transistors, on the other hand, are active devices

6-3
that are often operated in a nonlinear region.) The RC circuit therefore has an impuse response
h(t), which could be found by calculating the output y(t) when the input is x(t) = δ(t). That is,
the impulse response h(t) must satisfy δ(t) = h(t) + RC dh(t)
dt . This equation is not easy to solve.
Fourier methods present a way forward. Since the system is LTI it has a frequency response H(ω)
that equals the Fourier transform of h(t). Now, if we take the Fourier transform of both sides of
the expression δ(t) = h(t) + RC dh(t)
dt we get

1 = H(ω) + RCjωH(ω),

which is an algebraic equation rather than a differential equation. We can factorise it as 1 =

H(ω)(1 + jωRC), and solve for H(ω):
1
1 RC
H(ω) = = 1 .
1 + jωRC RC + jω
The inverse transform then yields the impulse response
1 − 1 t
h(t) = e RC u(t),
RC
which is plotted below:

1
RC h(t)

t
0

Mathematically, the RC lowpass filter given takes the input x(t) and convolves it with this function
to give the output y(t). The quantity τ = RC is called the time constant of the circuit, and can
quite easily be shown to be the time required for the impulse response voltage to fall to 1/e of its
initial value.
Recall that the impulse response is the output of the circuit when the input is δ(t). While we
can’t actually generate a delta function, and therefore can’t put one into a real circuit, we certainly
can put one into a mathematical model of our circuit. The interpretation is as follows. Initially
there is zero voltage at the input and the circuit is at rest — the capacitor is discharged and the
output voltage is zero. At time t = 0 the circuit is ”kicked” by an impulse, which instantaneously
1
transfers a charge to the capacitor. The output voltage immediately jumps to RC . After this
time the input voltage returns to zero, so the voltage source is effectively a short circuit. The
capacitor discharges through the resistor, so the output voltage gradually drops. If R is big then
the discharge is slow, so the impulse response falls off gradually. Similarly, if the capacitor is big
then the discharging is also slow. The product τ = RC is a parameter that takes both of these
effects into accout: when large the rate of decay of h(t) is low, and when small it is high.
If one doesn’t explicitly need the impulse response or the frequency response then the original
differential equation linking input and output can be used directly: the equation x(t) = y(t) +
RC dy(t)
dt can be Fourier transformed to give an input-output relation in the frequency domain:

X(ω) = Y (ω) + RCjωY (ω) = Y (ω)(1 + jωRC).

We can express any input x(t) in terms of its Fourier transform X(ω), in which case the Fourier
transform of the output is
1
Y (ω) = X(ω).
1 + jωRC
Note that this is just a statement of the expression Y (ω) = H(ω)X(ω), which is the frequency-
domain counterpart of the time-domain input-output relationship defined by convolution: y(t) =
h(t) ∗ x(t).

6-4
6.3 Frequency-domain analysis of RC lowpass filter
We normally visualise filters (and signals) in the frequency domain. We showed that the RC
lowpass filter in the previous section has a transfer function or frequency response of
1
H(ω) = .
1 + jωRC
To understand this filter it helps to plot the transfer function — both magnitude and phase,
since it is generally complex. To express this transfer function in polar coordinates we consider
1 + jωRC. For each value of ω this is a point in the Argand diagram, [draw] and by consdiering
the length and angle of this vector we see that it can be written as
p
1 + jωRC = 1 + ω 2 (RC)2 ej arctan(ωRC) ,

The required frequency response is therefore

1 1
H(ω) = = p e−j arctan(ωRC) ,
1 + jωRC 1 + ω 2 (RC)2
which is plotted below for R = C = 1:

1
|H(ω)|

0.5

0
-6 -4 -2 0 2 4 6

1
∠H(ω)

-1

-2
-6 -4 -2 0 2 4 6
ω

This is called the Bode plot of the system (although typically Bode plots are displayed on loga-
rithmic axes, as will be shown shortly).
The definition of the frequency response states that an input of x(t) = ejω0 t to the system produces
an output y(t) = H(ω0 )ejω0 t . Therefore an input of x(t) = ej2t will produce the output
1
y(t) = H(2)ej2t = p e−j arctan(2RC) ej2t
1 + 22 (RC)2
1
= p ej(2t−arctan(2RC)) = |H(2)|ej(2t+∠H(2)) ,
2
1 + 2 (RC)2

which for the R = C = 1 case plotted above gives

y(t) = 0.4472ej(2t−1.1071).

Note that the values in this result can be read directly off of the graph at ω = 2. Thus the plot
of the frequency response immediately carries all of the useful information about what happens

6-5
to any given frequency at the input as it passes through the system. We see that this RC lowpass
filter tends to reduce the magnitude of high frequencies: the effectively get multiplied by a number
with a small magnitude, and their amplitude decreases accordingly.
The magnitude |H(ω)| of the frequency response is called the gain of the system for frequency ω.
If the gain is greater than one then the amplitude of a complex exponential input at that frequency
will be amplified by the corresponding gain value; otherwise it will be reduced. However, engineers
and physicists often prefer to express the gain on a logarithmic scale, also called the decibel (dB)
scale, which for a voltage signal is defined as
GdB (ω) = 20 log10 |H(ω)|.
1
To investigate this further we define ωc = 1/τ = RC , and from the magnitude response calculated
earlier for the RC lowpass system we see that
1 1
GdB (ω) = 20 log10 p = 20 log10 p
1 + (ωRC) )2 1 + (ω/ωc )2 )
= 20 log10 ( 1 + (ω/ωc )2 )−1 = −20 log10 ( 1 + (ω/ωc )2 )
p p

A plot of the gain in dB versus frequency on a log axis for the case of ωc = 1 is shown as the solid
(red) curve below:

0
-3.0103

-10
Gain GdB (ω)

-20

-30

-40
-1 0 1 2
10 10 10 10
ω

Also shown are dotted√(blue) asymptotes to the curve. Firstly, for ω ≪ ωc the gain is approximately
GdB (ω) ≈ −20 log10 ( 1)p= 0. This is the p
horizontal asymptote in the plot. Secondly, for ω ≫ ωc
the ratio ω/ωc ≫ 1 and 1 + (ω/ωc )2 ≈ (ω/ωc )2 = ω/ωc , so the gain is approximately
GdB (ω) ≈ −20 log10 (ω/ωc ) = −20 log10 (ω) + 20 log10 (ωc ).
This is the asymptote in the plot that matches the curve for high frequencies, and defines what is
called the roll-off of the filter. We observe that in this region an increase in frequency by a factor
of 10 (e.g. from 100 to 101 , or from 101 to 102 ) always reduces the gain by 20dB. We therefore
say that the filter has a roll-off of 20dB/decade, which is a measure of how effectively the lowpass
filter attentuates or reduces high frequencies. Equivalently, it is quite easy to show that when the
frequency doubles in this region the gain drops by 6dB. Since going up an octave corresponds to
a doubling of frequency (also in music!), we could equivalently say that the filter has a roll-off of
6dB/octave. Note that this is the roll-off rate of any first-order lowpass filter.
The two asymptotes cross at ω = ωc , which is called the cut-off, breakpoint, or knee of the filter
— effectively it can be considered the frequency at which the filter transitions from the low-
frequency
p pass band
p to the high-frequency
√ stop band. Noting that at this frequency we have
1 + (ω/ω0 )2 = 1 + (ω0 /ω0 )2 = 2, we see that
√
GdB (ω0 ) = −20 log10 ( 2) = −3.0103.

6-6
 The cut-off frequency ωc of the filter is therefore also called the −3dB point. It can be expressed
1
in Hertz by noting that ωc = 2πfc and ωc = RC , so
1
fc =.
2πRC
The cut-off frequency of the RC lowpass filter can therefore be varied by changing either R or
C, and for mathematical purposes only the product RC (equal to the circuit time constant τ )
matters.
Exercise: Repeat the whole analysis for a first-order highpass filter, where the resistor and the capacitor in
the circuit are swapped.

6.4 Response to a real-valued sinusoid

The frequency response of an LTI system can easily be used to find the output y(t) when the input
is the sinusoidal signal x(t) = A cos(ω0 t + φ). The result follows from the input-output pair
ejω0 t −→ H(ω0 )ejω0 t .
Since we can write
x(t) = A/2ejω0 t+φ + A/2e−jω0 t+φ = (A/2ejφ )ejω0 t + (A/2e−jφ )e−jω0 t
the required output is
y(t) = (A/2ejφ )H(ω0 )ejω0 t + (A/2e−jφ )H(−ω0 )e−jω0 t .
If the impulse response h(t) is real valued then H(ω) is conjugate symmetric, so
H(ω0 ) = |H(ω0 )|ej∠H(ω0 ) and H(−ω0 ) = H ∗ (ω0 ) = |H(ω0 )|e−j∠H(ω0 ) .
Substituting and rearranging gives the following:
y(t) = (A/2ejφ )|H(ω0 )|ej∠H(ω0 ) ejω0 t + (A/2e−jφ )|H(ω0 )|e−j∠H(ω0 ) e−jω0 t
= A/2|H(ω0 )|ej(ω0 t+φ+∠H(ω0 )) + A/2|H(ω0 )|e−j(ω0 t+φ+∠H(ω0 ))
= A|H(ω0 )| cos(ω0 t + φ + ∠H(ω0 )).
Thus the following input-output pair results:
A cos(ω0 t + φ) −→ A|H(ω0 )| cos(ω0 t + φ + ∠H(ω0 ).
A sinusoid of frequency ω0 at the input results in a sinusoid of the same frequency at the output,
but with magnitude scaled by the factor |H(ω0 )| and phase shifted by ∠H(ω0 ). These two values
can be obtained from the complex frequency response H(ω) evaluated at the frequency ω0 of the
input.
LTI systems do not generate frequencies at the output that are not present at the input. For AC
analysis at a fixed frequency this makes it possible to use phasor notation to perform the required
calculations. With fixed ω0 the sinusoidal signal x(t) = A cos(ω0 t + φ) has two free parameters, A
and φ, and we can represent it by the quantity X̃ in the phasor domain:
x(t) = A cos(ω0 t + φ) −→ X̃ = Aejφ .
The output signal above can be expressed in the form
y(t) = A|H(ω0 )| cos(ω0 t + φ + ∠H(ω0 )) −→ Ỹ = A|H(ω0 )|eφ+∠H(ω0 ) = H̃ X̃,
where the frequency response for the frequency of interest is represented by the phasor H̃ =
|H(ω0 )|ej∠H(ω0 ) . For the single-frequency case, convolution in time is equivalent to multiplication
in the phasor domain.
Exercise: The signals x1 (t) = A1 cos(ω0 t+φ1 ) and x2 (t) = A2 cos(ω0 t+φ2 ) can be represented by the phasors
X̃1 = A1 ejφ1 and X̃2 = A2 ejφ2 . Show that x1 (t) + x2 (t) is also a sinusoid at frequency ω0 and
corresponds to the phasor X̃1 + X̃2 .

6-7
6.5 Circuit analysis in the frequency domain
Linear constant coefficient differential equations in time always become algebraic equations in the
frequency domain.
Consider a capacitor. The principles of electrostatics demand that the voltage signal vc (t) across
d
the capacitor must always satisfy the equation ic (t) = C dt vc (t), where ic (t) is the signal that
represents the current through the capacitor. This differential equation is difficult to work with.
Now take the Fourier transform of the voltage-current relationship: Ic (ω) = (jωC)Vc (ω). If we
define the ratio of voltage to current in the frequency domain to be the impedance of the device,
then for the capacitor we have
Vc (ω) 1
= zC = .
Ic (ω) jωC
d
Similarly, for an inductor the time-domain equation is vL (t) = L dt iL (t), so in the frequency
domain we have VL (ω) = (jωL)IL (ω). Finally, for a resistor vr (t) = RiR (t), so Vr (ω) = RIR (ω).
Thus for these latter two devices
VL (ω) VR (ω)
= ZL = jωL and = ZR = R.
IL (ω) IR (ω)
Note that in all three cases we have voltage = impedance × current as long as voltages and currents
are expressed in the frequency domain. The resistor is a special case where this relationship also
holds in the time domain, but the frequency domain works for all three devices. The net result is
that circuit analysis using these three passive devices is simple as long as it is done in the frequency
domain.
Example: Consider the circuit below, which is a current source driving a combination of a resistor,
a capacitor, and an inductor:

R
vR (t)

x(t) = i(t) i1 (t) i2 (t) vC (t) = y(t)

C
L
vL (t)

We could find a differential equation linking the input x(t) and the output y(t), but it is simpler
to just derive the relationship in the frequency domain. The required quantities are

R
VR (ω)

X(ω) = I(ω) I1 (ω) I2 (ω) VC (ω) = Y (ω)

C
L
VL (ω)

Now the basic voltage-current relationships for the three devices are

VR (ω) = zR I1 (ω), VL (ω) = zL I1 (ω), VC (ω) = zC I2 (ω).

In the first branch the voltage-current relationship is

Y (ω) = VR (ω) + VL (ω) = zR I1 (ω) + zL I1 (ω) = (zR + zL )I1 (ω).

6-8
The impedances therefore add for the series combination. Now
 
1 1 1 1
I(ω) = I1 (ω) + I2 (ω) = Y (ω) + Y (ω) = + Y (ω),
zR + zL zc zR + zL zc

so the voltage-current relationship for the overall LRC combination is Y (ω) = zeff I(ω), where the
effective impedance satisfies
1 1 1
= + .
zeff zR + zL zc
This is the usual formula for combining impedances in parallel. The overall input-output relation-
ship satisfies
   
1 1 + jωC(R + jωL)
X(ω) = + jωC Y (ω) = Y (ω)
R + jωL R + jωL
and in terms of a system the transfer function of the circuit relating the input current to the
output voltage is
Y (ω) R + jωL
H(ω) = = .
X(ω) 1 + jωRC + (jω)2 LC
This transfer function is plotted below for L = C = 1 and three different values of R.

20
Gain GdB (ω)

R=0.2
0 R=2
R=10
-20

-40
-2 -1 0 1 2
10 10 10 10 10

2
∠G(ω)

-2
10-2 10-1 100 101 102
ω

The impulse response h(t) is the inverse transform of this quantity.

Finally, note that it is possible to obtain a differential equation for the system from this transfer
function. Multiplying out gives

(R + jωL)X(ω) = (1 + jωRC + (jω)2 LC)Y (ω),

and inverting gives the second-order differential equation

d d d2
Rx(t) + L x(t) = y(t) + RC y(t) + LC 2 y(t).
dt dt dt
The same expression could have been obtained using first principles in the time domain.

6.6 General second-order response

RLC circuit (lowpass, bandpass, highpass).

6-9
6.7 Sampling
All the signals considered up to this point are analog signals such as x(t), expressed as a function
of a time variable t that can take on any real value. While such continuous-time signals are fine
up to a point, they are quite limited. Most importantly, they don’t really work in the modern
world, where almost all processing is done using digital systems.
We can sample a continuous-time signal at regular intervals to obtain a discrete-time signal:

x[n] = x(t)|t=nT = x(nT ).

Here n is an integer index, so the spacing of the samples is T seconds. The square-bracket notation
[·] is often used instead of (·) to emphasise the fact that the signal is only defined for integer n. It
is not permitted, for example, to ask for the value of x[3/2] — it is not defined.
Let p(t) = ∞
P
n=−∞ δ(t − kT ) be an impulse train with spacing T . For a given continuous-time
signal x(t) we can form the sampled signal
∞ ∞ ∞
!
X X X
xs (t) = p(t)x(t) = δ(t − nT ) x(t) = x(nT )δ(t − nT ) = x[n]δ(t − nT ).
n=−∞ k=−∞ k=−∞

This signal xs (t) forms the link between the continuous-time domain and the discrete-time domain.
We can clearly obtain xs (t) from x(t), just by using the modulation operation P described, but we
can also construct xs (t) from the discrete sampled signal x[n] using xs (t) = ∞ k=−∞ x[n]δ(t − nT ).
In short, xs (t) contains only that information in x(t) that is also present x[n]. The picture in the
time domain is as follows:

x(t)

t
0
(1) p(t)

t
0 T

xs (t)

t
0

The interesting fact is that x(t) can be recovered from xs (t) as long as x(t) is appropriately
bandlimited. Suppose B is the highest frequency present in x(t), so X(ω) = 0 for |ω| > B. An
impulse train in time corresponds to an impulse train in frequency, so
∞
X
P (ω) = ωs δ(ω − kωs )
k=−∞

2π
where the sampling frequency is ωs = T . Since multiplication in time is convolution in frequency
we can find Xs (ω):
∞ ∞
1 1 X X 1
Xs (ω) = P (ω) ∗ X(ω) = δ(ω − kωs ) ∗ X(ω) = X(ω − kωs ).
2π T T
k=−∞ k=−∞

The picture in the frequency domain is as follows:

6-10
 A X(ω)

ω
0 B
( 2π
T ) P (ω)

ω
0 ωs

A/T Xs (ω)
ω
0 B

The signal Xs (ω) can be seen to contain replicas of the signal X(ω) at integer multiples of the
sampling frequency ωs = 2πT . In the case drawn above we have ωs > 2B, and the replicas are not
overlapping. Evidently we can then recover X(ω) from Xs (ω) by passing it through a filter with
the following frequency response:

T
Hr (ω)
ωs ω
0 2

The transfer function of this filter can be expressed as Hr (ω) = T Pωs (ω).
Since X(ω) = Hr (ω)Xs (ω) in the frequency domain, we must have x(t) = hr (t) ∗ xs (t) in time.
We can therefore reconstruct x(t) from the sampled signal x[n] as follows: use the samples to form
the signal
∞
X
xs (t) = x[n]δ(t − nT ),
k=−∞

and then put this signal through an ideal lowpass filter with impulse response
 
−1 t
hr (t) = F {Hr (ω)} = sinc
T

to recover the original x(t). Equivalently we can write the reconstruction formula as
∞ ∞  
X X 1
x(t) = xs (t) ∗ hr (t) = x[n]hr (t − nT ) = x[n]sinc (t − nT ) .
T
k=−∞ k=−∞

The reconstruction process works as long as the spectral replicas in Xs (ω) are not overlapping.
Note, though, that these replicas occur at spacings ωs = 2π T . Thus if we choose T such that
ωs ≥ 2B is satisfied then the signal x(t) can be reconstructed from x[n]. The signal must be
sampled at twice its highest frequency for this to be possible. This is called the Nyquist sampling
criterion, and ωs = 2B is called the Nyquist sampling rate.
In the case shown above the sampling frequency ωs is bigger than it needs to be for perfect recon-
struction to be possible, since the replicas are far from overlapping. Equivalently, the sampling
period T is smaller than it has to be — we have more samples than are required. The signal is
said to be over-sampled, and this is a good thing.
If we choose ωs = 2B then the signal x(t) is critically sampled. The frequency plots then look like
this:

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 A X(ω)

ω
0 B
( 2π
T ) P (ω)

ω
0 ωs

A/T Xs (ω)
ω
0 B

The signal x(t) can just be reconstructed from the samples x[n], although to do this we do need
a perfect reconstruction filter.
If we sample at a spacing larger than the spacing for this critical frequency — in other words,
if we take too few samples — then the spectral replicas will overlap and we say that aliasing
has occurred. It is important to note that the signal Xs (ω) contains the sum of all the shifted
replicas: if they overlap then it is impossible to recover each individual replica, in the same way
that it is impossible to determine x and y given that x + y = 10. In this case the signal has been
destroyed by the discretisation process: there is no way to recover x(t) from x[n]. The signal has
been under-sampled, aliasing has occurred, and x(t) has been lost.

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