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Module-4
Ideal Characteristics of filters
Objective: To understand the magnitude response characteristics of ideal filters and concept of
causality and physical reliazability.
Introduction:
The simplest ideal filters aim at retaining a portion of spectrum of the input in some pre-defined
region of the frequency axis and removing the rest .
Description:
Ideal filter characteristics:
A filter is a frequency selective network. It allows transmission of signals of certain
frequencies with no attenuation or with very little attenuation, and it rejects or heavily attenuates
signals of all other frequencies.
An ideal filter has very sharp cutoff characteristics, and it passes signals of certain
specified band of frequencies exactly and totally rejects signals of frequencies outside this band.
Its phase spectrum is linear.
Filters are usually classified according to their frequency response characteristics as low-
pass filter (LPF), high- pass filter (HPF), band-pass filter (BPF) and band-elimination or band
stop or band reject filter (BEF, BSF, BRF). Ideal version of these filters are described below and
their magnitude response are shown in Figure
Ideal LPF
An ideal low-pass filter transmits, without any distortion, all of the signals of frequencies
below a certain frequency c radians per second. The signals of frequencies above
cradians/second are completely attenuated. c is called thecutoff frequency. The corresponding
phase function for distortion less transmission is -td.
the transfer function of an ideal LPF is given by
H() = 1, ˂c
0, >c
The frequency response characteristics of an ideal LPF are shown in figure (a). It is a gate
function.
Ideal HPF
An ideal high-pass filter transmits, without any distortion, all of the signals of frequencies
above a certain frequency c radians per second and attenuates completely the signals of
frequencies below c radians per second, where c is called the cutoff frequency.
The corresponding phase function for distortion less transmission is -td.
the transfer function of an ideal LPF is given by
H() = 0, ˂c 1, >c
The frequency response characteristics of an ideal HPF are shown in figure (b).
Ideal BPF
An ideal band-pass filter transmits, without any distortion, all of the signals of
frequencies within a certain frequency band 2-1 radians per second and attenuates completely
the signals of frequencies outside this band. (2-1) is the bandwidth of the band-pass filter. The
corresponding phase function for distortion less transmission is -td.
An ideal BPF is given by
H() = 1, 1˂˂2
0, ˂1and >2
The frequency response characteristics of an ideal BPF are shown in figure (c).
Ideal BRF
An ideal band-rejection filter rejects totally all of the signals of frequencies within a certain
frequency band 2-1 radians per second and transmits without any distortion all signals of
frequencies outside this band. (2-1) is the rejection band. The corresponding phase function
for distortion less transmission is -td.
An ideal BRF is given by
H() = 0, 1˂˂2
1, ˂1and >2
The frequency response characteristics of an ideal BRF are shown in figure (d).
In addition to these filters, there is one more filter called an all pass filter.
An all pass filter transmits signals of all frequencies without any distortion, that is, its bandwidth
is ∞ as shown in Figure (e).
An ideal all pass filter is specified by
H()=1 (for all frequencies)
The corresponding phase function for distortion less transmission is -td.
All ideal filters are non-causal systems. Hence none of them is physically realizable.
Causality and Paley-Wiener Criterion For Physical Realization:
A system is said to be causal if it does not produce an output
before the input is applied .For an LTI system to be causal, the condition to be satisfied is its
impulse response must be zero for t less than zero, i.e.
h(t)=0 for t<0
Physical realizability implies that it is physically possible to
construct that system in real time. A physically realizable system cannot have a response before
the input is applied. This is known as causality condition. It means the unit impulse response h(t)
of a physically realizable system must be causal. This is the time domain criterion of physical
realizability. In the frequency domain, this criterion implies that a necessary and sufficient
condition for a magnitude function H(ω) to be physically realizable is:
ln 𝐻 𝜔)
1+𝜔2𝑑𝜔
∞
−∞<∞
The magnitude function H(ω)must, however, be square-
integrable before the paley-wiener criterion valid, that is,
𝐻 𝜔)2𝑑𝜔∞
−∞<∞
A system whose magnitude function violets the paley-wiener
creation has non-causal impulse response, the response exists prior to the application of the
driving function.
The following conclusions can be drawn from the paley- wiener criterion:
1. The magnitude function 𝐻 𝜔) may be zero at some discrete frequencies, but it
cannot be zero over a finite band of frequencies since this will cause the integral in
the equation of paley-wiener creation to become infinite. That means ideal filters
are not physically realizable.
2. The magnitude function 𝐻 𝜔) cannot fall off to zero faster than a function of
exponential order. It implies, a realizable magnitude characteristic cannot have too
great to total attenuation.
Relationship between bandwidth and rise time:
Risetime is an easily measured parameter that provides considerable insight into the
potential pitfalls in performing a measurement or designing a circuit. Risetime is defined as the
time it takes for a signal to rise (or fall for falltime) from 10% to 90% of its final value.
We know that the transfer function of an ideal LPF is given by
H()=H()=𝑒−𝑗𝑡𝑑
where H() = 1, <c
0, >c
c is called the cutoff frequency.
H() = e-jt
-c≤≤c , i.e., <c
= 0 >c The impulse response h(t) of the LPF is obtained by taking the inverse Fourier transform of the
transfer function H().
h(t) = F-1
[H()] = 1
2𝜋 𝑒−𝑗𝑡𝑑
𝑐
−𝑐𝑒−𝑗𝑡 d
= 1
2𝜋 𝑒−𝑗(𝑡−𝑡𝑑 )𝑐
−𝑐 d
= 1
2𝜋
e−j(t−td )
j(t−td ) −c
c
=1
2𝜋
ejc (t−td )−e−jc (t−td )
j(t−td )
= 1
𝜋(t−td ) sin𝑐(𝑡 − 𝑡𝑑)
= 𝑐
𝜋
sin 𝑐(𝑡−𝑡𝑑 )
𝑐(𝑡−𝑡𝑑 )
The impulse response of the ideal LPF is shown in below figure. The impulse response has a
peak value at t=td. This value c/𝜋 is proportional to cutoff frequency c. The width of the main
lobe is 2𝜋/c. As c⟶∞, the LPF becomes an all pass filter. As td⟶0, the output response peak
⟶∞, that is, the output response approaches input.
Further, the impulse response h(t) is non-zero for t<0, even though the input δ(t) is applied
at t=0. That is, the impulse response begins before the input is applied. In real life, no system
exhibits such type of characteristics. Hence we can conclude that ideal LPF is physically not
realizable.
If the impulse response is known, the step response can be obtained by convolution.
The step response y(t) = h(t)*u(t) = ℎ 𝜏)𝑑𝜏𝑡
−∞
We have h(t) = c
π
sin c (t−td )
c (t−td )
y(t) = c
π
sin c (τ−td )
c (τ−td ) dτ
𝑡
−∞
Let x= c(τ − td)
dx =cdτ or dτ = dx
c
therefore y(t) = c
π
sinx
x
dx
c
c (t−td )
−∞ =
1
π
sinx
x dx
c (t−td )
−∞
=1
𝜋 𝑆𝑖(𝑥) −∞
c (t−td )
Where Si is the sine integral function.
The properties of sine integral functions are :
1. Si(x) is an odd function, that is Si(-x) = -Si(x)
2. Si(0) =0
3. Si(∞) = 𝜋
2 and Si(-∞) = - (
𝜋
2)
A sketch of Si(x) is shown figure (a).
The step response can be expressed as:
Y(t) =1
𝜋Si[ωc(t-td)]-Si(-∞)
=1
𝜋Si[ωc(t-td)]-
𝜋
2
=1
2+
1
𝜋Si[ωc(t-td)]
If ωc→∞, then the response is:
Y(t) = 1
2+
1
𝜋Si(∞) =
1
2+
1
𝜋(𝜋
2) = 1
If ωc→- ∞, then the response is:
Y(t) = 1
2+
1
𝜋Si(−∞) =
1
2+
1
𝜋(−
𝜋
2) = 0
The step response of LPF is shown in figure (b).
Figure (a) Figure (b)
From figure(b), we can observe that y(t) approaches a delayed unit step u(t-td). But the abrupt
rise of input corresponds to more gradual rise of the output.
The rise time tr is defined as the time required for the response to reach from 0% to 100% of the
final value. To find it, draw a tangent at t= td with the line y(t) = 0 and y(t) =1. From figure (b),
we have
𝑑𝑦 (𝑡)
𝑑𝑡 𝑡=𝑡𝑑
= 1
𝑡𝑟 =
𝜔𝑐
𝜋
sin 𝜔𝑐(𝑡−𝑡𝑑 )
𝜔𝑐(𝑡−𝑡𝑑 ) 𝑡=𝑡𝑑
= 𝜔𝑐
𝜋
tr= 𝜋
𝜔𝑐
For a low-pass filter
Cut off frequency = Bandwidth
So the rise time is inversely proportional to the bandwidth.
Bandwidth * Rise time = constant
Worked out Problems:
Example 1: Find the output voltage of the RC low pass filter shown in bellow figure, for an
input voltage of te-t/RC
.
Solution:
The output of the circuit can be obtained very easily by Laplace transforming the
given network as shown in below figure.
The transfer function of the circuit is given by
H(s) = 𝑌(𝑠)
𝑋(𝑠)=
1/Cs
R+(1
Cs) =
1
1+RCs
Then output Y(s) = 1
1+RCsX(s) =
1
RC[
1
s+(1Cs
)]X(s)
Given Input voltage x(t)= te-t/RC
X(s)= l[te-t/RC
] = 1
s+ 1Cs
2
Y(s) = 1
𝑅𝐶[
1
s+(1Cs
)]
1
s+ 1Cs
2=
1
𝑅𝐶
1
s+ 1Cs
3
Taking inverse Laplace transform on both sides, we get
y(t) = L-1
1
𝑅𝐶[
1
s+ 1Cs
3]=
1
RCe
-t/RCt2
2 =
t2
2RCe
-t/RC u(t)
Output voltage y(t) = t2
2RCe
-t/RC u(t)
Example 2: A system produces an output of y(t) = e-t𝑢(t) for an input of x(t) = e
-2t𝑢(t). Determine the impulse response and frequency response of the system.
Solution: Given Output y(t) = e-t𝑢(t)
Y(s) = L[e-t𝑢(t)] =
1
s+1
Input x(t) = e-2t𝑢(t)
X(s) = L[e-2t𝑢(t) ] =1
s+2
The transfer function of the system
H(s) = Y(s)
X(s) =
1
s+11
s+2
=s+2
s+1
H(s) = s+2
s+1 =
(s+1)+1
s+1 = 1 +
1
s+1
The impulse response of the system
h(t) = L-1
[H(s)] = L-1
(1 + 1
s+1 ) = δ(t) + e
-t𝑢(t)
the frequency response of the system
H(ω) = H(s)s=jω = s+2
s+1s=jω =
2+jω
1+jω
Magnitude response
H(ω) = 22+𝜔2
1+𝜔2
Phase response
𝐻(𝜔) = tan−1 𝜔
2 -tan−1 𝜔
Example 3: Determine the maximum bandwidth of signals that can be transmitted through the
low-pass RC filter shown in figure, if over this bandwidth, the gain variation is to be within 10%
and the phase variation is to be within 7% of the ideal charecteristics.
Solution:
Taking 20 kΩ = R and nF = C, the given RC network transformed into frequency-
domain can be represented as shown in figure (a).Combining the parallel R and C at the output
side, the equivalent circuit is shown in figure (b)
From the basic circuit theory, the input/output relationship [transfer function ] of the given
circuit is:
𝑌(𝜔)
𝑋(𝜔) = H(ω) =
𝑅/(1+𝑗𝜔𝑅𝐶 )
𝑅+[𝑅
1+𝑗𝜔𝑅𝐶]
=R
2R+jωR2C =
1
2+jωRC
But, given R=20kΩ and C=10nF, we get
H(ω)= 1
2+jω 20∗103)(10∗10−9) =
1
2+jω(2∗10−4)
= 1
2[1+jω 10−4)] =
104
2[jω+ 10−4)] =
5000
jω+ 10−4)
H(ω) = 5000
jω+10000
Magnitude response 𝐻(𝜔) = 5000
𝜔2+10000 2 =
5000
𝜔2+108
Phase response θ(ω) = 𝐻 𝜔) = tan−1(𝜔
10000)
At ω = 0
𝐻(𝜔) ω=0 = 5000
10000 = 0.5
But there is 10% variation in gain over the bandwidth B.
𝐻(𝜔) = 0.5-0.5*10% = 0.45
But H(ω) = 5000
𝐵2+108
B2 +10
8 = (
5000
0.45)2 = 123.46*10
6
B2=123.46*10
6-100*10
6 = 23.46*10
6
B=4.84*103 = 4.84 kHz
But B=2𝜋𝑓
𝑓 =𝐵
2𝜋 =
4.84∗103
2𝜋 = 770.8 Hz
Phase at frequency 𝑓 = 770.8 Hz is:
Θ(ω) = -tan−1(𝜔
10000) = -tan−1(
4.84∗103
10∗103 )
= - 25.83 (7% of ideal value)
Example 4: Consider a stable LTI system characterized by the differential equation dy (t)
dt+2y(t)
=x(t). Find its impulse response.
Solution: The given differential equation is:
dy (t)
dt+2y(t) =x(t)
Taking Laplace transform on both sides, we have
sY(s)-y(0)+2Y(s) =X(s)
Neglecting initial conditions, we have
sY(s)+2Y(s) =X(s) i.e. Y(s)(s+2) =X(s)
The transfer function of the system
H() = Y(s)
X(s) =
1
s+2
The impulse response of the system
h(t) = L-1
[H(s)] = L-1
1
s+2 = e
-2t u(t)
Example 5: The input voltage to an RC circuit is given as x(t)= te-3t
u(t), and the impulse
response of this circuit is given as 2e-4t
u(t). Determine the output y(t).
Solution: Input x(t)= te
-3tu(t)
X(s)=1/(s+3)2
Impulse response h(t)=2e-4t
u(t)
H(s)=2/(s+4)
We know that Output y(t)=x(t)*h(t)
Y(s)=X(s)H(s)
Output y(t)=L-1
[X(s) H(s)]
Y(s)=X(s)H(s)= [1/(s+3)2][2/(s+4)] =
2
s+3)2(s+4)
Taking partial fractions, we have
Y(s)=2
s+3)2(s+4) =
A
s+3)2+
B
s+3+
C
(s+4)
Where A=(s+3)2Y(s)s = -3 =
2
(s+4)s=-3 = 2
B=d[ s+3)2Y s)]
dss=-3 =
d
ds
2
(s+4) s=-3 =
2
(s+4)²s=-3 = -2
C=(s+4)Y(s)s =-4 = 2
(s+3)²s=-4= 2
Y(s)= 2
s+3)2−
2
s+3+
2
(s+4)
Taking inverse Laplace transform on both sides, we have the output
y(t) = L-1
[Y(s)] = L-1
2
s+3)2 − L-1 2
s+3 +L
-1 2
s+4
∴ y(t)= 2te-3t
u(t)-2e-3t
u(t)+2e-4t
u(t)
Example 6: Consider a causal LTI system with frequency response
H(ω) = 1
4+jω
For a particular input x(t), the system is observed to produce the output
y(t) = 𝑒−2𝑡u(t) - 𝑒−4𝑡u(t)
Find the input x(t).
Solution: Given frequency response H(ω) = 1
4+jω
And output y(t) = 𝑒−2𝑡u(t) - 𝑒−4𝑡u(t)
Y(ω) =F[y(t)] = F[𝑒−2𝑡u(t)] - F[𝑒−4𝑡u(t)]
= 1
2+jω−
1
4+jω =
2
(2+jω)(4+jω)
Input X(ω) = Y(ω)
H(ω) =
2
(2+jω)(4+jω)1
4+jω
= 2
2+jω
Taking inverse Fourier transform, we have
Input x(t) = 𝐹−1[X1(ω)] = 𝐹−1 2
2+jω = 2𝑒−2𝑡u(t)
Example 7: The impulse response of a continuous-time system is expressed as:
h(t) = 1
RCe
-t/RCu(t)
find the frequency response of the system.
Solution: Given impulse response of the system
h(t) = 1
RCe
-t/RCu(t)
The frequency response of the system H() is the Fourier transform of the impulse response of
the system
H() =F[h(t)] = F 1
RCe−
1
RC t u(t)
= 1
RC
1
j+(1/RC ) =
1
1+jRC
Magnitude response H() = 1
1+(RC )²
Phase response H() = - tan-1RC.
Example 8. For the system shown below:
f(t) y(t)
f(t) = e−at , t≥0.a≥1
0 , otherwise
Filter
Y() = 1
a+j, -a<<a
0 , otherwise
Find the transfer function and impulse response of the system.
Solution:
f(t)= e-at
for t≥0
f(t)= e-at
u(t)
∴ F() = F[e-at
u(t)] = 1
j+a
Y() =1
j+a
The transfer function of the system
H() =Y()
F()
1/(j+a)
1/(j+a)=1
The impulse response of the system
h(t)= F-1
[H()] = F-1
[1] =δ(t)
Example 9: Find the impulse response for the RL filter shown in fig below. Also find the
transfer function. What would be its frequency response?
Solution: The transfer function of the RL filter can be easily determined by transforming the
network into frequency domain as shown in below figure.
From the Fourier transformed network of above figure , the transfer function is
H(ω) =Y(ω)
X(ω) = X(ω) =
jωL
R+jωL =
jω
jω+(R/L)
The impulse response of the system is:
h(t) = 𝐹−1[H(ω)] = 𝐹−1 jω
𝑗ω+(RL
)
Magnitude response 𝐻(𝜔) =𝜔
𝜔2+(𝑅/𝐿)2
Phase response 𝐻(𝜔 ) = 𝜋
2 - tan−1 𝐿
𝑅ω
Example 10:Find the Magnitude response of the system whose impulse response is given by
ℎ(𝑡) = 𝑒−2𝑡𝑢(𝑡)
Solution:
Given impulse response ℎ(𝑡) = 𝑒−2𝑡𝑢(𝑡)
Applying Fourier Transform 𝐻(𝜔) =1
2+𝑗 𝜔
Magnitude response is 𝐻(𝜔) =1
4+ 𝜔2
AssignmentProblems
1. The impulse response of a continuous-time system is expressed as: h(t)= e-2t
u(t). Find the
frequency response of the system. Plot the frequency response.
2. Determine the maximum bandwidth of signals that can be transmitted through the low
pass RC filter shown in below figure, if over this bandwidth, the gain variation is to be
within 8% and the phase variation is to be within 8% of the ideal characteristics.
3. There are several possible ways of estimating an essential bandwidth of non-band limited
signal. For a low-pass signal, for example, the essential bandwidth may be chosen as a
frequency where the amplitude spectrum of the signal decays to K% of its peak value.
The choice of K depends on the nature of application. Choosing K=10, determine the
essential bandwidth of g(t) = e-2at
u(t).
4. The input voltage to an RC circuit is given as x(t) = te-3t
u(t), and the impulse response of
this circuit is given as 2e-3t
u(t). Determine the output y(t).
5. Consider a causal LTI system with frequency response H() = 1/(3+j). for a particular
input x(t), the system is observed to produce the output y(t) = e-t
u(t) – e-3t
u(t). find the
input x(t).
6. The transfer function of a system is given by H() = K, where K is a constant. Sketch the
magnitude and phase function of this transfer function, Evaluate the impulse response of
this filter. Sketch this response and state whether the filter is physically realizable.
7. For a system excited by x(t) = e-3t
u(t), the impulse response is h(t) = e-2t
u(t)+e-2t
u(-t).
Find the output for the system.
8. Determine the maximum bandwidth of signals that can be transmitted through the low
pass RC filter shown in below figure, if over this bandwidth, the gain variation is to be
within 5% and the phase variation is to be within 5% of the ideal characteristics.
9. There are several possible ways of estimating an essential bandwidth of non-band limited
signal. For a low-pass signal, for example, the essential bandwidth may be chosen as a
frequency where the amplitude spectrum of the signal decays to K% of its peak value. The
choice of K depends on the nature of application. Choosing K=8, determine the essential
bandwidth of g(t) = e-5at
u(t).
10. Consider a causal LTI system with frequency response H() = 1/(8+j). for a particular
input x(t), the system is observed to produce the output y(t) = e-4t
u(t) – e-5t
u(t). Find the
input x(t).
Simulation:
% Given system
%y(n)=-(1/3)y(n-1)+(1/3)y(n-2)+x(n)+(1/4)x(n-1)
% To verify stability and physical realizability of the System
num = input (' type the numerator vector '); den = input (' type the denominator vector '); [z,p,k] = tf2zp(num,den); disp ('Gain constant is '); disp(k); disp (' Zeros are at '); disp(z) disp ('radius of Zeros ') ; radzero = abs(z) disp ('Poles are at '); disp(p) disp ('radius of Poles ') ; radpole = abs(p) if max(radpole) >= 1 disp (' ALL the POLES do not lie within the Unit Circle '); disp (' The given LTI system is NOT a stable system '); else disp (' ALL the POLES lie WITHIN the Unit Circle '); disp (' The given LTI system is a REALIZABLE and STABLE system
'); end; zplane(num,den) title ( ' Pole-Zero plot of the LTI system ' );
INPUT
type the numerator vector [1 1/4]
type the denominator vector [1 1/3 -1/3]
Gain constant is
1
Zeros are at
-0.2500
radius of Zeros
radzero =
0.2500
Poles are at
-0.7676
0.4343
radius of Poles
radpole =
0.7676
0.4343
ALL the POLES lie WITHIN the Unit Circle
The given LTI system is a REALIZABLE and STABLE system
References:
[1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, “Signals & Systems”, Second
edition, Pearson Education, 8th
Indian Reprint, 2005.
[2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”,
Second edition,McGraw-Hill Education,2011
[3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals
and Systems using MATLAB”,Prentice Hall Signal Processing Series
[4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008
[5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011
[6] A.Anand Kumar, “Signals and Systems” , PHI Learning Private Limited ,2011