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