DT287/3

Transmission line questions

Q1
Measurements on a 50 Ω transmission line produced a maximum voltage Vmax = 10
mV and a minimum voltage Vmin 2 mV at an operating frequency f = 1 GHZ. The first
voltage minimum from the load was measured at 25.5 mm. If εr = 2, determine
(i)The VSWR,
(ii)The wavelength,
(iii)The magnitude of the reflection coefficient , and
(iv)The distance in wavelengths between the load and the first minimum.
[8 marks]
It is required to match the load to the line using a short-circuited single stub having
the same characteristics as the line. Determine the location and length of the stub for
correct matching.
[12 marks]
Q2
(a)A 10 km length of screened telephone cable, operating at a frequency of 10 kHz,
has the following, primary transmission line parameters:
L = 700 mH per km,
C = 0.05 µF per km,
R = 28 Ω per km, and
G = 1 µS per km
Determine the phase and attenuation constants for this cable and hence calculate
the characteristic impedance.
[8 marks]
(b) An expression for the input impedance of a transmission line, terminated in a
resistance ZR, = 35 Ω is:
 ZR 
 tanβ l + jZ o 
Z in = Z o  
 Z o + jZ 
 tanβ l R 
 
Where β is the phase change coefficient and Zo is the characteristic impedance
equal to 75 Ω. Determine the input impedance of a section of line whose length is
λ/4 metres.
[7 marks]
(c) Describe one graphical technique for matching a transmission line to a load,
whose value is not equal to the characteristic impedance . Illustrate your answer
using a Smith chart.
[10 marks]
Q3
(a)A 1 km length of screened telephone cable, operating at a frequency of 10 kHz,
has the following parameters:

L = 700 mH per km,

C = 0.05 µF per km,

1
 R = 28 Ω per km, and
G = 1 µS per km
Determine the characteristic impedance, phase and attenuation constants for this
cable.
(b)A transmission cable is terminated by an impedance which has a value twice the
characteristic impedance value. Calculate the position and length of a short-circuited
stub required to achieve matched conditions on the line. The stub is constructed from
the same cable type as the line and the frequency of operation is 1 MHz. The velocity
of propagation is 0.67c, where c is the velocity of light.
[ 8 marks]
If this cable is terminated by an impedance ZL = 123.69- j36.01 Ω, calculate the
position and length of a short-circuited stub required to achieve matched conditions
on the line. The stub is constructed from the same cable type as the line.
Q4
(a) Define the term voltage reflection coefficient. Use the solution for the voltage
along a transmission line V = V1 e -(α + jβ) z + V2 e(α + jβ) z in your answer.
[4 marks]
(b)A transmission line with a 75 Ω characteristic impedance Zo, is terminated in an
impedance ZL = 40 + j20 Ω. Calculate
(i) The reflection coefficient,
(ii) The voltage standing ratio, and
(iii) The input impedance.
[6 marks]
(c) Hence use a Smith chart to verify approximately, the three values calculated in (i)
to (iii).
[12 marks]
Q5
(a) Explain the transmission line terms: Voltage standing wave ratio and voltage
reflection coefficient.
[6marks]
(b) A twin-lead transmission line whose characteristic impedance Zo = 300 + j0 Ω
has an aerial of impedance 225 - j175 Ω connected as a load. Matching by means of a
single stub connected a distanced metres from the load is used. Estimate the length in
metres of the stub and the distance d if the operating frequency f = 500 MHz assume
that the stub is formed from a section of the same air-spaced transmission line.
[14 marks]
Solutions
a) Definitions
b) The procedure for using a single stub as a matching device is:
i) Normalise the aerial impedance as ZL = 0.75 - j0.583 Ω

ii) A circle of constant VSWR is then constructed through the point L.

iii) Since we are connecting a short circuit stub in parallel with the line, then it makes
sense to work with admittance. The admittance of the load is

2
 Figure 1

iv) Going around the constant VSWR circle to the point where the intersection with
the constant conductance circle of 1(i.e. through the circle G/Y = 1.0 + j0.72).

v) A stub is attached whose admittance ys = -j0.72s will cancel out the admittance
j0.72 s. The points M and B. which are read from the chart as M = 0.131 λ towards
the generator and B = 0.153 λ. The difference between these two measurements is d
= (0.153 - 0.131) λ = 0.022 λ metres towards the generator. The relationship v = λ/f
gives us a way of calculating the distance d in metres i.e. λ = v/f = 300.108/500.108 =
0.6m or 60 cm so that the distance d = 0.022 x 60 cm = 1.32 cm

vi) The length of the stub is found from the distance = (0.25-0.099) λ = 0.151 λ
so that the length of the stub is = 0.151 x 60 cm = 9.06 cm
[5 marks]
Q6
A transmission line whose characteristic impedance Zo is (75 + j 0) Ω, has a velocity
factor of 0.6 and negligible attenuation. Determine, using a Smith chart, the VSWR on
the line given that the transmission line is terminated in a impedance ZR = 187.5 +
j187.5 Ω and the operating frequency is f = 90 MHz,
[4 marks]
On the chart, determine the length and location of a single stub which will achieve
correct matching. You may assume the stub is made from a line with the same
transmission characteristics.
[16 marks]
Determine, with the aid of a Smith chart, the characteristic impedance of a quarter
wave transmission line to achieve correct matching. You may assume the transformer
is inserted at a point nearest the load.
[8 marks]
Q7
a) A load Z = 100 - j50 Ω is connected to a transmission line whose characteristic
impedance is 75 Ω. Determine, with the aid of a Smith chart, the characteristic

3
 impedance of a quarter wave transmission line to achieve correct matching. You
may assume the transformer is inserted at a point nearest the load.
[8 marks]
b) A transmission line with characteristic impedance Z = 300 + j0 Ω, is connected to
aerial of impedance ZL = 225 - j175 Ω A single -stub is connected a distance d
metres from the load. If the frequency of operation is 500 MHz, estimate the
length of stub required for correct matching. (You may assume the stub is formed
from the same section of transmission line and the velocity of propagation is
2.01x108 m/s).
[12 marks]
Q8
A transmission line whose characteristic impedance is 75 + j0 Ω, a velocity factor of
0.6 c (Where c is the velocity of light) and negligible attenuation is used to connect a
transmitter to a load .The operating frequency of the transmitter is 90 MHz and the
impedance of the load is 187.5 + j187.5 Ω. Determine using a Smith chart
a) The VSWR on the line,
[2marks]
b) the physical length and position of a single short-circuited stub which will achieve
correct matching. Two solutions are possible. State with reasons the preferred
solution.
[12marks]
(c) If the transmitter frequency is reduced by 10 %, determine using a separate Smith
chart, the VSWR for the solution in b)
[6marks]
Solution
The line wavelength is calculated as :
wavelength λ = 2 m
The normalised impedance is Z/Z i.e. = 2.5 + j 2.5. This is located at point A on the
chart. The normalised admittance is 0.2 + j 0.2 and is at point B.

a) Position of stub relative to the load :

d (l) = 0.0033 + 0.184 = 0.217l

so that the length is 0.217x2 metres = 0.434 m

The normalised input of the line at the stub position is

y = 1 + j1.85

The normalised susceptance of the stub is s1. The required length of the stub s1
d (l ) = (0.033 + 0.316)l= 0.349 x 2 metres

The normalised input admittance of the line at the stub position :

The first solution is preferable since the stub is shorter and is positioned closer to the
load. For a frequency change of - 10 % f = 0.9 f = 81 MHz so that the normalised
impedance becomes 2.5 + j 2.25 see point A chart 2 and y ' = 0.23 + j 0.2 see point B
chart 2. Electrical length of the line between the load and the stub is :

4
 d ' = 0.217 x 0.9 = 0.195λ
Normalised input admittance of the line at the stub position :

y '= 0.675 + j 1.36 see point C chart 2

Electrical length of the stub for the changed frequency is

l ' = 0.785 x 0.9 = 0.07λ

Normalised susceptance of the line :

y ' = -j 2.15 see point F on chart 2

Normalised admittance of line and stub at stub position

y ' = y ' + y ' = 0.675 + j 1.36 - j 2.15

The resulting VSWR on the line is given by the intersection at E

VSWR = 2.8

The VSWR of the unmatched line is 5

Q9

Explain by means of a Smith chart the single stub method of matching a transmission
line to a load.
[10 marks]
A 15 km length of loss-free transmission line is terminated by a 400 Ω resistor. A
leakage fault develops across the transmission line at a point p km from the sending
end. To locate the fault the method of TDR is used. A record of the transmitted pulse
(sent at t = 0 s ) and the first two reflections is shown in Fig.1. Calculate
a) The inductance and the capacitance of the line per unit length,
b) The position of the fault and the magnitude of its conductance.

[10 marks]

Explanation of single stub method

[6 marks]

5
 In a loss-free line the velocity of propagation is 3 x 10 m/s so in 20 us a pulse travels
a distance of 6 km. Hence the fault is 3 km down the line and the -3 V is the reflected
voltage from the fault point.
Solution
Consider the Bewley diagram below.

At the fault let the total impedance Z be the parallel combination of the fault
resistance R and the characteristic impedance . We can then write
RZ 0
Z=
R + Zo
The reflection coefficient is defined as the ratio of the incident voltage to the
reflected voltage. So we can then write
RZ 0
=
R + Zo
The transmitted voltage beyond P is calculated by considering the fundamental
relationship for the total voltage V at a point on line where due to a mismatch there is
a reflection so But the reflection coefficient is given as where Z is the terminating
impedance at the point in question. Substituting c) into b) yields The transmitted
beyond P is This wave travels to Q where it causes an unknown reflection x volts
back towards the source. and beyond P we can write and substituting from a) solving
for R or a conductance of For a loss-free line we can write The velocity of
propagation V\2p From these two equations we can write for C as and hence the
inductance L is calculated as L = 379,456 x 5.41 x 10 = 2.051 uH per metre

Q10
(a) Standing waves were observed on a 50 Ω transmission line with a maximum
voltage Vmax = 10 mV. The minimum voltage Vmin = 2 mV was measured 50.8
mm from the load. If the wavelength for the operating frequency is 212 mm,
determine:
(i) The VSWR,
(ii) The magnitude of the reflection coefficient ρ,
(iii)The distance in wavelengths between the load and the first minimum, and

6
 (iv)The mismatch loss in dB.
[8 marks]
(b) Determine the location and the length of a single stub (same characteristics as
the mismatched line) for correct matching.
[10 marks]
(c) Describe the technique of TDR for measuring faults on transmission lines.
[7 marks]
Q11
(a) Define the following transmission line terms: Voltage standing wave ratio and
voltage reflection coefficient.
[5 marks]
(b) A transmission line, whose characteristic impedance Zo = 300 + j0 Ω, has an
antenna of impedance 225 - j175 Ω connected as a load. Matching by means of
a single stub connected, at a distance d metres from the load, is used. Estimate
the length in metres of the stub and the distance d if the operating frequency f =
500 MHz. Assume that the stub is formed from a section of the same air-spaced
transmission line.
[12 marks]
(c)Calculate the characteristic impedance, phase coefficient and attenuation constant
for a 10 km length of telephone cable, operating at a frequency of 10 kHz, if the cable
has the following, primary transmission line parameters:
L = 700 mH per km, C = 0.05 µF per km, R = 28 Ω per km, G = 1 µS per km
[8 marks]
Q12
A 50 Ω transmission line is connected to an antennae with Zant = 100 – j60 Ω.
Determine the following for a single stub tuner:
a) Shortest distance to the stub (from the antennae load),
b) Value and type of reactance needed to tune(resonant) the line, and
c) Length of short-circuited stub.

Solution
a) 0.126λ,
b) Inductive reactance = j44.25Ω, and
0.115λ
Q13
A load of Zant = 100 – j60 Ω is to be matched to a 50 Ω line with a quarter wave
transformer.
(a) Where should the transformer be located,
(b) Determine the characteristic impedance of the transformer, and
(c) If f = 1 GHz, determine the length of the transformer in metres if the
transmission line dielectric constant, is e = 2.25.S

Solution
0.21λ from the load
(b)29.6 Ω, and
(c)0.05λ

7
 Active Filters Questions
Q14

(a)
Obtain a transfer function for a high-pass filter, which uses a Butterworth loss
function, and which meets the following specification:
The maximum passband loss Amax = 3 dB
The minimum stopband loss Amin = 28 dB
The passband edge frequency ωp = 6000 rs-1
The stopband edge frequency ωs = 2000 rs-1
[10 marks]
(b)Show how the transfer function, for the second stage of the Sallen and Key,
high- pass filter shown in figure 4, is:

E3 s2
=
E2 2 1
s2 + s + 2
CR2 C R1 R2
[8 marks]
Hence calculate component values for this circuit, which could implement the
transfer function obtained in part (a).
R = 20 kΩ.
[7 marks]

Figure 4
Q15
(a) Discuss the effect, component tolerance, has on the performance of active filters.
[5 marks]
(b) The frequency spectrum of a 1 kHz squarewave is shown in figure 5. It is
desired to extract, from this squarewave, a 1 kHz sinusoidal signal. Obtain a transfer
function for a low-pass filter, which will produce a maximum attenuation of the
fundamental component of 1 dB. The third harmonic (3 kHz) should be attenuated by
12 dB (Butterworth loss functions tables are available for use in your analysis).
[12 marks]

8
 Figure 5
(c) Show how you could implement the transfer function obtained in part (a) using
a Sallen and Key VCVS active filter.
[8 marks]
Q16
(a) It is desired to produce a 1 kHz sinusoidal signal from a 1 kHz squarewave.
Obtain the transfer function for a low-pass filter that will produce a maximum
attenuation of the fundamental signal by 1 dB. The third harmonic should be
attenuated by 12 dB (Use a Butterworth loss function in the analysis).
[10 marks]
(b) Discuss the technique of frequency transformation applied to low-pass loss
functions in the design of high-pass filters. Illustrate the answer using a second-order
low-pass Butterworth loss function to produce a high-pass transfer function with a
desired passband edge frequency of 1 kHz and A max equal to 3 dB.
[10 marks]
Q17
(a)It is desired to produce a 1 kHz sinusoidal signal from a 1 kHz squarewave. Obtain
the transfer function for a low-pass filter, which will produce a maximum
attenuation of the fundamental signal of 1 dB. The third harmonic (3 kHz) should
be attenuated by 12 dB (Butterworth loss functions tables are available for use in
your analysis).
[10 marks]
(b) Identify the circuit shown in figure 6. State one possible application for this
circuit. Obtain the high-pass transfer function, assuming equal value
resistances.
[10 marks]

9
 Figure 6

Q18
1.(a)Identify the order of the active filter circuit in Figure 7. Hence obtain a
Butterworth approximation function, using the tables supplied, which will
have a passband edge frequency ωp = 1000 rs-1. The frequency correction
factor ε = 1.
[8 marks]
(b)Show that the transfer function for this circuit is:
 1 
E3  1   2
R C2C3 
=   
E1 1 + sC1 R   s 2 + s 2 + 1 
 C2 R R 2C2C3 
Hint: (Apply nodal analysis separately to each circuit isolated by the buffer
amplifier).
[10 marks]
c) Calculate suitable values for the three capacitors if the resistance R = 100 kΩ.
Sketch the form of the frequency response for the two outputs at node 6.
[7 marks]

10
 Figure 7
Q19
a)Indicate how the order of a filter may be obtained using the expression for A(ω) and
the following terms:
Maximum passband attenuation Amax, = 3 dB,
Minimum stopband attenuation Amin, = 28 dB,
Passband edge frequency ωp, = 1k rs-1, and
Stopband edge frequency ωs =10 k rs-1.
A ( ω ) = 10 log 10 [1 + ε 2 ( ω ω p ) 2 n
[7 marks]
(b)Hence obtain a transfer function to meet the above specification.
[10 marks]
(c) Discuss the method of frequency transformation used in designing bandpass
filters. Use the first order Butterworth approximation function A($) = $ + 1 to
illustrate your answer. Include in your discussion any relevant formulae.
[8 marks]
Q20
(a) Obtain a transfer function for a third-order high-pass filter, which uses a
Butterworth loss function, and which meets the following specification:
The maximum passband loss Amax = 3 dB
The minimum stopband loss Amin = 28 dB
The passband edge frequency ωp = 6000 rs-1
The stopband edge frequency ωs = 2000 rs-1
[10 marks]
(b) A second-order IGMF bandpass active filter is shown in Figure 8. Show, by
means of nodal analysis, how the transfer function is:
1
s
Vout R1C
=−
Vin 2 1
s2 + s + 2
CR2 C R1 R2
[8 marks]
Calculate a value for the centre frequency ωo, the -3 dB bandwidth and the
passband gain for the circuit values given.

11
 R1 = 1 kΩ, R2 = 100 kΩ, C =15 nF.
[7 marks]

Figure 8

Q22
(a) A low-pass filter is required to meet the filter specification:
The maximum passband loss Amax = 0.5 dB
The minimum stopband loss Amin = 12 dB
The passband edge frequency ωp = 100 rs-1
The stopband edge frequency ωs = 400 rs-1

Determine the filter order n using the relationship

 10 0.1 Amin − 1
log10  0.1 A 
 10 max − 1 
n= .
ωs
2 log10 ( )
ωp
Sketch the pole-zero diagram using the relationship between n and the angle θk,
between the poles
3600
θk =
2n
Hence obtain a transfer function, which meets the specification.
Note: Butterworth tables are not supplied.
[15 marks]
(b) Obtain circuit values for an equal component value Sallen and Key VCVS
circuit configuration that would meet the specification in part (a).
[10 marks]
Q23
(a) State one advantage and one disadvantage of using Chebychev polynomials in
approximation loss function analysis.
[5 marks]

12
 (b) A bandpass filter is required for a particular application, which will tolerate ripple
in the passband equal to 0.5 dB. The specification for the bandpass filter is:
ω bs1 = the lower stopband edge frequency = 688 rs-1
ω bs2 = the upper stopband edge frequency = 1930 rs-1
ω bp1 = the lower passband edge frequency = 970 rs-1
ω bp2 = the upper passband edge frequency = 1370 rs-1
The maximum passband attenuation Amax = 0.5 dB.
The minimum stopband attenuation Amin = 15 dB.
Obtain the bandpass transfer function, which will meet this specification. (Use the
available Chebychev tables).
[20 marks]
Q24
Draw the circuit diagram of a second-order bi-quad filter, which utilises three
operational amplifiers.
[4marks]
Obtain an expression for the bandpass voltage transfer function for this
configuration.
[10marks]
Such a transfer function is defined as
G(s) =

Select suitable resistive component values for this filter if equal value 100 nF
capacitors are to be used in the design. Obtain an expression for the bandpass voltage
transfer function for this configuration.
[10marks]
Q25

The loss-function for a low-pass Butterworth filter, expressed in dB is

ω 2n
A ( ω ) = 10 log 10 [1 + ε 2 ( ) ]
ω p
Using the above function, obtain expressions for the frequency correction factor and
the order in terms of Amax, Amin, the passband and stopband edge frequencies.
[8 marks]
Determine a value for ε and η if the specification for an active low-pass filter is:
Maximum passband attenuation Amax = 2 dB,
Minimum stopband attenuation Amin = 12 dB,
Passband edge frequency ωp = 1 krs-1,
Stopband edge frequency ωs = 2 krs-1.
[12 marks]
Q26
The bandpass circuit in Fig. 10 has a transfer function. Determine suitable
component values for C, R and R which will. The required bandpass frequency
transformation is: where B is the -3 dB bandwidth that can be expressed as

The normalised low pass stopband frequency is defined as

13
 Sketch a Sallen and Key equal-value component circuit-diagram that could be used to
implement the transfer function so derived (circuit values not required).
[4 marks]

Then the corresponding 3rd order normalised loss function is

A(s) = (s + 1)(s2 + s + 1)

1 1
H(s) = = )
A( s ) (s + 1)(s2 + s + 1)

To find the attenuation use the following

A(s) = 10 log [1 + 0.585.3]

A(s) = 10 log [1 + 0.585x729] = 26.3 dB

The angle is calculated as

Q27
Show that the transfer function H(s) for the circuit shown in Fig 11 is:
1
H(s) = 2
( s + 3s + 3)
Sketch the pole-zero plot. Hence evaluate the magnitude of the transfer function at a
frequency of 4 rs −1 .
[4 marks]
A unit impulse signal is applied to the second order active filter shown in Fig 11.
Obtain an expression for the output voltage and hence evaluate the output voltage
after an elapsed time of one second.
R1 = 0.5 Ω, R2 = 1 Ω, C1 = C2 = 1 F.
[6 marks]

Figure 11
To find the attenuation at ωs use the following

1 1 1
H ( s) = = = = −28 dB
A( s ) ω 6000 6
10 log[1 + ( s ) 2 n ] 10 log[1 + ( ) ]
ωp 2000

14
Q28
A state-variable filter is shown in Fig. 12. Obtain the bandpass transfer function
Vo1(s)/Vin(s). Assuming a value for R = 10 kΩ, calculate a value for C which will
set the centre frequency to a value f0 = 10 krs-1. Calculate suitable values for Ra and
Rb, which will produce a Q-factor of 50.
[14marks]
th
A 4 order bandpass is required whose Butterworth transfer function is:
Calculate suitable component values for two cascaded state-variable bandpass
circuits whose overall transfer function is
[14marks]
Solution
For the gain determining elements:

Q = 50 = 1/3(1 + R4/R3) or 150 = 1 + R4/R3 or R4/R3 = 149

i. e. R4 = 149 R3. Assume a value for R3 = 1 kΩ so that R4 = 149 kΩ Using these

values the transfer function is (s) = BW 31. 83 Hz Since the transfer function in
factored form indicates a BP with centre frequency = 1, then cascading two such
circuits as in the first section will suffice. The damping value for each section is
determined by the resistors R3, R4, R3' and R4'.
The circuit values can be calculated as follows: Since ωo = 1 and selecting a value for
C = 10 uF and R = 100 kΩ and comparing coefficients of the s terms gives then from
R3 = R4 i. e. 2. 92R3 = R4. Choose R3 = 1 kΩ 6 R4 = 2. 92 kΩ similarly for the
second stage or suitable values R4' = 623 Ω and R3' = 1 kΩ R4. So for each other.
The transfer function for the Bandpass output is

For ωo = 1/CR = 10 krs-1 6 C = 10 nF.

A state-variable filter, which uses three operational amplifiers is shown in Fig.13.

Obtain the low-pass transfer function H(s).
[10 marks]
for the gain determining elements as

Q = 50 = 1/3(1+R4/R3) i.e.
150 = 1 + R4/R3 or
R4/R3 = 149 i.e.R4 = 149 R3

Assume a value for R3 =1 kΩ so that R4 = 149 kΩ

Using these values the transfer function is

H(s) =
BW =31.83 Hz

Since the transfer function in factored form indicates a BP with centre frequency = 1,
then cascading two such circuits as in the first section will suffice. The damping of
each section will be determined by R3 and R4 and R3'and 4'.The circuit values can be
calculated as follows:

15
 Since ωo = 1 and selecting a value for C= 10 uF and R =100 kΩ and comparing
coefficients of the s terms gives
then from (1)
R3 = R4
i.e.2.92R3 = R4
Choose R3 = 1 kΩ 6 R4 = 2.92 kΩ
similarly for the second stage or suitable values R4' = 623 Ω and R3' = 1 kΩ R4 of
each other. The transfer function for the Bandpass output is For ωo =1/CR = 10 krs
=10 nF.
Q29
A low -pass Butterworth filter is required to meet the following specifications:
Maximum passband attenuation Amax = 2 dB,
Minimum stopband attenuation Amin = 12 dB,
passband edge frequency fp = 1 kHz, and
stopband edge frequency fs = 2 kHz.
Determine
(a) The order n of the filter,
(b) ε, the frequency scaling factor, and
(c) The transfer function.

Sketch the pole zero pattern for this transfer function.

[6marks]
Determine the transfer function for a high-pass filter which would have Amax = 2 dB,
Amin = 12 dB and a passband edge frequency = 1 kHz.(Table 1 gives the Butterworth
normalised loss functions)
[8 marks]
Q30
Discuss the technique of frequency transformation in obtaining a transfer function.
for a high-pass filter to meet desired hp specifications and using available low-pass
loss function tables.
[6marks]
The specifications desired for a high-pass filter are :
Maximum passband edge attenuation Amax = 2 dB
Minimum stopband edge attenuation Amin = 20dB
Passband edge frequency ωp = 3 krs-1
Stopband edge frequency ω s = 1.5 krs-1 By using an appropriate frequency transform
, obtain a Butterworth high-pass transfer function which will meet these
specifications.
[12 marks]
Q31
A low-pass filter transfer function is given as:
1
H ( s) =
s +1
Using frequency transformation techniques on this transfer function, demonstrate
how to obtain:

16
 i) A Chebychev high-pass filter with ωp =10 rs-1, and
ii)A Chebychev bandpass filter with ωo =1 krs-1 and -3 dB bandwidth = 100 rs -1.
[10 marks]
Q32
Sketch the circuit diagram of a second-order filter circuit that could be used in the
equalisation section of an audio mixing desk. Such a circuit would have low-pass,
high-pass and bandpass outputs. Indicate which components determine the frequency
characteristics of the filter. State briefly how you would change the configuration
from a Butterworth to a Chebychev filter.
[10 marks]
Q33
a) Indicate how an expression for the order of a filter may be obtained using the
following terms:
Maximum passband attenuation Amax,
Minimum stopband attenuation Amin,
Passband edge frequency ωp, and
Stopband edge frequency ωs.
You may use a Butterworth or a Chebychev function in your analysis.
[6 marks]

b) Obtain an appropriate loss -function A($) which will meet these specifications.
Hence calculate the actual loss in dB produced by the filter at the stopband edge
frequency.
Show how this transfer function could be implemented using an active filter. Circuit
values are not required but indicate which components determine the passband
frequency. Discuss briefly which components determine whether the circuit
configuration is a Chebychev or a Butterworth type of filter.
[8 marks]
Q34
Discuss a graphical technique for locating the poles of a Chebychev approximation
loss-function on an s-plane plot.
[4 marks]
The bandpass circuit configuration in Fig. 14 has a transfer function. Determine
suitable component values for C ,R and R which will The required bandpass
frequency transformation is :where B is the -3dB bandwidth which can be expressed
as
The normalised low pass stopband frequency is defined as

Q35
An anti-aliasing low-pass filter is required to meet the following filter specifications:
The maximum passband loss Amax = 2 dB
The minimum stopband loss Amin = 23 dB
The passband edge frequency ωp = 3.10 rs-1
The stopband edge frequency ωs = 9.10 rs-1

17
 Obtain an appropriate loss -function A($) which will meet these specifications. Hence
calculate the actual loss in dB produced by the filter in dB the stopband edge
frequency. [12 marks]
From the transfer function H($)= 1/A($), calculate the phase shift produced at the
passband edge frequency .
[4 marks]

Sketch a Sallen and Key equal-value component circuit diagram which could be used
to implement the transfer function so derived.(circuit values not required)
[4 marks]
Solution
[7 marks]

Then the corresponding 3rd order normalised loss function is

A($) = (s + 1)(s2 + s + 1)

H(S) =

To find the attenuation use the following

A($) = 10 log[1 + 0.585x3 ]

A($) = 10 log[1 + 0.585x729] = 26.3 dB

The angle is calculated as

Q36
Discuss a graphical technique for locating the poles of a Chebychev approximation
loss-function on an s-plane plot.
[4 marks]
The bandpass circuit configuration in Fig. 15 has a transfer function. Determine
suitable component values for C, R and R which will The required bandpass
frequency transformation is :where B is the –3 dB bandwidth which can be expressed
as bp bp1. The normalised low pass stopband frequency is defined as

POLE-ZERO PLOT OF H(s)

THE STEP RESPONSE
The output voltage V is V(s) = H(s)V (s) so for a step i/p
Let s + 1/ C R
1 - t/C (R +R )

Obtain the voltage transfer function H(s) and plot the pole-zero constellation for H(s)
. Hence using this diagram, show that the network in Fig.2 is an ALL-PASS filter [12
marks]
Explain the terms minimum- phase function and non - minimum -phase function .
Show how the network in Fig.16 is a non-minimum phase network and state one
application for this network.

18
 The Laplace Transform of a function f(t) is defined as:

∫0
∝
L[f(t] = f(t)e −st dt

Show that a capacitor charged to V volts, can be represented as a capacitive reactance

1/sC in series with a voltage source V/s.
[6 marks]
A 10 H coil with a measured resistance of 1600 Ω is connected across the terminals
of a capacitor, which had been previously charged to 100 V. Obtain an expression (as
a function of time) for the voltage across the coil after the capacitor has been
connected. Calculate a value for this voltage at a time t = 1 second has elapsed.
[14 marks]
Q37
(a)Show that the unit step response G(t), of a simple CR low-pass filter, is the time
integral of its impulse response.
[6marks]
(b) For the circuit shown in Fig.17, steady state conditions are established when S is
opened. S is then closed. Derive an expression for the voltage V (s) across C. Apply
the initial and final value theorems to obtain a value for V (t) at time t = 0 and
seconds.
[14marks]
E = 1 V; R = R = 1; L = 1 H; C = 1 F.

Draw a block diagram of a PLL.

Obtain a value for the open loop gain k for the circuit parameters given. Hence show,
from first principles, how the closed loop gain is For the circuit in Fig 1 and using
mesh analysis show that current in R = 0 (the balance condition ) obtains for the
condition
[8 marks]

POLE-ZERO PLOT OF H(s)

THE STEP RESPONSE

The output voltage V is V(s) = H(s)V (s) so for a step i/p
Let s + 1/ C R
1 - t/C (R +R )

b)A 1 volt step is applied to this circuit .Sketch to scale the output voltage over the
period 0 t 5 seconds for the following circuit values :
R = 10 kΩ, R = 100 kΩ C = 10 nF [12 marks]

a)Derive the voltage transfer function H(s) and sketch the pole-zero plot for the
network in Fig.2
[6 marks]

19
b)Sketch to scale the output voltage over the period 0 9< 3t 9< 15 s when the
following signals are applied to the input :i)a unit step , and
ii)a unit impulse [12 marks]
c)Show that the differential of the step response from i) is the impulse response
obtained in ii). [2 marks]
R = 10 kΩ 1,C = 10 nF
To find values for A and B use the partial fraction method 1

The transfer function H(s) = The required transfer function is :

1/C RH(s) =1s + 1/CR
Let us consider the STEP response for the CR low-pass filter . The Laplace transform
of a step is 1/s so that we can write the output response V(s) :V (s) = H(s).1/s 22u

POLE-ZERO PLOT OF H(s)

IMPULSE RESPONSE

1/C R H(s) = 1s + 1/CR

The impulse response is the transfer function since the transform of an impulse
function is one so we can write V221(s)= H(s) We can write that -3t1/CR
V221(3t1)= 1/CR e 8uFig 7 3Impulse response of Low-pass CR net1If we
differentiate the step response V221(3t1) we get the result : -3t1/CR V (3t1) = 1/CR e
221Which is the result we obtained for the impulse response i.e. differentiate a step
3response 1and you get the impulse response/+,

Obtain the voltage transfer function H(s) and plot the pole-zero constellation for H(s)
Hence using this diagram, show that the network in Fig.2 is an ALL-PASS filter [12
marks]
Explain the terms minimum- phase function and non - minimum -phase function.
Show how the network in Fig.2 is a non-minimum phase network and state one
application for this network.

MINIMUM AND NON-MINIMUM PHASE FUNCTIONS

A system with left -hand zeros only are called minimum - phase functions. It can be
seen from a pole -zero plot of such functions that the angle of the transfer function
will vary over a much angle than a system which has right -hand zeros. Considering
the all-pass network in Fig.2 the angle for the transfer function H(s)is given as One
application for this network is where a signal is suffering from an unwanted phase lag
then the network can supply enough additional phase lag to bring the total angle to 0.

From the table of Butterworth loss functions we obtain a 4th order polynomial.

A($) = ($ 1+ 0.765$ + 1)($ + 1.848$ + 1)

De-normalising and LP to HP transforming substituting for 1 as b= ωhp/1/s =

(0.764) = 2800/s
substituting this into the required high-pass approximating function which, when
inverted, gives the required H-P transfer function. 0( )

20
A(s) =0{1[(2800 /s) +0.765(2800 /s) + 1][(2800/s) + 1.848(2800/s) +1]0}

Inverting this to obtain the required transfer function.

H(s)=(s + 2142s + 2800)(s + 5174s + 2800)

The passband frequency is determined by C211,R21 1.R2a 1R2b 1R2c 1R2d 1will
determine whether we have a Butterworth or a Chebychev response . The damping is
set by k = 1 + Ra/R2b 1and 1 + Rc/Rd
Draw the circuit configuration for an equal component high-pass Sallen and Key
active filter. Obtain the transfer function H(s) for this circuit and show that passband
edge frequency wp 1and damping (wp/Q)are 1/CR and (3 - K)/CR resp. You may
assume the operational amplifier is ideal .
[20 marks]
HIGH-PASS SPECIFICATIONS = 0 1.5krs 3 krs EQUIVALENT LOW-PASS
substituting values From the table of Butterworth loss functions we obtain a 4th order
polynomial. De-normalising and LP to HP transforming substituting for as
substituting this into the required high-pass approximating function which, when
inverted, gives the required H-P transfer function.

A(s) ={[(2800 /s) +0.765(2800 /s) + 1][(2800/s) + 1.848(2800/s) +1]}

Inverting this to obtain the required transfer function.

H(s) =(s + 2142s + 2800)(s + 5174s + 2800)

The passband frequency is determined by C,Rc will determine whether we have a

Butterworth or a Chebychev response. The damping is set by k = 1 + Ra/Rb and 1 +
Rc/Rd.
[20 marks]
Draw equivalent circuits for a capacitor C with an initial voltage volts and an
inductor L with an initial current I = 20 amps
[4 marks]
Steady state conditions are established for the circuit in Fig.1 with opened. S is then
closed. Derive an expression for the voltage V2o across R 1Obtain a value for V at
time = 0 and 98 seconds by applying the Initial and Final value theorems
[16marks]
Applying Thévenin's theorem to the circuit yields The initial conditions are
calculated from Fig 1b above by considering the capacitor C as an o/c and L as a s/c.
Thus The initial voltage on the capacitor is The initial condition for the inductor is
calculated: The equivalent s-plane circuit for t > 0 is Thus applying KVL to the
circuit in Fig.2 Applying the initial value to this equation gives a value for V\2R2 and
the Final value theorem Value at \98 \1is = Lim f(t) = Lim s F(s) a)
Derive the voltage transfer function H(s) and sketch the pole-zero plot for the
network in Fig.2
[6 marks]
b)Sketch to scale the output voltage over the period 0< 3t < 15 s when the following
signals are applied to the input : i)a unit step and ii)a unit impulse [12 marks]
c)Show that the differential of the step response from i) is the impulse response

21
obtained in ii).

[2 marks]
= 10 kΩ, C = 10 nF Fig.2 To find values for A and B use the partial fraction method
Fig 4 Step response of low-pass CR net Fig 5 \3low-pass CR filter The transfer
function H(s) = The required transfer function is: Let us consider the STEP response
for the CR low-pass filter. The Laplace transform of a step is 1/s so that we can write
the output response V1(s):

IMPULSE RESPONSE The impulse response is the transfer function since the
transform of an impulse function is one so we can write We can write that Fig 7
Impulse response of Low-pass CR net If we differentiate the step response we get the
result : Which is the result we obtained for the impulse response i.e. differentiate a
step response and you get the impulse response
b) A 1 volt step is applied to this circuit .Sketch to scale the output voltage over the
period 0 t 5 seconds for the following circuit values :
R = 10 k R = 100 k C = 10 nF
[12 marks]
POLE-ZERO PLOT OF H(s)
THE STEP RESPONSE
The output voltage V is V(s) = H(s)V (s) so for a step i/p
Let s + 1/ C R
1 - t/C (R +R )
Obtain the voltage transfer function H(s) and plot the pole-zero constellation for H(s)
. Hence using this diagram, show that the network in Fig.2 is an ALL-PASS filter [12
marks]. Explain the terms minimum- phase function and non - minimum -phase
function . Show how the network in Fig.2 is a non-minimum phase network and state
one application for this network.

22