UNIT 2 G.C, AND D.C. CIRCUITS

Structure

2.1 Introduction Objective

2.2 Recall Complex Impedance

2.3 Resonant Circuits Series Resonant. Circuit Impedance end Phase Angle of Series Resonant Citcuit Voltages and '"urrent in a Series Resonant Circuit Bandwidth of RLC Circuit Parallel Resonance' Q Factor of a Parallel Resonant Circuit

2.4 Impedance Matching

Theory of Passive Filters Constant K Low Pass Piltcr Constant K High Pass Filter Band Pass Filters

Attenuators 7'-type Attenuator Z-type Attenuator Lattice Attenuator

Terminal Questions

2.9 Solutions and Answers

2.1 INTRODUCTION

In this unit you will study the practical application of some circuits of passive elements. The most important among these are resonant circuits. The series resonant circuits we used to offer a low impedance and the parallel resonant circuit to offer high impedance at a desired frequency. In every day life we see that by moving a knob in a trmsistor radio we can get the programe from a desired station. This is achieved by filters, which allows to select desired frequencies from a signal (source) having wide range of frequencies. In order to reduce or stop power at the output at desired frequencies without loosing power in the circuit (reactive elements) we use attanuators. We have learnt in maximum power trnnsfer theorem, that the maximu~n power from a source to a load is transfetred when certain impedance conditions are satisfied (matched impedance). It is aot always possible to have a source and a load to wGch power has to be delivered satisfy the matched impedance conditions. This is achieved by usin; special impedance matching circuit so that the source at its output sees a matched impedance and the load to which power has to be delivered sccs a matched impedance at its input. Resonant circuits, fiIters, attenuators and impedance matching circuits fitad applications in Instrumentation, telemetering equipment etc,, where it is necessary to transmit or attenhte a limited range of frequencies. The topics of this unit, find wide applications in communication world and hence are very important.

' I t Objcctivcs

After going thrdugh this unit, you will be able to

32 e expbdn cornpiex impedance, . . ' ..

e describe resonance phenomenon in series and parallel resonant circuits,

r explain concept of impedance matching,

explain characterisics' of low pass, high pass and band pass filters and their designs, and

describe attenuators.

2.2 RECALL COMPLEX IMPEDANCE

You have already learned that in the resistive element, there k no phase difference between the voltage and current whereas in case of pure inductors, the voltage leads over cursent by n/2 and in pure capacitors, the voltage lags behind the current by z/2. Almost all the electric circuits offer impedance to the flow of curtent. Impedance is the vector sum of inductive reactance, capacitative reactance and resistance.

Consider the RL series circuit shown in Rg.2.1. If we apply the real function V, cos ot to the circuit, the response may be I, cos at. Similarly, if we apply

Fig. 2.11 Series RL leiw work.

imaginary function jV, sin at to the same circuit, the response will be j I , sin at. If

we apply a complex function, which is the combination of real and imaginnry

functions, we will get the complex response. The complex function is V, eiat - V, (cos ot + j sin at). Applyhg KVL (Kirchofrs Voltage Law) to the circuit shown in Fig. 2.1, we get

This is a first order linear differential equation with constan! coefficient and can be easily, solved. The simplest way to solve is to choose a trial solution for i ( t ) as :

If this happens to be a solution of the differential equation, then i t must satisfy i t completely. By substituting, we gct

Impedance is defined a s the ratio of voltage to current as follows :

.AX. nnd D.C. Circui1.r

-----

Network Analyals and &vices

Complex impedance (2) is the total opposition offered by the circuit elements to ac '

current and can be displayed on the complex plane. In the expression for 2, the resistance R is the real part of the impedance, and the reactance XL is the imaginary part of the impedance. The resultant of R and XL is called the complex impedance as shown jn Pig. 2.2. This is called impedance diagram. From Figure, i t is.-clear that

Fis 2.2: Impedance dlagram of RL network.

I Z I - and angle 0 = tan-' OL . Here, impedance is the vector sum of [ R l

resistance and inductive reactance.

Where 121 = @G?F

and

This can be deduced as follows :

Let R-Acos8

and wL-Asin0

Squaring & adding, we get

P + w2 (cos2 + sid 0)

-- .

also, A s h e L _ _tm-t [%] -=- AcosO R

I Clearly, ~ = ~ ( c o s 0 t j s i n 8 ) = ~ e I ~

- Similarly, 'if we consider RC series circuit, as shown in Fig. 2.3, h d apply complex

voltage v ( t ) = V, el? we get, complex response as. shown below : Applying KVL in the circuit shown we get,

I v ( 1 ) = v,cJN1

FYg 2.3: Serlea RC network

L C . and D.C. Clrct~lts

, As v - i relation for capacitor is given by

d v, ( t ) i ( t ) = C- d t

Taking trial solution of the above equation as

i ( t ) - I,,, ejm* we get

vm ~9 i ( t ) - (R - j/"Q

jut, Y.(t> ( R - j/oQ

I

I :. The impedance Z = a - R - j/wC i 0 )

I I Here, the impedance Z consists of resktance (R), which is real part, and capacidve

which i s imaginary part of the impedance. The impedance reactance Xc = - wc

/ diagram is shown in Fig. 2.4. From this diagram, the impedance I

(- j ) J. Imaginary

Ap. 2.4: Impedance dlapam for -lee RC network

1 a d angle 0 = tantan1 i I j Also, we can write,

1 I 2 - R - - 1 0 C !

Example 1. In a circuit, consisting of 1 kQ resistor connected in series with 50 mH coil, a loV. m., 10 kHz signal is applied. Find impedance 2, current I, phase angle 0, VR and VL.

I

I

Network Annlymlr and De- Solution. R = 1 WZ = 1000 P

. x,- wL-2zfZ

Current I = - = - = vms lo 3 . 0 3 m ~ Z 3295.4

phase angle (0) = tan-

Example 2. A sine wave generator suppliea a 500 Hz, 10 V,, dgnal to a 2 kQ resistor in series with a 0.1 f l capacitor as shown in Pip. 2.5. Determine the total impedance 2, m e n t 2, phase angle 0, capacitive voltage VC , d mdstfve voltage VR .

1 Solution. Xc - - - 1 2 s 2 x 3.14 x 14 x 500 x 0.1 x 10+

- 3184.752

Total impedance i 1 R - - z - R + ~ ~ = 2nf - R - j X c

Phase angle -xc

v,, 10 Current ( I ) - - = - = 2.66mA 12 ( 3760.0

Capacitive voltaga, Vc = I X c - 2.66 x x 3 184.7 = 8.47 V I

Remnrks : You must have noted 'that the arithmetic sum of Vc and VR does not give , 36 the applied voltage (1OV). In fact, the total applied voltage is a' complex quantity

- .- - - _ _ _I __ - -

I I given by

A.C. and D.C. Clrcuits

I

SAQ 1

Determine the source voltage and the phase angle, if the voltage across the resistance is 70V and the voltage across the inductive reactance is 20V as shown in Fig. 2.6.

vs Fig. 2.6;

SAQ 2

Determine the source voltage and phase angle when the voltage across the resistor is 20V and the voltage across the capacitor is 30V as shown in Fig. 2.7.

Fig. 2.78

2.3 RESONANT CIRCUITS

We call a circuit a resonant circuit if at a particular frequency called resoaant frequency the capacitive reactance Xc and inductive reactance XL are equal i-e.,

XL - XC. In many of the electrical circuit, resonance is a very important phenomenon. The study of re'sonance is very useful, particularly in the area of communications. For example, the ability of a radio receiver to select a certain frequency, transmitted by a station and to eliminate frequencies from other stations is based on the principle of resonance. In this section we will study the essential features of series and parallel resonant circuit.

2.3.1 Series Resonant Circuit

In a series RLC circuit, the current lags behind, or leads the applied voltage depending upon the values of XL andXc. XL causes the total current to lag behind the applied voltage, while Xc causes the tdtal current to lead over the applied voltage. When XL > XC, the circuit is predominantly inductive and when Xc > XL the circuit i~ predominantly capacitive. However, if one of the parameters of the series RLC circuit is varied in such a way that the current is in phase with the applied voltago, then the circuit is said to be ih ltresonancetl. consider a series RLC circuit shown in Rg. 2.8. The total impedance for the series RLC circuit h :

- Y~twork Annlysla and Devices

vs Fig. 1.8: Seriea RLC clrcuit.

The circuit is said to b: .n resonance if the current is in phase with the applied voitage. In a series RLC circuit, 'resonance" occurs when XL = XC. The frequency at

which the resonance occurs is called the 'resonant frequency". When XL - Xc, the impedance in a series RLC circuit is purely resistive. At resonant frequeocy (fo), the

voltages across capacitance and inductance are equal in magnitude. Since they are 180" out of phase with each other, they cancel each other, hence, zero voltage appears across the LC combination.

At resonance, XL = X c

In a series RLC circuit, the resonance may be produced by varying the frequency, keeping L and C constant, otherwise resonance may be produced by varying either L or C for a fmed frequencey.

2.3.2 Impedance and Phase Angle of a Series Resonant Circuit

As we have already seen in previous section, the impedance of a series RLC circuit i s given by :

The variation of XL, Xc and Z with frequency is shown in Fig. 2.9. At zero \ frequency, both Xc and 2 are infinitely latge and XL is zerd- because the capacitor

acts as open circuit at zero frequency. As the frequency &creases, Xc decreases and

XL increases. Since XC is larger than XL at frequencies below the resonant frequency

&), Z decreases along with Xc. At resonant frequency VS), Xc = XL and Z5 R. At

frequencies above the respnant frequency, XL is larger than Xc causing Z to increase.

F¶5 2 3 1 Variation of XL ,Xcmad Z with frequency.

The phase angle associated with total impedance in series RLC circuit is given by :

= tan-' R

The variation of phase angle with frequency Is shown in Fig. 2 . 1 0 At a frequency

Fie ZlOt VarlsUon of phamm angle wllh frequency.

below the pcsonant frequency, the current l ads over source voltagzbecause the capacitive reactance 1s greater than the inductive reactance. The phase angle

decrtasea PB the f*quency approachas the resonant value, and is O0 at resonance. At frequencies above resonance, the current lags behind the source voltage, because the inductive reactance ia greater than capacitive reactance. As the' frequency goes higher, the phase angle approaches + 90".

2.3.3 Voltages and Current in a Series Resonant Circuit

We know, in a series RLC network

--

I ~ r t w o r k Annlysls and Devices

Where Vo and I are amplitude of voltage and current respectively. The variation of

impedance and current with frequency is shown in Fig. 2.11. As we have learned in

Flg. 2.11: Variation of impedance and current with frequency.

, I~l~pcdancc (2)

t I

previous sections, at resonant frequency, the capacitive reactance is equal to inductive reactance, and hence the impedance is minimum. Because of the minimum impedance, maximum current flows through the circuit. This value of current is given by

1~iipcntl:mcc or

c u r e n

vo lo - current at resonant frequency = - R

Current ( I )

I I I I

The voltage drop across resistance, inductance and capacitance also varies with frequency. At f =0, the capacitor acts as an open circuit and blocks current. Hence, complete source voltage appears across the capacitor. As the frequency increases, Xc decreases and XL increases, causing total reactance (XC - XL ) to decrease. As a

result, the impedance decreases and the current increases. As current increases, voltage drop across resistance, inductance and capacitance (VR, VL & VC) also

increases. When the frequency becomes equal to the resonant frequency ( fo), the

impedence is equal to resistance (R) and hence current reaches its maximum value and VR is also of maximum value. At this frequency, the voltage drop across

capacitance and inductance are equal in magnitude and opposite in phase. As the frequency is increased further, X; continues to increase and XC continues to

decrease, causing the total reactance (XL - XC) to increase. As a result, there is

decrease in current. If the frequency is increased still further, the current approaches zero, both VR and Vc approaches zero and VL approaches V@ The response of

different voltage with frequency is shown in Fig. 2.12.

0 .L> *f

Fig. 2.12: Yarlatlon of Vc, VL and VR wlth frequency.

I 2.3.4 Band Width of a PULC Circuit

.. -

A.C. nnd D.C.

i The band width of any-*<tern is defined as the range of frequencies for which the current or output voltage is equal to 70.7% of its peak value at the resonant

I frequency. Fig. 2.13 shows the frequency response of a series RLC circuit. &re the

1 Fig. 2.131 Frequency response of RLC aetwark.

frequency fl is the frequency at which the current is 0.707 times the currcnt at

resonant value, and i t is called the lower cut off frequency. The frequency f2 is the frequency at which the current is 0.707 times the current nt resonant valuc and is called the upper cut off frequency. The band width (BW ) is defined as :

If the current at PI is 0.707 lo, the impedance of the circuit nt this point is ER, and hence

I Equating these equation, we obtain

We know, 2 1 a()=- LC

Therefore, at= at w2

Adding equations for R, we get I

Circuits

I

' Network Analyllb mnd Drvica - ( a 2 - w l ) L + 0 2 . 0 1 C

1 1 Since a:=-&c=-

LC O ~ L

Substituting these values, we obtain

R :. BW @and width) = 2 x ~

From Fig.2.13 we have

R The lower frequency limit fi fo - - 4s L

R The upper frequency limit fi = fo + -

4n; L

Rearranging Eq. (2.1) (dividing both sides by fo ), we get

Here an important property of a coil is defined:It is the ratio of the. reactance of the coil to its resistance. This ratio is defined as the Q of the coil. Q is known as Quality factor or figure of merit, and is an indication of the 'quality of the coil.

.

Clearly, 1 f2 -f1 A f -a=-=-

Q fo fo'

' C

fo Resonant frequency * Band. width

I

Clearly, a higher value of Q results in a smaller ba6d width and a lower value of Q -- causes a latger band width.

I 2.3.5 Parallel Resonance

A parallel RLC network is shown in Fig. 2.14. Parallel resonance occurs when

Xc XL and the frequency-at which this occup, is called-remnant frequency. When

Xc = XL, the two branch currents are equal (provided Rc - RL ) in magnitude and

180" out of pdase with each other. Therefore, the two current caticel each other and the total cumnt is zero. Prom Fig. 2.14. the total admittance ( Y ) is:

--4

Y-' ..-A R L + l j a L l y '

%-,c ,

42

Hg. 2.14: Pumllcl U C network.

1 At resonance (o = COO) the imaginary paa becomes zero, i.e., I

I M s is general condition for resonance. In special case, when & = RL, we get

You are advised to deternine the resonant circuit for a 'tank circuit" &own in -.

...

i Fig. 2.15: Tank clrcult.

Fig. 2.15 a d show that the resonant Frequency turns out to be

I -

LC. and D.C. Clrcults

&, you will observe from the expression for impedana (reciprocal of admittance which has been derived), i t becomes maximum at the resonant frequency and decreases at lower and higher frequencies as shown in Fig. 2.16 At lower frequencies, XL is very small and Xc is very large and so the total impedance is

essentially inductive. As the frequency increases, the impedance also increases and inductive reactance dominates till the remnant frequency is reached. At this point XL = X C B d the impedance is at its maximum value. As the frequency goes above the resonance, the capacitive reactance dominates and the impedance decreases as shown in Fig. 2.16.

Jo

Fir. 2.161 Variellon of impdnece with frequency.

2.3.6 Q factor of a Parallel Resonant Circuit

Now, let us consider a parallel resonant circuit shown in Pig. 2.17. The total admittance (Y) is

At resonant frequency ( f x f o ) , the imaginary part is zero i.e.,

The voltage and current variation with frequency is shown in Fig. 2.17. Clearly, at resonant frequency, the current is minimum because in parallel resonant circuit, the impedance is maximum.

The band width (BW) fi - fi

For, parallel circuit, to obtain the lower cut off frcqucncy (A),

Similarly, to obtain upper cut off ftequency, (fi).

I 1 BW - 0 2 - 01- - RC

I

1 Also, fi - f i g

fo 0 0 The Quality factor Q = - = ---- f2 - f 1 0 2 - 0 1

I

I I I- I 1

00 = mo RC 1/RC

, Example 3. Determine the resonant frequancey for the circuit shown in Pig. 2,18(a)

i

E %ohtiion : 73e resonant frequency

1

LC. and D.C' CIrcvltn , . ,

;sH , ;:, 1 4 ; y , CL -

-4

(4 I Fig. 2.18 (a) Example 3 (b) Example 4 (b)

Nciwork Analylsis and Devicea

= 1 = 2.25 KHz

2x 410 x lok6 x 0.5 x lom3 . Eraaargle 4: For the circuit shown in 2.18 @) determine the impedance at resonant frequency, 10 Hz above resonant frequency' and 10 Hz below resonant frequency.

Solution : At resonant frequency, Xc = XL and hence total impedance in a series

RLC resonant circuit is purely resistive.

:. Z = R = 10 !2 at resonance

1 Resonant frequency ( fo, = - =

1 = 159.2Hz

2 n m 2x 40.1 x 10 x 10-6

Given, fi - fo - lOHz - 159.2 - 10 = 149.2Hz

fi =fo+ lo&= 159.2+ l Q = 169.2Hz

1 Now, Xc, - - = 106.6P

w,C 2 x 4 C

i 94.0652 Also, x q = ~ = - =

Clearly, impedance at fl = 149.2Ha

Here, Xcl > XLI , so ZI is capacitive.

And impodance at 2 = 1692 Hz

Clearly X4 > XC, at this frequency, so Z is inductive in nature.

SAQ 3 In the circuit shown in Fig. 2.19 the inductance of 0.1 H having a Q of 5 is in parallel with capacitor. Determine the value of capacitance and coil

- resistance at resonant frequency of 500 Hz ?

SAQ 4 A series RLC circuit consists of 50 l-2 resistance, 0.2 H inductance and 10 pH capacitor with the applied voltage of 20 V. Determine the resonant frequency. Find the Q factor of the circuit. Compute the lower and upper

1, frequency limits and also find the band width of the circuit.

2.4 IMPEDANCE MATCHING .

You hive already learnt maximum power transfer theorem which states that the muimum power is delivered from source to load when load resistance is equal to

the source resistance. It is for this reason that the impedance matching between circuits is so important. As an example, the apdio output transformer must match the high impedance of the audio power amplifier and its output to the low inpa jmpadulce of the speaker. But, maximum power transfer is not always desirable, dme the transfer occurs i t a 50% efficiency. For many a situations, a maximum voltage transfer is desired, which means that unmatched impedances are necessary. If maximum power transfer la required, the load resistance should be equal to the liven source msistnnce. Maximum power transfer theorem can be applied to complex impedance circuits. If the source impedance is complex then the maximum power transfer occurs when the load impedance is complex conjugate of the source impedance.

Comidcr a circuit shown in Fig. 2.20 consisting of source impedance delivering powtr to a complex load. Clearly, the current passing through the circuit is

vs i (t) = (4, + j X,) + (RL f ~ X L )

I 4 Flg. 2.208 Maximum power transfer for complex hd.

'Ihc magitude of current I. = I i(t) 1 is given by

m e power delivered to the load is given by

Clearly, ifi the ahve equation, if RL is kept fixed, the power becomes maximum when

under this conditfon, the power Is given by

Now, let us assume that RL is variable. In this case, the maximum power is

t t ans fe~d when the load resistance is equal to source rksistkce. If R . a Rs and Xs = - XL then,

I ZL ' RL ?.jXL

A.C. and D.C Clrcults '

I Network Anelysle .ad Devlew I This means hat the mum power tamfer occurs when the load ce is I

equal to the complex conjugate of the source impedance Zs . I

Exarnplc 5 : For the circuit shown Jn Fig. 2.21, detem.int the value of load impedance for which mwcd delivers maximum power. Cdculatc the value of the

maximum power. (25 + j30) R

50 LO o

Flg. 1 2 1 8 Example 5.

Solution : As we how, the maximurn power will be transferred to the load when load impedance Is complex conjugate of the source ce i.e.

when ZL = 25 - j 30, the cuPrent passing through the circuit is

Maximum power delivered to the load is

P -I:& - ( 1 F x 25- 2 5 ~

SAQ 5

For the circuit shown in Fig. 2.22, the resistance Rs is variable from 2 SZ to 25 5L.

Determine what value of Rs results in maximum power transfer across the terminals XY.

2.5 THEORY OF PASSIVE FILTERS /

The wnve filters were first invented by G.A. Campbell and 0.1. Lobe1 of the Bell Telephone laboratories. A filter b essentially a reactive network that freely passes the desired bands of frequencies while almost totally suppressing all other bands. A filter is constructed from purely reactive elements, for otherwise the attenuation would never become zero in the pass bond of the filter network. Filters differ from simple resonant circuits ($iscussed in section 2.3) in providing a substantially constant transmission over the bnhd which they accept. Ideally, a filter should produce no attenuation in the desired band called the -pass bmd" and should provide infinite attenuation in all other frequencies called the 'Stop band*. The frequency which separates pass band and stop band is defined as the 'cut off frequency* (f,). A filter may, in principle, have any number of pass bands separated

by attenuation bands. However, they are classified into four common types : low

p-, ~ g h ps., band pnr a d band reject. in caning sections, we will discuss a b u t these ypa of filters ia detail. Before we get into tbcsc filters, we would like to study the cqundom of filter naworb. Ibc study of the behavior of any filter

the calculation of its popagatioa constant h), attenution(d, phase shift (p) and fa chamcted~dc lmpwna (&>

T-Neh~orlr

Consider a symmetrical T-netwo* M showtl In Fig. 213. AS we h o w , if the image i m w c e at port 11' and pt 22' uo qd a each o k , the i m q e impdance is then called "ChurcWfstic" or "iterative hpdnne" (a. Thus, If the network shown in Ffg. 2.23 b termhated in & , ita input Impedance will also bc & . Clearly, the input fmpebnce for thb network will be :

Also, = i n = %

The characteriedc impedance of a symmetricel T-network is,

The characteristic impedance can be expressed in terms of the open circuit impedance Zoc and short circuit impedance Zsc of the T-hetwork.

)

"

Clearly, ZSC =- + - - - - -

2 z~ 2Zi+45 z+Z2

A.C. and D.C. circuit^

Network Analysis %nd Devices

and ZI Z O C " ~ + Z Z

By definition, the propagation constanst (7) of the T-network is given by (See Fig. 2.24)

~'aking .the mesh equation, we get

The characteristic impedance of T-network is

Putting this in abwe equation and simplifying, we obtain :

Dividing both side by 2, we get

x-network

Consider a ~~mrnetrical x-section shown in Pig. 2.25. When the network is terminated in Zo at port 22', its input impedance is given by : .

Fig. 2.25: Symmetrical x-network. 2'

By definition of characterstic impedance, -Zin = Zo

The propagation constant of a symmetrical A- network is same as that for a symmetrical T-network and is given by

2.5.1 Constant-K Low Pass Filter

By definition, a low pass (LP) filter is one which passes without attenuation all frequencies upto the cut off frequency (f, ) and attenuates all other frequencies greater then f,. This attenuation characteristic of an ideal LP filter is shown in

Pig. 2.26. A network, eilher T or n , is said to be of the constant K-type if ZI andZ2 of the network satisfy the relation :

where Z1 andZ2 arc impedance in the T and n section as shown in Fig. 2.27 (a) and

A.C. and D.C. Circuits

Network Analysls and Devices

,A& 2.26: Attenuation chnracterlstic of an Ideal LP nlter,

(b) rcspectively and K is a real constant impedance or nomlnal impedance of the Constant-K filter.

The constant K, T or rc type filter is also known as 'prototype' because other more complex networks can be derived from it. A prototype T and n sections are shown in Rg. 2.27 (a) end (b) 'and where ZI = j m L and & - l / j w C

- (4

Fig, 2.271 Constant K type LP fllter (a) T-sectlon (b) rr Sectlon.

Since product of Z1 && is constant (independent of frequency), the filter is a constant -K-type, As we know, the cut off frequencies are determined by the equation given by

z 4 f = O and -&.-l

Also,

1 f, = cut off frequency = - x m

The pass band can be determined graphically. The reactances of ZI and4 Zz will vary

with frequency as shown in Fig. 2.28. The cut off frequency at the intersection of the curves ZI and - 4 & is indicated as f,. On the X-axis, as Z1 = - 4 & at cut off

frequency, the pass band lies between the frequencies at which Zi - 0 and ZI - -4 &. All frequencies above f, lie in a stop or attenuation band. We h o w ,

t Keac tance

0 f

Fis 2.28: 'Vorinlton or mctanee with frequency.

0 jo m

2

1 - ,m?=- For LP filter, fc = - x m fc

We know that in the pass band, -

and p = Phase angle - 2 sin-' ; a - 0 [fJ In the attenuation band,

and P-n

Example 6 : Design a low pass filter having a cut off frequency of 2KHz to operate with a terminated load resistance of 500 ohm.

Solution : W e know for constant K LP filter

T and n section of Constant-K tP filter is shown in Fig. 2.29.

1

I Network Anelydn end Devlcen . L/2 = 39.8 mH 39.8 mH L ='79.8 mH

-, 0.318 yF

T C12 = C12 = 0.159 pF

(T-section) (rr-section)

Fig. 2.293 T and x section or low pass filter (Answer to Example 6)

2.512 Constant K-High Pass Filter

By definition, a high pass (HP) filter attenuates all frequencies below a designated cut off frequency f, , and pass frequencies above f, . l h u s , the pass band of tbis

fdter is.the frequency range above f, and stop band is the frequency range below f,. The attenuation characteristic of an ideal HP filter is shown in Fig. 2.30. Constant K-high pass filter can be obtained by changing the position of series and shunt arms of the network shown in previous section. The prototype high pass filters are shown

1 in Fig. 2.31. (a) and (b) where ZI - - and & - joL.

P C

z , q - L = K 2 C

Ng. 2.31: Constant K-type HP fllter (a) T sedlon (b) x e l o n .

I

The cut off frequ cies are given by ZI = 0 and21 - -4%.

, = O d - = o d 0 -+ 0( i lac Zl=-42 ,

' I I .

-+ - +jmL m c

- -

1 f==m

5 54 ' The variation of reactances (Z1 and&) with frequency are shown in Fig. 2.32. As , I

L. I

aeen from the figure, the filter transmita all frequencies between f = f, and f -. The point f, from the graph L a point at which Za -422.. We know,

kc. and D.C. CimuIts '

Irlk 2.32: Vnrlmtbn of rrectaaoca wlth kquenq.

; 21 f, In the pass band, - 1 < - < 0, a = 0 or the region in which - < 1 is a pass band

4% f

with p = phase shift - 2 sin- ' [$I. 21 In the attenuation band, - < - 1 - > 1 422 f

Here, a = 2 cash-I [ ] - 2 rnSh- 14 f

and fl--n .

Example 7 : Dedgn a high pass filter having a cut off frequency of 1 KH2: with a load resistance of 600 ohm.

Solution : Here, f, - 100 Hz and K - 600 hL

Network Analyals and Devdem c=-= 0.133pP I 4 m - c

T and n sections are shown in Fig. 2.33.

2.5.3 Band Pass filter

By definition, a band pass filter b one which passes frequencies between two designated cut off frequencies ahad attenuates all other frequencies. Fig. 2.34 shows an ideal attenuation characteristic of a band pass filter. A band pass filter may be o b h h d by using a low pass filter followed by a high pass filter in which the cut off freqiency of the LP filter is above the cut off frequency of HP filter, the

Plg, 254: Idcnl sltenuntlon charnctedslie of a bnnd puss filler.

overlap thus allowing only a band of frequencies to pass. This is not economical in practice, it is more economical to combine the low and high pass functions into a single filtcr section.

Fig. 2.35: Bnhd pnse fllter.

Consider, the circuit shown in Fig. 2.35 (a) and @), each arm has a resonant circuit with same resonant frequency i.e., the resonant frequency of the series arm and the resonant frequency of the shunt ann are made equal to obtain band pass cbnracterlstics.

Clearly, for condition of equal resonant frequencies

ZI - Impedance of scriee arm - j a22:1 -11 & = Impedance of shunt arm =

[I -::c2 ]

% where K is a constant. %us the filter is a constant K-type.

i.e., the value of Z1 at lower cut off fkequency is equal to thc aegative of the ZI at uppet cut off frequency.

It can be shown, fo = resonant frequencya

-

SAQ 6

Determine cut off frequency for the LIP film shown In Pig. 2.36 (a) and @).

SAQ 7 I

Determine the cut bff frequency for the HP filter shown Fig 2.37 (a) and @).

dlPF , 0.2 ].IF

it- -pil+-

2.6 ATTENUATQRS

An attenuator is a two port resistive network and is used to reduce the signal level by a given amount. In number of application, it is necessary to introduce a specified loss bctween source and a matched load without altering the impedance relationship. 'Attenuators may be symmetrical or asymmetrical, and can be either fixed or variable. A fixed attenuator of constant attenuation is called a Pad". Variable attenuators are used as volume controls in radio broadcasting, speed control of farm etc. Attenuators are also used in laboratory to obtain a smaU value of voltage or current for testing circuits. Attenuation is generally expressed in decibels (dB) or in nepers. Attenuation offered by a network in dB is given as follows:

Attenuation in dB - 10loglo - . [2) Where PI is input power and P2 is output power,

kc. and D.C. Circuit.

-. N- antilog [g] We will now discuss various types of amnutors vjz T-type, %-type and lattice auenuatm.

2.6.1 T-type Attenuatqr

Fig. 2.38 shows the symmetrical T-type attenuator. An attenuator is to be desigaedt for desired values of characteristic resistance 4 and attenuation. .-

The network shown in the figure is a symmctical resistive circuit, hence & = & md y - a . 'Ihe design equation can be obtained by applying Kirchoff a law to @ network.

Tbt chamterisdc impedance of the attenuator is R,, when it is terminated in a load

LC. and D.C. Clrcuits (N- 1) I, R l = % ( ~ + 1)

c ~ p r d o a r for R1 and R2 are called the design equntion for a symmetrical

T-rmu1tor.

m p h 8 r Deaign a T-pad attenuator to give an attenuation of 60 dB and to work . ia r line of 100 Q impedance.

fJdmtioa : We Know,

Now, each of the d e s arm (RID are given by

2.6.2 A-Type Attenuator I

Ibe c h i t shown in Fig. 2.39 shows a symmctri&l x-type attcnuatar.;T& serics

, / :

r- Ro

2.391 Symmetrical ktype sitemstar.

4 dut.arm of the attenuator can be specified in terms of 4 and propagation ,

constant y. In this network also, it is fonned by resistance, and is symmetrical, hmfoh, Zo ' Ro and y - a. From fundamental equations

Network Andy& and Dcdwa

5-4-[;]

By definition of propagation conshat, we get

Here, y m a c a - N

using this, we get

The expressions fot RI andRz are called the design equations of the symmetrical mattenuator.

Example 9 : Design a z-type attenuator to give 40 dB attenuation and to have a characteristic impedance of 100 a. Solution : We know,

- 102 Q

2.6.3 Lattice Attenuator 1

A symmetrical resistance lattice is shown in Pig. 2.40 (a). The series and the I

diagonal arms of the network can be specified in terms of the characteristic impedance &, and propagation constant, y. We know that the characteristic impedance of symmetrical network is geometric mean of the open and short circuit impedance of the network.

'Zbe &cult of Pig. 2.40 (a) is redrawn in Fig. 2AO @) to calculate open and shod circuit impedances. - I

I

1

T

,- - 1'

11 F!ig. 2.401 (a) Symmetrical lattic~ stteausbr (b) Red- d.EuY d mymdrbJ UUa *UclluJor.

Hence,

==m \ I.Q Fig. 2*.40 (b), applying kirchoffs laws, we get

.I, a = Propagatinn Constt.

Rl 1 + --7

4 Also, from NP- RI 1 -- Ro

The above equations are called design equations for, lattice attenuator.

Network Annlysis Devices Example 10: Design a symmetrical lattice attenuator to have characteriptic impedance of 10052 and attenuation of 20 dB.

Solution: Given 4 = 100 SZ and attenuation - 20 dB

We know, N = antilog - e, antilog - = 10 [ [3 Prom design equations for symmetrical lattice network

The resulting lattice attenuator is shown h Fig. 2.41.

SAQ 8

Design a T-type attenuator to give an attenuation of 20 dB and to work in a line of 800 Q

Design a n-type attenuator to give an attenuation of 20 dB and to have characteristic impedance of 500 SZ.

SAQ 10

Design a symmetrical lattice type attenuator to give attenuation of 100 dB and to have charactcristic impedance of 1000 Sl.

2.7 SUMMARY

0 The frequency at which the resonance occurs is called resonant frequency.

At resonant frequency the capacitive reactance and inductive reactance satisfy .

the following relation, XL = Xc .

R e Band width of RLC series circuit is - 2 x L

I XL I I

0 For series resomat circuit quality factor (or figure of Merit) is given by Q = - . R i

e For parallel circuit, Q-factor is given by Q = oo RC.

e Maximum power transfer occurs when the load impedance is equal to the 1

4 62 compIex conjugate of the source impedance. I

I I i

0 Filter can be classfitd as : Low pass, High pass and Band pass filters.

An nttenuator is n network which is used to reduce the gigan1 level by a given amaunt i

a Attenuation offered by n network in dB is given by

PI Attenuation in dB = 10 log - pz

Where PI is input power and P2 is output power.

2.8 TERMINAL QUESTIONS

1. A signal generator supplies a 30V, 100 Hz signal to a series circuit shown in Fig. 2.42. Detenuinc tbt impedance, the line current and phase angle of the given circuit.

2. For the circuit shown in Fig. 2.43. determlne the value of R, for which the

3. Deduce . I'

2.9 SOLUTIONS AND ANSWERS

Phase angle 8 = tan-' - [:I

m CLC. and D.C. Clrculb

Network Analysis and ~ e i c w

BW = Band width =A-fi a 39.79Hz

5. In the circuit shown, the resistance RL is fmed. Hence the maximum power transfer theorem does not apply. Maxinium cutrent flows the circuit whcn Rs Is minimum. For, the 'maximum cumeat,

Total impedance of circuit 2 = Rs - j 5 + RL

Maximum power = (2.221~ x 20 - 98.6 W

L 6. (a) T - network *: - - 40 mH, C = 0.5 pP 2

i2 (b) x - network - = 0.5 pF : L = 10 rnH 2

(a) T-network : 2C - 0.2 pP - C=0.1@, L = 30 mH

@) 1~ - network 2L - 100 mH -, L = 50 mH

C = 4 y F

, Network Analysls and Devices vs 30L0° I = - = ZT 40+j43.98

= 0.56 L- 47.7'.

Tho current lags behind voltago by 47 .7

The phase angle 6 = 47.7

1 2. Use, oo =m

Ans. RC - 1.77 0

3. (i) See Text

(11) See text

'Y 4-= ~ ~ R o v e d . use, s i n h ~ a 1 + -

225-1