INTRODUCTION OF WAVEGUIDES Under guidance of Joydeep Sengupta sir VNIT BT14ECE031 CHARAN SAI KATAKAM 1
INTRODUCTION TO WAVEGUIDES In a waveguide energy is transmitted in the form of electromagnetic waves whereas in transmission line, it is transmitted in the form of voltage and current. Conduction of energy takes place not through the walls whose function is only to confine this energy but through the dielectric filling the waveguide which is usually air. ADVANTAGES OF WAVEGUIDE OVER A TRANSMISSION LINE : In a transmission line high frequency signals cannot be transmitted because Large mutual induction. Length is very large. Multiplexing the signals is not possible in transmission lines. Flashover is less in case of waveguide. Minimum loss of power in case of waveguide. Power handling capability is 10 times than coaxial cable. VNIT BT14ECE031 CHARAN SAI KATAKAM 2
INTRODUCTION TO WAVEGUIDES Mechanical simplicity. Higher maximum operating frequency.(3ghz to 100GHZ). Normally waveguides are of two types. Rectangular waveguide. Circular waveguide. a b Rectangular waveguide d Circular waveguide REFLECTION OF WAVEFORM IN CONDUCTING PLANE : V I V n V g V R V g is velocity guided by wall (group velocity) V n is the normal velocity to wall VNIT BT14ECE031 CHARAN SAI KATAKAM 3
WAVEGUIDES V n and V g are components of V R. V I incident velocity V R reflected velocity TEM Transverse Electro Magnetic In TEM, electric,magnetic and propagation of wave should be perpendicular to each other. In rectangular waveguide, if TEM strikes the wall, as it is a conductor it get absorbed by wall whereas in rectangular waveguide, reflection takes place when TEM strikes wall. Therefore TEM cannot be propagated through rectangular waveguide but in circular waveguide. Only TE and TM are possible in rectangular waveguide. TE- Transverse Electric: Electric field is perpendicular to propagation of wave. TM-Transverse Magnetic: Magnetic field is perpendicular to propagation of wave. VNIT BT14ECE031 CHARAN SAI KATAKAM 4
DOMINANT MODE OF OPERATION DOMINANT MODE : The natural mode of oper- -ation for a waveguide is called dominant mode. This mode is the lowest possible frequency that can be propagated in a waveguide. a L L m=no. of half wavelength across waveguide. n=no. of half wavelength along the waveguide height. a MAGNETIC FIELD > > > a L ELECTRIC FIELD VNIT BT14ECE031 CHARAN SAI KATAKAM 5
DOMINANT MODE OF OPERATION cont n=1 => 1 circle n=2 => 2 circles m- electric field n magnetic field The signal of maximum wavelength that can pass through a waveguide is λ 0 =2a/m Magnetic field BASIC BEHAVIOUR m=1 m=2 An electromagnetic plane wave in space is transverse electromagnetic or TEM. The electric field, the magnetic field and direction of propagation are mutually perpendicular. If such a wave were sent straight down a waveguide, it would not propagate in it. This is because the electric field would be short-circuited by the walls since the walls are assumed to be VNIT BT14ECE031 CHARAN SAI KATAKAM 6
BASIC BEHAVIOUR cont. Perfect conductors and a potential cannot exist across them. What must be found in some method of propagation which does t e ui e a ele t i field to e ist ea a all a d simultaneously the parallel to it. This is achieved by sending the wave down the waveguide in a zigzag fashion bouncing it off the walls and setting of a field i.e., maxima at the centre of waveguide and zero at the walls. In this case, the walls are nothing to be short circuit and they do t i te fe e ith the a e patte setup et ee the. To measure consequences of zigzag propagation are apparent. The first is that velocity of propagation in a waveguide must be less than that of free space. VNIT BT14ECE031 CHARAN SAI KATAKAM 7
PLANE WAVE OF A CONDUCTING SURFACE If actual velocity of wave is V c, then the simple trigonometry shows that the velocity of the wave in the direction parallel to conducting is V g and velocity normal to the wall is V n. V g λ P INCIDENT WAVE V n θ V C REFLECTED WAVE λ n λ PARALLEL AND NORMAL WAVE LENGTH: Distance between 2 successive identical points i.e., successive crests or successive troughs. In the figure, it is seen that the wavelength in direction of propagation of wave is λ being the distance between 2 successive crests in this direction. VNIT BT14ECE031 CHARAN SAI KATAKAM 8
PARALLEL AND NORMAL WAVELENGTH cont. So the distance between 2 consecutive crests in the direction parallel to conducting plane is λ p and wavelength perpendicular to surface is λ n. λ p = λ sinθ λ n = λ cosθ V C = f λ V g = V C sinθ V p = fλ p = f λ sinθ = V Eq 1 C sinθ V p phase velocity velocity with which wave changes its phase. V g group velocity velocity parallel to wall. PARALLEL PLANE WAVEGUIDE In order to wave to propagate in a waveguide there should be no voltage at the walls because walls are purely conductive if there exists some voltage at walls the wave get shorted and there will be no propagation of wave. VNIT BT14ECE031 CHARAN SAI KATAKAM 9
PARALLEL PLANE WAVEGUIDE cont.. Voltage variation in waveguide is almost similar to the transmission lines. Waveform of voltage in transmission lines is as follows λ/2 2λ/2 So dimension of waveguide can have following values such that voltage at walls will be zero. a=3 λ n / 2 3λ/2 a=2 λ n / 2 a= λ n / 2 VNIT BT14ECE031 CHARAN SAI KATAKAM 10
DERIVATION OF CUT OFF WAVELENGTH a is the dista e et ee the alls λ n is the wavelength in the direction normal to both walls. is the o.of half a ele gth of ele t i i te sit to e established between walls which is nothing but integer. a = mλ n mλ 2 = From Eq 1 2cosθ cosθ = mλ 2a λ P = λ λ sinθ = 1- cos 2 θ λ 1- (mλ) 2 (2a) 2 From equation, it is easy to say that as the free space wave length is increased, there comes a point beyond which, the a e a o li ge p opagate i a a eguide ith fi ed a &. The f ee spa e a ele gth at hi h this takes pla e is called cut off wavelength and it defines as the smallest free VNIT BT14ECE031 CHARAN SAI KATAKAM 11
DERIVATION OF CUT OFF WAVELENGTH Space wavelength that is just unable to propagate in waveguide under such condition. mλ 1- ( o ) 2 2a = 0 λ o = 2a m λ o = cut off wavelength for dominant mode, m=1, λ o =2a λ λ p = 1 - λ ( 2a m ) 2 λ p = guided wavelength λ = f ee spa e a ele gth λ p = λ 1 - ( λ λ o ) 2 So a waveguide allows a signal having a frequency more than cut off frequency so a waveguide acts as high pass filter. The lowest cut off frequency can be calculated through fc =1.5*10 8 ( m a ) 2 + ( b n ) 2 fc - lower cut off frequency a,b waveguide measurements m,n integers indicating the modes VNIT BT14ECE031 CHARAN SAI KATAKAM 12
PROBLEM Calculate the lowest frequency and determine the mode closest to the dominant mode of rectangular waveguide 5.1cm*2.4cm.Calculate the cut off frequency of dominant mode? Ans) From the formula in previous slide for dominant mode is TE 1,0 so m=1,n=0,a=5.1*10-2,b=2.4*10-2 we get f c = 2.94GHz Mode closest to dominant mode can be determined by substituting m, n values. for m=0, n=1, we get f c = 6.25GHz for m=0, n=2, we get f c = 12.5GHz for m=2, n=0, we get f c = 5.8GHz As f c = 5.8GHz in TE 2,0 is close to f c = 2.94GHz in TE 2,0, therefore mode closest to dominant mode is TE 2,0. VNIT BT14ECE031 CHARAN SAI KATAKAM 13
GROUP AND PHASE VELOCITY IN WAVEGUIDE V g = V c sinθ V p = sinθ V p V g = (V c ) 2 ------Eq 2 V p =f λ p V p =f V p = λ 1 - ( λ 1 - V c λ o ) 2 λ ( ) λ o 2 Substituting above equation in Eq 2 we get V c ( 1- ( λ ) 2 λo ) θ V n V g = V c The above equation represents that velocity of propagation or group velocity in a waveguide is lower than in free space. Group velocity decreases as the free space wavelength approaches the cut off wavelength and is zero when the two wavelength are equal. VNIT BT14ECE031 CHARAN SAI KATAKAM 14 V g V c θ λ n λ λ p
PROBLEM It is necessary to propagate a 10GHz signal in a waveguide whose wall separation is 6cm.What is the greatest no.of halfwaves of electric intensity which it will possible to establish between 2 walls. Calculate the guide wavelength for this mode of operation? Solution: a = 6 cm, f = 10GHz => λ = f/c = 3cm λ 0 = (2*a)/m for m=1 => λ 0 =12/1 =12 cm for m=2 => λ 0 =12/2 =6 cm for m=3 => λ 0 =12/3 =4 cm for m=4 => λ 0 =12/4 =3 cm So maximum value of m is 4. VNIT BT14ECE031 CHARAN SAI KATAKAM 15
RECTANGULAR WAVEGUIDE y a x All equations are same for parallel plane waveguide and rect- -angular waveguide are same except characteristic impedance. Z o = z 1 -( λ ) TE m,0 MODE λo 2 where Z 0 = characteristic impedance Z = 120π = 377ohms (characteristic impedance of free space). If λ= λ o in particular mode then Z o =. i.e., a e p opagatio will have infinite resistance i.e., V g = 0. b VNIT BT14ECE031 CHARAN SAI KATAKAM 16
TE m,n MODES The obvious difference between TM m,n mode mode and those described thus foe is that magnetic field here is transverse only and the electric field has a component in the direction of propagation. Although most of the behaviour of thus modes is same for TE modes, a no.of differences do exist. The first such difference is due to the fact that lines of magnetic field lines are closed loops consequently if magnetic field exists and is changing in x- direction it must also exist and be changing in Y- direction. Hence TM m,0 modes cannot exist in rectangular wave guide. If we want to propagate TM mode, if a wave is propagating in z-direction it may propagate in x-direction and y-direction. In TE m,0 mode if λ = λ 0 then Z 0 = 0, if potential difference between two sides is zero there will be no variation of field. VNIT BT14ECE031 CHARAN SAI KATAKAM 17
CIRCULAR WAVEGUIDE Let e i te al adius of i ula a eguide then cut of wavelength of waveguide is 2πr λ 0 = k r solution of a vessel function equation. k r TE MODE TM MODE MODE Kr MODE Kr TE O,1 3.83 TM O,1 2.40 TE 1,1 1.84 TM 1,1 3.83 TE 2,1 3.05 TM 2,1 5.14 TE 0,2 7.02 TM 0,2 0.52 TE 1,2 5.33 TM 1,2 7.02 TE 2,2 6.71 TM 2,2 8.42 The are values of Kr for different modes of transverse electric and transverse magnetic modes of circular wave guide. VNIT BT14ECE031 CHARAN SAI KATAKAM 18
PROBLEM Calculate the ratio of cross section of a circular waveguide to that of rectangular waveguide if each is to have same cut off wavelength for its dominant mode. Solution: for circular for λ 0 = 2πr/(kr) = 2πr/1.84 = 3.14r TE 1,1 is dominant mode r = (λ 0 * 1.84)/ 2π = λ 0 /3.41 A c = π*r 2 = π*(λ 0 /3.41) 2 for rectangular waveguide λ 0 = 2a/m TE 1,0 is dominant mode => a = (λ 0 *1)/2 (since m=1) Ratio of a : b = 2 :1,area = a *b = a*(a/2) = a 2 /2=(λ 0 /2) 2 /2 Ratio of cross sections is π*(λ 0 /3.41) 2 : (λ 0 /2) 2 /2 2.17 : 1 VNIT BT14ECE031 CHARAN SAI KATAKAM 19
CIRCULAR WAVEGUIDE Smallest value of Kr, the TE 1,1 mode is dominant in circular waveguide. The cut off wavelength of this mode is λ 0 = (2πr)/(1.84) λ 0 =1.7d d inner diameter of circular waveguide. Another difference lies in the different method of mode levelling, which must be used because of the circular cross se tio. The i tege de otes the no.of full wave intensity a iatio alo g the i u fe e e a d ep ese ts the no.of half wave intensity changes radially out from the centre to the wall. It is seen that cylindrical co-ordinates are used here. DISADVANTAGES OF CIRCULAR WAVEGUIDE OVER RECTANGULAR WAVEGUIDE: The first drawback associated with the circular waveguide is that its cross section will be much bigger in area than that of a corresponding rectangular waveguide used to carry the same signal. VNIT BT14ECE031 CHARAN SAI KATAKAM 20
CIRCULAR WAVEGUIDE Another problem with circular waveguide is that it is possible for the plane of polarisation to rotate during the waves travel through waveguide. This may happen because of roughness and discontinuities in the wall at departure from circular cross section. ADVANTAGES: It is easier to manufacture than rectangular waveguide. They are also easier to join together. VNIT BT14ECE031 CHARAN SAI KATAKAM 21