IIT Bombay Course Code : EE 611 Department: Electrical Engineering - PowerPoint PPT Presentation

Page 0 IIT Bombay Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in Lecture 6 EE 611 Lecture 6 Jayanta Mukherjee

Page 1 IIT Bombay Subtopics - Binomial and Chebyshev matching networks EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 2 Maximally Flat (Binomial) Transformer • The maximally flat transformer is a N section transformer • It achieves broad-band matching by setting the reflection coefficient and its N-1 derivatives to 0 at the center frequency ( θ = π /2) 0.4 N=1 π   0.35 N=2 Γ θ = =   N=3 0 0.3 N=4 in   2 0.25 Γ k in d 0.2 Γ = in 0 θ k 0.15 d π θ = 0.1 2 ≤ < − with 1 k N 1 0.05 0 0 0.2 0.4 0.6 0.8 1 θ / π EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 3 Binomial Solution Many functions can be implemented For a N-section maximally flat transformer we choose ( ) ( ) − θ N Γ θ = + 2 j A 1 e in • The input reflection coefficient verifies Γ in ( π /2)=0 • All the derivatives of Γ in ( θ ) upto order N-1 vanish at θ = π /2 EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 4 Binomial Solution ( ) Γ θ ( ) d − N k ∝ + − θ < < 2 j 1 e for 1 k N - 1 θ k d θ = We can determine the constant A by letting 0 : − R Z Γ = = = Γ N L 0 (0) A2 + in L R Z L 0 such that we have R - Z = = − Γ - N N L 0 A 2 2 + L R Z L 0 EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 5 Binomial Transformer Bandwidth ( ) Γ θ We can write the magnitude of as N θ − θ + j j e e ( ) N − θ Γ θ = Γ j in e L N 2 = Γ θ N cos L ( ) Γ θ = Γ If we decide on a maximum tolerable value m m θ = θ for the reflection coefficien t at the point , we have the m Γ θ following relationsh ip between and : m m EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 6 Binomial Transformer Bandwidth θ Solving for gives m 1 / N   Γ θ = - 1   m cos Γ m     L The Fractional Bandwidh is given by ∆ θ f 4 = − m 2 π f 0 EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 7 Fractional Bandwidth vs Number of Sections The Fractional Bandwidh is given by 1 / N   ∆ Γ 0.4 f 4 = − - 1 m   cos N=1 2 π Γ 0.35   N=2 f   0 L N=3 0.3 N=4 The maximally flat transforme r 0.25 will have wider bandwidth in 0.2 Γ as the number of sections 0.15 is increased and the reflection 0.1 coefficien t becomes more 0.05 flat around the center frequency 0 0 0.2 0.4 0.6 0.8 1 θ / π EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 8 Design of the transformer: finding the Γ k ’ s So far we have just chosen a “maximally flat” function we would like to have the reflection coefficient of our transformer to produce. The next step is to find how the transformer can actually produce this maximally flat reflection coefficient. The binomial expansion formula is given by: N ∑ ( ) + = N N k 1 x C x k = 0 k N! = N C where, is a " binomial coefficien t" which can also be ( ) k − N n !n! rapidly calculated using Pascal' s triangle EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 9 Design of the transformer: finding the Γ k ’ s Pascal’s Triangle N/k 0 1 2 3 4 0 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 10 Design of the transformer: finding the Γ k ’ s • Applying the binomial expansion to our maximally flat function we have ( ) N ∑ ( ) N − − θ − − θ Γ θ = Γ + = Γ N 2 j N N 2 jk 2 1 e 2 C e in L L k = k 0 • This is exactly of the same form as the input reflection coefficient Γ in of the multi-section network we analyzed in the previous lecture using the small reflection approximation N = ∑ ( ) − θ Γ θ Γ 2 jk e in k = k 0 Γ which permits us to identify as : k = Γ -N N Γ 2 C k k L EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 11 Attribute of the Reflection Coefficients − Γ = 2 Γ N N C k k L • Note that the binomial transformer will produce symmetric Γ k ‘s because C k N =C N N-k • Notice that the C k N are always positive real numbers • However Γ L can be positive or negative depending on whether R L >Z 0 or vice-versa • Thus the Γ k ‘s are all either positive or negative and we should have uniformly decreasing or increasing line impedances • Now that we know the Γ k ‘s we need to determine the characteristic impedances of the lines to obtain these Γ k ‘s EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 12 Approximate Method for Finding the Z k ’s • Since the Γ k are given by Γ k =(Z k+1 -Z k )/(Z k+1 +Z k ) we could write a system of equations and solve it to find Z k values. • Note that in this approximate theory we have actually more equations than variables which is a reminder that our design is approximate. • However, if Z k+1 is not too different from Z k (required to keep our theory of small reflections valid), we have seen that we can approximate Γ k by : − Z Z Z 1 + + Γ = ≈ k 1 k k 1 ln k + Z Z Z 2 + k 1 k k • This is the recommended approach as it yields self-consistent solution EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 13 Approximate Method for Finding the Z k ’s • Since we know the Γ k ’s we can obtain the following expression   − Z R Z R + − −   = Γ = ≈ N N N N k 1 L 0 L ln 2 2 2 C 2 C ln   k + k k   Z R Z Z k L 0 0 • where the last expression is valid provided R L and Z 0 are within around a factor of 2 or ½. • We can step through the above equation starting at k=0 to find all the impedance values in the transformer. EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 14 2 Section Example Let us consider a two section example N=2 • Starting from   − Z R Z R + − −   = Γ = ≈ k 1 N L 0 N N N L ln 2 2 2 C 2 C ln   k + k k   Z R Z Z k L 0 0 we have : 1 / 4   Z R R 1   = = = L L 1 ln ln ln for k 0     Z 4 Z Z 0 0 0 1 / 2   Z 1 R R   = = = 2 L L for k ln ln ln 1     Z 2 Z Z 1 0 0 EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 15 2 Section Example The resulting characteristic impedances are then = = 1 / 4 1 / 4 3 / 4 1 / 4 Z R Z Z R Z , L L 1 0 2 0 • Note that for k=2 solution using this method is not accurate. This is due to the approximate nature of the design equations we have derived. Pozar (pg 281) provides a table with more accurate values. EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 16 Design Example Design a three section binomial transformer to match a load R L =100 ohms to a 50 ohm source. If a reflection coefficient of 0.1 is considered tolerable, what is the relative bandwidth obtained? Since Γ m =0.1, 1 / 3   Γ + R Z − m L θ = 0 = 1   cos 0 . 8374 rads m −   R Z   L 0 θ π = The relative bandwidth is then 2 - 4 / 0.9339 or 93.39% m = The binomial coefficien ts for N 3(expandin g a cubic) are = = = = 0 1 2 3 C 1 C 3 C 3 C 1 3 3 3 3 Γ The ' s are thus i Γ = = Γ Γ = = Γ 0.0417 3(0.0417) 0 3 1 2 EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 17 Design Example Γ Instead of using the exact we use : k   − Z R Z R + − −   = Γ = ≈ N N N N k 1 L 0 L ln 2 2 2 C 2 C ln   k + k k   Z R Z Z k L 0 0 Find Z' s from Z R − = = 1 3 3 L ln 2 C ln 0 . 0866 0 Z Z 0 0 Z R − = = 3 3 L 2 ln 2 C ln 0 . 2599 1 Z Z 1 0 Z R − = = 3 3 3 L C ln 2 ln 0 . 2599 2 Z Z 2 0 EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 18 Design Example = = = Solving gives Z 54.523, Z 70.7055, Z 91.6909 1 3 2 These are uniformly increasing values between 50 ohm and 100 ohm. These values are very close to those obtained using the table in Pozar. EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 19 Result 0.35 Exact Binomial 0.3 Approximate Design 0.25 0.2 in Γ 0.15 0.1 0.05 0 0 0.2 0.4 0.6 0.8 1 θ / π EE 611 Lecture 6 Jayanta Mukherjee

IIT Bombay Page 20 Chebyshev Transformer • To improve on the maximally flat transformer we consider next the Chebyshev transformer. • We will get more bandwidth for a given Γ in if we allowed some ripples inside the passband (i.e. between θ = θ m and θ = π - θ m ) 0.14 N=1 N=2 • No perfect match at center 0.12 N=3 N=4 Frequency for N even 0.1 0.08 Γ (in) • Γ in has N zeros 0.06 0.04 0.02 0 0 0.2 0.4 0.6 0.8 1 θ / π EE 611 Lecture 6 Jayanta Mukherjee