djoint networks and other elements of circuit theory
One-port reciprocal networks one-port network is reciprocal if: V I I V = Where and are two different tests on the element Example: a linear impedance (R,L,C) is reciprocal, since V V = ZI V I I V = ZI I ZI I = = ZI Note: non-linear element will usually be non-reciprocal, since V = f ( I) nd there is every reason to expect that at two different operating points: ( ) ( ) V I I V = f I I f I I
3 statements: -port reciprocal networks u I I u I =I u V V u V =V u I V u V =I
n-port reciprocal networks Can be proven that an n-port network is reciprocal if: n ( V ) n In InVn = The summation taken over the n-ports If the network is linear we can write the vector I in terms of V and the admittance matrix: V V T Vn In InVn = V V n Y V V n Y n V n V n We conclude that a n-port is reciprocal if it is linear and possesses a symmetric admittance matrix.
More on reciprocal networks Networks made out of resistors, capacitors, inductors, and mutual inductors are reciprocal Networks which include controlled sources and gyrators are not reciprocal We focus on two-port networks. Reciprocal networks have: y = y, z = z, h = h, g = g, D C =
Graphs and incidence matrices bstract a circuit as a directed graph Each circuit element becomes a directed line 4 3 3 5 5 6 4 Write the incidence matrix: # nodes as rows # branches as columns Entries: if branch leaves node, - if branch arrives at node, if no connection
The incidence matrix 4 3 3 5 5 6 4 The incidence matrix of the above is 6 branches 3 4 5 6 = 5 nodes 3 4 5
CL and Tellegen s theorem Let I be the column vector of the branch currents irchhoff's current law is: (!) I T =[] (row vector of zeroes, one for each node) I=[] (column vector of zeroes, one for each node) Tellegen s Theorem: If two networks, have the same incidence matrix, then: n branches V I If, refer to the same network this is the statement of conservation of power: s much power as generated by sources will be dissipated by the rest of the network. Exercise: show that any network with internal RLC components only is reciprocal. (hint: apply Tellegen s theorem with an input applied only to P or to P and argue that the rest of the terms in the sum are identical for both configurations; after all it IS the same network, so it does have the same graph!) n n =
djoint networks Two n-port networks, are inter-reciprocal or adjoint if: n ( V ) n In InVn = The previous observations apply: Two networks are adjoint if: y = y, z = z, g = g, h = h Note that the reverse gain of a voltage amplifier (g parameters) is a current gain lso, the reverse gain of a current amplifier (h parameters) is a voltage gain We expect, therefore the adjoint of a voltage amplifier to be a reversely connected voltage amplifier which acts like a current amplifier.
The adjoint of an ideal voltage amplifier From the reciprocity theorem it is evident that a current amplifier of the same gain will be the adjoint of a voltage amplifier. V u u I V =I Note that the thevenin impedances of sources and loads remain the same! lso note that the current gain is the negative of the voltage gain
Constructing the adjoint of a circuit Leave the resistors, capacitors, and inductors as they are. Replace voltage sources by short circuits (i.e. current sinks the inputs become outputs) (and vice-versa) Replace voltage meters by current sources (i.e. the outputs become inputs (and vice-versa) The following circuits have the same function, one in voltage, the other in current mode. otha are Sallen-ey filters
nd order filter transfer functions ( iquads ): Review Second order filter transfer functions are all of the following form: ( ) ( s ) ω ζs ω C s/ ω + ζs/ ω + H( s) = H, Q= / + / + ζ H is the overall amplitude, ω the break (or peak) frequency, and ζ the damping factor ζ isrelated to the quality factor Q by: Q=/ζ The 3d bandwidth of an underdamped nd order filter is approx /Q times the peak frequency. Function Low Pass High Pass and Pass C The coefficients,, C determine the function of the filter: and Stop ll Pass -
nd order low pass passive RC filter R R Vin C C Vout ( ) H s = = s RCRC + s RC + RC + RC + s + s + + + ( ) ττ ( τ τ τ ) τ τ τ = /, = = + + > Q τ τ τ ω ττ ζ Since the minimum value of x+/x is It follows that passive RC nd order filters are OVERDMPED The passive band pass filter transfer function calculation is part of experiment Y in the lab.
The Sallen ey Low Pass Filter () Vin R C R C Vout + H y superposition, there are: n RC LPF in the forward signal path, of gain: + + + + = srcrc s RC RC RC ( ) n RC PF in the (positive) feedback path, reinforcing Q sr C = s RCRC + s ( RC + RC + RC ) +
The Sallen ey Low Pass filter () + From the block diagram it follows that H = and are both rational functions, with the same denominator: Q s, sr C = = ( ) Q( s) H = = Q RC srcrc + s RC + RC + RC + H (( ) )
Vin R The Sallen ey Low Pass filter (3) C R C = RCRC ω = ω H = RCRC Vout H H = = s / + s / + srcrc + s RC + RC + RC + ω ζ ω ζ RC RC RC = = ( ) RC + RC + R C ζ = = ( ) + + ω Qω Q RC RC RC For large enough the circuit will have Q< and will become dynamically unstable, i.e. it will become an oscillator + H (( ) )
The Sallen ey High pass filter Vin C R C R Vout + H y superposition, there are: n RC HPF in the forward signal path n RC PF in the (positive) feedback path, reinforcing Q nalysis very similar to that of the S-LPF Detailed calculation left as a homework problem
The Sallen ey and pass filter R3 Vin R C C R Vout + H This has identical in form passive band pass filters in the forward and feedback paths, shown on the middle. The block diagram in the right is the same form as the other S filters. If R =R 3 then the two filters are identical and =. The transfer function of each path filter is: sτ = =, τ = RC, τ = RC, τ = RC s ττ + ( τ+ τ+ τ) s+ The entire S filter has a transfer function: H H = = sτ H s ττ /+ τ + τ + τ s/+ / (( ) )
The Sallen ey Notch filter Vin R C C R C R/ Vout + H Networks, may be solved by nodal analysis or any other suitable method.
Tee Pi transformations dual networks When analysing active band pass or band stop active filters we often encounter the twin-tee passive notch filter topology This requires quite a bit of algebra to compute, so we prove a, useful for simplifying networks, lemma: Y3 Z Z is equivalent to Y Y Z3 Proof: write the z matrix of the Tee< and the y matrix of the Pi and require that the two circuits are representations of same network: Z+ Z3 Z3 Y+ Y3 Y3 ZTee =, Pi, Pi Tee Z3 Z Z Y = 3 Y3 Y Y Y = Z + + 3 Y Y Y3 Z =, Z =, Z3 = YY + YY 3+ YY 3 YY + YY 3+ YY 3 YY + YY 3+ YY 3 Z Z Z3 Y =, Y =, Y3 = ZZ + ZZ3+ ZZ3 ZZ + ZZ3+ ZZ3 ZZ + ZZ3+ ZZ3