# Signal flow graphs

Signal flow graphs solved examples

Consider the following two-port network

A signal flow graph is a diagram depicting the relationships between signals in a network. It can also be used to solve for ratios of these signals. Signal flow graphs are used in control systems, power systems and other fields besides microwave engineering. Key elements of a signal flow graph are:
1. The network must be linear,
2. Nodes represent the system variables,
3. Branches represent paths for signal flow.
For example, referring to the two-port above, the nodes and branches are

4. A signal yk traveling along a branch between nodes ak and bjis multiplied by the gain of that branch:

That is, bj = Sjk ak
5. Signals travel along branches only in the direction of the arrows.
This restriction exists so that a branch from ak to bj denotes a proportional dependence of bj on ak , but not the reverse.
Solving Signal Flow Graphs
Signal flow graphs (SFGs) can form an intuitive picture of the signal flow in a network. As an application, we will develop SFGs in the next lecture to help us calibrate out systematic errors present when we make measurements with a VNA. Another useful characteristic is that we can solve for ratios of signals directly from a SFG using a simple algebra. There are four rules that form the algebra of SFGs:
1. Series Rule. Given the two proportional relations then
V2 = S21V1 and V3 = S32 V2
V3= (S32S21)V1
In a SFG, this is represented as

In other words, two series paths are equivalent to a single path with a transmission factor equal to a product of the two original transmission factors.
2. Parallel Rule. Consider the relation:
V2 = Sa V1 + Sb V1 =(Sa +Sb) V1
In a SFG, this is represented as

In other words, two parallel paths are equivalent to a single path with a transmission factor equal to the sum of the original transmission coefficients.
3. Self-Loop Rule. Consider the relations
V2 = S21 V1 + S22 V2
and V3 = S32V2 We will choose to eliminate V2 . From
V2(1-S22)=S11V1 ÞV2=(S21/1-S22)V1
Substituting this into (4) gives
V3=(S32S21/1-S22)V1
In a SFG, this is represented as

In other words, a feedback loop may be eliminated by dividing the input transmission factor by one minus the transmission factor around the loop.
4. Splitting Rule. Consider the relationships:
V4 = S42 V2 V2 = S21 V1 and V3=S32V2
The SFG is

From (6) and (7) we find that V4 = S42S21V1. Hence, if we use the Series Rule “in reverse” we can define:
V4 S42 V4= and V4= S21 V1
In a SFG, this is represented as

In other words, a node can be split such that the product of transmission factors from input to output is unchanged.
Example N21.1. Construct a signal flow graph for the network shown below. Determine tin and VL using only SFG algebra.

The signal flow graph is:

Notice the arrow directions for Ts and TL . These are the correct orientations since

And

We will systematically apply the four rules above to reduce this diagram to a form that will directly allow us to determine both Tin and VL .
Starting this solution process is probably the most difficult part. The rest is fairly systematic.
Step 1. Start by splitting node a2 using Rule 4:

Why split TL and not S22 ? Because the TL arrow is in the same direction as S12 .
Step 2. Eliminate nodes a2 and a2 using Rule 1:

Step 3. Eliminate the self loop at b2 using Rule 3:

Step 4. Eliminate node b2 using Rule 1:

Step 5. Apply Rule 2:

From this last diagram we can directly solve for Tin :

Next, we will determine the load voltage VL . The voltage on this
TL can be expressed as V2( z) = b2 (e-jbz + TL e-jbz) . At the terminal plane z = 0, then
V2(0) =VL = b2(1+TL)
In this expression, V2(0) =VL because the TL is very short. So, we see from (9) that to find VL we need to determine b2 . In Step 4, however, we eliminated that node. Let’s start again from Step 3, but now split node b1 :
Step 4?. Split node b1 using Rule 4:

Step 5?. Next, we can use Rule 1, then Rule 3 (Self-Loop Rule) to all branches feeding node a1 :

Step 6?. Split node b2 using Rule 4:

Step 7?. Apply Rule 3 one last time to remove the self loop:

Step 8?. Using the Series Rule, we can now find b2 as:

Or

Using this b2 , we can determine VL from (9) to be