## Equivalence of Flowgraphs and Difference Equations The following figure shows a flowgraph of a filter which
contains one zero and one pole. This flowgraph can be converted
to a The first path of the signal is direct from left to right. There
are no delays or gains affecting the direct signal through the filter
on this path,
so the algebraic term for this piece is very simple: The second path is through the feedforward delay at the left
of the flowgraph. Going along this path, notice two element: a
delay and a scale (by a x[n-1] which means that the previous sample,
since x[n] can be thought of as the current sample. n
is the unit of time, so n-1 is the previous instant of time.
The value a is multiplied by the previous input
sample value since it is a constant gain on the signal going through
the feedforward delay.
_{1} The last signal path is a feedback from the output of the filter
and is not directly associated with the signal input -- only the
signal output. The algebraic expression for the feedback portion
of the filter is:
y[n]. Thus,
the value y[n-1] is the previous output from the filter.
The feedback path contains a gain of b which
is multiplied by the previous output to generate the third signal
component: _{1}b.
_{1} y[n-1]Now you should be able to figure out how to convert a flowgraph (in its generalize form) into a difference equation which can be implemented in a program written in a language such as C. |