Minimum-Phase
Systems
The phase response of a digital filter is dependent
upon the poles and zeroes of the transfer function.
Example: Find the phase response for the following
filter:
1
H ( z)  1  z .
Rather than simply plugging-in z=ejw, let us take a
geometrical approach to this problem.
z 1
H ( z) 
.
z
We have one zero at z=-1 and one pole at z=0.
The pole-zero plot is shown on the following slide.
Im{z}
Re{z}
We can treat the numerator and denominator of H(z)
as vectors.
z+1 = z-(-1) is a vector whose tip is at z and whose
tail is at –1.
z = z-(0) is a vector whose tip is at z and whose tail
is at 0.
For magnitude and frequency responses, we take
z=ejw – along the unit circle.
Im{z}
z
w
Re{z}
The phase response of the transfer function is
computed by taking the angle of the numerator (z+1)
and subtracting the angle of the denominator (z).
H ( z)  ( z  1)  ( z).
Now, plugging-in z=ejw, we see that
jw
H (e )  ( z  1)  ( z).
H (e )  0  0  0.
j0
j 2
H (e )  4  2   4 .
j



H (e )  2     2 .
Example: Find the phase response for the following
filter:
1
H ( z)  1  z .
1
2
Proceeding as we have before,
z  12
H ( z) 
.
z
Im{z}
z
w
Re{z}
jw
H (e )  ( z  )  ( z).
1
2
H (e )  0  0  0.
j0
j 2
H (e )  1.107 2  0.4636.

j
H (e )      0.
Example: Find the phase response for the following
filter:
1
H ( z)  1  2z .
Again, proceeding as before,
z2
H ( z) 
.
z
Im{z}
z
w
Re{z}
jw
H (e )  ( z  2)  ( z).
H (e )  0  0  0.
j0
j 2
H (e )  0.4636 2  1.1071.

j
H (e )  0     .
In the example
1
H ( z)  1  z .
1
2
the phase increased to a maximum value (at w =
/2), and then decreased to zero.
In the example
1
H ( z)  1  2z .
the phase increased monotonically to -.
This first case was an example of a minimumphase system.
This second case is an example of a maximumphase system.
Minimum-phase systems have zeroes inside the
unit circle.
Maximum-phase systems have zeroes outside the
unit circle.
It is often desirable to have a system whose phase
response is minimal (the phase lag or phase delay is
minimal).
In the last example, we clearly had a response
which was not minimal. It is possible, however, to
transform this filter into one with a minimal response.
Suppose we take our transfer function
H ( z)  1  2 z
1
and multiply and divide it by H(z-1) = 1 + 2z:
H ( z)
H ( z)  H ( z )
1
H (z )
1
1
1 2z
 (1  2 z )
.
1 2z
The first factor
1
1  2z  2z( z 1)
1
2
is minimum phase.
We denote the remaining factor
1
1 2z
H ap ( z ) 
1  2z
Let us examine the magnitude response of Hap(z):
w
ejw
0
1
/2
j

-1
Hap(ejw) |Hap(ejw)|
1  2(1)
1
1  2(1)
1  2( j )
1  2( j )
1  2(1)
1  2(1)
1
1
We see that Hap(z) is an all-pass filter.
The original transfer function can be represented as

H ( z)  2z 1  z
1
2
1
H
ap
( z)
or,
H ( z)  H min ( z) H ap ( z)
As it turns out, any filter (corresponding to a causal
system) can be decomposed into the following
product:
H ( z)  Hmin ( z) Hap ( z).
Example: Decompose the following transfer function
into the product of a minimum-phase transfer
function and an all-pass transfer function.

H ( z)  1  3z
1
1
1
4
z
1
.
Solution: to eliminate the zero at z=-3, we multiply
and divide by (1+3z):
1
1  3z
1 1
H ( z )  1  3z 
1 4 z
1  3z
1
1 1 1  3 z
 1  3z  1  4 z
.
 1  3 z







H min ( z )
H ap ( z )

H min ( z )  1  3z  1  z
1
1  3z
H ap ( z ) 
.
1  3z
1
4
1
.