Hilbert Transforms
Hilbert transforms are used to change the phase of
a signal by 90°. For example
H cos  0 t   sin  0 t .
What is the process by which this operation is
performed?
Let us consider this operation in frequency domain.
We start with the Fourier transforms of cosine and
sine:
d(+0)
F {cos 0t}
d(-0)

j d(+0)
F {sin 0t}
-j d(-0)

What frequency operation is performed to go from
sine to cosine?
A useful function in formulating this operation is the
signum or sign function:
1 (   0 ),
sgn(  )  
  1 (   0 ).
This function is like a unit step function with a nonzero negative part.
sgn ()

We readily see that, in frequency domain, the Hilbert
transform is equivalent to multiplication by
 j sgn(  )
as can be seen on the following slide.
F {cos 0t}

-j sgn ()

F {sin 0t}
j

-j
So, a Hilbert transform can be reduced to a filtering
operation, where
H (  )   j sgn(  ).
We can design a digital filter with this frequency
response:
H d ( )   j sgn(  ).
The coefficients cn for the digital filter can be found
using the Fourier series method:
cn 

1
2
1
2




H d ( ) e


 jn 
j sgn(  ) e
d
 jn 
j  0
 jn 

(1) e
d 

2   

d

0
(  1) e
 jn 
d 


j  1
1
jn 
 jn 

1 e

e
1 

2   jn
 jn





1
2 n
1
2 n
1
n
 2  e
)
jn 
 2  2 cos
  1  cos
n
e
n
)

 jn 

)
)
The plot of the magnitude and phase of this filter is
shown on the following slide.
Frequency Response
2
j
|H(e )|
1.5
r = 20
r = 50
1
r=5
0.5
0
-0.5
-1.5
-1
-0.5
-1
-0.5
0
0.5
1
1.5
0
0.5
1
1.5
, x 
j
H(e ), x 
1
0.5
0
-0.5
-1
-1.5
, x 