3-Coefficient Differentiator
Suppose we sample a time sinusoid at discrete time intervals. The angular frequency of the sinusoid is ω and the sampling time interval is T . The sampled signal looks like:
y k = A sin ( ω k T )
In the above equation, k is an integer denoting the sample number. Suppose we define two discrete differentiators: one with two coefficients (the classic difference quotient), and another with three coefficients.
f 2 = T 1 y k − T 1 y k − 1 f 3 = T 1 . 5 y k − T 2 y k − 1 + T 0 . 5 y k − 2
If we take the z-transform of both of the above equations, we obtain the following transfer functions (the " z " in parentheses denotes functions of z ):
Y ( z ) F 2 ( z ) = H 2 ( z ) Y ( z ) F 3 ( z ) = H 3 ( z ) z = e j ω T j = − 1
For a sinusoid, the ideal transfer function is H i d e a l = j ω . We can therefore say that the error of the differentiator is H a c t u a l − j ω . As it turns out, the three-coefficient differentiator has a better response for some frequencies (given some sampling rate).
Calculate the following error ratio:
∣ H 3 − j ω H 2 − j ω ∣
Details and Assumptions:
-
ω
=
1
2
0
π
-
T
=
8
0
0
0
1
-
∣
⋅
∣
denotes the modulus of a complex number
-
Hint:
Use the time-shift property of the z-transform to get the transfer functions
The answer is 31.834.
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2 solutions
Interesting observations! Thank you for the problem and for the insights. My takeaway from this problem is that while differentiating high frequency and sampled signals, it is always better to use the 3 coefficient differentiator as opposed to a simple (first order) Taylor series approximation (2 coefficient).
If your application is sensitive to both magnitude and phase, then you are probably better off with the 3-coefficient version. The downside though is that it has a 50% greater buffer depth, and takes correspondingly longer to "charge". It also requires more computation to implement. So it depends on the desired balance between accuracy, speed, and computational burden.
Consider the first difference equation:
f 2 ( k ) = T y ( k ) − y ( k − 1 )
Taking the Z-Transform on both sides assuming no initial conditions gives us:
F 2 ( z ) = T 1 − z − 1 Y ( Z )
This Implies:
H 2 ( z ) = T 1 − z − 1
Doing the same for the second difference equation:
f 3 ( k ) = T 1 . 5 y ( k ) − 2 y ( k − 1 ) + 0 . 5 y ( k − 2 )
Taking the Z-Transform on both sides, assuming no initial conditions and dividing by Y(z) gives us:
H 3 ( z ) = T 1 . 5 − 2 z − 1 + 0 . 5 z − 2
Now, it is already given that
z = e j ω T = cos ( ω T ) + j sin ( ω T )
This means:
z − 1 = e − j ω T = cos ( ω T ) − j sin ( ω T )
And:
z − 2 = e − 2 j ω T = cos ( 2 ω T ) − j sin ( 2 ω T )
The values of T and ω are known.
Plugging this into each of the transfer functions H 2 ( z ) , H 3 ( z ) gives us two complex numbers in terms of T and ω .
These complex numbers are, in turn, plugged into the required expression and the modulus of the resulting complex number is computed. The substitutions and simplifications are straightforward, but are a bit tedious. The answer comes out to be 31.8339
Nice solution, thanks. The 2-coefficient differentiator has a slightly better magnitude response, but the 3-coefficient differentiator has a much better phase response. Maybe I'll upload some plots showing that.
See the solution by @Karan Chatrath . I will just add some plots here. The ideal normalized magnitude response is 1 (after normalizing by ω ), and the ideal phase response is 90 degrees. We can see from the plots that the 2-coefficient differentiator has a slightly better magnitude response, and that the 3-coefficient differentiator has a substantially better phase response.