# partial fractions expansion inverse Z-transform, help

I have the correct solution from teacher's solution guide, but I was slightly confused by some algebra about the partial fractions expansion evidently

difference equation is as follows

$$y[n] = \frac{1}{4} \cdot \left( x[n] +x[n-1]+x[n-2]+x[n-3] \right)$$

where $$x[n]$$ will be unit-step sequence $$u[n]$$

PROBLEM STATEMENT:

Find what will the output sequence y[n] when input x[n] = u[n] = unit step

We can use tactic to get X(z) and multiply X(z)*H(z) so we get Y(z) Then inverse Z-transform to get y[k]

we know from earlier calculation that the transfer function for this system willbe as follows

$$H(z) = \frac{z^3 + z^2 + z +1}{4z^3} = \frac{1}{4} * \frac{z^3 + z^2 + z +1}{z^3}$$

because unit step input sequence

$$X(z) = \frac{z}{z-1}$$

We can find out $$Y(z) = X(z) * H(z)$$

I had some difficulty with the partial fraction expansion because of the $$\frac{1}{4}$$ multiplier, how does that constant multiplier term become accounted into the partial fraction expansion?

We were tipped by our teacher to find out $$Y(z)/z$$ , so that we can find good matches in our z-transforms-table. So, I try to make Y(z)/z into the partial fraction expansion...

$$Y(z)/z = 1/4 * \frac{z^3 + z^2 + z +1}{z^3*(z-1)}$$

So, do we just "ignore" the constant multiplier $$\frac{1}{4}$$ on the r.h.s of the partial fraction as follows???

$$Y(z)/z = 1/4 * \frac{z^3 + z^2 + z +1}{z^3*(z-1)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z^3} + \frac{D}{z-1}$$

So, after the "equating coefficients technique" I obtain followins system of equations

4A + 4D = 1

-4A + 4B = 1

-4B +4C = 1

1 = -4C

from these I obtianed coefficients as follows

A= -3/2

B= -1/2

C= -1/4

D= 1

With these results we should multiply both sides by z, so we get Y(z) in partial fractions format

$$Y(z) = A + B/z + \frac{C}{z^2} + \frac{D*z}{z-1}$$

From here, we should be able to find inverse-Z-transforms from the table as follows

$$Z^{-1}(Y(z)) = -3/4 * Z^{-1}(1) - 1/2 * Z^{-1}(1*z^{-1}) - 1/4 * Z^{-1}( 1 * z^{ -2 } ) + Z^{-1}( \frac{ z }{ z-1 } ))$$

$$y[k] = -3/4 * \delta[k] - 1/2 * \delta[k-1] - 1/4 * \delta[k-2] +u[k]$$

where $$\delta[k]$$, is the unit impulse, aka the deltafunction

I put that into my Texas instruments calculator and the function seemed to hold true for some initial values

expected values are that for negatives indexes output will be zero

then for k=0: output = 1/4

k=1: output = 2/4

k=2: output = 3/4

k>=3: output = 4/4

• Are you required to use partial fractions? Because I think the solution can be obtained much more easily. – Matt L. Nov 21 '18 at 16:50
• not required but what did u have in mind?, just plug-in u[n] into x[n]? i dont know what else could be done... – Late347 Nov 21 '18 at 17:07
• Yes, sure, as simple as that. – Matt L. Nov 21 '18 at 17:11
• yea, I suppose my teacher had that inverse-Z-transformed sequence as the correct answer... but I suppose both styles would in actual fact produce the same output sequence y[k]... It's just that the plug-in technique gives output as sums of unit step terms, where as the laborious technique has those delta terms and a unit step term... – Late347 Nov 21 '18 at 17:14
• It's just a matter of taste how you express the solution. You can easily express the sum of shifted steps in terms of deltas for the first 3 samples. – Matt L. Nov 21 '18 at 17:16

First convert your input-output relation to Z-domain to see:

$$Y(z) = 0.25 \left( 1 + z^{-1} + z^{-2} + z^{-3} \right) X(z)$$

From which you find the transfer function of the LTI system as:

$$H(z) = \frac{Y(z)}{X(z)} = 0.25 \left( 1 + z^{-1} + z^{-2} + z^{-3} \right)$$

Now this is an FIR system with 4 taps. As MattL. tried (without luck) to convince you that for such a system a partial fraction expansion is not the proper way to compute the output. Indeed for any FIR system, the transfer function is of the form of a $$H(z) = \sum h[k]z^{-k}$$ and for such a system the output is defined through a convolution as the following:

$$Y(z) = H(z) X(z) = \left( \sum h[k]z^{-k} \right) X(z) \implies y[n] = \sum h[k]x[n-k]$$

so for your example the proper way to describe the output for the input $$x[n]= u[n]$$ is

$$\boxed{ y[n] = 0.25 ~( u[n] + u[n-1] + u[n-2] + u[n-3]) }$$

Nevertheless, assuming you want to insist on a partial fraction expansion to compute the output $$y[n]$$ then you would do the following:

Given $$x[n] = u[n]$$ and its Z-transform $$X(z) = \frac{1}{1-z^{-1}}$$, then the output is:

$$Y(z) = \frac{1}{4} ~\frac{ 1 + z^{-1} + z^{-2} + z^{-3} } {1 -z^{-1} }$$

Now, ignoring the linear scale factor $$1/4$$ to the end, since the degree of numerator polynomial is larger (in negative) than that of the denominator polynomial, you should first apply a long division of numerator polynomial into denominator polynomial to get :

$$\frac{ 1 + z^{-1} + z^{-2} + z^{-3} } {1 -z^{-1} } = -(3 + 2 z^{-1} + z^{-2}) + \frac{ 4} {1 -z^{-1} }$$

hence you have this (including the scale factor): $$Y(z) = -(\frac{3}{4} + \frac{2}{4} z^{-1} + \frac{1}{4} z^{-2}) + \frac{1} {1 -z^{-1} }$$

And finally you get the output from the PFE method via inverse Z-transformas: $$\boxed{ y[n] = u[n] - \frac{3}{4} \delta[n] - \frac{2}{4} \delta[n-1] - \frac{1}{4} \delta[n-2] }$$

• This was interesting but is there some easy algebra to show that the two answers are equivalent. I think I get why they are ( 1-3/4 = 0.25 so you lose part of the step of initial step response ) but I don't know the general way to do it. Thanks. – mark leeds May 3 at 0:06