partial fractions expansion inverse Z-transform, help - Signal Processing Stack Exchange most recent 30 from dsp.stackexchange.com 2019-07-22T22:07:51Z https://dsp.stackexchange.com/feeds/question/53526 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://dsp.stackexchange.com/q/53526 1 partial fractions expansion inverse Z-transform, help Late347 https://dsp.stackexchange.com/users/38885 2018-11-21T16:03:05Z 2019-05-02T22:01:55Z <p>I have the correct solution from teacher's solution guide, but I was slightly confused by some algebra about the partial fractions expansion evidently</p> <p>difference equation is as follows</p> <p><span class="math-container">$y[n] = \frac{1}{4} \cdot \left( x[n] +x[n-1]+x[n-2]+x[n-3] \right)$</span></p> <p>where <span class="math-container">$x[n]$</span> will be unit-step sequence <span class="math-container">$u[n]$</span></p> <p><strong>PROBLEM STATEMENT:</strong></p> <p>Find what will the output sequence y[n] when input x[n] = u[n] = unit step</p> <p>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]</p> <p>we know from earlier calculation that the transfer function for this system willbe as follows</p> <p><span class="math-container">$H(z) = \frac{z^3 + z^2 + z +1}{4z^3} = \frac{1}{4} * \frac{z^3 + z^2 + z +1}{z^3}$</span></p> <p>because unit step input sequence</p> <p><span class="math-container">$X(z) = \frac{z}{z-1}$</span></p> <p>We can find out <span class="math-container">$Y(z) = X(z) * H(z)$</span></p> <p>I had some difficulty with the partial fraction expansion because of the <span class="math-container">$\frac{1}{4}$</span> multiplier, how does that constant multiplier term become accounted into the partial fraction expansion?</p> <p>We were tipped by our teacher to find out <span class="math-container">$Y(z)/z$</span> , 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...</p> <p><span class="math-container">$Y(z)/z = 1/4 * \frac{z^3 + z^2 + z +1}{z^3*(z-1)}$</span></p> <p>So, do we just "ignore" the constant multiplier <span class="math-container">$\frac{1}{4}$</span> on the r.h.s of the partial fraction as follows???</p> <p><span class="math-container">$Y(z)/z = 1/4 * \frac{z^3 + z^2 + z +1}{z^3*(z-1)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z^3} + \frac{D}{z-1}$</span></p> <p>So, after the "equating coefficients technique" I obtain followins system of equations</p> <p>4A + 4D = 1</p> <p>-4A + 4B = 1</p> <p>-4B +4C = 1</p> <p>1 = -4C</p> <p>from these I obtianed coefficients as follows</p> <p>A= -3/2</p> <p>B= -1/2</p> <p>C= -1/4</p> <p>D= 1</p> <p>With these results we should multiply both sides by z, so we get Y(z) in partial fractions format</p> <p><span class="math-container">$Y(z) = A + B/z + \frac{C}{z^2} + \frac{D*z}{z-1}$</span></p> <p>From here, we should be able to find inverse-Z-transforms from the table as follows</p> <p><span class="math-container">$Z^{-1}(Y(z)) = -3/4 * Z^{-1}(1) - 1/2 * Z^{-1}(1*z^{-1}) - 1/4 * Z^{-1}( 1 * z^{ -2 } ) + Z^{-1}( \frac{ z }{ z-1 } ))$</span></p> <p><span class="math-container">$y[k] = -3/4 * \delta[k] - 1/2 * \delta[k-1] - 1/4 * \delta[k-2] +u[k]$</span></p> <p>where <span class="math-container">$\delta[k]$</span>, is the unit impulse, aka the deltafunction</p> <p>I put that into my Texas instruments calculator and the function seemed to hold true for some initial values</p> <p>expected values are that for negatives indexes output will be zero</p> <p>then for k=0: output = 1/4</p> <p>k=1: output = 2/4</p> <p>k=2: output = 3/4</p> <p>k>=3: output = 4/4</p> https://dsp.stackexchange.com/questions/53526/-/53533#53533 1 Answer by Fat32 for partial fractions expansion inverse Z-transform, help Fat32 https://dsp.stackexchange.com/users/13309 2018-11-21T21:17:17Z 2018-11-21T21:17:17Z <p>First convert your input-output relation to Z-domain to see:</p> <p><span class="math-container">$$Y(z) = 0.25 \left( 1 + z^{-1} + z^{-2} + z^{-3} \right) X(z)$$</span></p> <p>From which you find the transfer function of the LTI system as:</p> <p><span class="math-container">$$H(z) = \frac{Y(z)}{X(z)} = 0.25 \left( 1 + z^{-1} + z^{-2} + z^{-3} \right)$$</span></p> <p>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 <span class="math-container">$H(z) = \sum h[k]z^{-k}$</span> and for such a system the output is defined through a <strong>convolution</strong> as the following: </p> <p><span class="math-container">$$Y(z) = H(z) X(z) = \left( \sum h[k]z^{-k} \right) X(z) \implies y[n] = \sum h[k]x[n-k]$$</span></p> <p>so for your example the proper way to describe the output for the input <span class="math-container">$x[n]= u[n]$</span> is</p> <p><span class="math-container">$$\boxed{ y[n] = 0.25 ~( u[n] + u[n-1] + u[n-2] + u[n-3]) }$$</span></p> <p>Nevertheless, assuming you want to insist on a partial fraction expansion to compute the output <span class="math-container">$y[n]$</span> then you would do the following:</p> <p>Given <span class="math-container">$x[n] = u[n]$</span> and its Z-transform <span class="math-container">$X(z) = \frac{1}{1-z^{-1}}$</span>, then the output is:</p> <p><span class="math-container">$$Y(z) = \frac{1}{4} ~\frac{ 1 + z^{-1} + z^{-2} + z^{-3} } {1 -z^{-1} }$$</span></p> <p>Now, <strong>ignoring</strong> the linear scale factor <span class="math-container">$1/4$</span> to the end, since the degree of numerator polynomial is larger (in negative) than that of the denominator polynomial, you should first apply a <strong>long division</strong> of numerator polynomial into denominator polynomial to get :</p> <p><span class="math-container">$$\frac{ 1 + z^{-1} + z^{-2} + z^{-3} } {1 -z^{-1} } = -(3 + 2 z^{-1} + z^{-2}) + \frac{ 4} {1 -z^{-1} }$$</span></p> <p>hence you have this (including the scale factor): <span class="math-container">$$Y(z) = -(\frac{3}{4} + \frac{2}{4} z^{-1} + \frac{1}{4} z^{-2}) + \frac{1} {1 -z^{-1} }$$</span></p> <p>And finally you get the output from the PFE method via inverse Z-transformas: <span class="math-container">$$\boxed{ y[n] = u[n] - \frac{3}{4} \delta[n] - \frac{2}{4} \delta[n-1] - \frac{1}{4} \delta[n-2] }$$</span></p>