$\mathcal Z$-transform if the output is given

Question

A impulse response for a LTI system is given by:

$$h[n]=\left(\frac{2}{3}\right)^n u[n]+2 \left(\frac{1}{5}\right)^n u [n]$$

and if the putput for the system is given by:

$$y[n]= \left(\frac{1}{3}\right)^n u [n] $$

• What is the input in the $z$-domain ($X(z)$)?
• And what is the output in the time-domain ?

I'm not sure what is the correct procedure for solving this type of problem. Can you just simply transform $y[n]$ into $Y(z)$ and then solve for $X(z)$?

$$\frac{Y(z)}{X(z)}=\frac{3-\frac{23}{12}z^{-1} }{\left(1-\frac 23z^{-1}\right)\left(1-\frac 15z^{-1}\right) }$$

Then I replace $ Y(z)$ with $\frac{1}{1-1/3z^{-1}}$ and then I isolate $ X(z)$.

$$\frac{\frac{1}{1-\frac 13z^{-1}}}{X(z)}=\frac{3-\frac{23}{12}z^{-1} }{\left(1-\frac 23z^{-1}\right)\left(1-\frac 15z^{-1}\right)}.$$

Then I get $X(z)=\frac{6z^-2-39z^{-1}+45}{23z^{-1}-11z^{-1}+135}$ and then I transform it into a sequence in the time domain.

Is it right? or is there a better method for solving this.

Answer 1

Since

$$Y(z) = H(z)X(z)$$ therefore $$X(z)=\frac{Y(z)}{H(z)}$$

Now, $$y[n]=\big(\frac{1}{3}\big)^n u[n] \longleftrightarrow Y(z)=\frac{1}{1-\frac{1}{3}z^{-1}}$$ and $$h[n]=\big(\frac{2}{3}\big)^n u[n] + 2\big(\frac{1}{5}\big)^n u[n] \longleftrightarrow H(z)=\frac{1}{1-\frac{2}{3}z^{-1}}+\frac{2}{1-\frac{1}{5}z^{-1}}$$

will you please perform the necessary arithmetic... to find $$X(z) = \frac{\frac{1}{1-\frac{1}{3}z^{-1}}}{\frac{1}{1-\frac{2}{3}z^{-1}}+\frac{2}{1-\frac{1}{5}z^{-1}}}$$

Lets also display the further steps: \begin{align} H(z) &=\frac{1}{1-\frac{2}{3}z^{-1}}+\frac{2}{1-\frac{1}{5}z^{-1}} = \frac{3-\frac{23}{15}z^{-1}}{(1 - \frac{2}{3}z^{-1})(1-\frac{1}{5}z^{-1})} = \frac{3(1-\frac{23}{45}z^{-1})}{(1 - \frac{2}{3}z^{-1})(1-\frac{1}{5}z^{-1})}\\ Y(z) &=\frac{1}{1-\frac{1}{3}z^{-1}}\\ \\ X(z) &= Y(z)/H(z) = \frac{1}{1-\frac{1}{3}z^{-1}} \frac{(1 - \frac{2}{3}z^{-1})(1-\frac{1}{5}z^{-1})}{3(1-\frac{23}{45}z^{-1})}\\ \\ X(z) &= \frac{\frac{1}{3}(1 - \frac{2}{3}z^{-1})(1-\frac{1}{5}z^{-1})}{(1-\frac{1}{3}z^{-1})(1-\frac{23}{45}z^{-1})} ~~~~~~\scriptstyle{\text{Performing partial fraction expansion on X(z), results in:}}\\ \\ X(z) &= 0.26087 + \frac{0.25}{1-\frac{1}{3}z^{-1}} + \frac{-0.17754}{1-\frac{23}{45}z^{-1}} ~~~~\scriptstyle{\text{Transform each term back into time by a table...}}\\ \\ x[n] &= 0.26087 \cdot \delta[n] + 0.25 \big(\frac{1}{3}\big)^n u[n] - 0.17754 \big(\frac{23}{45}\big)^n u[n] \\ \end{align}

Sorry that the partial fraction expansion resulted in numeric format, rather than algebraic...

Answer 2

I haven't checked your solution, but the procedure appears correct:

$$x[n]=\mathcal{Z}^{-1}\left\{\frac{\mathcal{Z}\{y[n]\}}{\mathcal{Z}\{h[n]\}}\right\}$$