# Output of a discrete-time LTI system different form than input?

This question is related to this one. I'm going through old exams for a 2nd year systems and transforms course, and came across this question. I'm posting this question just in case my other question doesn't actually represent this issue properly...

The question states:

Suppose that we are given the following information about an LTI system: if the input to the system is $x_1[n]=(1/6)^n u[n]$, then the output is $y_1[n]=[a(1/2)^n +10(1/3)^n]u[n]$. If $x_2[n]=(-1)^n$, then the output is $y_2[n]=(7/4)(-1)^n$ where $a$ is a real number. Determine the transfer function of this system and the value of the number $a$.

The solution proceeds by determining the z-transform of $x_1$ and $y_1$ and stating the transfer function as $H(z)=Y_1(z)/X_1(z)$.

However, it seems to me that it's impossible for this system to generate an output of $(1/2)^n+(1/3)^n$ if the input is $(1/6)^n$,and hence the question is somewhat contrived if not outright bogus - see the discussion pertaining to my other question.

What am I missing here?

• Your confusion stems from your ignoring of the unit step. Aug 6, 2015 at 16:46
• Could you please elaborate? Aug 6, 2015 at 16:54
• @Jazzmaniac means the $u[n]$ part of the input signal. Your input is not $(1/6)^n$. It's that, multiplied by the unit step.
– Peter K.
Aug 6, 2015 at 16:55
• Sorry - that part I got; what I meant was what effect that has on the question... But now I wonder if what the question is indicating is that $y_1[n]$ is the transient response of the system? Aug 6, 2015 at 16:56

If you compute the $\mathcal{Z}$-transforms of $x_1[n]$ and of $y_1[n]$ you get (as already shown in Brian's answer)

$$X_1(z)=\frac{1}{1-\frac16z^{-1}}\\ Y_1(z)=\frac{a}{1-\frac12z^{-1}}+\frac{10}{1-\frac13z^{-1}}=(a+10)\frac{1-\frac{a+15}{3(a+10)}z^{-1}}{(1-\frac12z^{-1})(1-\frac13z^{-1})}$$

The system's transfer function must be

$$H(z)=\frac{Y_1(z)}{X_1(z)}=(a+10)\frac{\left(1-\frac{a+15}{3(a+10)}z^{-1}\right)(1-\frac16z^{-1})}{(1-\frac12z^{-1})(1-\frac13z^{-1})}\tag{1}$$

From the relation between $x_2[n]$ and $y_2[n]$ we know that $H(-1)=7/4$, because for an input signal $b^n$, the output signal of a discrete-time LTI system with transfer function $H(z)$ is $H(b)\cdot b^n$ (see also this answer). If you plug that into $(1)$ you indeed get $a=-9$. With this value for $a$, the transfer function becomes

$$H(z)=\frac{(1-2z^{-1})(1-\frac16z^{-1})}{(1-\frac12z^{-1})(1-\frac13z^{-1})}\tag{2}$$

Note that the output signal $y_1[n]$ only reflects the system's poles, i.e. the transient response, because the steady state response is zero due to the system's zero at $z=1/6$. Note that the system's reaction to the input signal $x_3[n]=(\frac16)^n$ (without the step function!) is $y_3[n]=H(\frac16)\cdot(\frac16)^n=0$. Since your input signal is switched on at $n=0$ (due to the step function) you see a decaying transient response (because the system is stable), but no other contribution from the input signal.

It is important to realize that you can't determine a transfer function from the relation between $x_2[n]$ and $y_2[n]$ (this was the mistake in Brian's answer, see also my comment there). That relation only determines the transfer function for one value of $z$, namely $z=-1$. Note that the sequence $x_2[n]=(-1)^n$ has no $\mathcal{Z}$-transform. The function $1/(1+z^{-1})$ is the $\mathcal{Z}$-transform of $(-1)^n\cdot u[n]$!

1. The output of an LTI system just has the same form as the input for the eigenfunctions $a^n$ (without the step function!), not for other functions. As mentioned before, the corresponding output is $H(a)\cdot a^n$, where $H(z)$ is the system's transfer function.

2. A sinusoidal input signal only excites the system at one single frequency, so from the output signal you can only determine the system's response at that single frequency. The signal $(-1)^n$ is such a (discrete-time) sinusoidal signal. So from the corresponding output, only the value $H(-1)$ can be determined, nothing else.

3. If the input signal contains all frequencies, which is the case for a signal that is switched on at a certain time (i.e., multiplied by a step function), the system's transfer function can be determined from the output signal and the given input signal. This is the case for the input $x_1[n]$, where you can simply determine $H(z)$ from $X(z)$ and $Y(z)$. Since the step function contains all frequencies, all signals multiplied by it also contain all frequencies. If you excite a system with such a signal, you can determine the system's response at all frequencies. In general, a system's response can be determined only at those frequencies that are present in the input signal. If the input signal contains a single frequency (a sinusoid), you only get information about that single frequency. In order to determine the complete transfer function, you need an input signal containing all frequencies, such as an impulse or a step (or any signal multiplied by a step).

• I generally follow what you've written; how you (and @brian) arrived at the transfer function makes sense. However, I'm still not seeing how to 'reconcile' the explanation here with the notion discussed in my other question of the output being of the same form as the input. I suspect the 'notion' is too simplistic... Aug 7, 2015 at 6:18
• Could you kindly elaborate on why you can't determine transfer function from the relation between $x_2[n]$ and $y_2[n]$? The qualitative difference betwen $x_1[n] -> y_1[n]$ and $x_2[n] -> y_2[n]$ seems to be the step function... Aug 7, 2015 at 6:20
• On that note, why does the step function's presence/absence in the input make such a big difference? Aug 7, 2015 at 6:21

NOTE: The answer is not complete; however, it seems the problem cannot be solved.

We have the following z-transforms of the input signals: $$x_1[n]=(1/6)^n\cdot u[n]\quad\longleftrightarrow\quad X_1(z)=\frac{1}{1-(1/6)\cdot z^{-1}}$$ $$x_2[n] = (-1)^n\quad\longleftrightarrow\quad X_2(z)=\frac{1}{1+z^{-1}}$$

For the output signals we have the following z-transforms: $$y_1[n] = [a(1/2)^2 + 10(1/3)^n]u[n] \quad\longleftrightarrow\quad Y_1(z) = a\frac{1}{1-(1/2)\cdot z^{-1}} + 10\frac{1}{1-(1/3)\cdot z^{-1}}$$ $$y_2[n] = (7/4)(-1)^n\quad\longleftrightarrow\quad Y_2(z) = (7/4)\frac{1}{1+z^{-1}}$$

For both pairs of signals we know that the transfer function equals $H(z)=\frac{Y(z)}{X(z)}$. From $\{X_2(z),Y_2(z)\}$ we know that $H(z) = 7/4$ (Which is trivial due to linearity, i.e., it follows also directly from $\{x_2[n],y_2[n]\}$).

What remains is to solve $Y_1(z) = H(z)\cdot X_1(z)$ for $a$; however, I don't get a proper result.

EDIT: Another way to appraoch this last step is to solve the following: $$a\frac{2}{2-z^{-1}} + b\frac{3}{3-z^{-1}} = c\frac{6}{6-z^{-1}}$$ where you may choose $b=10$ if desired.

In matrix representation: $$\mathbf{A}\left[\begin{array}{1}a\\b\\c\end{array}\right] = \mathbf{b}$$ the matrix $\mathbf{A}$ is a full rank size 3x3 matrix, which shows that no scalar solution for $H(z)=c$ with $b=10$ exists.

EDIT2: The solution seems to be that $H(-1)=7/4$ and $a=-9$. We already found that $H(z)=7/4$ for any $z$, based on the signals $\{X_2(z), Y_2(z)\}$. When evaluating $\{X_1(z),Y_1(z)\}$ for $z=-1$, the solution $a=-9$ follows easily. However, for another value of $z$, say $z=-2$ we obtain $a\approx -9.375$.

Conclusion, the decomposition only holds for $z=-1$.

• I get exactly the same thing... I think something got incorrectly copied over somewhere. Aug 6, 2015 at 18:12
• I've edited to state exactly what the question says; the changes (insignificant, in my mind) are in italics... Aug 6, 2015 at 18:24
• What I still don't get is how multiplying by $u[n]$ allows this system to generate an output involving $(1/2)^n + (1/3)^n$ with an input of $(1/6)^n$. Why does the discussion from the related question not apply? Aug 6, 2015 at 18:25
• To determine the value of $a$, the solution I have sets $(7/4)=H(-1)$ and proceeds from there, resulting in a value of $a=-9$. But again, I'm trying to understand the underlying concepts... Aug 6, 2015 at 18:28
• The signals $a^n$ and $a^n u[n]$ are different. You can use signals of the form $a^n$ to decompose a signal $x[n]$, leading to the representation $X(a)=\sum_n x[n]a^n$ ; however, you cannot do this when a step function is involved. One way go get this is that the signal $x[n]$ may be non-zero for negative time-indices, say for $n=-1$. With the signal $a^n$ you can represent such signals; however, with $a^nu[n]$ you cannot since for any $a$ the signal $a^{-1}u[-1]$ is zero. Aug 6, 2015 at 18:32