Let a linear, time-invariant, causal discrete linear system be described by the following difference equation:

$$y[n] = 0.9y[n-1] + 0.1x[n]$$

Assuming that $y[-1]=2$ and $x[n]=20\cos(\Omega n)u[n]$, $\Omega = \omega T_s = 0.2\pi$, find the total response, identifiying the terms related to the zero state and zero input responses.

I've found the transfer function of the system to be: $$H(z) = 0.1\frac{z}{z-0.9}$$

and from it I've determined the magnitude and phase of the steady-state response: $$|H \left(e^{j \Omega}\right)| = \frac{0.1}{\sqrt{1.81-1.8\cos(\Omega)}}$$ $$\angle H \left(e^{j \Omega}\right) = -\arctan\left[\frac{-\sin(\Omega)}{1-0.9\cos(\Omega)} \right]$$

Therefore the steady-state response of the system is: $$y_{ss}[n] = 20 \frac{0.1}{\sqrt{1.81-1.8\cos(\Omega)}} \cos\left(\Omega n -\arctan\left[\frac{-\sin(\Omega)}{1-0.9\cos(\Omega)} \right] \right)$$

$$\therefore \boxed{y_{ss}[n] = 3.363\cos(0.2\pi n + 1.138)}$$

But how do I find the total response (zero state and zero input) from the steady-state response?

  • $\begingroup$ why are you assuming $y[-1]=2$? $\endgroup$ Jul 4, 2019 at 5:38
  • $\begingroup$ I don't think he is assuming it, that's part of the problem description. $\endgroup$
    – Max
    Jul 4, 2019 at 5:40
  • $\begingroup$ Max is correct. The condition is given by the problem description. $\endgroup$
    – MCarsten
    Jul 4, 2019 at 5:53

1 Answer 1


Strictly speaking, an LTI system (characterized by an LCCDE) can have a zero-state response, but not a zero-input response. The latter requires nonzero initial conditions which conflicts with the requirement that an LTI system's LCCDE should have zero initial conditions, a.k.a. initial-rest.

However, this problem can be understood in the sense that the given LCCDE represents an LTI system under initial-rest assumptions, yet we still want to compute its response due to nonzero initial conditions as well, in which case the LCCDE actually represents a non LTI system.

Then you can proceed as follows: Lets use one-sided (unilateral) Z-transform to account for the initial conditions: $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} \tag{1} $$

and its useful property: $$ x[n-1] \longleftrightarrow z^{-1}X(z) + x[-1] \tag{2}$$

and for general shift of k: $$ x[n-k] \longleftrightarrow z^{-k} X(z) + \sum_{m=0}^{k-1} z^{-m}x[m-k] \tag{3} $$

Given an N-th order LCCDE: $$ \sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k] \tag{4} $$

Apply Eq.3 to Eq.4, assuming $x[n]=0, n< 0$:

$$ \sum_{k=0}^{N} a_k \left( z^{-k} Y(z) + \sum_{m=0}^{k-1} z^{-m} y[m-k] \right) = \sum_{k=0}^{M} b_k z^{-k} X(z) \tag{5} $$

$$ Y(z) \left( \sum_{k=0}^{N} a_k z^{-k} \right) + \sum_{k=0}^{N} a_k \sum_{m=0}^{k-1} z^{-m} y[m-k] = X(z) \left(\sum_{k=0}^{M} b_k z^{-k} \right) \tag{6} $$

Denote the double sum $\sum_{k=0}^{N} a_k \sum_{m=0}^{k-1} z^{-m} y[m-k]$ as $Y_{ic}(z)$ the initial-conditions part of output, then Eq.6 becomes: $$ Y(z) \left( \sum_{k=0}^{N} a_k z^{-k} \right) + Y_{ic}(z) = X(z) \left(\sum_{k=0}^{M} b_k z^{-k} \right) \tag{7} $$

Then express $Y(z)$ as: $$ Y(z) = \frac{ \sum_{k=0}^{M} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} X(z) - \frac{ Y_{ic}(z) }{\sum_{k=0}^{N} a_k z^{-k} } \tag{8} $$

Denoting $\frac{ \sum_{k=0}^{M} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}}$ as $H(z)$; the Transfer Function of the associated LTI system, we have:

$$ Y(z) = H(z) X(z) - \frac{ Y_{ic}(z) }{\sum_{k=0}^{N} a_k z^{-k} } \tag{9} $$

Eq.9 expresses the LCCDE output $Y(z)$ in terms of the input $X(z)$ and the initial-conditions $Y_{ic}(z)$.

The first term $H(z) X(z)$ is the zero-state response (or forced response) $Y_{zs}(z)$ corresponding to the input $X(z)$, and the second term is the zero-input response (or natural response) $Y_{zi}(z)$ corresponding to the initial-conditions $Y_{ic}(z)$ which is typically a transient response due to initial conditions.

Hence we have:

$$Y(z) = Y_{zs}(z) + Y_{zi}(z) \tag{10}$$

The zero-state response, $Y_{zs}(z) = H(z)X(z)$ corresponds to the LTI system that's characterized by the LCCDE, because it assumes zero initial conditions;i.e., $Y_{IC}(z) = 0$, and then the LCCDE output is given by :

$$ Y(z) = H(z)X(s) \longleftrightarrow y[n] = h[n] \star x[n] \tag{11} $$

Eq.11 displays the equivalence between the zero-state response of an LCCDE and the convolution output of the corresponding LTI system with impulse response $h[n]$.

For a suddenly applied sinudoidal $x[n] = K \cos(\omega_0 n)u[n]$, the zero-state response will include two terms: a switching transient part that corresponds to the sudden application of the input; i.e. $x[n] = 0 , n < 0$ , and a sinusoidal steady-state part that remains persistent after switching transients (and initial-condition transients as well, if that also exists) decay, and which corresponds to the response of the LTI system as if the input were $x[n] = \cos(\omega_0 n)$.

When applied to your LCCDE, you get the following:

$$ y[n] + a y[n-1] = b_0 x[n] ~~~,~~~ y[-1]=2 \tag{12} $$


$$x[n] = K \cos(\omega_0 n)u[n] \leftrightarrow \frac{K/2}{ 1 - e^{j \omega_0} z^{-1} } + \frac{K/2}{ 1 - e^{-j \omega_0} z^{-1} } \tag{13}$$

where $a=-0.9 ,~ b_0 = 0.1 ,~ \omega_0 = 0.2 \pi$ , and $K=20$.

Then the total output becomes:

$$ Y(z) = \frac{b_0}{1 + a z^{-1}} \left( \frac{K / 2} {1 - e^{j \omega_0} z^{-1}} + \frac{K / 2} {1 - e^{-j \omega_0} z^{-1}} + \right) - \frac{ a y[-1]}{1 + a z^{-1}} \tag{14}$$

Where the term in the paranthesis corresponds to the zero-state output, and the remaing last term is the zero-input response.

Apply partial fraction expansion, PFE, to the zero-state response part:

$$Y(z) = \frac{A_1}{1 - e^{j\omega_0} z^{-1}} + \frac{A_2}{1 - e^{-j\omega_0} z^{-1}} + \frac{B_1}{1 + a z^{-1}} + \frac{B_2}{1 + a z^{-1}} - \frac{a y[-1]}{1 + a z^{-1}}\tag{15} $$

It will turn out that $A_2 = A_1^*$ and $B_2 = B_1^*$, and we have $$A = \frac{b_0 K/2}{1 + a e^{-j \omega_0}} ~~ ,~~ B = \frac{b_0 K/2}{1 + a^{-1} e^{j\omega_0}} $$.

Finally in the time-domain, the following is obtained for the total output:

$$y[n] = 2|A| \cos(\omega_0 n + \phi_A) + 2 \mathcal{Re}\{B\} (-a)^n - a y[-1]( -a)^n \tag{16} $$

For the given values it happens to be:

$$y[n] = 3.3626 \cos(0.2\pi n - 1.0960) u[n] + 0.4629 (0.9)^n u[n] + 1.8 (0.9)^n u[n] \tag{17} $$

Finally we can denote the parts of total output as follows:

the sinusoidal steady-state response: $$y_{ss}[n] = 3.3626 \cos(0.2\pi n - 1.0960) u[n] $$

the input switching transient response: $$y_{is}[n] = 0.4629 (0.9)^n u[n] $$

the zero-state response: $$y_{zs}[n] = y_{ss}[n] + y_{is}[n] $$

the zero-input (transient) response: $$ y_{zi}[n] = 1.8 (0.9)^n u[n] $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.